Zebra puzzle
The Zebra puzzle, a.k.a. Einstein's Riddle, is a logic puzzle which is to be solved programmatically.
You are encouraged to solve this task according to the task description, using any language you may know.
It has several variants, one of them this:
- There are five houses.
- The English man lives in the red house.
- The Swede has a dog.
- The Dane drinks tea.
- The green house is immediately to the left of the white house.
- They drink coffee in the green house.
- The man who smokes Pall Mall has birds.
- In the yellow house they smoke Dunhill.
- In the middle house they drink milk.
- The Norwegian lives in the first house.
- The man who smokes Blend lives in the house next to the house with cats.
- In a house next to the house where they have a horse, they smoke Dunhill.
- The man who smokes Blue Master drinks beer.
- The German smokes Prince.
- The Norwegian lives next to the blue house.
- They drink water in a house next to the house where they smoke Blend.
The question is, who owns the zebra?
Additionally, list the solution for all the houses.
Optionally, show the solution is unique.
- Related tasks
Ada
Not the prettiest Ada, but it's simple and very fast. Similar to my Dinesman's code; uses enums to keep things readable. <lang Ada>with Ada.Text_IO; use Ada.Text_IO; procedure Zebra is
type Content is (Beer, Coffee, Milk, Tea, Water,
Danish, English, German, Norwegian, Swedish,
Blue, Green, Red, White, Yellow,
Blend, BlueMaster, Dunhill, PallMall, Prince,
Bird, Cat, Dog, Horse, Zebra);
type Test is (Drink, Person, Color, Smoke, Pet);
type House is (One, Two, Three, Four, Five);
type Street is array (Test'Range, House'Range) of Content;
type Alley is access all Street;
procedure Print (mat : Alley) is begin
for H in House'Range loop
Put(H'Img&": ");
for T in Test'Range loop
Put(T'Img&"="&mat(T,H)'Img&" ");
end loop; New_Line; end loop;
end Print;
function FinalChecks (mat : Alley) return Boolean is
function Diff (A, B : Content; CA , CB : Test) return Integer is begin
for H1 in House'Range loop for H2 in House'Range loop
if mat(CA,H1) = A and mat(CB,H2) = B then
return House'Pos(H1) - House'Pos(H2);
end if;
end loop; end loop;
end Diff;
begin
if abs(Diff(Norwegian, Blue, Person, Color)) = 1
and Diff(Green, White, Color, Color) = -1
and abs(Diff(Horse, Dunhill, Pet, Smoke)) = 1
and abs(Diff(Water, Blend, Drink, Smoke)) = 1
and abs(Diff(Blend, Cat, Smoke, Pet)) = 1
then return True;
end if;
return False;
end FinalChecks;
function Constrained (mat : Alley; atest : Natural) return Boolean is begin
-- Tests seperated into levels for speed, not strictly necessary
-- As such, the program finishes in around ~0.02s
case Test'Val (atest) is
when Drink => -- Drink
if mat (Drink, Three) /= Milk then return False; end if;
return True;
when Person => -- Drink+Person
for H in House'Range loop
if (mat(Person,H) = Norwegian and H /= One)
or (mat(Person,H) = Danish and mat(Drink,H) /= Tea)
then return False; end if;
end loop;
return True;
when Color => -- Drink+People+Color
for H in House'Range loop
if (mat(Person,H) = English and mat(Color,H) /= Red)
or (mat(Drink,H) = Coffee and mat(Color,H) /= Green)
then return False; end if;
end loop;
return True;
when Smoke => -- Drink+People+Color+Smoke
for H in House'Range loop
if (mat(Color,H) = Yellow and mat(Smoke,H) /= Dunhill)
or (mat(Smoke,H) = BlueMaster and mat(Drink,H) /= Beer)
or (mat(Person,H) = German and mat(Smoke,H) /= Prince)
then return False; end if;
end loop;
return True;
when Pet => -- Drink+People+Color+Smoke+Pet
for H in House'Range loop
if (mat(Person,H) = Swedish and mat(Pet,H) /= Dog)
or (mat(Smoke,H) = PallMall and mat(Pet,H) /= Bird)
then return False; end if;
end loop;
return FinalChecks(mat); -- Do the next-to checks
end case;
end Constrained;
procedure Solve (mat : Alley; t, n : Natural) is
procedure Swap (I, J : Natural) is
temp : constant Content := mat (Test'Val (t), House'Val (J));
begin
mat (Test'Val (t), House'Val (J)) := mat (Test'Val (t), House'Val (I));
mat (Test'Val (t), House'Val (I)) := temp;
end Swap;
begin
if n = 1 and Constrained (mat, t) then -- test t passed
if t < 4 then Solve (mat, t + 1, 5); -- Onto next test
else Print (mat); return; -- Passed and t=4 means a solution
end if;
end if;
for i in 0 .. n - 1 loop -- The permutations part
Solve (mat, t, n - 1);
if n mod 2 = 1 then Swap (0, n - 1);
else Swap (i, n - 1); end if;
end loop;
end Solve;
myStreet : aliased Street; myAlley : constant Alley := myStreet'Access;
begin
for i in Test'Range loop for j in House'Range loop -- Init Matrix
myStreet (i,j) := Content'Val(Test'Pos(i)*5 + House'Pos(j));
end loop; end loop;
Solve (myAlley, 0, 5); -- start at test 0 with 5 options
end Zebra;</lang>
- Output:
ONE: DRINK=WATER PERSON=NORWEGIAN COLOR=YELLOW SMOKE=DUNHILL PET=CAT TWO: DRINK=TEA PERSON=DANISH COLOR=BLUE SMOKE=BLEND PET=HORSE THREE: DRINK=MILK PERSON=ENGLISH COLOR=RED SMOKE=PALLMALL PET=BIRD FOUR: DRINK=COFFEE PERSON=GERMAN COLOR=GREEN SMOKE=PRINCE PET=ZEBRA FIVE: DRINK=BEER PERSON=SWEDISH COLOR=WHITE SMOKE=BLUEMASTER PET=DOG
ALGOL 68
Attempts to find solutions using the rules. <lang algol68>BEGIN
# attempt to solve Einstein's Riddle - the Zebra puzzle #
INT unknown = 0, same = -1;
INT english = 1, swede = 2, dane = 3, norwegian = 4, german = 5;
INT dog = 1, birds = 2, cats = 3, horse = 4, zebra = 5;
INT red = 1, green = 2, white = 3, yellow = 4, blue = 5;
INT tea = 1, coffee = 2, milk = 3, beer = 4, water = 5;
INT pall mall = 1, dunhill = 2, blend = 3, blue master = 4, prince = 5;
[]STRING nationality = ( "unknown", "english", "swede", "dane", "norwegian", "german" );
[]STRING animal = ( "unknown", "dog", "birds", "cats", "horse", "ZEBRA" );
[]STRING colour = ( "unknown", "red", "green", "white", "yellow", "blue" );
[]STRING drink = ( "unknown", "tea", "coffee", "milk", "beer", "water" );
[]STRING smoke = ( "unknown", "pall mall", "dunhill", "blend", "blue master", "prince" );
MODE HOUSE = STRUCT( INT nationality, animal, colour, drink, smoke );
# returns TRUE if a field in a house could be set to value, FALSE otherwise #
PROC can set = ( INT field, INT value )BOOL: field = unknown OR value = same;
# returns TRUE if the fields of house h could be set to those of #
# suggestion s, FALSE otherwise #
OP XOR = ( HOUSE h, HOUSE s )BOOL:
( can set( nationality OF h, nationality OF s ) AND can set( animal OF h, animal OF s )
AND can set( colour OF h, colour OF s ) AND can set( drink OF h, drink OF s )
AND can set( smoke OF h, smoke OF s )
) # XOR # ;
# sets a field in a house to value if it is unknown #
PROC set = ( REF INT field, INT value )VOID: IF field = unknown AND value /= same THEN field := value FI;
# sets the unknown fields in house h to the non-same fields of suggestion s #
OP +:= = ( REF HOUSE h, HOUSE s )VOID:
( set( nationality OF h, nationality OF s ); set( animal OF h, animal OF s )
; set( colour OF h, colour OF s ); set( drink OF h, drink OF s )
; set( smoke OF h, smoke OF s )
) # +:= # ;
# sets a field in a house to unknown if the value is not same #
PROC reset = ( REF INT field, INT value )VOID: IF value /= same THEN field := unknown FI;
# sets fields in house h to unknown if the suggestion s is not same #
OP -:= = ( REF HOUSE h, HOUSE s )VOID:
( reset( nationality OF h, nationality OF s ); reset( animal OF h, animal OF s )
; reset( colour OF h, colour OF s ); reset( drink OF h, drink OF s )
; reset( smoke OF h, smoke OF s )
) # -:= # ;
# attempts a partial solution for the house at pos #
PROC try = ( INT pos, HOUSE suggestion, PROC VOID continue )VOID:
IF pos >= LWB house AND pos <= UPB house THEN
IF house[ pos ] XOR suggestion THEN
house[ pos ] +:= suggestion; continue; house[ pos ] -:= suggestion
FI
FI # try # ;
# attempts a partial solution for the neighbours of a house #
PROC left or right = ( INT pos, BOOL left, BOOL right, HOUSE neighbour suggestion
, PROC VOID continue )VOID:
( IF left THEN try( pos - 1, neighbour suggestion, continue ) FI
; IF right THEN try( pos + 1, neighbour suggestion, continue ) FI
) # left or right # ;
# attempts a partial solution for all houses and possibly their neighbours #
PROC any2 = ( REF INT number, HOUSE suggestion
, BOOL left, BOOL right, HOUSE neighbour suggestion
, PROC VOID continue )VOID:
FOR pos TO UPB house DO
IF house[ pos ] XOR suggestion THEN
number := pos;
house[ number ] +:= suggestion;
IF NOT left AND NOT right THEN # neighbours not involved #
continue
ELSE # try one or both neighbours #
left or right( pos, left, right, neighbour suggestion, continue )
FI;
house[ number ] -:= suggestion
FI
OD # any2 # ;
# attempts a partial solution for all houses #
PROC any = ( HOUSE suggestion, PROC VOID continue )VOID:
any2( LOC INT, suggestion, FALSE, FALSE, SKIP, continue );
# find solution(s) #
INT blend pos;
INT solutions := 0;
# There are five houses. #
[ 1 : 5 ]HOUSE house;
FOR h TO UPB house DO house[ h ] := ( unknown, unknown, unknown, unknown, unknown ) OD;
# In the middle house they drink milk. #
drink OF house[ 3 ] := milk;
# The Norwegian lives in the first house. #
nationality OF house[ 1 ] := norwegian;
# The Norwegian lives next to the blue house. #
colour OF house[ 2 ] := blue;
# They drink coffee in the green house. #
# The green house is immediately to the left of the white house. #
any2( LOC INT, ( same, same, green, coffee, same )
, FALSE, TRUE, ( same, same, white, same, same ), VOID:
# In a house next to the house where they have a horse, #
# they smoke Dunhill. #
# In the yellow house they smoke Dunhill. #
any2( LOC INT, ( same, horse, same, same, same )
, TRUE, TRUE, ( same, same, yellow, same, dunhill ), VOID:
# The English man lives in the red house. #
any( ( english, same, red, same, same ), VOID:
# The man who smokes Blend lives in the house next to the #
# house with cats. #
any2( blend pos, ( same, same, same, same, blend )
, TRUE, TRUE, ( same, cats, same, same, same ), VOID:
# They drink water in a house next to the house where #
# they smoke Blend. #
left or right( blend pos, TRUE, TRUE, ( same, same, same, water, same ), VOID:
# The Dane drinks tea. #
any( ( dane, same, same, tea, same ), VOID:
# The man who smokes Blue Master drinks beer. #
any( ( same, same, same, beer, blue master ), VOID:
# The Swede has a dog. #
any( ( swede, dog, same, same, same ), VOID:
# The German smokes Prince. #
any( ( german, same, same, same, prince ), VOID:
# The man who smokes Pall Mall has birds. #
any( ( same, birds, same, same, pall mall ), VOID:
# if we can place the zebra, we have a solution #
any( ( same, zebra, same, same, same ), VOID:
( solutions +:= 1;
FOR h TO UPB house DO
print( ( whole( h, 0 )
, " ", nationality[ 1 + nationality OF house[ h ] ]
, ", ", animal [ 1 + animal OF house[ h ] ]
, ", ", colour [ 1 + colour OF house[ h ] ]
, ", ", drink [ 1 + drink OF house[ h ] ]
, ", ", smoke [ 1 + smoke OF house[ h ] ]
, newline
)
)
OD;
print( ( newline ) )
)
) # zebra #
) # pall mall #
) # german #
) # swede #
) # beer #
) # dane #
) # blend L/R #
) # blend #
) # red #
) # horse #
) # green # ;
print( ( "solutions: ", whole( solutions, 0 ), newline ) )
END </lang>
- Output:
1 norwegian, cats, yellow, water, dunhill 2 dane, horse, blue, tea, blend 3 english, birds, red, milk, pall mall 4 german, ZEBRA, green, coffee, prince 5 swede, dog, white, beer, blue master solutions: 1
AutoHotkey
BBC BASIC
<lang bbcbasic> REM The names (only used for printing the results):
DIM Drink$(4), Nation$(4), Colr$(4), Smoke$(4), Animal$(4)
Drink$() = "Beer", "Coffee", "Milk", "Tea", "Water"
Nation$() = "Denmark", "England", "Germany", "Norway", "Sweden"
Colr$() = "Blue", "Green", "Red", "White", "Yellow"
Smoke$() = "Blend", "BlueMaster", "Dunhill", "PallMall", "Prince"
Animal$() = "Birds", "Cats", "Dog", "Horse", "Zebra"
REM Some single-character tags:
a$ = "A" : b$ = "B" : c$ = "C" : d$ = "D" : e$ = "E"
REM BBC BASIC Doesn't have enumerations!
Beer$=a$ : Coffee$=b$ : Milk$=c$ : Tea$=d$ : Water$=e$
Denmark$=a$ : England$=b$ : Germany$=c$ : Norway$=d$ : Sweden$=e$
Blue$=a$ : Green$=b$ : Red$=c$ : White$=d$ : Yellow$=e$
Blend$=a$ : BlueMaster$=b$ : Dunhill$=c$ : PallMall$=d$ : Prince$=e$
Birds$=a$ : Cats$=b$ : Dog$=c$ : Horse$=d$ : Zebra$=e$
REM Create the 120 permutations of 5 objects:
DIM perm$(120), x$(4) : x$() = a$, b$, c$, d$, e$
REPEAT
p% += 1
perm$(p%) = x$(0)+x$(1)+x$(2)+x$(3)+x$(4)
UNTIL NOT FNperm(x$())
REM Express the statements as conditional expressions:
ex2$ = "INSTR(Nation$,England$) = INSTR(Colr$,Red$)"
ex3$ = "INSTR(Nation$,Sweden$) = INSTR(Animal$,Dog$)"
ex4$ = "INSTR(Nation$,Denmark$) = INSTR(Drink$,Tea$)"
ex5$ = "INSTR(Colr$,Green$+White$) <> 0"
ex6$ = "INSTR(Drink$,Coffee$) = INSTR(Colr$,Green$)"
ex7$ = "INSTR(Smoke$,PallMall$) = INSTR(Animal$,Birds$)"
ex8$ = "INSTR(Smoke$,Dunhill$) = INSTR(Colr$,Yellow$)"
ex9$ = "MID$(Drink$,3,1) = Milk$"
ex10$ = "LEFT$(Nation$,1) = Norway$"
ex11$ = "ABS(INSTR(Smoke$,Blend$)-INSTR(Animal$,Cats$)) = 1"
ex12$ = "ABS(INSTR(Smoke$,Dunhill$)-INSTR(Animal$,Horse$)) = 1"
ex13$ = "INSTR(Smoke$,BlueMaster$) = INSTR(Drink$,Beer$)"
ex14$ = "INSTR(Nation$,Germany$) = INSTR(Smoke$,Prince$)"
ex15$ = "ABS(INSTR(Nation$,Norway$)-INSTR(Colr$,Blue$)) = 1"
ex16$ = "ABS(INSTR(Smoke$,Blend$)-INSTR(Drink$,Water$)) = 1"
REM Solve:
solutions% = 0
TIME = 0
FOR nation% = 1 TO 120
Nation$ = perm$(nation%)
IF EVAL(ex10$) THEN
FOR colr% = 1 TO 120
Colr$ = perm$(colr%)
IF EVAL(ex5$) IF EVAL(ex2$) IF EVAL(ex15$) THEN
FOR drink% = 1 TO 120
Drink$ = perm$(drink%)
IF EVAL(ex9$) IF EVAL(ex4$) IF EVAL(ex6$) THEN
FOR smoke% = 1 TO 120
Smoke$ = perm$(smoke%)
IF EVAL(ex14$) IF EVAL(ex13$) IF EVAL(ex16$) IF EVAL(ex8$) THEN
FOR animal% = 1 TO 120
Animal$ = perm$(animal%)
IF EVAL(ex3$) IF EVAL(ex7$) IF EVAL(ex11$) IF EVAL(ex12$) THEN
PRINT "House Drink Nation Colour Smoke Animal"
FOR house% = 1 TO 5
PRINT ; house% ,;
PRINT Drink$(ASCMID$(Drink$,house%)-65),;
PRINT Nation$(ASCMID$(Nation$,house%)-65),;
PRINT Colr$(ASCMID$(Colr$,house%)-65),;
PRINT Smoke$(ASCMID$(Smoke$,house%)-65),;
PRINT Animal$(ASCMID$(Animal$,house%)-65)
NEXT
solutions% += 1
ENDIF
NEXT animal%
ENDIF
NEXT smoke%
ENDIF
NEXT drink%
ENDIF
NEXT colr%
ENDIF
NEXT nation%
PRINT '"Number of solutions = "; solutions%
PRINT "Solved in " ; TIME/100 " seconds"
END
DEF FNperm(x$())
LOCAL i%, j%
FOR i% = DIM(x$(),1)-1 TO 0 STEP -1
IF x$(i%) < x$(i%+1) EXIT FOR
NEXT
IF i% < 0 THEN = FALSE
j% = DIM(x$(),1)
WHILE x$(j%) <= x$(i%) j% -= 1 : ENDWHILE
SWAP x$(i%), x$(j%)
i% += 1
j% = DIM(x$(),1)
WHILE i% < j%
SWAP x$(i%), x$(j%)
i% += 1
j% -= 1
ENDWHILE
= TRUE</lang>
Output:
House Drink Nation Colour Smoke Animal 1 Water Norway Yellow Dunhill Cats 2 Tea Denmark Blue Blend Horse 3 Milk England Red PallMall Birds 4 Coffee Germany Green Prince Zebra 5 Beer Sweden White BlueMasterDog Number of solutions = 1 Solved in 0.12 seconds
Bracmat
<lang bracmat>( (English Swede Dane Norwegian German,)
(red green white yellow blue,(red.English.))
(dog birds cats horse zebra,(dog.?.Swede.))
( tea coffee milk beer water
, (tea.?.?.Dane.) (coffee.?.green.?.)
)
( "Pall Mall" Dunhill Blend "Blue Master" Prince
, ("Blue Master".beer.?.?.?.)
("Pall Mall".?.birds.?.?.)
(Dunhill.?.?.yellow.?.)
(Prince.?.?.?.German.)
)
( 1 2 3 4 5
, (3.?.milk.?.?.?.) (1.?.?.?.?.Norwegian.)
)
: ?properties
& ( relations
= next leftOf
. ( next
= a b A B
. !arg:(?S,?A,?B)
& !S:? (?a.!A) ?:? (?b.!B) ?
& (!a+1:!b|!b+1:!a)
)
& ( leftOf
= a b A B
. !arg:(?S,?A,?B)
& !S:? (?a.!A) ?:? (?b.!B) ?
& !a+1:!b
)
& leftOf
$ (!arg,(?.?.?.green.?.),(?.?.?.white.?.))
& next$(!arg,(Blend.?.?.?.?.),(?.?.cats.?.?.))
& next
$ (!arg,(?.?.horse.?.?.),(Dunhill.?.?.?.?.))
& next
$ (!arg,(?.?.?.?.Norwegian.),(?.?.?.blue.?.))
& next$(!arg,(?.water.?.?.?.),(Blend.?.?.?.?.))
)
& ( props
= a constraint constraints house houses
, remainingToDo shavedToDo toDo value values z
. !arg:(?toDo.?shavedToDo.?house.?houses)
& ( !toDo:(?values,?constraints) ?remainingToDo
& !values
: ( ?a
( %@?value
& !constraints
: ( ?
( !value
. ?constraint
& !house:!constraint
)
?
| ~( ?
( ?
. ?constraint
& !house:!constraint
)
?
| ? (!value.?) ?
)
)
)
( ?z
& props
$ ( !remainingToDo
. !shavedToDo (!a !z,!constraints)
. (!value.!house)
. !houses
)
)
|
& relations$!houses
& out$(Solution !houses)
)
| !toDo:
& props$(!shavedToDo...!house !houses)
)
)
& props$(!properties...) & done );</lang> Output:
Solution
(4.Prince.coffee.zebra.green.German.)
(1.Dunhill.water.cats.yellow.Norwegian.)
(2.Blend.tea.horse.blue.Dane.)
(5.Blue Master.beer.dog.white.Swede.)
(3.Pall Mall.milk.birds.red.English.)
{!} done
C
<lang c>#include <stdio.h>
- include <string.h>
enum HouseStatus { Invalid, Underfull, Valid };
enum Attrib { C, M, D, A, S };
// Unfilled attributes are represented by -1 enum Colors { Red, Green, White, Yellow, Blue }; enum Mans { English, Swede, Dane, German, Norwegian }; enum Drinks { Tea, Coffee, Milk, Beer, Water }; enum Animals { Dog, Birds, Cats, Horse, Zebra }; enum Smokes { PallMall, Dunhill, Blend, BlueMaster, Prince };
void printHouses(int ha[5][5]) {
const char *color[] = { "Red", "Green", "White", "Yellow", "Blue" };
const char *man[] = { "English", "Swede", "Dane", "German", "Norwegian" };
const char *drink[] = { "Tea", "Coffee", "Milk", "Beer", "Water" };
const char *animal[] = { "Dog", "Birds", "Cats", "Horse", "Zebra" };
const char *smoke[] = { "PallMall", "Dunhill", "Blend", "BlueMaster", "Prince" };
printf("%-10.10s%-10.10s%-10.10s%-10.10s%-10.10s%-10.10s\n",
"House", "Color", "Man", "Drink", "Animal", "Smoke");
for (int i = 0; i < 5; i++) {
printf("%-10d", i);
if (ha[i][C] >= 0)
printf("%-10.10s", color[ha[i][C]]);
else
printf("%-10.10s", "-");
if (ha[i][M] >= 0)
printf("%-10.10s", man[ha[i][M]]);
else
printf("%-10.10s", "-");
if (ha[i][D] >= 0)
printf("%-10.10s", drink[ha[i][D]]);
else
printf("%-10.10s", "-");
if (ha[i][A] >= 0)
printf("%-10.10s", animal[ha[i][A]]);
else
printf("%-10.10s", "-");
if (ha[i][S] >= 0)
printf("%-10.10s\n", smoke[ha[i][S]]);
else
printf("-\n");
}
}
int checkHouses(int ha[5][5]) {
int c_add = 0, c_or = 0; int m_add = 0, m_or = 0; int d_add = 0, d_or = 0; int a_add = 0, a_or = 0; int s_add = 0, s_or = 0;
// Cond 9: In the middle house they drink milk.
if (ha[2][D] >= 0 && ha[2][D] != Milk)
return Invalid;
// Cond 10: The Norwegian lives in the first house.
if (ha[0][M] >= 0 && ha[0][M] != Norwegian)
return Invalid;
for (int i = 0; i < 5; i++) {
// Uniqueness tests.
if (ha[i][C] >= 0) {
c_add += (1 << ha[i][C]);
c_or |= (1 << ha[i][C]);
}
if (ha[i][M] >= 0) {
m_add += (1 << ha[i][M]);
m_or |= (1 << ha[i][M]);
}
if (ha[i][D] >= 0) {
d_add += (1 << ha[i][D]);
d_or |= (1 << ha[i][D]);
}
if (ha[i][A] >= 0) {
a_add += (1 << ha[i][A]);
a_or |= (1 << ha[i][A]);
}
if (ha[i][S] >= 0) {
s_add += (1 << ha[i][S]);
s_or |= (1 << ha[i][S]);
}
// Cond 2: The English man lives in the red house.
if ((ha[i][M] >= 0 && ha[i][C] >= 0) &&
((ha[i][M] == English && ha[i][C] != Red) || // Checking both
(ha[i][M] != English && ha[i][C] == Red))) // to make things quicker.
return Invalid;
// Cond 3: The Swede has a dog.
if ((ha[i][M] >= 0 && ha[i][A] >= 0) &&
((ha[i][M] == Swede && ha[i][A] != Dog) ||
(ha[i][M] != Swede && ha[i][A] == Dog)))
return Invalid;
// Cond 4: The Dane drinks tea.
if ((ha[i][M] >= 0 && ha[i][D] >= 0) &&
((ha[i][M] == Dane && ha[i][D] != Tea) ||
(ha[i][M] != Dane && ha[i][D] == Tea)))
return Invalid;
// Cond 5: The green house is immediately to the left of the white house.
if ((i > 0 && ha[i][C] >= 0 /*&& ha[i-1][C] >= 0 */ ) &&
((ha[i - 1][C] == Green && ha[i][C] != White) ||
(ha[i - 1][C] != Green && ha[i][C] == White)))
return Invalid;
// Cond 6: drink coffee in the green house.
if ((ha[i][C] >= 0 && ha[i][D] >= 0) &&
((ha[i][C] == Green && ha[i][D] != Coffee) ||
(ha[i][C] != Green && ha[i][D] == Coffee)))
return Invalid;
// Cond 7: The man who smokes Pall Mall has birds.
if ((ha[i][S] >= 0 && ha[i][A] >= 0) &&
((ha[i][S] == PallMall && ha[i][A] != Birds) ||
(ha[i][S] != PallMall && ha[i][A] == Birds)))
return Invalid;
// Cond 8: In the yellow house they smoke Dunhill.
if ((ha[i][S] >= 0 && ha[i][C] >= 0) &&
((ha[i][S] == Dunhill && ha[i][C] != Yellow) ||
(ha[i][S] != Dunhill && ha[i][C] == Yellow)))
return Invalid;
// Cond 11: The man who smokes Blend lives in the house next to the house with cats.
if (ha[i][S] == Blend) {
if (i == 0 && ha[i + 1][A] >= 0 && ha[i + 1][A] != Cats)
return Invalid;
else if (i == 4 && ha[i - 1][A] != Cats)
return Invalid;
else if (ha[i + 1][A] >= 0 && ha[i + 1][A] != Cats && ha[i - 1][A] != Cats)
return Invalid;
}
// Cond 12: In a house next to the house where they have a horse, they smoke Dunhill.
if (ha[i][S] == Dunhill) {
if (i == 0 && ha[i + 1][A] >= 0 && ha[i + 1][A] != Horse)
return Invalid;
else if (i == 4 && ha[i - 1][A] != Horse)
return Invalid;
else if (ha[i + 1][A] >= 0 && ha[i + 1][A] != Horse && ha[i - 1][A] != Horse)
return Invalid;
}
// Cond 13: The man who smokes Blue Master drinks beer.
if ((ha[i][S] >= 0 && ha[i][D] >= 0) &&
((ha[i][S] == BlueMaster && ha[i][D] != Beer) ||
(ha[i][S] != BlueMaster && ha[i][D] == Beer)))
return Invalid;
// Cond 14: The German smokes Prince
if ((ha[i][M] >= 0 && ha[i][S] >= 0) &&
((ha[i][M] == German && ha[i][S] != Prince) ||
(ha[i][M] != German && ha[i][S] == Prince)))
return Invalid;
// Cond 15: The Norwegian lives next to the blue house.
if (ha[i][M] == Norwegian &&
((i < 4 && ha[i + 1][C] >= 0 && ha[i + 1][C] != Blue) ||
(i > 0 && ha[i - 1][C] != Blue)))
return Invalid;
// Cond 16: They drink water in a house next to the house where they smoke Blend.
if (ha[i][S] == Blend) {
if (i == 0 && ha[i + 1][D] >= 0 && ha[i + 1][D] != Water)
return Invalid;
else if (i == 4 && ha[i - 1][D] != Water)
return Invalid;
else if (ha[i + 1][D] >= 0 && ha[i + 1][D] != Water && ha[i - 1][D] != Water)
return Invalid;
}
}
if ((c_add != c_or) || (m_add != m_or) || (d_add != d_or)
|| (a_add != a_or) || (s_add != s_or)) {
return Invalid;
}
if ((c_add != 0b11111) || (m_add != 0b11111) || (d_add != 0b11111)
|| (a_add != 0b11111) || (s_add != 0b11111)) {
return Underfull;
}
return Valid;
}
int bruteFill(int ha[5][5], int hno, int attr) {
int stat = checkHouses(ha);
if ((stat == Valid) || (stat == Invalid))
return stat;
int hb[5][5];
memcpy(hb, ha, sizeof(int) * 5 * 5);
for (int i = 0; i < 5; i++) {
hb[hno][attr] = i;
stat = checkHouses(hb);
if (stat != Invalid) {
int nexthno, nextattr;
if (attr < 4) {
nextattr = attr + 1;
nexthno = hno;
} else {
nextattr = 0;
nexthno = hno + 1;
}
stat = bruteFill(hb, nexthno, nextattr);
if (stat != Invalid) {
memcpy(ha, hb, sizeof(int) * 5 * 5);
return stat;
}
}
}
// We only come here if none of the attr values assigned were valid. return Invalid;
}
int main() {
int ha[5][5] = {{-1, -1, -1, -1, -1}, {-1, -1, -1, -1, -1},
{-1, -1, -1, -1, -1}, {-1, -1, -1, -1, -1},
{-1, -1, -1, -1, -1}};
bruteFill(ha, 0, 0); printHouses(ha);
return 0;
}</lang>
- Output:
% gcc -Wall -O3 -std=c99 zebra.c -o zebra && time ./zebra House Color Man Drink Animal Smoke 0 Yellow Norwegian Water Cats Dunhill 1 Blue Dane Tea Horse Blend 2 Red English Milk Birds PallMall 3 Green German Coffee Zebra Prince 4 White Swede Beer Dog BlueMaster ./zebra 0.00s user 0.00s system 0% cpu 0.002 total
The execution time is too small to be reliably measured on my machine.
C Generated from Perl
I'll be the first to admit the following doesn't quite look like a C program. It's in fact in Perl, which outputs a C source, which in turn solves the puzzle. If you think this is long, wait till you see the C it writes. <lang perl>#!/usr/bin/perl
use utf8; no strict;
my (%props, %name, @pre, @conds, @works, $find_all_solutions);
sub do_consts { local $"; for my $p (keys %props) { my @s = @{ $props{$p} };
$" = ", "; print "enum { ${p}_none = 0, @s };\n";
$" = '", "'; print "const char *string_$p [] = { \"###\", \"@s\" };\n\n"; } print "#define FIND_BY(p) \\ int find_by_##p(int v) { \\ int i; \\ for (i = 0; i < N_ITEMS; i++) \\ if (house[i].p == v) return i; \\ return -1; }\n";
print "FIND_BY($_)" for (keys %props);
local $" = ", "; my @k = keys %props;
my $sl = 0; for (keys %name) { if (length > $sl) { $sl = length } }
my $fmt = ("%".($sl + 1)."s ") x @k; my @arg = map { "string_$_"."[house[i].$_]" } @k; print << "SNIPPET"; int work0(void) { int i; for (i = 0; i < N_ITEMS; i++) printf("%d $fmt\\n", i, @arg); puts(\"\"); return 1; } SNIPPET
}
sub setprops { %props = @_; my $l = 0; my @k = keys %props; for my $p (@k) { my @s = @{ $props{$p} };
if ($l && $l != @s) { die "bad length @s"; } $l = @s; $name{$_} = $p for @s; } local $" = ", "; print "#include <stdio.h>
- define N_ITEMS $l
struct item_t { int @k; } house[N_ITEMS] = Template:0;\n"; }
sub pair {NB. h =.~.&> compose&.>~/y,<h
my ($c1, $c2, $diff) = @_; $diff //= [0]; $diff = [$diff] unless ref $diff;
push @conds, [$c1, $c2, $diff]; }
sub make_conditions { my $idx = 0; my $return1 = $find_all_solutions ? "" : "return 1"; print "
- define TRY(a, b, c, d, p, n) \\
if ((b = a d) >= 0 && b < N_ITEMS) { \\ if (!house[b].p) { \\ house[b].p = c; \\ if (n()) $return1; \\ house[b].p = 0; \\ }} ";
while (@conds) { my ($c1, $c2, $diff) = @{ pop @conds }; my $p2 = $name{$c2} or die "bad prop $c2";
if ($c1 =~ /^\d+$/) { push @pre, "house[$c1].$p2 = $c2;"; next; }
my $p1 = $name{$c1} or die "bad prop $c1"; my $next = "work$idx"; my $this = "work".++$idx;
print " /* condition pair($c1, $c2, [@$diff]) */ int $this(void) { int a = find_by_$p1($c1); int b = find_by_$p2($c2); if (a != -1 && b != -1) { switch(b - a) { "; print "case $_: " for @$diff; print "return $next(); default: return 0; }\n } if (a != -1) {"; print "TRY(a, b, $c2, +($_), $p2, $next);" for @$diff; print " return 0; } if (b != -1) {"; print "TRY(b, a, $c1, -($_), $p1, $next);" for @$diff; print " return 0; } /* neither condition is set; try all possibles */ for (a = 0; a < N_ITEMS; a++) { if (house[a].$p1) continue; house[a].$p1 = $c1; ";
print "TRY(a, b, $c2, +($_), $p2, $next);" for @$diff; print " house[a].$p1 = 0; } return 0; }"; }
print "int main() { @pre return !work$idx(); }"; }
sub make_c { do_consts; make_conditions; }
- ---- above should be generic for all similar puzzles ---- #
- ---- below: per puzzle setup ---- #
- property names and values
setprops ( 'nationality' # Svensk n. a Swede, not a swede (kålrot). # AEnglisk (from middle Viking "Æŋløsåksen") n. a Brit. => [ qw(Deutsch Svensk Norske Danske AEnglisk) ], 'pet' => [ qw(birds dog horse zebra cats) ], 'drink' => [ qw(water tea milk beer coffee) ], 'smoke' => [ qw(dunhill blue_master prince blend pall_mall) ], 'color' => [ qw(red green yellow white blue) ] );
- constraints
pair(AEnglisk, red); pair(Svensk, dog); pair(Danske, tea); pair(green, white, 1); # "to the left of" can mean either 1 or -1: ambiguous pair(coffee, green); pair(pall_mall, birds); pair(yellow, dunhill); pair(2, milk); pair(0, Norske); pair(blend, cats, [-1, 1]); pair(horse, dunhill, [-1, 1]); pair(blue_master, beer); # Nicht das Deutsche Bier trinken? Huh. pair(Deutsch, prince); pair(Norske, blue, [-1, 1]); pair(water, blend, [-1, 1]);
- "zebra lives *somewhere* relative to the Brit". It has no effect on
- the logic. It's here just to make sure the code will insert a zebra
- somewhere in the table (after all other conditions are met) so the
- final print-out shows it. (the C code can be better structured, but
- meh, I ain't reading it, so who cares).
pair(zebra, AEnglisk, [ -4 .. 4 ]);
- write C code. If it's ugly to you: I didn't write; Perl did.
make_c;</lang>
output (ran as perl test.pl | gcc -Wall -x c -; ./a.out):
0 dunhill cats yellow water Norske 1 blend horse blue tea Danske 2 pall_mall birds red milk AEnglisk 3 prince zebra green coffee Deutsch 4 blue_master dog white beer Svensk
C#
"Manual" solution (Norvig-style)
This is adapted from a solution to a similar problem by Peter Norvig in his Udacity course CS212, originally written in Python. This is translated from example python solution on exercism. This is a Generate-and-Prune Constraint Programming algorithm written with Linq. (See Benchmarks below) <lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text; using static System.Console;
public enum Colour { Red, Green, White, Yellow, Blue } public enum Nationality { Englishman, Swede, Dane, Norwegian,German } public enum Pet { Dog, Birds, Cats, Horse, Zebra } public enum Drink { Coffee, Tea, Milk, Beer, Water } public enum Smoke { PallMall, Dunhill, Blend, BlueMaster, Prince}
public static class ZebraPuzzle {
private static (Colour[] colours, Drink[] drinks, Smoke[] smokes, Pet[] pets, Nationality[] nations) _solved;
static ZebraPuzzle()
{
var solve = from colours in Permute<Colour>() //r1 5 range
where (colours,Colour.White).IsRightOf(colours, Colour.Green) // r5
from nations in Permute<Nationality>()
where nations[0] == Nationality.Norwegian // r10
where (nations, Nationality.Englishman).IsSameIndex(colours, Colour.Red) //r2
where (nations,Nationality.Norwegian).IsNextTo(colours,Colour.Blue) // r15
from drinks in Permute<Drink>()
where drinks[2] == Drink.Milk //r9
where (drinks, Drink.Coffee).IsSameIndex(colours, Colour.Green) // r6
where (drinks, Drink.Tea).IsSameIndex(nations, Nationality.Dane) //r4
from pets in Permute<Pet>()
where (pets, Pet.Dog).IsSameIndex(nations, Nationality.Swede) // r3
from smokes in Permute<Smoke>()
where (smokes, Smoke.PallMall).IsSameIndex(pets, Pet.Birds) // r7
where (smokes, Smoke.Dunhill).IsSameIndex(colours, Colour.Yellow) // r8
where (smokes, Smoke.Blend).IsNextTo(pets, Pet.Cats) // r11
where (smokes, Smoke.Dunhill).IsNextTo(pets, Pet.Horse) //r12
where (smokes, Smoke.BlueMaster).IsSameIndex(drinks, Drink.Beer) //r13
where (smokes, Smoke.Prince).IsSameIndex(nations, Nationality.German) // r14
where (drinks,Drink.Water).IsNextTo(smokes,Smoke.Blend) // r16
select (colours, drinks, smokes, pets, nations);
_solved = solve.First(); } private static int IndexOf<T>(this T[] arr, T obj) => Array.IndexOf(arr, obj);
private static bool IsRightOf<T, U>(this (T[] a, T v) right, U[] a, U v) => right.a.IndexOf(right.v) == a.IndexOf(v) + 1;
private static bool IsSameIndex<T, U>(this (T[] a, T v)x, U[] a, U v) => x.a.IndexOf(x.v) == a.IndexOf(v);
private static bool IsNextTo<T, U>(this (T[] a, T v)x, U[] a, U v) => (x.a,x.v).IsRightOf(a, v) || (a,v).IsRightOf(x.a,x.v);
// made more generic from https://codereview.stackexchange.com/questions/91808/permutations-in-c public static IEnumerable<IEnumerable<T>> Permutations<T>(this IEnumerable<T> values) { if (values.Count() == 1) return values.ToSingleton();
return values.SelectMany(v => Permutations(values.Except(v.ToSingleton())),(v, p) => p.Prepend(v)); }
public static IEnumerable<T[]> Permute<T>() => ToEnumerable<T>().Permutations().Select(p=>p.ToArray());
private static IEnumerable<T> ToSingleton<T>(this T item){ yield return item; }
private static IEnumerable<T> ToEnumerable<T>() => Enum.GetValues(typeof(T)).Cast<T>();
public static new String ToString()
{
var sb = new StringBuilder();
sb.AppendLine("House Colour Drink Nationality Smokes Pet");
sb.AppendLine("───── ────── ──────── ─────────── ────────── ─────");
var (colours, drinks, smokes, pets, nations) = _solved;
for (var i = 0; i < 5; i++)
sb.AppendLine($"{i+1,5} {colours[i],-6} {drinks[i],-8} {nations[i],-11} {smokes[i],-10} {pets[i],-10}");
return sb.ToString();
}
public static void Main(string[] arguments)
{
var owner = _solved.nations[_solved.pets.IndexOf(Pet.Zebra)];
WriteLine($"The zebra owner is {owner}");
Write(ToString());
Read();
}
}</lang> Produces:
The zebra owner is German
House Colour Drink Nationality Smokes Pet
───── ────── ──────── ─────────── ────────── ─────
1 Yellow Water Norwegian Dunhill Cats
2 Blue Tea Dane Blend Horse
3 Red Milk Englishman PallMall Birds
4 Green Coffee German Prince Zebra
5 White Beer Swede BlueMaster Dog
"Manual" solution (Combining Houses)
This is similar to the Scala solution although there are differences in how the rules are calculated and it keeps all the original constraints/rules rather than does any simplification of them.
This is a different type of generate-and-prune compared to Norvig. The Norvig solution generates each attribute for 5 houses, then prunes and repeats with the next attribute. Here all houses with possible attributes are first generated and pruned to 78 candidates. The second phase proceeds over the combination of 5 houses from that 78, generating and pruning 1 house at a time. (See Benchmarks below) <lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text;
using static System.Console;
namespace ZebraPuzzleSolver {
public enum Colour { Red, Green, White, Yellow, Blue }
public enum Nationality { Englishman, Swede, Dane, Norwegian, German }
public enum Pet { Dog, Birds, Cats, Horse, Zebra }
public enum Drink { Coffee, Tea, Milk, Beer, Water }
public enum Smoke { PallMall, Dunhill, Blend, BlueMaster, Prince }
public struct House
{
public Drink D { get; }
public Colour C { get; }
public Pet P { get; }
public Nationality N { get; }
public Smoke S { get; }
House(Drink d, Colour c, Pet p, Nationality n, Smoke s) => (D, C, P, N, S) = (d, c, p, n, s);
public static House Create(Drink d, Colour c, Pet p, Nationality n, Smoke s) => new House(d, c, p, n, s);
public bool AllUnequal(House other) => D != other.D && C != other.C && P != other.P && N != other.N && S != other.S;
public override string ToString() =>$"{C,-6} {D,-8} {N,-11} {S,-10} {P,-10}";
}
public static class LinqNoPerm
{
public static IEnumerable<T> ToEnumerable<T>() => Enum.GetValues(typeof(T)).Cast<T>();
public static IEnumerable<House> FreeCandidates(this IEnumerable<House> houses, IEnumerable<House> picked) =>
houses.Where(house => picked.All(house.AllUnequal));
static Dictionary<Type, Func<House, dynamic, bool>> _eFn = new Dictionary<Type, Func<House, dynamic, bool>>
{ {typeof(Drink),(h,e)=>h.D==e},
{typeof(Nationality),(h,e)=>h.N==e},
{typeof(Colour),(h,e)=>h.C==e},
{typeof(Pet),(h,e)=>h.P==e},
{typeof(Smoke),(h, e)=>h.S==e}
};
public static bool IsNextTo<T, U>(this IEnumerable<House> hs,T t, U u) => hs.IsLeftOf(t,u) || hs.IsLeftOf(u, t);
public static bool IsLeftOf<T, U>(this IEnumerable<House> hs, T left, U right) =>
hs.Zip(hs.Skip(1), (l, r) => (_eFn[left.GetType()](l, left) && _eFn[right.GetType()](r, right))).Any(l => l);
static House[] _solved;
static LinqNoPerm()
{
var candidates =
from colours in ToEnumerable<Colour>()
from nations in ToEnumerable<Nationality>()
from drinks in ToEnumerable<Drink>()
from pets in ToEnumerable<Pet>()
from smokes in ToEnumerable<Smoke>()
where (colours == Colour.Red) == (nations == Nationality.Englishman) //r2
where (nations == Nationality.Swede) == (pets == Pet.Dog) //r3
where (nations == Nationality.Dane) == (drinks == Drink.Tea) //r4
where (colours == Colour.Green) == (drinks == Drink.Coffee) //r6
where (smokes == Smoke.PallMall) == (pets == Pet.Birds) //r7
where (smokes == Smoke.Dunhill) == (colours == Colour.Yellow) // r8
where (smokes == Smoke.BlueMaster) == (drinks == Drink.Beer) //r13
where (smokes == Smoke.Prince) == (nations == Nationality.German) // r14
select House.Create(drinks,colours,pets,nations, smokes);
var members =
from h1 in candidates
where h1.N == Nationality.Norwegian //r10
from h3 in candidates.FreeCandidates(new[] { h1 })
where h3.D == Drink.Milk //r9
from h2 in candidates.FreeCandidates(new[] { h1, h3 })
let h123 = new[] { h1, h2, h3 }
where h123.IsNextTo(Nationality.Norwegian, Colour.Blue) //r15
where h123.IsNextTo(Smoke.Blend, Pet.Cats)//r11
where h123.IsNextTo(Smoke.Dunhill, Pet.Horse) //r12
from h4 in candidates.FreeCandidates(h123)
from h5 in candidates.FreeCandidates(new[] { h1, h3, h2, h4 })
let houses = new[] { h1, h2, h3, h4, h5 }
where houses.IsLeftOf(Colour.Green, Colour.White) //r5
select houses;
_solved = members.First();
}
public static new String ToString()
{
var sb = new StringBuilder();
sb.AppendLine("House Colour Drink Nationality Smokes Pet");
sb.AppendLine("───── ────── ──────── ─────────── ────────── ─────");
for (var i = 0; i < 5; i++)
sb.AppendLine($"{i + 1,5} {_solved[i].ToString()}");
return sb.ToString();
}
public static void Main(string[] arguments)
{
var owner = _solved.Where(h=>h.P==Pet.Zebra).Single().N;
WriteLine($"The zebra owner is {owner}");
Write(ToString());
Read();
}
}
}</lang> Produces
The zebra owner is German
House Colour Drink Nationality Smokes Pet
───── ────── ──────── ─────────── ────────── ─────
1 Yellow Water Norwegian Dunhill Cats
2 Blue Tea Dane Blend Horse
3 Red Milk Englishman PallMall Birds
4 Green Coffee German Prince Zebra
5 White Beer Swede BlueMaster Dog
"Amb" solution
This uses the second version of the Amb C# class in the Amb challenge
<lang csharp>using Amb; using System; using System.Collections.Generic; using System.Linq; using static System.Console;
static class ZebraProgram {
public static void Main()
{
var amb = new Amb.Amb();
var domain = new[] { 1, 2, 3, 4, 5 };
var terms = new Dictionary<IValue<int>, string>();
IValue<int> Term(string name)
{
var x = amb.Choose(domain);
terms.Add(x, name);
return x;
};
void IsUnequal(params IValue<int>[] values) =>amb.Require(() => values.Select(v => v.Value).Distinct().Count() == 5);
void IsSame(IValue<int> left, IValue<int> right) => amb.Require(() => left.Value == right.Value);
void IsLeftOf(IValue<int> left, IValue<int> right) => amb.Require(() => right.Value - left.Value == 1);
void IsIn(IValue<int> attrib, int house) => amb.Require(() => attrib.Value == house);
void IsNextTo(IValue<int> left, IValue<int> right) => amb.Require(() => Math.Abs(left.Value - right.Value) == 1);
IValue<int> english = Term("Englishman"), swede = Term("Swede"), dane = Term("Dane"), norwegian = Term("Norwegian"), german = Term("German");
IsIn(norwegian, 1);
IsUnequal(english, swede, german, dane, norwegian);
IValue<int> red = Term("red"), green = Term("green"), white = Term("white"), blue = Term("blue"), yellow = Term("yellow");
IsUnequal(red, green, white, blue, yellow);
IsNextTo(norwegian, blue);
IsLeftOf(green, white);
IsSame(english, red);
IValue<int> tea = Term("tea"), coffee = Term("coffee"), milk = Term("milk"), beer = Term("beer"), water = Term("water");
IsIn(milk, 3);
IsUnequal(tea, coffee, milk, beer, water);
IsSame(dane, tea);
IsSame(green, coffee);
IValue<int> dog = Term("dog"), birds = Term("birds"), cats = Term("cats"), horse = Term("horse"), zebra = Term("zebra");
IsUnequal(dog, cats, birds, horse, zebra);
IsSame(swede, dog);
IValue<int> pallmall = Term("pallmall"), dunhill = Term("dunhill"), blend = Term("blend"), bluemaster = Term("bluemaster"),prince = Term("prince");
IsUnequal(pallmall, dunhill, bluemaster, prince, blend);
IsSame(pallmall, birds);
IsSame(dunhill, yellow);
IsNextTo(blend, cats);
IsNextTo(horse, dunhill);
IsSame(bluemaster, beer);
IsSame(german, prince);
IsNextTo(water, blend);
if (!amb.Disambiguate())
{
WriteLine("No solution found.");
Read();
return;
}
var h = new List<string>[5];
for (int i = 0; i < 5; i++)
h[i] = new List<string>();
foreach (var (key, value) in terms.Select(kvp => (kvp.Key, kvp.Value)))
{
h[key.Value - 1].Add(value);
}
var owner = String.Concat(h.Where(l => l.Contains("zebra")).Select(l => l[0]));
WriteLine($"The {owner} owns the zebra");
foreach (var house in h)
{
Write("|");
foreach (var attrib in house)
Write($"{attrib,-10}|");
Write("\n");
}
Read();
}
}</lang> Produces
The zebra owner is German
House Colour Drink Nationality Smokes Pet
───── ────── ──────── ─────────── ────────── ─────
1 Yellow Water Norwegian Dunhill Cats
2 Blue Tea Dane Blend Horse
3 Red Milk Englishman PallMall Birds
4 Green Coffee German Prince Zebra
5 White Beer Swede BlueMaster Dog
"Automatic" solution
<lang csharp>using System; using System.Collections.Generic; using System.Linq; using System.Text; using Microsoft.SolverFoundation.Solvers;
using static System.Console;
static class ZebraProgram {
static ConstraintSystem _solver;
static CspTerm IsLeftOf(this CspTerm left, CspTerm right) => _solver.Equal(1, right - left); static CspTerm IsInSameHouseAs(this CspTerm left, CspTerm right) => _solver.Equal(left, right); static CspTerm IsNextTo(this CspTerm left, CspTerm right) => _solver.Equal(1,_solver.Abs(left-right)); static CspTerm IsInHouse(this CspTerm @this, int i) => _solver.Equal(i, @this);
static (ConstraintSystem, Dictionary<CspTerm, string>) BuildSolver()
{
var solver = ConstraintSystem.CreateSolver();
_solver = solver;
var terms = new Dictionary<CspTerm, string>();
CspTerm Term(string name)
{
CspTerm x = solver.CreateVariable(solver.CreateIntegerInterval(1, 5), name);
terms.Add(x, name);
return x;
};
CspTerm red = Term("red"), green = Term("green"), white = Term("white"), blue = Term("blue"), yellow = Term("yellow");
CspTerm tea = Term("tea"), coffee = Term("coffee"), milk = Term("milk"), beer = Term("beer"), water = Term("water");
CspTerm english = Term("Englishman"), swede = Term("Swede"), dane = Term("Dane"), norwegian = Term("Norwegian"),
german = Term("German");
CspTerm dog = Term("dog"), birds = Term("birds"), cats = Term("cats"), horse = Term("horse"), zebra = Term("zebra");
CspTerm pallmall = Term("pallmall"), dunhill = Term("dunhill"), blend = Term("blend"), bluemaster = Term("bluemaster"),
prince = Term("prince");
solver.AddConstraints(
solver.Unequal(english, swede, german, dane, norwegian),
solver.Unequal(red, green, white, blue, yellow),
solver.Unequal(dog, cats, birds, horse, zebra),
solver.Unequal(pallmall, dunhill, bluemaster, prince, blend),
solver.Unequal(tea, coffee, milk, beer, water),
english.IsInSameHouseAs(red), //r2
swede.IsInSameHouseAs(dog), //r3
dane.IsInSameHouseAs(tea), //r4
green.IsLeftOf(white), //r5
green.IsInSameHouseAs(coffee), //r6
pallmall.IsInSameHouseAs(birds), //r7
dunhill.IsInSameHouseAs(yellow), //r8
milk.IsInHouse(3), //r9
norwegian.IsInHouse(1), //r10
blend.IsNextTo(cats), //r11
horse.IsNextTo(dunhill),// r12
bluemaster.IsInSameHouseAs(beer), // r13
german.IsInSameHouseAs(prince), // r14
norwegian.IsNextTo(blue), //r15
water.IsNextTo(blend) //r16
);
return (solver, terms);
}
static List<string>[] TermsToString(ConstraintSolverSolution solved, Dictionary<CspTerm, string> terms)
{
var h = new List<string>[5];
for (int i = 0; i < 5; i++)
h[i] = new List<string>();
foreach (var (key, value) in terms.Select(kvp => (kvp.Key, kvp.Value)))
{
if (!solved.TryGetValue(key, out object house))
throw new InvalidProgramException("Can't find a term - {value} - in the solution");
h[(int)house - 1].Add(value);
}
return h; }
static new string ToString(List<string>[] houses)
{
var sb = new StringBuilder();
foreach (var house in houses)
{
sb.Append("|");
foreach (var attrib in house)
sb.Append($"{attrib,-10}|");
sb.Append("\n");
}
return sb.ToString();
}
public static void Main()
{
var (solver, terms) = BuildSolver();
var solved = solver.Solve();
if (solved.HasFoundSolution)
{
var h = TermsToString(solved, terms);
var owner = String.Concat(h.Where(l => l.Contains("zebra")).Select(l => l[2]));
WriteLine($"The {owner} owns the zebra");
WriteLine();
Write(ToString(h));
}
else
WriteLine("No solution found.");
Read();
}
}</lang> Produces:
The German owns the zebra |yellow |water |Norwegian |cats |dunhill | |blue |tea |Dane |horse |blend | |red |milk |Englishman|birds |pallmall | |green |coffee |German |zebra |prince | |white |beer |Swede |dog |bluemaster|
Benchmarking the 3 solutions:
BenchmarkDotNet=v0.10.12, OS=Windows 10 Redstone 3 [1709, Fall Creators Update] (10.0.16299.192) Intel Core i7-7500U CPU 2.70GHz (Kaby Lake), 1 CPU, 4 logical cores and 2 physical cores Frequency=2835943 Hz, Resolution=352.6164 ns, Timer=TSC DefaultJob : .NET Framework 4.6.1 (CLR 4.0.30319.42000), 64bit RyuJIT-v4.7.2600.0 Method Mean Error StdDev Norvig 65.32 ms 1.241 ms 1.328 ms Combin 93.62 ms 1.792 ms 1.918 ms Solver 148.7 us 2.962 us 6.248 us
I think that it is Enums (not the use of dynamic in a dictionary, which is only called 8 times in Combine), and Linq query comprehensions (versus for loops) slow down the 2 non_solver solutions. A non-type-safe non-Enum int version of Combine (not posted here) runs at ~21ms, which is a nearly 5x speed up for that algo. (Not tried with Norvig). Regardless, learning and using the Solver class (and that solution already uses ints rather than enums) provides a dramatic x 100 + performance increase compared to the best manual solutions.
C++
This is a modification of the C submission that uses rule classes and reduces the number of permutations evaluated.
<lang cpp>
- include <stdio.h>
- include <string.h>
- define defenum(name, val0, val1, val2, val3, val4) \
enum name { val0, val1, val2, val3, val4 }; \
const char *name ## _str[] = { # val0, # val1, # val2, # val3, # val4 }
defenum( Attrib, Color, Man, Drink, Animal, Smoke ); defenum( Colors, Red, Green, White, Yellow, Blue ); defenum( Mans, English, Swede, Dane, German, Norwegian ); defenum( Drinks, Tea, Coffee, Milk, Beer, Water ); defenum( Animals, Dog, Birds, Cats, Horse, Zebra ); defenum( Smokes, PallMall, Dunhill, Blend, BlueMaster, Prince );
void printHouses(int ha[5][5]) {
const char **attr_names[5] = {Colors_str, Mans_str, Drinks_str, Animals_str, Smokes_str};
printf("%-10s", "House");
for (const char *name : Attrib_str) printf("%-10s", name);
printf("\n");
for (int i = 0; i < 5; i++) {
printf("%-10d", i);
for (int j = 0; j < 5; j++) printf("%-10s", attr_names[j][ha[i][j]]);
printf("\n");
}
}
struct HouseNoRule {
int houseno; Attrib a; int v;
} housenos[] = {
{2, Drink, Milk}, // Cond 9: In the middle house they drink milk.
{0, Man, Norwegian} // Cond 10: The Norwegian lives in the first house.
};
struct AttrPairRule {
Attrib a1; int v1; Attrib a2; int v2;
bool invalid(int ha[5][5], int i) {
return (ha[i][a1] >= 0 && ha[i][a2] >= 0) &&
((ha[i][a1] == v1 && ha[i][a2] != v2) ||
(ha[i][a1] != v1 && ha[i][a2] == v2));
}
} pairs[] = {
{Man, English, Color, Red}, // Cond 2: The English man lives in the red house.
{Man, Swede, Animal, Dog}, // Cond 3: The Swede has a dog.
{Man, Dane, Drink, Tea}, // Cond 4: The Dane drinks tea.
{Color, Green, Drink, Coffee}, // Cond 6: drink coffee in the green house.
{Smoke, PallMall, Animal, Birds}, // Cond 7: The man who smokes Pall Mall has birds.
{Smoke, Dunhill, Color, Yellow}, // Cond 8: In the yellow house they smoke Dunhill.
{Smoke, BlueMaster, Drink, Beer}, // Cond 13: The man who smokes Blue Master drinks beer.
{Man, German, Smoke, Prince} // Cond 14: The German smokes Prince
};
struct NextToRule {
Attrib a1; int v1; Attrib a2; int v2;
bool invalid(int ha[5][5], int i) {
return (ha[i][a1] == v1) &&
((i == 0 && ha[i + 1][a2] >= 0 && ha[i + 1][a2] != v2) ||
(i == 4 && ha[i - 1][a2] != v2) ||
(ha[i + 1][a2] >= 0 && ha[i + 1][a2] != v2 && ha[i - 1][a2] != v2));
}
} nexttos[] = {
{Smoke, Blend, Animal, Cats}, // Cond 11: The man who smokes Blend lives in the house next to the house with cats.
{Smoke, Dunhill, Animal, Horse}, // Cond 12: In a house next to the house where they have a horse, they smoke Dunhill.
{Man, Norwegian, Color, Blue}, // Cond 15: The Norwegian lives next to the blue house.
{Smoke, Blend, Drink, Water} // Cond 16: They drink water in a house next to the house where they smoke Blend.
};
struct LeftOfRule {
Attrib a1; int v1; Attrib a2; int v2;
bool invalid(int ha[5][5]) {
return (ha[0][a2] == v2) || (ha[4][a1] == v1);
}
bool invalid(int ha[5][5], int i) {
return ((i > 0 && ha[i][a1] >= 0) &&
((ha[i - 1][a1] == v1 && ha[i][a2] != v2) ||
(ha[i - 1][a1] != v1 && ha[i][a2] == v2)));
}
} leftofs[] = {
{Color, Green, Color, White} // Cond 5: The green house is immediately to the left of the white house.
};
bool invalid(int ha[5][5]) {
for (auto &rule : leftofs) if (rule.invalid(ha)) return true;
for (int i = 0; i < 5; i++) {
- define eval_rules(rules) for (auto &rule : rules) if (rule.invalid(ha, i)) return true;
eval_rules(pairs);
eval_rules(nexttos);
eval_rules(leftofs);
}
return false;
}
void search(bool used[5][5], int ha[5][5], const int hno, const int attr) {
int nexthno, nextattr;
if (attr < 4) {
nextattr = attr + 1;
nexthno = hno;
} else {
nextattr = 0;
nexthno = hno + 1;
}
if (ha[hno][attr] != -1) {
search(used, ha, nexthno, nextattr);
} else {
for (int i = 0; i < 5; i++) {
if (used[attr][i]) continue;
used[attr][i] = true;
ha[hno][attr] = i;
if (!invalid(ha)) {
if ((hno == 4) && (attr == 4)) {
printHouses(ha);
} else {
search(used, ha, nexthno, nextattr);
}
}
used[attr][i] = false;
}
ha[hno][attr] = -1;
}
}
int main() {
bool used[5][5] = {};
int ha[5][5]; memset(ha, -1, sizeof(ha));
for (auto &rule : housenos) {
ha[rule.houseno][rule.a] = rule.v;
used[rule.a][rule.v] = true;
}
search(used, ha, 0, 0);
return 0;
} </lang>
- Output:
$ g++ -O3 -std=c++11 zebra.cpp -o zebracpp && time ./zebracpp House Color Man Drink Animal Smoke 0 Yellow Norwegian Water Cats Dunhill 1 Blue Dane Tea Horse Blend 2 Red English Milk Birds PallMall 3 Green German Coffee Zebra Prince 4 White Swede Beer Dog BlueMaster real 0m0.003s user 0m0.000s sys 0m0.000s
My measured time is slower than that posted for the original C code, but on my machine this C++ code is faster than the original C code.
Clojure
This solution uses the contributed package clojure.core.logic (with clojure.tools.macro), a mini-Kanren based logic solver. The solution is basically the one in Swannodette's logic tutorial, adapted to the problem statement here. <lang clojure>(ns zebra.core
(:refer-clojure :exclude [==])
(:use [clojure.core.logic]
[clojure.tools.macro :as macro]))
(defne lefto [x y l]
([_ _ [x y . ?r]])
([_ _ [_ . ?r]] (lefto x y ?r)))
(defn nexto [x y l]
(conde ((lefto x y l)) ((lefto y x l))))
(defn zebrao [hs]
(macro/symbol-macrolet [_ (lvar)]
(all
(== [_ _ _ _ _] hs)
(membero ['englishman _ _ _ 'red] hs)
(membero ['swede _ _ 'dog _] hs)
(membero ['dane _ 'tea _ _] hs)
(lefto [_ _ _ _ 'green] [_ _ _ _ 'white] hs)
(membero [_ _ 'coffee _ 'green] hs)
(membero [_ 'pallmall _ 'birds _] hs)
(membero [_ 'dunhill _ _ 'yellow] hs)
(== [_ _ [_ _ 'milk _ _] _ _ ] hs)
(firsto hs ['norwegian _ _ _ _])
(nexto [_ 'blend _ _ _] [_ _ _ 'cats _ ] hs)
(nexto [_ _ _ 'horse _] [_ 'dunhill _ _ _] hs)
(membero [_ 'bluemaster 'beer _ _] hs)
(membero ['german 'prince _ _ _] hs)
(nexto ['norwegian _ _ _ _] [_ _ _ _ 'blue] hs)
(nexto [_ _ 'water _ _] [_ 'blend _ _ _] hs)
(membero [_ _ _ 'zebra _] hs))))
(let [solns (run* [q] (zebrao q))
soln (first solns)
zebra-owner (->> soln (filter #(= 'zebra (% 3))) first (#(% 0)))]
(println "solution count:" (count solns))
(println "zebra owner is the" zebra-owner)
(println "full solution (in house order):")
(doseq [h soln] (println " " h)))
</lang>
- Output:
solution count: 1 zebra owner is the german full solution (in house order): [norwegian dunhill water cats yellow] [dane blend tea horse blue] [englishman pallmall milk birds red] [german prince coffee zebra green] [swede bluemaster beer dog white]
Alternate solution (Norvig-style)
This is adapted from a solution to a similar problem by Peter Norvig in his Udacity course CS212, originally written in Python but equally applicable in any language with for-comprehensions.
<lang clojure>(ns zebra
(:require [clojure.math.combinatorics :as c]))
(defn solve []
(let [arrangements (c/permutations (range 5))
before? #(= (inc %1) %2)
after? #(= (dec %1) %2)
next-to? #(or (before? %1 %2) (after? %1 %2))]
(for [[english swede dane norwegian german :as persons] arrangements
:when (zero? norwegian)
[red green white yellow blue :as colors] arrangements
:when (before? green white)
:when (= english red)
:when (after? blue norwegian)
[tea coffee milk beer water :as drinks] arrangements
:when (= 2 milk)
:when (= dane tea)
:when (= coffee green)
[pall-mall dunhill blend blue-master prince :as cigs] arrangements
:when (= german prince)
:when (= yellow dunhill)
:when (= blue-master beer)
:when (after? blend water)
[dog birds cats horse zebra :as pets] arrangements
:when (= swede dog)
:when (= pall-mall birds)
:when (next-to? blend cats)
:when (after? horse dunhill)]
(->> [[:english :swede :dane :norwegian :german]
[:red :green :white :yellow :blue]
[:tea :coffee :milk :beer :water]
[:pall-mall :dunhill :blend :blue-master :prince]
[:dog :birds :cats :horse :zebra]]
(map zipmap [persons colors drinks cigs pets])))))
(defn -main [& _]
(doseq [[[persons _ _ _ pets :as solution] i]
(map vector (solve) (iterate inc 1))
:let [zebra-house (some #(when (= :zebra (val %)) (key %)) pets)]]
(println "solution" i)
(println "The" (persons zebra-house) "owns the zebra.")
(println "house nationality color drink cig pet")
(println "----- ----------- ------- ------- ------------ ------")
(dotimes [i 5]
(println (apply format "%5s %-11s %-7s %-7s %-12s %-6s"
(map #(% i) (cons inc solution)))))))
</lang>
- Output:
user=> (time (zebra/-main))
solution 1
The :german owns the zebra.
house nationality color drink cig pet
----- ----------- ------- ------- ------------ ------
1 :norwegian :yellow :water :dunhill :cats
2 :dane :blue :tea :blend :horse
3 :english :red :milk :pall-mall :birds
4 :german :green :coffee :prince :zebra
5 :swede :white :beer :blue-master :dog
"Elapsed time: 10.555482 msecs"
nil
Crystal
<lang ruby>CONTENT = {House: [""],
Nationality: %i[English Swedish Danish Norwegian German],
Colour: %i[Red Green White Blue Yellow],
Pet: %i[Dog Birds Cats Horse Zebra],
Drink: %i[Tea Coffee Milk Beer Water],
Smoke: %i[PallMall Dunhill BlueMaster Prince Blend]}
def adjacent?(n, i, g, e)
(0..3).any? { |x| (n[x] == i && g[x + 1] == e) || (n[x + 1] == i && g[x] == e) }
end
def leftof?(n, i, g, e)
(0..3).any? { |x| n[x] == i && g[x + 1] == e }
end
def coincident?(n, i, g, e)
n.each_index.any? { |x| n[x] == i && g[x] == e }
end
def solve_zebra_puzzle
CONTENT[:Nationality].each_permutation { |nation|
next unless nation.first == :Norwegian # 10
CONTENT[:Colour].each_permutation { |colour|
next unless leftof?(colour, :Green, colour, :White) # 5
next unless coincident?(nation, :English, colour, :Red) # 2
next unless adjacent?(nation, :Norwegian, colour, :Blue) # 15
CONTENT[:Pet].each_permutation { |pet|
next unless coincident?(nation, :Swedish, pet, :Dog) # 3
CONTENT[:Drink].each_permutation { |drink|
next unless drink[2] == :Milk # 9
next unless coincident?(nation, :Danish, drink, :Tea) # 4
next unless coincident?(colour, :Green, drink, :Coffee) # 6
CONTENT[:Smoke].each_permutation { |smoke|
next unless coincident?(smoke, :PallMall, pet, :Birds) # 7
next unless coincident?(smoke, :Dunhill, colour, :Yellow) # 8
next unless coincident?(smoke, :BlueMaster, drink, :Beer) # 13
next unless coincident?(smoke, :Prince, nation, :German) # 14
next unless adjacent?(smoke, :Blend, pet, :Cats) # 11
next unless adjacent?(smoke, :Blend, drink, :Water) # 16
next unless adjacent?(smoke, :Dunhill, pet, :Horse) # 12
print_out(nation, colour, pet, drink, smoke)
}
}
}
}
}
end
def print_out(nation, colour, pet, drink, smoke)
width = CONTENT.map { |k, v| {k.to_s.size, v.max_of { |y| y.to_s.size }}.max }
fmt = width.map { |w| "%-#{w}s" }.join(" ")
national = nation[pet.index(:Zebra).not_nil!]
puts "The Zebra is owned by the man who is #{national}", ""
puts fmt % CONTENT.keys, fmt % width.map { |w| "-" * w }
[nation, colour, pet, drink, smoke].transpose.each.with_index(1) { |x, n| puts fmt % ([n] + x) }
end
solve_zebra_puzzle</lang>
Curry
<lang curry>import Constraint (allC, anyC) import Findall (findall)
data House = H Color Man Pet Drink Smoke
data Color = Red | Green | Blue | Yellow | White data Man = Eng | Swe | Dan | Nor | Ger data Pet = Dog | Birds | Cats | Horse | Zebra data Drink = Coffee | Tea | Milk | Beer | Water data Smoke = PM | DH | Blend | BM | Prince
houses :: [House] -> Success
houses hs@[H1,_,H3,_,_] = -- 1
H _ _ _ Milk _ =:= H3 -- 9
& H _ Nor _ _ _ =:= H1 -- 10
& allC (`member` hs)
[ H Red Eng _ _ _ -- 2
, H _ Swe Dog _ _ -- 3
, H _ Dan _ Tea _ -- 4
, H Green _ _ Coffee _ -- 6
, H _ _ Birds _ PM -- 7
, H Yellow _ _ _ DH -- 8
, H _ _ _ Beer BM -- 13
, H _ Ger _ _ Prince -- 14
]
& H Green _ _ _ _ `leftTo` H White _ _ _ _ -- 5
& H _ _ _ _ Blend `nextTo` H _ _ Cats _ _ -- 11
& H _ _ Horse _ _ `nextTo` H _ _ _ _ DH -- 12
& H _ Nor _ _ _ `nextTo` H Blue _ _ _ _ -- 15
& H _ _ _ Water _ `nextTo` H _ _ _ _ Blend -- 16
where
x `leftTo` y = _ ++ [x,y] ++ _ =:= hs
x `nextTo` y = x `leftTo` y
? y `leftTo` x
member :: a -> [a] -> Success
member = anyC . (=:=)
main = findall $ \(hs,who) -> houses hs & H _ who Zebra _ _ `member` hs</lang>
- Output:
Using web interface.
Execution time: 180 msec. / elapsed: 180 msec. [([H Yellow Nor Cats Water DH,H Blue Dan Horse Tea Blend,H Red Eng Birds Milk PM,H Green Ger Zebra Coffee Prince,H White Swe Dog Beer BM],Ger)]
D
Most foreach loops in this program are static. <lang d>import std.stdio, std.traits, std.algorithm, std.math;
enum Content { Beer, Coffee, Milk, Tea, Water,
Danish, English, German, Norwegian, Swedish,
Blue, Green, Red, White, Yellow,
Blend, BlueMaster, Dunhill, PallMall, Prince,
Bird, Cat, Dog, Horse, Zebra }
enum Test { Drink, Person, Color, Smoke, Pet } enum House { One, Two, Three, Four, Five }
alias TM = Content[EnumMembers!Test.length][EnumMembers!House.length];
bool finalChecks(in ref TM M) pure nothrow @safe @nogc {
int diff(in Content a, in Content b, in Test ca, in Test cb)
nothrow @safe @nogc {
foreach (immutable h1; EnumMembers!House)
foreach (immutable h2; EnumMembers!House)
if (M[ca][h1] == a && M[cb][h2] == b)
return h1 - h2;
assert(0); // Useless but required.
}
with (Content) with (Test)
return abs(diff(Norwegian, Blue, Person, Color)) == 1 &&
diff(Green, White, Color, Color) == -1 &&
abs(diff(Horse, Dunhill, Pet, Smoke)) == 1 &&
abs(diff(Water, Blend, Drink, Smoke)) == 1 &&
abs(diff(Blend, Cat, Smoke, Pet)) == 1;
}
bool constrained(in ref TM M, in Test atest) pure nothrow @safe @nogc {
with (Content) with (Test) with (House)
final switch (atest) {
case Drink:
return M[Drink][Three] == Milk;
case Person:
foreach (immutable h; EnumMembers!House)
if ((M[Person][h] == Norwegian && h != One) ||
(M[Person][h] == Danish && M[Drink][h] != Tea))
return false;
return true;
case Color:
foreach (immutable h; EnumMembers!House)
if ((M[Person][h] == English && M[Color][h] != Red) ||
(M[Drink][h] == Coffee && M[Color][h] != Green))
return false;
return true;
case Smoke:
foreach (immutable h; EnumMembers!House)
if ((M[Color][h] == Yellow && M[Smoke][h] != Dunhill) ||
(M[Smoke][h] == BlueMaster && M[Drink][h] != Beer) ||
(M[Person][h] == German && M[Smoke][h] != Prince))
return false;
return true;
case Pet:
foreach (immutable h; EnumMembers!House)
if ((M[Person][h] == Swedish && M[Pet][h] != Dog) ||
(M[Smoke][h] == PallMall && M[Pet][h] != Bird))
return false;
return finalChecks(M);
}
}
void show(in ref TM M) {
foreach (h; EnumMembers!House) {
writef("%5s: ", h);
foreach (immutable t; EnumMembers!Test)
writef("%10s ", M[t][h]);
writeln;
}
}
void solve(ref TM M, in Test t, in size_t n) {
if (n == 1 && constrained(M, t)) {
if (t < 4) {
solve(M, [EnumMembers!Test][t + 1], 5);
} else {
show(M);
return;
}
}
foreach (immutable i; 0 .. n) {
solve(M, t, n - 1);
swap(M[t][n % 2 ? 0 : i], M[t][n - 1]);
}
}
void main() {
TM M;
foreach (immutable t; EnumMembers!Test)
foreach (immutable h; EnumMembers!House)
M[t][h] = EnumMembers!Content[t * 5 + h];
solve(M, Test.Drink, 5);
}</lang>
- Output:
One: Water Norwegian Yellow Dunhill Cat Two: Tea Danish Blue Blend Horse Three: Milk English Red PallMall Bird Four: Coffee German Green Prince Zebra Five: Beer Swedish White BlueMaster Dog
Run-time about 0.04 seconds.
Alternative Version
This requires the module of the first D entry from the Permutations Task. <lang d>import std.stdio, std.math, std.traits, std.typecons, std.typetuple, permutations1;
uint factorial(in uint n) pure nothrow @nogc @safe in {
assert(n <= 12);
} body {
uint result = 1;
foreach (immutable i; 1 .. n + 1)
result *= i;
return result;
}
enum Number { One, Two, Three, Four, Five } enum Color { Red, Green, Blue, White, Yellow } enum Drink { Milk, Coffee, Water, Beer, Tea } enum Smoke { PallMall, Dunhill, Blend, BlueMaster, Prince } enum Pet { Dog, Cat, Zebra, Horse, Bird } enum Nation { British, Swedish, Danish, Norvegian, German }
enum size_t M = EnumMembers!Number.length;
auto nullableRef(T)(ref T item) pure nothrow @nogc {
return NullableRef!T(&item);
}
bool isPossible(NullableRef!(immutable Number[M]) number,
NullableRef!(immutable Color[M]) color=null,
NullableRef!(immutable Drink[M]) drink=null,
NullableRef!(immutable Smoke[M]) smoke=null,
NullableRef!(immutable Pet[M]) pet=null) pure nothrow @safe @nogc {
if ((!number.isNull && number[Nation.Norvegian] != Number.One) ||
(!color.isNull && color[Nation.British] != Color.Red) ||
(!drink.isNull && drink[Nation.Danish] != Drink.Tea) ||
(!smoke.isNull && smoke[Nation.German] != Smoke.Prince) ||
(!pet.isNull && pet[Nation.Swedish] != Pet.Dog))
return false;
if (number.isNull || color.isNull || drink.isNull || smoke.isNull ||
pet.isNull)
return true;
foreach (immutable i; 0 .. M) {
if ((color[i] == Color.Green && drink[i] != Drink.Coffee) ||
(smoke[i] == Smoke.PallMall && pet[i] != Pet.Bird) ||
(color[i] == Color.Yellow && smoke[i] != Smoke.Dunhill) ||
(number[i] == Number.Three && drink[i] != Drink.Milk) ||
(smoke[i] == Smoke.BlueMaster && drink[i] != Drink.Beer)||
(color[i] == Color.Blue && number[i] != Number.Two))
return false;
foreach (immutable j; 0 .. M) {
if (color[i] == Color.Green && color[j] == Color.White &&
number[j] - number[i] != 1)
return false;
immutable diff = abs(number[i] - number[j]);
if ((smoke[i] == Smoke.Blend && pet[j] == Pet.Cat && diff != 1) ||
(pet[i] == Pet.Horse && smoke[j] == Smoke.Dunhill && diff != 1) ||
(smoke[i] == Smoke.Blend && drink[j] == Drink.Water && diff != 1))
return false;
}
}
return true;
}
alias N = nullableRef; // At module level scope to be used with UFCS.
void main() {
enum size_t FM = M.factorial;
static immutable Number[M][FM] numberPerms = [EnumMembers!Number].permutations; static immutable Color[M][FM] colorPerms = [EnumMembers!Color].permutations; static immutable Drink[M][FM] drinkPerms = [EnumMembers!Drink].permutations; static immutable Smoke[M][FM] smokePerms = [EnumMembers!Smoke].permutations; static immutable Pet[M][FM] petPerms = [EnumMembers!Pet].permutations;
// You can reduce the compile-time computations using four casts like this: // static colorPerms = cast(immutable Color[M][FM])numberPerms;
static immutable Nation[M] nation = [EnumMembers!Nation];
foreach (immutable ref number; numberPerms)
if (isPossible(number.N))
foreach (immutable ref color; colorPerms)
if (isPossible(number.N, color.N))
foreach (immutable ref drink; drinkPerms)
if (isPossible(number.N, color.N, drink.N))
foreach (immutable ref smoke; smokePerms)
if (isPossible(number.N, color.N, drink.N, smoke.N))
foreach (immutable ref pet; petPerms)
if (isPossible(number.N, color.N, drink.N, smoke.N, pet.N)) {
writeln("Found a solution:");
foreach (x; TypeTuple!(nation, number, color, drink, smoke, pet))
writefln("%6s: %12s%12s%12s%12s%12s",
(Unqual!(typeof(x[0]))).stringof,
x[0], x[1], x[2], x[3], x[4]);
writeln;
}
}</lang>
- Output:
Found a solution: Nation: British Swedish Danish Norvegian German Number: Three Five Two One Four Color: Red White Blue Yellow Green Drink: Milk Beer Tea Water Coffee Smoke: PallMall BlueMaster Blend Dunhill Prince Pet: Bird Dog Horse Cat Zebra
Run-time about 0.76 seconds with the dmd compiler.
Short Version
This requires the module of the second D entry from the Permutations Task. <lang d>void main() {
import std.stdio, std.algorithm, permutations2;
enum E { Red, Green, Blue, White, Yellow,
Milk, Coffee, Water, Beer, Tea,
PallMall, Dunhill, Blend, BlueMaster, Prince,
Dog, Cat, Zebra, Horse, Birds,
British, Swedish, Danish, Norvegian, German }
enum has = (E[] a, E x, E[] b, E y) => a.countUntil(x) == b.countUntil(y); enum leftOf = (E[] a, E x, E[] b, E y) => a.countUntil(x) == b.countUntil(y) + 1; enum nextTo = (E[] a, E x, E[] b, E y) => leftOf(a, x, b, y) || leftOf(b, y, a, x);
with (E) foreach (houses; [Red, Blue, Green, Yellow, White].permutations)
if (leftOf(houses, White, houses, Green))
foreach (persons; [Norvegian, British, Swedish, German, Danish].permutations)
if (has(persons, British, houses, Red) && persons[0] == Norvegian &&
nextTo(persons, Norvegian, houses, Blue))
foreach (drinks; [Tea, Coffee, Milk, Beer, Water].permutations)
if (has(drinks, Tea, persons, Danish) &&
has(drinks, Coffee, houses, Green) && drinks[$ / 2] == Milk)
foreach (pets; [Dog, Birds, Cat, Horse, Zebra].permutations)
if (has(pets, Dog, persons, Swedish))
foreach (smokes; [PallMall, Dunhill, Blend, BlueMaster, Prince].permutations)
if (has(smokes, PallMall, pets, Birds) &&
has(smokes, Dunhill, houses, Yellow) &&
nextTo(smokes, Blend, pets, Cat) &&
nextTo(smokes, Dunhill, pets, Horse) &&
has(smokes, BlueMaster, drinks, Beer) &&
has(smokes, Prince, persons, German) &&
nextTo(drinks, Water, smokes, Blend))
writefln("%(%10s\n%)\n", [houses, persons, drinks, pets, smokes]);
}</lang>
- Output:
[ Yellow, Blue, Red, Green, White] [ Norvegian, Danish, British, German, Swedish] [ Water, Tea, Milk, Coffee, Beer] [ Cat, Horse, Birds, Zebra, Dog] [ Dunhill, Blend, PallMall, Prince, BlueMaster]
The run-time is 0.03 seconds or less.
EchoLisp
We use the amb library to solve the puzzle. The number of tries - calls to zebra-puzzle - is only 1900, before finding all solutions. Note that there are no declarations for things (cats, tea, ..) or categories (animals, drinks, ..) which are discovered when reading the constraints. <lang scheme> (lib 'hash) (lib 'amb)
- return #f or house# for thing/category
- houses
- = (0 1 2 3 4)
(define (house-get H category thing houses)
(for/or ((i houses)) #:continue (!equal? (hash-ref (vector-ref H i) category) thing)
i))
;; return house # for thing (eg cat) in category (eq animals)
;; add thing if not already here
(define-syntax-rule (house-set thing category)
(or
(house-get H 'category 'thing houses)
(dispatch H 'category 'thing context houses )))
- we know that thing/category is in a given house
(define-syntax-rule (house-force thing category house)
(dispatch H 'category 'thing context houses house))
- return house# or fail if impossible
(define (dispatch H category thing context houses (forced #f))
(define house (or forced (amb context houses))) ;; get a house number
(when (hash-ref (vector-ref H house) category) (amb-fail)) ;; fail if occupied
(hash-set (vector-ref H house) category thing) ;; else remember house contents
house)
(define (house-next h1 h2)
(amb-require (or (= h1 (1+ h2)) (= h1 (1- h2)))))
(define (zebra-puzzle context houses )
(define H (build-vector 5 make-hash)) ;; house[i] := hash(category) -> thing
- In the middle house they drink milk.
(house-force milk drinks 2)
- The Norwegian lives in the first house.
(house-force norvegian people 0)
- The English man lives in the red house.
(house-force red colors(house-set english people))
- The Swede has a dog.
(house-force dog animals (house-set swede people))
- The Dane drinks tea.
(house-force tea drinks (house-set dane people))
- The green house is immediately to the left of the white house.
(amb-require (= (house-set green colors) (1- (house-set white colors))))
- They drink coffee in the green house.
(house-force coffee drinks (house-set green colors))
- The man who smokes Pall Mall has birds.
(house-force birds animals (house-set pallmall smoke))
- In the yellow house they smoke Dunhill.
(house-force dunhill smoke (house-set yellow colors))
- The Norwegian lives next to the blue house.
(house-next (house-set norvegian people) (house-set blue colors))
- The man who smokes Blend lives in the house next to the house with cats.
(house-next (house-set blend smoke) (house-set cats animals))
- In a house next to the house where they have a horse, they smoke Dunhill.
(house-next (house-set horse animals) (house-set dunhill smoke))
- The man who smokes Blue Master drinks beer.
(house-force beer drinks (house-set bluemaster smoke))
- The German smokes Prince.
(house-force prince smoke (house-set german people))
- They drink water in a house next to the house where they smoke Blend.
(house-next (house-set water drinks) (house-set blend smoke))
- Finally .... the zebra 🐴
(house-set 🐴 animals) (for ((i houses)) (writeln i (hash-values (vector-ref H i)))) (writeln '----------) (amb-fail) ;; will ensure ALL solutions are printed
) </lang>
- Output:
<lang scheme> (define (task)
(amb-run zebra-puzzle (amb-make-context) (iota 5)))
(task)
→
0 (norvegian yellow dunhill cats water) 1 (dane tea blue blend horse) 2 (milk english red pallmall birds) 3 (green coffee german prince 🐴) 4 (swede dog white bluemaster beer)
→ #f
</lang>
Elixir
<lang elixir>defmodule ZebraPuzzle do
defp adjacent?(n,i,g,e) do
Enum.any?(0..3, fn x ->
(Enum.at(n,x)==i and Enum.at(g,x+1)==e) or (Enum.at(n,x+1)==i and Enum.at(g,x)==e)
end)
end
defp leftof?(n,i,g,e) do
Enum.any?(0..3, fn x -> Enum.at(n,x)==i and Enum.at(g,x+1)==e end)
end
defp coincident?(n,i,g,e) do
Enum.with_index(n) |> Enum.any?(fn {x,idx} -> x==i and Enum.at(g,idx)==e end)
end
def solve(content) do
colours = permutation(content[:Colour])
pets = permutation(content[:Pet])
drinks = permutation(content[:Drink])
smokes = permutation(content[:Smoke])
Enum.each(permutation(content[:Nationality]), fn nation ->
if hd(nation) == :Norwegian, do: # 10
Enum.each(colours, fn colour ->
if leftof?(colour, :Green, colour, :White) and # 5
coincident?(nation, :English, colour, :Red) and # 2
adjacent?(nation, :Norwegian, colour, :Blue), do: # 15
Enum.each(pets, fn pet ->
if coincident?(nation, :Swedish, pet, :Dog), do: # 3
Enum.each(drinks, fn drink ->
if Enum.at(drink,2) == :Milk and # 9
coincident?(nation, :Danish, drink, :Tea) and # 4
coincident?(colour, :Green, drink, :Coffee), do: # 6
Enum.each(smokes, fn smoke ->
if coincident?(smoke, :PallMall, pet, :Birds) and # 7
coincident?(smoke, :Dunhill, colour, :Yellow) and # 8
coincident?(smoke, :BlueMaster, drink, :Beer) and # 13
coincident?(smoke, :Prince, nation, :German) and # 14
adjacent?(smoke, :Blend, pet, :Cats) and # 11
adjacent?(smoke, :Blend, drink, :Water) and # 16
adjacent?(smoke, :Dunhill, pet, :Horse), do: # 12
print_out(content, transpose([nation, colour, pet, drink, smoke]))
end)end)end)end)end)
end
defp permutation([]), do: [[]]
defp permutation(list) do
for x <- list, y <- permutation(list -- [x]), do: [x|y]
end
defp transpose(lists) do
List.zip(lists) |> Enum.map(&Tuple.to_list/1)
end
defp print_out(content, result) do
width = for {k,v}<-content, do: Enum.map([k|v], &length(to_char_list &1)) |> Enum.max
fmt = Enum.map_join(width, " ", fn w -> "~-#{w}s" end) <> "~n"
nation = Enum.find(result, fn x -> :Zebra in x end) |> hd
IO.puts "The Zebra is owned by the man who is #{nation}\n"
:io.format fmt, Keyword.keys(content)
:io.format fmt, Enum.map(width, fn w -> String.duplicate("-", w) end)
fmt2 = String.replace(fmt, "s", "w", global: false)
Enum.with_index(result)
|> Enum.each(fn {x,i} -> :io.format fmt2, [i+1 | x] end)
end
end
content = [ House: ,
Nationality: ~w[English Swedish Danish Norwegian German]a,
Colour: ~w[Red Green White Blue Yellow]a,
Pet: ~w[Dog Birds Cats Horse Zebra]a,
Drink: ~w[Tea Coffee Milk Beer Water]a,
Smoke: ~w[PallMall Dunhill BlueMaster Prince Blend]a ]
ZebraPuzzle.solve(content)</lang>
- Output:
The Zebra is owned by the man who is German House Nationality Colour Pet Drink Smoke ----- ----------- ------ ----- ------ ---------- 1 Norwegian Yellow Cats Water Dunhill 2 Danish Blue Horse Tea Blend 3 English Red Birds Milk PallMall 4 German Green Zebra Coffee Prince 5 Swedish White Dog Beer BlueMaster
Erlang
This solution generates all houses that fits the rules for single houses, then it checks multi-house rules. It would be faster to check multi-house rules while generating the houses. I have not added this complexity since the current program takes just a few seconds. <lang Erlang> -module( zebra_puzzle ).
-export( [task/0] ).
-record( house, {colour, drink, nationality, number, pet, smoke} ). -record( sorted_houses, {house_1s=[], house_2s=[], house_3s=[], house_4s=[], house_5s=[]} ).
task() -> Houses = [#house{colour=C, drink=D, nationality=N, number=Nr, pet=P, smoke=S} || C <- all_colours(), D <- all_drinks(), N <- all_nationalities(), Nr <- all_numbers(), P <- all_pets(), S <- all_smokes(), is_all_single_house_rules_ok(C, D, N, Nr, P, S)], Sorted_houses = lists:foldl( fun house_number_sort/2, #sorted_houses{}, Houses ), Streets = [[H1, H2, H3, H4, H5] || H1 <- Sorted_houses#sorted_houses.house_1s, H2 <- Sorted_houses#sorted_houses.house_2s, H3 <- Sorted_houses#sorted_houses.house_3s, H4 <- Sorted_houses#sorted_houses.house_4s, H5 <- Sorted_houses#sorted_houses.house_5s, is_all_multi_house_rules_ok(H1, H2, H3, H4, H5)], [Nationality] = [N || #house{nationality=N, pet=zebra} <- lists:flatten(Streets)], io:fwrite( "~p owns the zebra~n", [Nationality] ), io:fwrite( "All solutions ~p~n", [Streets] ), io:fwrite( "Number of solutions ~p~n", [erlang:length(Streets)] ).
all_colours() -> [blue, green, red, white, yellow].
all_drinks() -> [beer, coffe, milk, tea, water].
all_nationalities() -> [danish, english, german, norveigan, swedish].
all_numbers() -> [1, 2, 3, 4, 5].
all_pets() -> [birds, cats, dog, horse, zebra].
all_smokes() -> [blend, 'blue master', dunhill, 'pall mall', prince].
house_number_sort( #house{number=1}=House, #sorted_houses{house_1s=Houses_1s}=Sorted_houses ) -> Sorted_houses#sorted_houses{house_1s=[House | Houses_1s]}; house_number_sort( #house{number=2}=House, #sorted_houses{house_2s=Houses_2s}=Sorted_houses ) -> Sorted_houses#sorted_houses{house_2s=[House | Houses_2s]}; house_number_sort( #house{number=3}=House, #sorted_houses{house_3s=Houses_3s}=Sorted_houses ) -> Sorted_houses#sorted_houses{house_3s=[House | Houses_3s]}; house_number_sort( #house{number=4}=House, #sorted_houses{house_4s=Houses_4s}=Sorted_houses ) -> Sorted_houses#sorted_houses{house_4s=[House | Houses_4s]}; house_number_sort( #house{number=5}=House, #sorted_houses{house_5s=Houses_5s}=Sorted_houses ) -> Sorted_houses#sorted_houses{house_5s=[House | Houses_5s]}.
is_all_different( [_H] ) -> true; is_all_different( [H | T] ) -> not lists:member( H, T ) andalso is_all_different( T ).
is_all_multi_house_rules_ok( House1, House2, House3, House4, House5 ) -> is_rule_1_ok( House1, House2, House3, House4, House5 ) andalso is_rule_5_ok( House1, House2, House3, House4, House5 ) andalso is_rule_11_ok( House1, House2, House3, House4, House5 ) andalso is_rule_12_ok( House1, House2, House3, House4, House5 ) andalso is_rule_15_ok( House1, House2, House3, House4, House5 ) andalso is_rule_16_ok( House1, House2, House3, House4, House5 ).
is_all_single_house_rules_ok( Colour, Drink, Nationality, Number, Pet, Smoke ) -> is_rule_ok( {rule_number, 2}, {Nationality, english}, {Colour, red}) andalso is_rule_ok( {rule_number, 3}, {Nationality, swedish}, {Pet, dog}) andalso is_rule_ok( {rule_number, 4}, {Nationality, danish}, {Drink, tea}) andalso is_rule_ok( {rule_number, 6}, {Drink, coffe}, {Colour, green}) andalso is_rule_ok( {rule_number, 7}, {Smoke, 'pall mall'}, {Pet, birds}) andalso is_rule_ok( {rule_number, 8}, {Colour, yellow}, {Smoke, dunhill}) andalso is_rule_ok( {rule_number, 9}, {Number, 3}, {Drink, milk}) andalso is_rule_ok( {rule_number, 10}, {Nationality, norveigan}, {Number, 1}) andalso is_rule_ok( {rule_number, 13}, {Smoke, 'blue master'}, {Drink, beer}) andalso is_rule_ok( {rule_number, 14}, {Nationality, german}, {Smoke, prince}).
is_rule_ok( _Rule_number, {A, A}, {B, B} ) -> true; is_rule_ok( _Rule_number, _A, {B, B} ) -> false; is_rule_ok( _Rule_number, {A, A}, _B ) -> false; is_rule_ok( _Rule_number, _A, _B ) -> true.
is_rule_1_ok( #house{number=1}=H1, #house{number=2}=H2, #house{number=3}=H3, #house{number=4}=H4, #house{number=5}=H5 ) -> is_all_different( [H1#house.colour, H2#house.colour, H3#house.colour, H4#house.colour, H5#house.colour] ) andalso is_all_different( [H1#house.drink, H2#house.drink, H3#house.drink, H4#house.drink, H5#house.drink] ) andalso is_all_different( [H1#house.nationality, H2#house.nationality, H3#house.nationality, H4#house.nationality, H5#house.nationality] ) andalso is_all_different( [H1#house.pet, H2#house.pet, H3#house.pet, H4#house.pet, H5#house.pet] ) andalso is_all_different( [H1#house.smoke, H2#house.smoke, H3#house.smoke, H4#house.smoke, H5#house.smoke] ); is_rule_1_ok( _House1, _House2, _House3, _House4, _House5 ) -> false.
is_rule_5_ok( #house{colour=green}, #house{colour=white}, _House3, _House4, _House5 ) -> true; is_rule_5_ok( _House1, #house{colour=green}, #house{colour=white}, _House4, _House5 ) -> true; is_rule_5_ok( _House1, _House2, #house{colour=green}, #house{colour=white}, _House5 ) -> true; is_rule_5_ok( _House1, _House2, _House3, #house{colour=green}, #house{colour=white} ) -> true; is_rule_5_ok( _House1, _House2, _House3, _House4, _House5 ) -> false.
is_rule_11_ok( #house{smoke=blend}, #house{pet=cats}, _House3, _House4, _House5 ) -> true; is_rule_11_ok( _House1, #house{smoke=blend}, #house{pet=cats}, _House4, _House5 ) -> true; is_rule_11_ok( _House1, _House2, #house{smoke=blend}, #house{pet=cats}, _House5 ) -> true; is_rule_11_ok( _House1, _House2, _House3, #house{smoke=blend}, #house{pet=cats} ) -> true; is_rule_11_ok( #house{pet=cats}, #house{smoke=blend}, _House3, _House4, _House5 ) -> true; is_rule_11_ok( _House1, #house{pet=cats}, #house{smoke=blend}, _House4, _House5 ) -> true; is_rule_11_ok( _House1, _House2, #house{pet=cats}, #house{smoke=blend}, _House5 ) -> true; is_rule_11_ok( _House1, _House2, _House3, #house{pet=cats}, #house{smoke=blend} ) -> true; is_rule_11_ok( _House1, _House2, _House3, _House4, _House5 ) -> false.
is_rule_12_ok( #house{smoke=dunhill}, #house{pet=horse}, _House3, _House4, _House5 ) -> true; is_rule_12_ok( _House1, #house{smoke=dunhill}, #house{pet=horse}, _House4, _House5 ) -> true; is_rule_12_ok( _House1, _House2, #house{smoke=dunhill}, #house{pet=horse}, _House5 ) -> true; is_rule_12_ok( _House1, _House2, _House3, #house{smoke=dunhill}, #house{pet=horse} ) -> true; is_rule_12_ok( #house{pet=horse}, #house{smoke=dunhill}, _House3, _House4, _House5 ) -> true; is_rule_12_ok( _House1, #house{pet=horse}, #house{smoke=dunhill}, _House4, _House5 ) -> true; is_rule_12_ok( _House1, _House2, #house{pet=horse}, #house{smoke=dunhill}, _House5 ) -> true; is_rule_12_ok( _House1, _House2, _House3, #house{pet=horse}, #house{smoke=dunhill} ) -> true; is_rule_12_ok( _House1, _House2, _House3, _House4, _House5 ) -> false.
is_rule_15_ok( #house{nationality=norveigan}, #house{colour=blue}, _House3, _House4, _House5 ) -> true; is_rule_15_ok( _House1, #house{nationality=norveigan}, #house{colour=blue}, _House4, _House5 ) -> true; is_rule_15_ok( _House1, _House2, #house{nationality=norveigan}, #house{colour=blue}, _House5 ) -> true; is_rule_15_ok( _House1, _House2, _House3, #house{nationality=norveigan}, #house{colour=blue} ) -> true; is_rule_15_ok( #house{colour=blue}, #house{nationality=norveigan}, _House3, _House4, _House5 ) -> true; is_rule_15_ok( _House1, #house{colour=blue}, #house{nationality=norveigan}, _House4, _House5 ) -> true; is_rule_15_ok( _House1, _House2, #house{drink=water}, #house{nationality=norveigan}, _House5 ) -> true; is_rule_15_ok( _House1, _House2, _House3, #house{drink=water}, #house{nationality=norveigan} ) -> true; is_rule_15_ok( _House1, _House2, _House3, _House4, _House5 ) -> false.
is_rule_16_ok( #house{smoke=blend}, #house{drink=water}, _House3, _House4, _House5 ) -> true; is_rule_16_ok( _House1, #house{smoke=blend}, #house{drink=water}, _House4, _House5 ) -> true; is_rule_16_ok( _House1, _House2, #house{smoke=blend}, #house{drink=water}, _House5 ) -> true; is_rule_16_ok( _House1, _House2, _House3, #house{smoke=blend}, #house{drink=water} ) -> true; is_rule_16_ok( #house{drink=water}, #house{smoke=blend}, _House3, _House4, _House5 ) -> true; is_rule_16_ok( _House1, #house{drink=water}, #house{smoke=blend}, _House4, _House5 ) -> true; is_rule_16_ok( _House1, _House2, #house{drink=water}, #house{smoke=blend}, _House5 ) -> true; is_rule_16_ok( _House1, _House2, _House3, #house{drink=water}, #house{smoke=blend} ) -> true; is_rule_16_ok( _House1, _House2, _House3, _House4, _House5 ) -> false. </lang>
- Output:
10> zebra_puzzle:task().
german owns the zebra
All solutions [[{house,yellow,water,norveigan,1,cats,dunhill},
{house,blue,tea,danish,2,horse,blend},
{house,red,milk,english,3,birds,'pall mall'},
{house,green,coffe,german,4,zebra,prince},
{house,white,beer,swedish,5,dog,'blue master'}]]
Number of solutions: 1
ERRE
<lang ERRE> PROGRAM ZEBRA_PUZZLE
DIM DRINK$[4],NATION$[4],COLR$[4],SMOKE$[4],ANIMAL$[4] DIM PERM$[120],X$[4]
PROCEDURE PERMUTATION(X$[]->X$[],OK)
LOCAL I%,J%
FOR I%=UBOUND(X$,1)-1 TO 0 STEP -1 DO
EXIT IF X$[I%]<X$[I%+1]
END FOR
IF I%<0 THEN OK=FALSE EXIT PROCEDURE END IF
J%=UBOUND(X$,1)
WHILE X$[J%]<=X$[I%] DO
J%=J%-1
END WHILE
SWAP(X$[I%],X$[J%])
I%=I%+1
J%=UBOUND(X$,1)
WHILE I%<J% DO
SWAP(X$[I%],X$[J%])
I%=I%+1
J%=J%-1
END WHILE
OK=TRUE
END PROCEDURE
BEGIN
! The names (only used for printing the results)
DATA("Beer","Coffee","Milk","Tea","Water")
DATA("Denmark","England","Germany","Norway","Sweden")
DATA("Blue","Green","Red","White","Yellow")
DATA("Blend","BlueMaster","Dunhill","PallMall","Prince")
DATA("Birds","Cats","Dog","Horse","Zebra")
FOR I%=0 TO 4 DO READ(DRINK$[I%]) END FOR FOR I%=0 TO 4 DO READ(NATION$[I%]) END FOR FOR I%=0 TO 4 DO READ(COLR$[I%]) END FOR FOR I%=0 TO 4 DO READ(SMOKE$[I%]) END FOR FOR I%=0 TO 4 DO READ(ANIMAL$[I%]) END FOR
! Some single-character tags:
A$="A" B$="B" c$="C" d$="D" e$="E"
! ERRE doesn't have enumerations!
Beer$=A$ Coffee$=B$ Milk$=c$ TeA$=d$ Water$=e$ Denmark$=A$ England$=B$ Germany$=c$ Norway$=d$ Sweden$=e$ Blue$=A$ Green$=B$ Red$=c$ White$=d$ Yellow$=e$ Blend$=A$ BlueMaster$=B$ Dunhill$=c$ PallMall$=d$ Prince$=e$ Birds$=A$ Cats$=B$ Dog$=c$ Horse$=d$ ZebrA$=e$
PRINT(CHR$(12);)
! Create the 120 permutations of 5 objects:
X$[0]=A$ X$[1]=B$ X$[2]=C$ X$[3]=D$ X$[4]=E$
REPEAT
P%=P%+1
PERM$[P%]=X$[0]+X$[1]+X$[2]+X$[3]+X$[4]
PERMUTATION(X$[]->X$[],OK)
UNTIL NOT OK
! Solve:
SOLUTIONS%=0
T1=TIMER
FOR NATION%=1 TO 120 DO
NATION$=PERM$[NATION%]
IF LEFT$(NATION$,1)=Norway$ THEN
FOR COLR%=1 TO 120 DO
COLR$=PERM$[COLR%]
IF INSTR(COLR$,Green$+White$)<>0 AND INSTR(NATION$,England$)=INSTR(COLR$,Red$) AND ABS(INSTR(NATION$,Norway$)-INSTR(COLR$,Blue$))=1 THEN
FOR DRINK%=1 TO 120 DO
DRINK$=PERM$[DRINK%]
IF MID$(DRINK$,3,1)=Milk$ AND INSTR(NATION$,Denmark$)=INSTR(DRINK$,TeA$) AND INSTR(DRINK$,Coffee$)=INSTR(COLR$,Green$) THEN
FOR SmOKe%=1 TO 120 DO
SmOKe$=PERM$[SMOKE%]
IF INSTR(NATION$,Germany$)=INSTR(SmOKe$,Prince$) AND INSTR(SmOKe$,BlueMaster$)=INSTR(DRINK$,Beer$) AND ABS(INSTR(SmOKe$,Blend$)-INSTR(DRINK$,Water$))=1 AND INSTR(SmOKe$,Dunhill$)=INSTR(COLR$,Yellow$) THEN
FOR ANIMAL%=1 TO 120 DO
ANIMAL$=PERM$[ANIMAL%]
IF INSTR(NATION$,Sweden$)=INSTR(ANIMAL$,Dog$) AND INSTR(SmOKe$,PallMall$)=INSTR(ANIMAL$,Birds$) AND ABS(INSTR(SmOKe$,Blend$)-INSTR(ANIMAL$,Cats$))=1 AND ABS(INSTR(SmOKe$,Dunhill$)-INSTR(ANIMAL$,Horse$))=1 THEN
PRINT("House Drink Nation Colour Smoke Animal")
PRINT("---------------------------------------------------------------------------")
FOR house%=1 TO 5 DO
PRINT(house%;)
PRINT(TAB(10);DRINK$[ASC(MID$(DRINK$,house%))-65];)
PRINT(TAB(25);NATION$[ASC(MID$(NATION$,house%))-65];)
PRINT(TAB(40);COLR$[ASC(MID$(COLR$,house%))-65];)
PRINT(TAB(55);SMOKE$[ASC(MID$(SmOKe$,house%))-65];)
PRINT(TAB(70);ANIMAL$[ASC(MID$(ANIMAL$,house%))-65])
END FOR
SOLUTIONS%=SOLUTIONS%+1
END IF
END FOR ! ANIMAL%
END IF
END FOR ! SmOKe%
END IF
END FOR ! DRINK%
END IF
END FOR ! COLR%
END IF
END FOR ! NATION%
PRINT("Number of solutions=";SOLUTIONS%)
PRINT("Solved in ";TIMER-T1;" seconds")
END PROGRAM</lang>
- Output:
House Drink Nation Colour Smoke Animal --------------------------------------------------------------------------- 1 Water Norway Yellow Dunhill Cats 2 Tea Denmark Blue Blend Horse 3 Milk England Red PallMall Birds 4 Coffee Germany Green Prince Zebra 5 Beer Sweden White BlueMaster Dog Number of solutions= 1 Solved in .109375 seconds
F#
This task uses Permutations_by_swapping#F.23 <lang fsharp> (*Here I solve the Zebra puzzle using Plain Changes, definitely a challenge to some campanoligist to solve it using Grandsire Doubles.
Nigel Galloway: January 27th., 2017 *)
type N = |English=0 |Swedish=1|Danish=2 |German=3|Norwegian=4 type I = |Tea=0 |Coffee=1 |Milk=2 |Beer=3 |Water=4 type G = |Dog=0 |Birds=1 |Cats=2 |Horse=3 |Zebra=4 type E = |Red=0 |Green=1 |White=2 |Blue=3 |Yellow=4 type L = |PallMall=0|Dunhill=1|BlueMaster=2|Prince=3|Blend=4 type NIGELz={Nz:N[];Iz:I[];Gz:G[];Ez:E[];Lz:L[]} let fn (i:'n[]) g (e:'g[]) l = //coincident?
let rec _fn = function |5 -> false |ig when (i.[ig]=g && e.[ig]=l) -> true |ig -> _fn (ig+1) _fn 0
let fi (i:'n[]) g (e:'g[]) l = //leftof?
let rec _fn = function |4 -> false |ig when (i.[ig]=g && e.[ig+1]=l) -> true |ig -> _fn (ig+1) _fn 0
let fg (i:'n[]) g (e:'g[]) l = (fi i g e l || fi e l i g) //adjacent? let n = Ring.PlainChanges [|for n in System.Enum.GetValues(typeof<N>)->n:?>N|]|>Seq.filter(fun n->n.[0]=N.Norwegian) //#10 let i = Ring.PlainChanges [|for n in System.Enum.GetValues(typeof)->n:?>I|]|>Seq.filter(fun n->n.[2]=I.Milk) //# 9 let g = Ring.PlainChanges [|for n in System.Enum.GetValues(typeof<G>)->n:?>G|] let e = Ring.PlainChanges [|for n in System.Enum.GetValues(typeof<E>)->n:?>E|]|>Seq.filter(fun n->fi n E.Green n E.White) //# 5 let l = Ring.PlainChanges [|for n in System.Enum.GetValues(typeof<L>)->n:?>L|] match n|>Seq.map(fun n->{Nz=n;Iz=[||];Gz=[||];Ez=[||];Lz=[||]})
|>Seq.collect(fun n->i|>Seq.map(fun i->{n with Iz=i}))|>Seq.filter(fun n-> fn n.Nz N.Danish n.Iz I.Tea) //# 4
|>Seq.collect(fun n->g|>Seq.map(fun i->{n with Gz=i}))|>Seq.filter(fun n-> fn n.Nz N.Swedish n.Gz G.Dog) //# 3
|>Seq.collect(fun n->e|>Seq.map(fun i->{n with Ez=i}))|>Seq.filter(fun n-> fn n.Nz N.English n.Ez E.Red && //# 2
fn n.Ez E.Green n.Iz I.Coffee&& //# 6
fg n.Nz N.Norwegian n.Ez E.Blue) //#15
|>Seq.collect(fun n->l|>Seq.map(fun i->{n with Lz=i}))|>Seq.tryFind(fun n->fn n.Lz L.PallMall n.Gz G.Birds && //# 7
fg n.Lz L.Blend n.Gz G.Cats && //#11
fn n.Lz L.Prince n.Nz N.German&& //#14
fg n.Lz L.Blend n.Iz I.Water && //#16
fg n.Lz L.Dunhill n.Gz G.Horse && //#12
fn n.Lz L.Dunhill n.Ez E.Yellow&& //# 8
fn n.Iz I.Beer n.Lz L.BlueMaster) with //#13
|Some(nn) -> nn.Gz |> Array.iteri(fun n g -> if (g = G.Zebra) then printfn "\nThe man who owns a zebra is %A\n" nn.Nz.[n]); printfn "%A" nn |None -> printfn "No solution found" </lang>
- Output:
The man who owns a zebra is German
{Nz = [|Norwegian; Danish; English; German; Swedish|];
Iz = [|Water; Tea; Milk; Coffee; Beer|];
Gz = [|Cats; Horse; Birds; Zebra; Dog|];
Ez = [|Yellow; Blue; Red; Green; White|];
Lz = [|Dunhill; Blend; PallMall; Prince; BlueMaster|];}
FormulaOne
<lang FormulaOne> // First, let's give some variables some values: Nationality = Englishman | Swede | Dane | Norwegian | German Colour = Red | Green | Yellow | Blue | White Cigarette = PallMall | Dunhill | BlueMaster | Blend | Prince Domestic = Dog | Bird | Cat | Zebra | Horse Beverage = Tea | Coffee | Milk | Beer | Water HouseOrder = First | Second | Third | Fourth | Fifth
{ We use injections to make the array-elements unique. Example: 'Pet' is an array of unique elements of type 'Domestic', indexed by 'Nationality'. In the predicate 'Zebra', we use this injection 'Pet' to define the array-variable 'pet'. The symbol used is the '->>'. 'Nationality->>Domestic' can be read as 'Domestic(Nationality)' in "plain array-speak"; the difference being that the elements are by definition unique.
So, in FormulaOne we use a formula like: 'pet(Swede) = Dog', which simply means that the 'Swede' (type 'Nationality') has a 'pet' (type 'Pet', which is of type 'Domestic', indexed by 'Nationality'), which appears to be a 'Dog' (type 'Domestic'). Or, one could say that the 'Swede' has been mapped to the 'Dog' (Oh, well...). }
Pet = Nationality->>Domestic Drink = Nationality->>Beverage HouseColour = Nationality->>Colour Smoke = Nationality->>Cigarette Order = HouseOrder->>Nationality
pred Zebra(houseColour::HouseColour, pet::Pet, smoke::Smoke, drink::Drink, order::Order) iff
// For convenience sake, some temporary place_holder variables are used. // An underscore distinguishes them:
houseColour(green_house) = Green &
houseColour(white_house) = White &
houseColour(yellow_house) = Yellow &
smoke(pallmall_smoker) = PallMall &
smoke(blend_smoker) = Blend &
smoke(dunhill_smoker) = Dunhill &
smoke(bluemaster_smoker) = BlueMaster &
pet(cat_keeper) = Cat &
pet(neighbour_dunhill_smoker) = Horse &
{ 2. The English man lives in the red house: }
houseColour(Englishman) = Red &
{ 3. The Swede has a dog: }
pet(Swede) = Dog &
{ 4. The Dane drinks tea: }
drink(Dane) = Tea &
{ 'smoke' and 'drink' are both nouns, like the other variables.
One could read the formulas like: 'the colour of the Englishman's house is Red' ->
'the Swede's pet is a dog' -> 'the Dane's drink is tea'.
}
{ 5. The green house is immediately to the left of the white house: }
{ The local predicate 'LeftOf' determines the order: }
LeftOf(green_house, white_house, order) &
{ 6. They drink coffee in the green house: }
drink(green_house) = Coffee &
{ 7. The man who smokes Pall Mall has birds: }
pet(pallmall_smoker) = Bird &
{ 8. In the yellow house they smoke Dunhill: }
smoke(yellow_house) = Dunhill &
{ 9. In the middle house they drink milk: }
drink(order(Third)) = Milk &
{10. The Norwegian lives in the first house: }
order(First) = Norwegian &
{11. The man who smokes Blend lives in the house next to the house with cats: }
{ Another local predicate 'Neighbour' makes them neighbours:}
Neighbour(blend_smoker, cat_keeper, order) &
{12. In a house next to the house where they have a horse, they smoke Dunhill: }
Neighbour(dunhill_smoker, neighbour_dunhill_smoker, order) &
{13. The man who smokes Blue Master drinks beer: }
drink(bluemaster_smoker) = Beer &
{14. The German smokes Prince: }
smoke(German) = Prince &
{15. The Norwegian lives next to the blue house: }
{10. The Norwegian lives in the first house,
so the blue house is the second house }
houseColour(order(Second)) = Blue &
{16. They drink water in a house next to the house where they smoke Blend: }
drink(neighbour_blend_smoker) = Water &
Neighbour(blend_smoker, neighbour_blend_smoker, order)
{ A simplified solution would number the houses 1, 2, 3, 4, 5
which makes it easier to order the houses. 'right in the center' would become 3; 'in the first house', 1 But we stick to the original puzzle and use some local predicates.
}
local pred Neighbour(neighbour1::Nationality, neighbour2::Nationality, order::Order)iff
neighbour1 <> neighbour2 &
order(house1) = neighbour1 &
order(house2) = neighbour2 &
( house1 = house2 + 1 |
house1 = house2 - 1 )
local pred LeftOf(neighbour1::Nationality, neighbour2::Nationality, order::Order) iff
neighbour1 <> neighbour2 & order(house1) = neighbour1 & order(house2) = neighbour2 & house1 = house2 - 1
{ The 'all'-query in FormulaOne:
all Zebra(houseColour, pet, smokes, drinks, order)
gives, of course, only one solution, so it can be replaced by:
one Zebra(houseColour, pet, smokes, drinks, order)
}
// The compacted version:
Nationality = Englishman | Swede | Dane | Norwegian | German Colour = Red | Green | Yellow | Blue | White Cigarette = PallMall | Dunhill | BlueMaster | Blend | Prince Domestic = Dog | Bird | Cat | Zebra | Horse Beverage = Tea | Coffee | Milk | Beer | Water HouseOrder = First | Second | Third | Fourth | Fifth
Pet = Nationality->>Domestic Drink = Nationality->>Beverage HouseColour = Nationality->>Colour Smoke = Nationality->>Cigarette Order = HouseOrder->>Nationality
pred Zebra(houseColour::HouseColour, pet::Pet, smoke::Smoke, drink::Drink, order::Order) iff
houseColour(green_house) = Green & houseColour(white_house) = White & houseColour(yellow_house) = Yellow & smoke(pallmall_smoker) = PallMall & smoke(blend_smoker) = Blend & smoke(dunhill_smoker) = Dunhill & smoke(bluemaster_smoker) = BlueMaster & pet(cat_keeper) = Cat & pet(neighbour_dunhill_smoker) = Horse &
houseColour(Englishman) = Red & pet(Swede) = Dog & drink(Dane) = Tea & LeftOf(green_house, white_house, order) & drink(green_house) = Coffee & pet(pallmall_smoker) = Bird & smoke(yellow_house) = Dunhill & drink(order(Third)) = Milk & order(First) = Norwegian & Neighbour(blend_smoker, cat_keeper, order) & Neighbour(dunhill_smoker, neighbour_dunhill_smoker, order) & drink(bluemaster_smoker) = Beer & smoke(German) = Prince & houseColour(order(Second)) = Blue & drink(neighbour_blend_smoker) = Water & Neighbour(blend_smoker, neighbour_blend_smoker, order)
local pred Neighbour(neighbour1::Nationality, neighbour2::Nationality, order::Order)iff
neighbour1 <> neighbour2 & order(house1) = neighbour1 & order(house2) = neighbour2 & ( house1 = house2 + 1 | house1 = house2 - 1 )
local pred LeftOf(neighbour1::Nationality, neighbour2::Nationality, order::Order) iff
neighbour1 <> neighbour2 & order(house1) = neighbour1 & order(house2) = neighbour2 & house1 = house2 - 1
</lang>
- Output:
houseColour = [ {Englishman} Red, {Swede} White, {Dane} Blue, {Norwegian} Yellow, {German} Green ]
pet = [ {Englishman} Birds, {Swede} Dog, {Dane} Horse, {Norwegian} Cats, {German} Zebra ]
smokes = [ {Englishman} PallMall, {Swede} BlueMaster, {Dane} Blend, {Norwegian} Dunhill, {German} Prince ]
drinks = [ {Englishman} Milk, {Swede} Beer, {Dane} Tea, {Norwegian} Water, {German} Coffee ]
order = [ {First} Norwegian, {Second} Dane, {Third} Englishman, {Fourth} German, {Fifth}, Swede ]
GAP
<lang gap>leftOf :=function(setA, vA, setB, vB) local i; for i in [1..4] do
if ( setA[i] = vA) and (setB[i+1] = vB) then return true ;fi;
od;
return false;
end;
nextTo :=function(setA, vA, setB, vB) local i; for i in [1..4] do
if ( setA[i] = vA) and (setB[i+1] = vB) then return true ;fi; if ( setB[i] = vB) and (setA[i+1] = vA) then return true ;fi;
od; return false; end;
requires := function(setA, vA, setB, vB)
local i;
for i in [1..5] do
if ( setA[i] = vA) and (setB[i] = vB) then return true ;fi;
od;
return false;
end;
pcolors :=PermutationsList(["white" ,"yellow" ,"blue" ,"red" ,"green"]);
pcigars :=PermutationsList(["blends", "pall_mall", "prince", "bluemasters", "dunhill"]);
pnats:=PermutationsList(["german", "swedish", "british", "norwegian", "danish"]);
pdrinks :=PermutationsList(["beer", "water", "tea", "milk", "coffee"]);
ppets :=PermutationsList(["birds", "cats", "horses", "fish", "dogs"]);
for colors in pcolors do
if not (leftOf(colors,"green",colors,"white")) then continue ;fi;
for nats in pnats do
if not (requires(nats,"british",colors,"red")) then continue ;fi;
if not (nats[1]="norwegian") then continue ;fi;
if not (nextTo(nats,"norwegian",colors,"blue")) then continue ;fi;
for pets in ppets do
if not (requires(nats,"swedish",pets,"dogs")) then continue ;fi;
for drinks in pdrinks do
if not (drinks[3]="milk") then continue ;fi;
if not (requires(colors,"green",drinks,"coffee")) then continue ;fi;
if not (requires(nats,"danish",drinks,"tea")) then continue ;fi;
for cigars in pcigars do
if not (nextTo(pets,"horses",cigars,"dunhill")) then continue ;fi;
if not (requires(cigars,"pall_mall",pets,"birds")) then continue ;fi;
if not (nextTo(cigars,"blends",drinks,"water")) then continue ;fi;
if not (nextTo(cigars,"blends",pets,"cats")) then continue ;fi;
if not (requires(nats,"german",cigars,"prince")) then continue ;fi;
if not (requires(colors,"yellow",cigars,"dunhill")) then continue ;fi;
if not (requires(cigars,"bluemasters",drinks,"beer")) then continue ;fi;
Print(colors,"\n");
Print(nats,"\n");
Print(drinks,"\n");
Print(pets,"\n");
Print(cigars,"\n");
od;od;od;od;od;
</lang>
Go
<lang go>package main
import (
"fmt"
"log"
"strings"
)
// Define some types
type HouseSet [5]*House type House struct {
n Nationality
c Colour
a Animal
d Drink
s Smoke
} type Nationality int8 type Colour int8 type Animal int8 type Drink int8 type Smoke int8
// Define the possible values
const (
English Nationality = iota
Swede
Dane
Norwegian
German
) const (
Red Colour = iota
Green
White
Yellow
Blue
) const (
Dog Animal = iota
Birds
Cats
Horse
Zebra
) const (
Tea Drink = iota
Coffee
Milk
Beer
Water
) const (
PallMall Smoke = iota
Dunhill
Blend
BlueMaster
Prince
)
// And how to print them
var nationalities = [...]string{"English", "Swede", "Dane", "Norwegian", "German"} var colours = [...]string{"red", "green", "white", "yellow", "blue"} var animals = [...]string{"dog", "birds", "cats", "horse", "zebra"} var drinks = [...]string{"tea", "coffee", "milk", "beer", "water"} var smokes = [...]string{"Pall Mall", "Dunhill", "Blend", "Blue Master", "Prince"}
func (n Nationality) String() string { return nationalities[n] } func (c Colour) String() string { return colours[c] } func (a Animal) String() string { return animals[a] } func (d Drink) String() string { return drinks[d] } func (s Smoke) String() string { return smokes[s] } func (h House) String() string {
return fmt.Sprintf("%-9s %-6s %-5s %-6s %s", h.n, h.c, h.a, h.d, h.s)
} func (hs HouseSet) String() string {
lines := make([]string, 0, len(hs))
for i, h := range hs {
s := fmt.Sprintf("%d %s", i, h)
lines = append(lines, s)
}
return strings.Join(lines, "\n")
}
// Simple brute force solution
func simpleBruteForce() (int, HouseSet) {
var v []House
for n := range nationalities {
for c := range colours {
for a := range animals {
for d := range drinks {
for s := range smokes {
h := House{
n: Nationality(n),
c: Colour(c),
a: Animal(a),
d: Drink(d),
s: Smoke(s),
}
if !h.Valid() {
continue
}
v = append(v, h)
}
}
}
}
}
n := len(v)
log.Println("Generated", n, "valid houses")
combos := 0
first := 0
valid := 0
var validSet HouseSet
for a := 0; a < n; a++ {
if v[a].n != Norwegian { // Condition 10:
continue
}
for b := 0; b < n; b++ {
if b == a {
continue
}
if v[b].anyDups(&v[a]) {
continue
}
for c := 0; c < n; c++ {
if c == b || c == a {
continue
}
if v[c].d != Milk { // Condition 9:
continue
}
if v[c].anyDups(&v[b], &v[a]) {
continue
}
for d := 0; d < n; d++ {
if d == c || d == b || d == a {
continue
}
if v[d].anyDups(&v[c], &v[b], &v[a]) {
continue
}
for e := 0; e < n; e++ {
if e == d || e == c || e == b || e == a {
continue
}
if v[e].anyDups(&v[d], &v[c], &v[b], &v[a]) {
continue
}
combos++
set := HouseSet{&v[a], &v[b], &v[c], &v[d], &v[e]}
if set.Valid() {
valid++
if valid == 1 {
first = combos
}
validSet = set
//return set
}
}
}
}
}
}
log.Println("Tested", first, "different combinations of valid houses before finding solution")
log.Println("Tested", combos, "different combinations of valid houses in total")
return valid, validSet
}
// anyDups returns true if h as any duplicate attributes with any of the specified houses func (h *House) anyDups(list ...*House) bool {
for _, b := range list {
if h.n == b.n || h.c == b.c || h.a == b.a || h.d == b.d || h.s == b.s {
return true
}
}
return false
}
func (h *House) Valid() bool {
// Condition 2:
if h.n == English && h.c != Red || h.n != English && h.c == Red {
return false
}
// Condition 3:
if h.n == Swede && h.a != Dog || h.n != Swede && h.a == Dog {
return false
}
// Condition 4:
if h.n == Dane && h.d != Tea || h.n != Dane && h.d == Tea {
return false
}
// Condition 6:
if h.c == Green && h.d != Coffee || h.c != Green && h.d == Coffee {
return false
}
// Condition 7:
if h.a == Birds && h.s != PallMall || h.a != Birds && h.s == PallMall {
return false
}
// Condition 8:
if h.c == Yellow && h.s != Dunhill || h.c != Yellow && h.s == Dunhill {
return false
}
// Condition 11:
if h.a == Cats && h.s == Blend {
return false
}
// Condition 12:
if h.a == Horse && h.s == Dunhill {
return false
}
// Condition 13:
if h.d == Beer && h.s != BlueMaster || h.d != Beer && h.s == BlueMaster {
return false
}
// Condition 14:
if h.n == German && h.s != Prince || h.n != German && h.s == Prince {
return false
}
// Condition 15:
if h.n == Norwegian && h.c == Blue {
return false
}
// Condition 16:
if h.d == Water && h.s == Blend {
return false
}
return true
}
func (hs *HouseSet) Valid() bool {
ni := make(map[Nationality]int, 5)
ci := make(map[Colour]int, 5)
ai := make(map[Animal]int, 5)
di := make(map[Drink]int, 5)
si := make(map[Smoke]int, 5)
for i, h := range hs {
ni[h.n] = i
ci[h.c] = i
ai[h.a] = i
di[h.d] = i
si[h.s] = i
}
// Condition 5:
if ci[Green]+1 != ci[White] {
return false
}
// Condition 11:
if dist(ai[Cats], si[Blend]) != 1 {
return false
}
// Condition 12:
if dist(ai[Horse], si[Dunhill]) != 1 {
return false
}
// Condition 15:
if dist(ni[Norwegian], ci[Blue]) != 1 {
return false
}
// Condition 16:
if dist(di[Water], si[Blend]) != 1 {
return false
}
// Condition 9: (already tested elsewhere)
if hs[2].d != Milk {
return false
}
// Condition 10: (already tested elsewhere)
if hs[0].n != Norwegian {
return false
}
return true
}
func dist(a, b int) int {
if a > b {
return a - b
}
return b - a
}
func main() {
log.SetFlags(0)
n, sol := simpleBruteForce()
fmt.Println(n, "solution found")
fmt.Println(sol)
}</lang>
- Output:
Generated 51 valid houses Tested 652 different combinations of valid houses before finding solution Tested 750 different combinations of valid houses in total 1 solution found 0 Norwegian yellow cats water Dunhill 1 Dane blue horse tea Blend 2 English red birds milk Pall Mall 3 German green zebra coffee Prince 4 Swede white dog beer Blue Master
Benchmark (not included but just calling simpleBruteForce):
BenchmarkBruteForce 1000 2687946 ns/op 550568 B/op 7560 allocs/op
Haskell
<lang haskell>module Main where
import Control.Applicative ((<$>), (<*>)) import Control.Monad (foldM, forM_) import Data.List ((\\))
-- types data House = House
{ color :: Color -- <trait> :: House -> <Trait>
, man :: Man
, pet :: Pet
, drink :: Drink
, smoke :: Smoke
}
deriving (Eq, Show)
data Color = Red | Green | Blue | Yellow | White
deriving (Eq, Show, Enum, Bounded)
data Man = Eng | Swe | Dan | Nor | Ger
deriving (Eq, Show, Enum, Bounded)
data Pet = Dog | Birds | Cats | Horse | Zebra
deriving (Eq, Show, Enum, Bounded)
data Drink = Coffee | Tea | Milk | Beer | Water
deriving (Eq, Show, Enum, Bounded)
data Smoke = PallMall | Dunhill | Blend | BlueMaster | Prince
deriving (Eq, Show, Enum, Bounded)
type Solution = [House]
main :: IO () main = do
forM_ solutions $ \sol -> mapM_ print sol
>> putStrLn "----"
putStrLn "No More Solutions"
solutions :: [Solution]
solutions = filter finalCheck . map reverse $ foldM next [] [1..5]
where
-- NOTE: list of houses is generated in reversed order
next :: Solution -> Int -> [Solution]
next sol pos = [h:sol | h <- newHouses sol, consistent h pos]
newHouses :: Solution -> Solution
newHouses sol = -- all combinations of traits not yet used
House <$> new color <*> new man <*> new pet <*> new drink <*> new smoke
where
new trait = [minBound ..] \\ map trait sol -- :: [<Trait>]
consistent :: House -> Int -> Bool
consistent house pos = and -- consistent with the rules:
[ man `is` Eng <=> color `is` Red -- 2
, man `is` Swe <=> pet `is` Dog -- 3
, man `is` Dan <=> drink `is` Tea -- 4
, color `is` Green <=> drink `is` Coffee -- 6
, pet `is` Birds <=> smoke `is` PallMall -- 7
, color `is` Yellow <=> smoke `is` Dunhill -- 8
, const (pos == 3) <=> drink `is` Milk -- 9
, const (pos == 1) <=> man `is` Nor -- 10
, drink `is` Beer <=> smoke `is` BlueMaster -- 13
, man `is` Ger <=> smoke `is` Prince -- 14
]
where
infix 4 <=>
p <=> q = p house == q house -- both True or both False
is :: Eq a => (House -> a) -> a -> House -> Bool
(trait `is` value) house = trait house == value
finalCheck :: [House] -> Bool
finalCheck solution = and -- fulfills the rules:
[ (color `is` Green) `leftOf` (color `is` White) -- 5
, (smoke `is` Blend ) `nextTo` (pet `is` Cats ) -- 11
, (smoke `is` Dunhill) `nextTo` (pet `is` Horse) -- 12
, (color `is` Blue ) `nextTo` (man `is` Nor ) -- 15
, (smoke `is` Blend ) `nextTo` (drink `is` Water) -- 16
]
where
nextTo :: (House -> Bool) -> (House -> Bool) -> Bool
nextTo p q = leftOf p q || leftOf q p
leftOf p q
| (_:h:_) <- dropWhile (not . p) solution = q h
| otherwise = False</lang>
- Output:
House {color = Yellow, man = Nor, pet = Cats, drink = Water, smoke = Dunhill}
House {color = Blue, man = Dan, pet = Horse, drink = Tea, smoke = Blend}
House {color = Red, man = Eng, pet = Birds, drink = Milk, smoke = PallMall}
House {color = Green, man = Ger, pet = Zebra, drink = Coffee, smoke = Prince}
House {color = White, man = Swe, pet = Dog, drink = Beer, smoke = BlueMaster}
----
No More SolutionsLP-like version
(a little faster version)
<lang haskell>import Control.Monad import Data.List
values :: (Bounded a, Enum a) => a values = permutations [minBound..maxBound]
data Nation = English | Swede | Dane | Norwegian | German
deriving (Bounded, Enum, Eq, Show)
data Color = Red | Green | White | Yellow | Blue
deriving (Bounded, Enum, Eq, Show)
data Pet = Dog | Birds | Cats | Horse | Zebra
deriving (Bounded, Enum, Eq, Show)
data Drink = Tea | Coffee | Milk | Beer | Water
deriving (Bounded, Enum, Eq, Show)
data Smoke = PallMall | Dunhill | Blend | BlueMaster | Prince
deriving (Bounded, Enum, Eq, Show)
answers = do
color <- values leftOf color Green color White -- 5
nation <- values first nation Norwegian -- 10 same nation English color Red -- 2 nextTo nation Norwegian color Blue -- 15
drink <- values middle drink Milk -- 9 same nation Dane drink Tea -- 4 same drink Coffee color Green -- 6
pet <- values same nation Swede pet Dog -- 3
smoke <- values same smoke PallMall pet Birds -- 7 same color Yellow smoke Dunhill -- 8 nextTo smoke Blend pet Cats -- 11 nextTo pet Horse smoke Dunhill -- 12 same nation German smoke Prince -- 14 same smoke BlueMaster drink Beer -- 13 nextTo drink Water smoke Blend -- 16
return $ zip5 nation color pet drink smoke
where
same xs x ys y = guard $ (x, y) `elem` zip xs ys
leftOf xs x ys y = same xs x (tail ys) y
nextTo xs x ys y = leftOf xs x ys y `mplus`
leftOf ys y xs x
middle xs x = guard $ xs !! 2 == x
first xs x = guard $ head xs == x
main = do
forM_ answers $ (\answer -> -- for answer in answers:
do
mapM_ print answer
print [nation | (nation, _, Zebra, _, _) <- answer]
putStrLn "" )
putStrLn "No more solutions!"</lang>
Output:
(Norwegian,Yellow,Cats,Water,Dunhill) (Dane,Blue,Horse,Tea,Blend) (English,Red,Birds,Milk,PallMall) (German,Green,Zebra,Coffee,Prince) (Swede,White,Dog,Beer,BlueMaster) [German] No more solutions!
J
Propositions 1 .. 16 without 9,10 and15 <lang j>ehs=: 5$a:
cr=: (('English';'red') 0 3} ehs);<('Dane';'tea') 0 2}ehs cr=: cr, (('German';'Prince') 0 4}ehs);<('Swede';'dog') 0 1 }ehs
cs=: <('PallMall';'birds') 4 1}ehs cs=: cs, (('yellow';'Dunhill') 3 4}ehs);<('BlueMaster';'beer') 4 2}ehs
lof=: (('coffee';'green')2 3}ehs);<(<'white')3}ehs
next=: <((<'Blend') 4 }ehs);<(<'water')2}ehs next=: next,<((<'Blend') 4 }ehs);<(<'cats')1}ehs next=: next,<((<'Dunhill') 4}ehs);<(<'horse')1}ehs</lang> Example <lang j> lof ┌─────────────────┬───────────┐ │┌┬┬──────┬─────┬┐│┌┬┬┬─────┬┐│ ││││coffee│green│││││││white│││ │└┴┴──────┴─────┴┘│└┴┴┴─────┴┘│ └─────────────────┴───────────┘</lang>
Collections of all variants of the propositions: <lang j>hcr=: (<ehs),. (A.~i.@!@#)cr hcs=:~. (A.~i.@!@#)cs,2$<ehs hlof=:(-i.4) |."0 1 lof,3$<ehs hnext=: ,/((i.4) |."0 1 (3$<ehs)&,)"1 ;(,,:|.)&.> next</lang>
We start the row of houses with fixed properties 9, 10 and 15. <lang j>houses=: ((<'Norwegian') 0}ehs);((<'blue') 3 }ehs);((<'milk') 2}ehs);ehs;<ehs</lang>
- Output:
<lang j> houses ┌───────────────┬──────────┬──────────┬──────┬──────┐ │┌─────────┬┬┬┬┐│┌┬┬┬────┬┐│┌┬┬────┬┬┐│┌┬┬┬┬┐│┌┬┬┬┬┐│ ││Norwegian││││││││││blue││││││milk││││││││││││││││││ │└─────────┴┴┴┴┘│└┴┴┴────┴┘│└┴┴────┴┴┘│└┴┴┴┴┘│└┴┴┴┴┘│ └───────────────┴──────────┴──────────┴──────┴──────┘</lang> Set of proposition variants: <lang j>constraints=: hcr;hcs;hlof;<hnext</lang> The worker and its helper verbs <lang j>select=: ~.@(,: #~ ,&(0~:#)) filter=: #~*./@:(2>#S:0)"1 compose=: [: filter f. [: ,/ select f. L:0"1"1 _
solve=: 4 :0 h=. ,:x whilst. 0=# z do.
for_e. y do. h=. h compose > e end. z=.(#~1=[:+/"1 (0=#)S:0"1) h=.~. h
end. )</lang>
- Output:
<lang j> >"0 houses solve constraints ┌─────────┬─────┬──────┬──────┬──────────┐ │Norwegian│cats │water │yellow│Dunhill │ ├─────────┼─────┼──────┼──────┼──────────┤ │Dane │horse│tea │blue │Blend │ ├─────────┼─────┼──────┼──────┼──────────┤ │English │birds│milk │red │PallMall │ ├─────────┼─────┼──────┼──────┼──────────┤ │German │ │coffee│green │Prince │ ├─────────┼─────┼──────┼──────┼──────────┤ │Swede │dog │beer │white │BlueMaster│ └─────────┴─────┴──────┴──────┴──────────┘</lang> So, the German owns the zebra.
Alternative
A longer running solver by adding the zebra variants.
<lang j>zebra=: (-i.5)|."0 1 (<(<'zebra') 1}ehs),4$<ehs
solve3=: 4 :0 p=. *./@:((0~:#)S:0) f=. [:~.&.> [: compose&.>~/y&, f. z=. f^:(3>[:#(#~p"1)&>)^:_ <,:x >"0 (#~([:*./[:;[:<@({.~:}.)\.;)"1)(#~p"1); z )</lang>
- Output:
<lang j> houses solve3 constraints,<zebra ┌─────────┬─────┬──────┬──────┬──────────┐ │Norwegian│cats │water │yellow│Dunhill │ ├─────────┼─────┼──────┼──────┼──────────┤ │Dane │horse│tea │blue │Blend │ ├─────────┼─────┼──────┼──────┼──────────┤ │English │birds│milk │red │PallMall │ ├─────────┼─────┼──────┼──────┼──────────┤ │German │zebra│coffee│green │Prince │ ├─────────┼─────┼──────┼──────┼──────────┤ │Swede │dog │beer │white │BlueMaster│ └─────────┴─────┴──────┴──────┴──────────┘</lang>
Java
This Java solution includes 3 classes. I developed and tested on a Win7 machine with JDK 1.8 platform, but it should work with earlier version 1.7 also.
First, create a class which express the puzzle line.
<lang Java> package zebra;
public class LineOfPuzzle implements Cloneable{
private Integer order;
private String nation;
private String color;
private String animal;
private String drink;
private String cigarette;
private LineOfPuzzle rightNeighbor;
private LineOfPuzzle leftNeighbor;
private PuzzleSet<LineOfPuzzle> undefNeighbors;
public LineOfPuzzle (Integer order, String nation, String color,
String animal, String drink, String cigarette){
this.animal=animal;
this.cigarette=cigarette;
this.color=color;
this.drink=drink;
this.nation=nation;
this.order=order;
}
public Integer getOrder() {
return order;
}
public void setOrder(Integer order) {
this.order = order;
}
public String getNation() {
return nation;
}
public void setNation(String nation) {
this.nation = nation;
}
public String getColor() {
return color;
}
public void setColor(String color) {
this.color = color;
}
public String getAnimal() {
return animal;
}
public void setAnimal(String animal) {
this.animal = animal;
}
public String getDrink() {
return drink;
}
public void setDrink(String drink) {
this.drink = drink;
}
public String getCigarette() {
return cigarette;
}
public void setCigarette(String cigarette) {
this.cigarette = cigarette;
}
/**
* Overrides object equal method
* @param obj
* @return
*/
@Override
public boolean equals(Object obj) {
if (obj instanceof LineOfPuzzle){
LineOfPuzzle searchLine = (LineOfPuzzle)obj;
return this.getWholeLine().equalsIgnoreCase(searchLine.getWholeLine());
}
else
return false;
}
public int getFactsCount(){
int facts = 0;
facts+=this.getOrder()!=null?1:0;
facts+=this.getNation()!=null?1:0;
facts+=this.getColor()!=null?1:0;
facts+=this.getAnimal()!=null?1:0;
facts+=this.getCigarette()!=null?1:0;
facts+=this.getDrink()!=null?1:0;
return facts;
}
public int getCommonFactsCount(LineOfPuzzle lineOfFacts){
int ordrCmp = (this.order!=null && lineOfFacts.getOrder()!= null &&
this.order.intValue()== lineOfFacts.getOrder().intValue())?1:0;
int natnCmp = (this.nation!=null && lineOfFacts.getNation()!= null &&
this.nation.equalsIgnoreCase(lineOfFacts.getNation()))?1:0;
int colrCmp = (this.color!=null && lineOfFacts.getColor()!= null &&
this.color.equalsIgnoreCase(lineOfFacts.getColor()))?1:0;
int petsCmp = (this.animal!=null && (lineOfFacts.getAnimal()!= null &&
this.animal.equalsIgnoreCase(lineOfFacts.getAnimal())))?1:0;
int cigrCmp = (this.cigarette!=null && lineOfFacts.getCigarette()!= null &&
this.cigarette.equalsIgnoreCase(lineOfFacts.getCigarette()))?1:0;
int drnkCmp = (this.drink!=null && lineOfFacts.getDrink()!= null &&
this.drink.equalsIgnoreCase(lineOfFacts.getDrink()))?1:0;
int result = (ordrCmp + natnCmp + colrCmp + petsCmp + cigrCmp + drnkCmp);
return result;
}
public void addUndefindedNeighbor(LineOfPuzzle newNeighbor){
if (this.undefNeighbors==null)
this.undefNeighbors = new PuzzleSet<>();
this.undefNeighbors.add(newNeighbor);
}
public boolean hasUndefNeighbors(){
return (this.undefNeighbors!=null);
}
public PuzzleSet<LineOfPuzzle> getUndefNeighbors(){
return this.undefNeighbors;
}
public void setLeftNeighbor(LineOfPuzzle leftNeighbor){
this.leftNeighbor = leftNeighbor;
this.leftNeighbor.setOrder(this.order - 1);
}
public void setRightNeighbor(LineOfPuzzle rightNeighbor){
this.rightNeighbor=rightNeighbor;
this.rightNeighbor.setOrder(this.order + 1);
}
public boolean hasLeftNeighbor(){
return (leftNeighbor!=null);
}
public LineOfPuzzle getLeftNeighbor(){
return this.leftNeighbor;
}
public boolean hasNeighbor(int direction){
if (direction < 0)
return (leftNeighbor!=null);
else
return (rightNeighbor!=null);
}
public boolean hasRightNeighbor(){
return (rightNeighbor!=null);
}
public LineOfPuzzle getRightNeighbor(){
return this.rightNeighbor;
}
public LineOfPuzzle getNeighbor(int direction){
if (direction < 0)
return this.leftNeighbor;
else
return this.rightNeighbor;
}
public String getWholeLine() {
String sLine = this.order + " - " +
this.nation + " - " +
this.color + " - " +
this.animal + " - " +
this.drink + " - " +
this.cigarette;
return sLine;
}
@Override
public int hashCode() {
int sLine = (this.order + " - " +
this.nation + " - " +
this.color + " - " +
this.animal + " - " +
this.drink + " - " +
this.cigarette
).hashCode();
return sLine;
}
public void merge(LineOfPuzzle mergedLine){
if (this.order == null) this.order = mergedLine.order;
if (this.nation == null) this.nation = mergedLine.nation;
if (this.color == null) this.color = mergedLine.color;
if (this.animal == null) this.animal = mergedLine.animal;
if (this.drink == null) this.drink = mergedLine.drink;
if (this.cigarette == null) this.cigarette = mergedLine.cigarette;
}
public LineOfPuzzle clone() {
try {
return (LineOfPuzzle) super.clone();
} catch (CloneNotSupportedException e) {
e.printStackTrace();
throw new RuntimeException();
}
}
} </lang>
Second, create a class which express a set of puzzle lines.
<lang Java> package zebra;
import java.util.Iterator; import java.util.LinkedHashSet;
public class PuzzleSet<T extends LineOfPuzzle> extends LinkedHashSet{
private T t;
private int countOfOne=0;
private int countOfTwo=0;
private int countOfThree=0;
private int countOfFour=0;
private int countOfFive=0;
PuzzleSet() {
super();
}
public void set(T t) { this.t = t; }
public T get(int index) {
return ((T)this.toArray()[index]);
}
public PuzzleSet<T> getSimilarLines(T searchLine) {
PuzzleSet<T> puzzleSubSet = new PuzzleSet<>();
for (Iterator<T> it = this.iterator(); it.hasNext();) {
T lineOfPuzzle = it.next();
if(lineOfPuzzle.getCommonFactsCount(searchLine) == searchLine.getFactsCount())
puzzleSubSet.add(lineOfPuzzle);
}
if (puzzleSubSet.isEmpty())
return null;
return puzzleSubSet;
}
public boolean contains(T searchLine) {
for (Iterator<T> it = this.iterator(); it.hasNext();) {
T puzzleLine = it.next();
if(puzzleLine.getCommonFactsCount(searchLine) == searchLine.getFactsCount())
return true;
}
return false;
}
public boolean accepts(T searchLine) {
int passed=0;
int notpassed=0;
for (Iterator<T> it = this.iterator(); it.hasNext();) {
T puzzleSetLine = it.next();
int lineFactsCnt = puzzleSetLine.getFactsCount();
int comnFactsCnt = puzzleSetLine.getCommonFactsCount(searchLine);
if( lineFactsCnt != comnFactsCnt && lineFactsCnt !=0 && comnFactsCnt !=0){
notpassed++;
}
if( lineFactsCnt == comnFactsCnt)
passed++;
}
return (passed >= 0 && notpassed == 0);
}
public void riseLineCountFlags(int lineOrderId){
switch (lineOrderId){
case 1: countOfOne++; break;
case 2: countOfTwo++; break;
case 3: countOfThree++; break;
case 4: countOfFour++; break;
case 5: countOfFive++; break;
default:;
}
}
public void clearLineCountFlags(){
countOfOne=0;
countOfTwo=0;
countOfThree=0;
countOfFour=0;
countOfFive=0;
}
public int getLineCountByOrderId(int lineOrderId){
switch (lineOrderId){
case 1: return countOfOne;
case 2: return countOfTwo;
case 3: return countOfThree;
case 4: return countOfFour;
case 5: return countOfFive;
default:return -1;
}
}
} </lang>
Third, create the main run class where there is a recursive call to check for rule sets. <lang Java> package zebra;
import java.util.ArrayList; import java.util.Iterator;
public class Puzzle {
private static final ArrayList<Integer> orders = new ArrayList<>(5);
private static final ArrayList<String> nations = new ArrayList<>(5);
private static final ArrayList<String> animals = new ArrayList<>(5);
private static final ArrayList<String> drinks = new ArrayList<>(5);
private static final ArrayList<String> cigarettes = new ArrayList<>(5);
private static final ArrayList<String> colors = new ArrayList<>(5);
private static PuzzleSet<LineOfPuzzle> puzzleTable;
static
{
// House orders
orders.add(1);
orders.add(2);
orders.add(3);
orders.add(4);
orders.add(5);
// Man nations
nations.add("English");
nations.add("Danish");
nations.add("German");
nations.add("Swedesh");
nations.add("Norwegian");
//Animals
animals.add("Zebra");
animals.add("Horse");
animals.add("Birds");
animals.add("Dog");
animals.add("Cats");
//Drinks
drinks.add("Coffee");
drinks.add("Tea");
drinks.add("Beer");
drinks.add("Water");
drinks.add("Milk");
//Smokes
cigarettes.add("Pall Mall");
cigarettes.add("Blend");
cigarettes.add("Blue Master");
cigarettes.add("Prince");
cigarettes.add("Dunhill");
//Colors
colors.add("Red");
colors.add("Green");
colors.add("White");
colors.add("Blue");
colors.add("Yellow");
}
public static void main (String[] args){
boolean validLine=true;
puzzleTable = new PuzzleSet<>();
//Rules
LineOfPuzzle rule2 = new LineOfPuzzle(null, "English", "Red", null, null, null);
LineOfPuzzle rule3 = new LineOfPuzzle(null, "Swedesh", null, "Dog", null, null);
LineOfPuzzle rule4 = new LineOfPuzzle(null, "Danish", null, null, "Tea", null);
LineOfPuzzle rule6 = new LineOfPuzzle(null, null, "Green", null, "Coffee", null);
LineOfPuzzle rule7 = new LineOfPuzzle(null, null, null, "Birds", null, "Pall Mall");
LineOfPuzzle rule8 = new LineOfPuzzle(null, null, "Yellow", null, null, "Dunhill");
LineOfPuzzle rule9 = new LineOfPuzzle(3, null, null, null, "Milk", null);
LineOfPuzzle rule10 = new LineOfPuzzle(1, "Norwegian", null, null, null, null);
LineOfPuzzle rule13 = new LineOfPuzzle(null, null, null, null, "Beer", "Blue Master");
LineOfPuzzle rule14 = new LineOfPuzzle(null, "German", null, null, null, "Prince");
LineOfPuzzle rule15 = new LineOfPuzzle(2, null, "Blue", null, null, null);
PuzzleSet<LineOfPuzzle> ruleSet = new PuzzleSet<>();
ruleSet.add(rule2);
ruleSet.add(rule3);
ruleSet.add(rule4);
ruleSet.add(rule6);
ruleSet.add(rule7);
ruleSet.add(rule8);
ruleSet.add(rule9);
ruleSet.add(rule10);
ruleSet.add(rule13);
ruleSet.add(rule14);
ruleSet.add(rule15);
//Creating all possible combination of a puzzle line.
//The maximum number of lines is 5^^6 (15625).
//Each combination line is checked against a set of knowing facts, thus
//only a small number of line result at the end.
for (Integer orderId : Puzzle.orders) {
for (String nation : Puzzle.nations) {
for (String color : Puzzle.colors) {
for (String animal : Puzzle.animals) {
for (String drink : Puzzle.drinks) {
for (String cigarette : Puzzle.cigarettes) {
LineOfPuzzle pzlLine = new LineOfPuzzle(orderId,
nation,
color,
animal,
drink,
cigarette);
// Checking against a set of knowing facts
if (ruleSet.accepts(pzlLine)){
// Adding rules of neighbors
if (cigarette.equalsIgnoreCase("Blend")
&& (animal.equalsIgnoreCase("Cats")
|| drink.equalsIgnoreCase("Water")))
validLine = false;
if (cigarette.equalsIgnoreCase("Dunhill")
&& animal.equalsIgnoreCase("Horse"))
validLine = false;
if (validLine){
puzzleTable.add(pzlLine);
//set neighbors constraints
if (color.equalsIgnoreCase("Green")){
pzlLine.setRightNeighbor(new LineOfPuzzle(null, null, "White", null, null, null));
}
if (color.equalsIgnoreCase("White")){
pzlLine.setLeftNeighbor(new LineOfPuzzle(null, null, "Green", null, null, null));
}
//
if (animal.equalsIgnoreCase("Cats")
&& !cigarette.equalsIgnoreCase("Blend") ){
pzlLine.addUndefindedNeighbor(new LineOfPuzzle(null, null, null, null, null, "Blend"));
}
if (cigarette.equalsIgnoreCase("Blend")
&& !animal.equalsIgnoreCase("Cats")){
pzlLine.addUndefindedNeighbor(new LineOfPuzzle(null, null, null, "Cats", null, null));
}
//
if (drink.equalsIgnoreCase("Water")
&& !animal.equalsIgnoreCase("Cats")
&& !cigarette.equalsIgnoreCase("Blend")){
pzlLine.addUndefindedNeighbor(new LineOfPuzzle(null, null, null, null, null, "Blend"));
}
if (cigarette.equalsIgnoreCase("Blend")
&& !drink.equalsIgnoreCase("Water")){
pzlLine.addUndefindedNeighbor(new LineOfPuzzle(null, null, null, null, "Water", null));
}
//
if (animal.equalsIgnoreCase("Horse")
&& !cigarette.equalsIgnoreCase("Dunhill")){
pzlLine.addUndefindedNeighbor(new LineOfPuzzle(null, null, null, null, null, "Dunhill"));
}
if (cigarette.equalsIgnoreCase("Dunhill")
&& !animal.equalsIgnoreCase("Horse")){
pzlLine.addUndefindedNeighbor(new LineOfPuzzle(null, null, null, "Horse", null, null));
}
}
validLine = true;
}
} //cigarette end
} //drinks end
} //animal end
} //color end
} //nations end
} //order end
System.out.println("After general rule set validation, remains "+
puzzleTable.size() + " lines.");
for (Iterator<LineOfPuzzle> it = puzzleTable.iterator(); it.hasNext();){
validLine=true;
LineOfPuzzle lineOfPuzzle = it.next();
if (lineOfPuzzle.hasLeftNeighbor()){
LineOfPuzzle neighbor = lineOfPuzzle.getLeftNeighbor();
if (neighbor.getOrder()<1 || neighbor.getOrder()>5){
validLine=false;
it.remove();
}
}
if (validLine && lineOfPuzzle.hasRightNeighbor()){
LineOfPuzzle neighbor = lineOfPuzzle.getRightNeighbor();
if (neighbor.getOrder()<1 || neighbor.getOrder()>5){
it.remove();
}
}
}
System.out.println("After removing out of bound neighbors, remains " +
puzzleTable.size() + " lines.");
//Setting left and right neighbors
for (Iterator<LineOfPuzzle> it = puzzleTable.iterator(); it.hasNext();) {
LineOfPuzzle puzzleLine = it.next();
if (puzzleLine.hasUndefNeighbors()){
for (Iterator<LineOfPuzzle> it1 = puzzleLine.getUndefNeighbors().iterator(); it1.hasNext();) {
LineOfPuzzle leftNeighbor = it1.next();
LineOfPuzzle rightNeighbor = leftNeighbor.clone();
//make it left neighbor
leftNeighbor.setOrder(puzzleLine.getOrder()-1);
if (puzzleTable.contains(leftNeighbor)){
if (puzzleLine.hasLeftNeighbor())
puzzleLine.getLeftNeighbor().merge(leftNeighbor);
else
puzzleLine.setLeftNeighbor(leftNeighbor);
}
rightNeighbor.setOrder(puzzleLine.getOrder()+1);
if (puzzleTable.contains(rightNeighbor)){
if (puzzleLine.hasRightNeighbor())
puzzleLine.getRightNeighbor().merge(rightNeighbor);
else
puzzleLine.setRightNeighbor(rightNeighbor);
}
}
}
}
int iteration=1;
int lastSize=0;
//Recursively validate against neighbor rules
while (puzzleTable.size()>5 && lastSize != puzzleTable.size()) {
lastSize = puzzleTable.size();
puzzleTable.clearLineCountFlags();
recursiveSearch(null, puzzleTable, -1);
ruleSet.clear();
// Assuming we'll get at leas one valid line each iteration, we create
// a set of new rules with lines which have no more then one instance of same OrderId.
for (int i = 1; i < 6; i++) {
if (puzzleTable.getLineCountByOrderId(i)==1)
ruleSet.addAll(puzzleTable.getSimilarLines(new LineOfPuzzle(i, null, null, null, null, null)));
}
for (Iterator<LineOfPuzzle> it = puzzleTable.iterator(); it.hasNext();) {
LineOfPuzzle puzzleLine = it.next();
if (!ruleSet.accepts(puzzleLine))
it.remove();
}
//
System.out.println("After " + iteration + " recursive iteration, remains "
+ puzzleTable.size() + " lines");
iteration+=1;
}
// Print the results
System.out.println("-------------------------------------------");
if (puzzleTable.size()==5){
for (Iterator<LineOfPuzzle> it = puzzleTable.iterator(); it.hasNext();) {
LineOfPuzzle puzzleLine = it.next();
System.out.println(puzzleLine.getWholeLine());
}
}else
System.out.println("Sorry, solution not found!");
}
// Recursively checks the input set to ensure each line has right neighbor.
// Neighbors can be of three type, left, right or undefined.
// Direction: -1 left, 0 undefined, 1 right
private static boolean recursiveSearch(LineOfPuzzle pzzlNodeLine,
PuzzleSet puzzleSet, int direction){
boolean validLeaf = false;
boolean hasNeighbor = false;
PuzzleSet<LineOfPuzzle> puzzleSubSet = null;
for (Iterator<LineOfPuzzle> it = puzzleSet.iterator(); it.hasNext();) {
LineOfPuzzle pzzlLeafLine = it.next();
validLeaf = false;
hasNeighbor = pzzlLeafLine.hasNeighbor(direction);
if (hasNeighbor){
puzzleSubSet = puzzleTable.getSimilarLines(pzzlLeafLine.getNeighbor(direction));
if (puzzleSubSet != null){
if (pzzlNodeLine !=null)
validLeaf = puzzleSubSet.contains(pzzlNodeLine);
else
validLeaf = recursiveSearch(pzzlLeafLine, puzzleSubSet, -1*direction);
}
else
validLeaf = false;
}
if (!validLeaf && pzzlLeafLine.hasNeighbor(-1*direction)){
hasNeighbor = true;
if (hasNeighbor){
puzzleSubSet = puzzleTable.getSimilarLines(pzzlLeafLine.getNeighbor(-1*direction));
if (puzzleSubSet != null){
if (pzzlNodeLine !=null)
validLeaf = puzzleSubSet.contains(pzzlNodeLine);
else
validLeaf = recursiveSearch(pzzlLeafLine, puzzleSubSet, direction);
}
else
validLeaf = false;
}
}
if (pzzlNodeLine != null && validLeaf)
return validLeaf;
if (pzzlNodeLine == null && hasNeighbor && !validLeaf){
it.remove();
}
if (pzzlNodeLine == null){
if (hasNeighbor && validLeaf){
puzzleSet.riseLineCountFlags(pzzlLeafLine.getOrder());
}
if (!hasNeighbor){
puzzleSet.riseLineCountFlags(pzzlLeafLine.getOrder());
}
}
}
return validLeaf;
}
} </lang>
- Output:
After general rule set validation, remains 60 lines. After removing out of bound neighbors, remains 52 lines. After 1 recursive iteration, remains 17 lines After 2 recursive iteration, remains 6 lines After 3 recursive iteration, remains 5 lines ------------------------------------------- 1 - Norwegian - Yellow - Cats - Water - Dunhill 2 - Danish - Blue - Horse - Tea - Blend 3 - English - Red - Birds - Milk - Pall Mall 4 - German - Green - Zebra - Coffee - Prince 5 - Swedesh - White - Dog - Beer - Blue Master
--VValache (talk) 14:14, 3 October 2014 (UTC)
jq
The main function presented here, zebra, generates all possible solutions. By letting it run to completion, we can see that there is only one solution to the puzzle. The program is fast because pruning takes place. That is, the program implements a "generate-and-prune" strategy.
Part 1: Generic filters for unification and matching <lang jq>
- Attempt to unify the input object with the specified object
def unify( object ):
# Attempt to unify the input object with the specified tag:value def unify2(tag; value): if . == null then null elif .[tag] == value then . elif .[tag] == null then .[tag] = value else null end; reduce (object|keys[]) as $key (.; unify2($key; object[$key]) );
- Input: an array
- Output: if the i-th element can be made to satisfy the condition,
- then the updated array, otherwise empty.
def enforce(i; cond):
if 0 <= i and i < length then (.[i] | cond) as $ans | if $ans then .[i] = $ans else empty end else empty end ;</lang>
Part 2: Zebra Puzzle <lang jq># Each house is a JSON object of the form:
- { "number": _, "nation": _, "owns": _, "color": _, "drinks": _, "smokes": _}
- The list of houses is represented by an array of five such objects.
- Input: an array of objects representing houses.
- Output: [i, solution] where i is the entity unified with obj
- and solution is the updated array
def solve_with_index( obj ):
. as $Houses | range(0; length) as $i | ($Houses[$i] | unify(obj)) as $H | if $H then $Houses[$i] = $H else empty end | [ $i, .] ;
def solve( object ):
solve_with_index( object )[1];
def adjacent( obj1; obj2 ):
solve_with_index(obj1) as $H
| $H[1]
| (enforce( $H[0] - 1; unify(obj2) ),
enforce( $H[0] + 1; unify(obj2) )) ;
def left_right( obj1; obj2 ):
solve_with_index(obj1) as $H | $H[1] | enforce( $H[0] + 1; unify(obj2) ) ;
- All solutions by generate-and-test
def zebra:
[range(0;5)] | map({"number": .}) # Five houses
| enforce( 0; unify( {"nation": "norwegian"} ) )
| enforce( 2; unify( {"drinks": "milk"} ) )
| solve( {"nation": "englishman", "color": "red"} )
| solve( {"nation": "swede", "owns": "dog"} )
| solve( {"nation": "dane", "drinks": "tea"} )
| left_right( {"color": "green"}; {"color": "white"})
| solve( {"drinks": "coffee", "color": "green"} )
| solve( {"smokes": "Pall Mall", "owns": "birds"} )
| solve( {"color": "yellow", "smokes": "Dunhill"} )
| adjacent( {"smokes": "Blend" }; {"owns": "cats"} )
| adjacent( {"owns": "horse"}; {"smokes": "Dunhill"})
| solve( {"drinks": "beer", "smokes": "Blue Master"} )
| solve( {"nation": "german", "smokes": "Prince"})
| adjacent( {"nation": "norwegian"}; {"color": "blue"})
| adjacent( {"drinks": "water"}; {"smokes": "Blend"})
| solve( {"owns": "zebra"} )
zebra</lang>
- Output:
<lang sh>$ time jq -n -f zebra.jq [
{
"number": 0,
"nation": "norwegian",
"color": "yellow",
"smokes": "Dunhill",
"owns": "cats",
"drinks": "water"
},
{
"number": 1,
"drinks": "tea",
"nation": "dane",
"smokes": "Blend",
"owns": "horse",
"color": "blue"
},
{
"number": 2,
"drinks": "milk",
"color": "red",
"nation": "englishman",
"owns": "birds",
"smokes": "Pall Mall"
},
{
"number": 3,
"color": "green",
"drinks": "coffee",
"nation": "german",
"smokes": "Prince",
"owns": "zebra"
},
{
"number": 4,
"nation": "swede",
"owns": "dog",
"color": "white",
"drinks": "beer",
"smokes": "Blue Master"
}
]
- Times include compilation:
real 0m0.284s user 0m0.260s
sys 0m0.005s</lang>Julia
<lang Julia>
- Julia 1.0
using Combinatorics function make(str, test )
filter(test, collect( permutations(split(str))) )
end
men = make("danish english german norwegian swedish",
x -> "norwegian" == x[1])
drinks = make("beer coffee milk tea water", x -> "milk" == x[3])
- Julia 1.0 compatible
colors = make("blue green red white yellow",
x -> 1 == findfirst(c -> c == "white", x) - findfirst(c -> c == "green",x))
pets = make("birds cats dog horse zebra")
smokes = make("blend blue-master dunhill pall-mall prince")
function eq(x, xs, y, ys)
findfirst(xs, x) == findfirst(ys, y)
end
function adj(x, xs, y, ys)
1 == abs(findfirst(xs, x) - findfirst(ys, y))
end
function print_houses(n, pet, nationality, colors, drink, smokes)
println("$n, $pet, $nationality $colors $drink $smokes")
end
for m = men, c = colors
if eq("red",c, "english",m) && adj("norwegian",m, "blue",c)
for d = drinks
if eq("danish",m, "tea",d) && eq("coffee",d,"green",c)
for s = smokes
if eq("yellow",c,"dunhill",s) &&
eq("blue-master",s,"beer",d) &&
eq("german",m,"prince",s)
for p = pets
if eq("birds",p,"pall-mall",s) &&
eq("swedish",m,"dog",p) &&
adj("blend",s,"cats",p) &&
adj("horse",p,"dunhill",s)
println("Zebra is owned by ", m[findfirst(c -> c == "zebra", p)])
println("Houses:")
for house_num in 1:5
print_houses(house_num,p[house_num],m[house_num],c[house_num],d[house_num],s[house_num])
end
end
end
end
end
end
end
end
end
</lang>
- Output:
Zebra is owned by german Houses: 1, cats, norwegian yellow water dunhill 2, horse, danish blue tea blend 3, birds, english red milk pall-mall 4, zebra, german green coffee prince 5, dog, swedish white beer blue-master
Constraint Programming Version
Using the rules from https://github.com/SWI-Prolog/bench/blob/master/zebra.pl <lang Julia># Julia 1.4 using JuMP using GLPK
c = Dict(s => i for (i, s) in enumerate(split("blue green ivory red yellow"))) n = Dict(s => i for (i, s) in enumerate(split("english japanese norwegian spanish ukrainian"))) p = Dict(s => i for (i, s) in enumerate(split("dog fox horse snails zebra"))) d = Dict(s => i for (i, s) in enumerate(split("coffee milk orangejuice tea water"))) s = Dict(s => i for (i, s) in enumerate(split("chesterfields kools luckystrikes parliaments winstons")))
model = Model(GLPK.Optimizer)
@variable(model, colors[1:5, 1:5], Bin) @constraints(model, begin
[h in 1:5], sum(colors[h, :]) == 1 [c in 1:5], sum(colors[:, c]) == 1
end)
@variable(model, nations[1:5, 1:5], Bin) @constraints(model, begin
[h in 1:5], sum(nations[h, :]) == 1 [n in 1:5], sum(nations[:, n]) == 1
end)
@variable(model, pets[1:5, 1:5], Bin) @constraints(model, begin
[h in 1:5], sum(pets[h, :]) == 1 [p in 1:5], sum(pets[:, p]) == 1
end)
@variable(model, drinks[1:5, 1:5], Bin) @constraints(model, begin
[h in 1:5], sum(drinks[h, :]) == 1 [d in 1:5], sum(drinks[:, d]) == 1
end)
@variable(model, smokes[1:5, 1:5], Bin) @constraints(model, begin
[h in 1:5], sum(smokes[h, :]) == 1 [s in 1:5], sum(smokes[:, s]) == 1
end)
@constraint(model, [h=1:5], colors[h, c["red"]] == nations[h, n["english"]]) @constraint(model, [h=1:5], nations[h, n["spanish"]] == pets[h, p["dog"]]) @constraint(model, [h=1:5], colors[h, c["green"]] == drinks[h, d["coffee"]]) @constraint(model, [h=1:5], nations[h, n["ukrainian"]] == drinks[h, d["tea"]]) @constraint(model, [h=1:5], colors[h, c["ivory"]] == get(colors, (h+1, c["green"]), 0)) @constraint(model, [h=1:5], pets[h, p["snails"]] == smokes[h, s["winstons"]]) @constraint(model, [h=1:5], colors[h, c["yellow"]] == smokes[h, s["kools"]]) @constraint(model, drinks[3, d["milk"]] == 1) @constraint(model, nations[1, n["norwegian"]] == 1) @constraint(model, [h=1:5], (1-pets[h, p["fox"]]) + get(smokes,(h-1, s["chesterfields"]), 0) + get(smokes, (h+1, s["chesterfields"]), 0) >= 1) @constraint(model, [h=1:5], (1-pets[h, p["horse"]]) + get(smokes,(h-1, s["kools"]), 0) + get(smokes, (h+1, s["kools"]), 0) >= 1) @constraint(model, [h=1:5], drinks[h, d["orangejuice"]] == smokes[h, s["luckystrikes"]]) @constraint(model, [h=1:5], nations[h, n["japanese"]] == smokes[h, s["parliaments"]]) @constraint(model, [h=1:5], (1-nations[h, n["norwegian"]]) + get(colors, (h-1, c["blue"]), 0) + get(colors, (h+1, c["blue"]), 0) >= 1)
optimize!(model)
if termination_status(model) == MOI.OPTIMAL && primal_status(model) == MOI.FEASIBLE_POINT
m = map(1:5) do h
[Dict(values(c) .=> keys(c))[findfirst(value.(colors)[h, :] .≈ 1.0)],
Dict(values(n) .=> keys(n))[findfirst(value.(nations)[h, :] .≈ 1.0)],
Dict(values(p) .=> keys(p))[findfirst(value.(pets)[h, :] .≈ 1.0)],
Dict(values(d) .=> keys(d))[findfirst(value.(drinks)[h, :] .≈ 1.0)],
Dict(values(s) .=> keys(s))[findfirst(value.(smokes)[h, :] .≈ 1.0)]]
end
end
using DataFrames DataFrame(colors=getindex.(m, 1),
nations=getindex.(m, 2),
pets=getindex.(m, 3),
drinks=getindex.(m, 4),
smokes=getindex.(m, 5))
</lang>
- Output:
: 5×5 DataFrame : │ Row │ colors │ nations │ pets │ drinks │ smokes │ : ├─────┼──────────┼───────────┼──────────┼─────────────┼───────────────┤ : │ 1 │ yellow │ norwegian │ fox │ water │ kools │ : │ 2 │ blue │ ukrainian │ horse │ tea │ chesterfields │ : │ 3 │ red │ english │ snails │ milk │ winstons │ : │ 4 │ ivory │ spanish │ dog │ orangejuice │ luckystrikes │ : │ 5 │ green │ japanese │ zebra │ coffee │ parliaments │
Kotlin
<lang scala>// version 1.1.3
fun nextPerm(perm: IntArray): Boolean {
val size = perm.size
var k = -1
for (i in size - 2 downTo 0) {
if (perm[i] < perm[i + 1]) {
k = i
break
}
}
if (k == -1) return false // last permutation
for (l in size - 1 downTo k) {
if (perm[k] < perm[l]) {
val temp = perm[k]
perm[k] = perm[l]
perm[l] = temp
var m = k + 1
var n = size - 1
while (m < n) {
val temp2 = perm[m]
perm[m++] = perm[n]
perm[n--] = temp2
}
break
}
}
return true
}
fun check(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
for (i in 0..4)
if (p[a1][i] == v1) return p[a2][i] == v2
return false
}
fun checkLeft(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
for (i in 0..3)
if (p[a1][i] == v1) return p[a2][i + 1] == v2
return false
}
fun checkRight(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
for (i in 1..4)
if (p[a1][i] == v1) return p[a2][i - 1] == v2
return false
}
fun checkAdjacent(a1: Int, a2: Int, v1: Int, v2: Int): Boolean {
return checkLeft(a1, a2, v1, v2) || checkRight(a1, a2, v1, v2)
}
val colors = listOf("Red", "Green", "White", "Yellow", "Blue") val nations = listOf("English", "Swede", "Danish", "Norwegian", "German") val animals = listOf("Dog", "Birds", "Cats", "Horse", "Zebra") val drinks = listOf("Tea", "Coffee", "Milk", "Beer", "Water") val smokes = listOf("Pall Mall", "Dunhill", "Blend", "Blue Master", "Prince")
val p = Array(120) { IntArray(5) { -1 } } // stores all permutations of numbers 0..4
fun fillHouses(): Int {
var solutions = 0
for (c in 0..119) {
if (!checkLeft(c, c, 1, 2)) continue // C5 : Green left of white
for (n in 0..119) {
if (p[n][0] != 3) continue // C10: Norwegian in First
if (!check(n, c, 0, 0)) continue // C2 : English in Red
if (!checkAdjacent(n, c, 3, 4)) continue // C15: Norwegian next to Blue
for (a in 0..119) {
if (!check(a, n, 0, 1)) continue // C3 : Swede has Dog
for (d in 0..119) {
if (p[d][2] != 2) continue // C9 : Middle drinks Milk
if (!check(d, n, 0, 2)) continue // C4 : Dane drinks Tea
if (!check(d, c, 1, 1)) continue // C6 : Green drinks Coffee
for (s in 0..119) {
if (!check(s, a, 0, 1)) continue // C7 : Pall Mall has Birds
if (!check(s, c, 1, 3)) continue // C8 : Yellow smokes Dunhill
if (!check(s, d, 3, 3)) continue // C13: Blue Master drinks Beer
if (!check(s, n, 4, 4)) continue // C14: German smokes Prince
if (!checkAdjacent(s, a, 2, 2)) continue // C11: Blend next to Cats
if (!checkAdjacent(s, a, 1, 3)) continue // C12: Dunhill next to Horse
if (!checkAdjacent(s, d, 2, 4)) continue // C16: Blend next to Water
solutions++
printHouses(c, n, a, d, s)
}
}
}
}
}
return solutions
}
fun printHouses(c: Int, n: Int, a: Int, d: Int, s: Int) {
var owner: String = ""
println("House Color Nation Animal Drink Smokes")
println("===== ====== ========= ====== ====== ===========")
for (i in 0..4) {
val f = "%3d %-6s %-9s %-6s %-6s %-11s\n"
System.out.printf(f, i + 1, colors[p[c][i]], nations[p[n][i]], animals[p[a][i]], drinks[p[d][i]], smokes[p[s][i]])
if (animals[p[a][i]] == "Zebra") owner = nations[p[n][i]]
}
println("\nThe $owner owns the Zebra\n")
}
fun main(args: Array<String>) {
val perm = IntArray(5) { it }
for (i in 0..119) {
for (j in 0..4) p[i][j] = perm[j]
nextPerm(perm)
}
val solutions = fillHouses()
val plural = if (solutions == 1) "" else "s"
println("$solutions solution$plural found")
}</lang>
- Output:
House Color Nation Animal Drink Smokes ===== ====== ========= ====== ====== =========== 1 Yellow Norwegian Cats Water Dunhill 2 Blue Danish Horse Tea Blend 3 Red English Birds Milk Pall Mall 4 Green German Zebra Coffee Prince 5 White Swede Dog Beer Blue Master The German owns the Zebra 1 solution found
Logtalk
The Logtalk distribution includes a solution for a variant of this puzzle (here reproduced with permission):
<lang logtalk> /* Houses logical puzzle: who owns the zebra and who drinks water?
1) Five colored houses in a row, each with an owner, a pet, cigarettes, and a drink.
2) The English lives in the red house.
3) The Spanish has a dog.
4) They drink coffee in the green house.
5) The Ukrainian drinks tea.
6) The green house is next to the white house.
7) The Winston smoker has a serpent.
8) In the yellow house they smoke Kool.
9) In the middle house they drink milk.
10) The Norwegian lives in the first house from the left.
11) The Chesterfield smoker lives near the man with the fox.
12) In the house near the house with the horse they smoke Kool.
13) The Lucky Strike smoker drinks juice.
14) The Japanese smokes Kent.
15) The Norwegian lives near the blue house.
Who owns the zebra and who drinks water?
- /
- - object(houses).
:- public(houses/1).
:- mode(houses(-list), one).
:- info(houses/1, [
comment is 'Solution to the puzzle.',
argnames is ['Solution']
]).
:- public(print/1).
:- mode(print(+list), one).
:- info(print/1, [
comment is 'Pretty print solution to the puzzle.',
argnames is ['Solution']
]).
houses(Solution) :-
template(Solution), % 1
member(h(english, _, _, _, red), Solution), % 2
member(h(spanish, dog, _, _, _), Solution), % 3
member(h(_, _, _, coffee, green), Solution), % 4
member(h(ukrainian, _, _, tea, _), Solution), % 5
next(h(_, _, _, _, green), h(_, _, _, _, white), Solution), % 6
member(h(_, snake, winston, _, _), Solution), % 7
member(h(_, _, kool, _, yellow), Solution), % 8
Solution = [_, _, h(_, _, _, milk, _), _, _], % 9
Solution = [h(norwegian, _, _, _, _)| _], % 10
next(h(_, fox, _, _, _), h(_, _, chesterfield, _, _), Solution), % 11
next(h(_, _, kool, _, _), h(_, horse, _, _, _), Solution), % 12
member(h(_, _, lucky, juice, _), Solution), % 13
member(h(japonese, _, kent, _, _), Solution), % 14
next(h(norwegian, _, _, _, _), h(_, _, _, _, blue), Solution), % 15
member(h(_, _, _, water, _), Solution), % one of them drinks water
member(h(_, zebra, _, _, _), Solution). % one of them owns a zebra
print([]).
print([House| Houses]) :-
write(House), nl,
print(Houses).
% h(Nationality, Pet, Cigarette, Drink, Color) template([h(_, _, _, _, _), h(_, _, _, _, _), h(_, _, _, _, _), h(_, _, _, _, _), h(_, _, _, _, _)]).
member(A, [A, _, _, _, _]). member(B, [_, B, _, _, _]). member(C, [_, _, C, _, _]). member(D, [_, _, _, D, _]). member(E, [_, _, _, _, E]).
next(A, B, [A, B, _, _, _]). next(B, C, [_, B, C, _, _]). next(C, D, [_, _, C, D, _]). next(D, E, [_, _, _, D, E]).
- - end_object.
</lang>
Sample query: <lang text> | ?- houses::(houses(S), print(S)). h(norwegian,fox,kool,water,yellow) h(ukrainian,horse,chesterfield,tea,blue) h(english,snake,winston,milk,red) h(japonese,zebra,kent,coffee,green) h(spanish,dog,lucky,juice,white)
S = [h(norwegian,fox,kool,water,yellow),h(ukrainian,horse,chesterfield,tea,blue),h(english,snake,winston,milk,red),h(japonese,zebra,kent,coffee,green),h(spanish,dog,lucky,juice,white)] </lang>
Mathematica
This creates a table that has 5 columns, and 25 rows. We fill the table; each column is the same and equal to all the options joined together in blocks of 5:
1 2 3 4 5 colors Blue Blue Blue Blue Blue colors Green Green Green Green Green colors Red Red Red Red Red colors White White White White White colors Yellow Yellow Yellow Yellow Yellow nationality Dane Dane Dane Dane Dane nationality English English English English English nationality German German German German German nationality Norwegian Norwegian Norwegian Norwegian Norwegian nationality Swede Swede Swede Swede Swede beverage Beer Beer Beer Beer Beer beverage Coffee Coffee Coffee Coffee Coffee beverage Milk Milk Milk Milk Milk beverage Tea Tea Tea Tea Tea beverage Water Water Water Water Water animal Birds Birds Birds Birds Birds animal Cats Cats Cats Cats Cats animal Dog Dog Dog Dog Dog animal Horse Horse Horse Horse Horse animal Zebra Zebra Zebra Zebra Zebra smoke Blend Blend Blend Blend Blend smoke Blue Master Blue Master Blue Master Blue Master Blue Master smoke Dunhill Dunhill Dunhill Dunhill Dunhill smoke Pall Mall Pall Mall Pall Mall Pall Mall Pall Mall smoke Prince Prince Prince Prince Prince
This should be read as follows: Each column shows (in blocks of 5) the possible candidates of each kind (beverage, animal, smoke...) We solve it now in a 'sudoku' way: We remove candidates iteratively until we are left with 1 candidate of each kind for each house. <lang Mathematica>ClearAll[EliminatePoss, FilterPuzzle] EliminatePoss[ct_, key1_, key2_] := Module[{t = ct, poss1, poss2, poss, notposs},
poss1 = Position[t, key1]; poss2 = Position[t, key2]; poss = Intersection[Last /@ poss1, Last /@ poss2]; notposs = Complement[Range[5], poss]; poss1 = Select[poss1, MemberQ[notposs, Last[#]] &]; poss2 = Select[poss2, MemberQ[notposs, Last[#]] &]; t = ReplacePart[t, poss1 -> Null]; t = ReplacePart[t, poss2 -> Null]; t
] FilterPuzzle[tbl_] := Module[{t = tbl, poss1, poss2, poss, notposs, rows, columns, vals, sets, delpos},
t = EliminatePoss[t, "English", "Red"]; (*2. The English man lives in the red house. *) t = EliminatePoss[t, "Swede", "Dog"]; (* 3. The Swede has a dog. *) t = EliminatePoss[t, "Dane", "Tea"]; (* 4. The Dane drinks tea. *) t = EliminatePoss[t, "Green", "Coffee"]; (* 6. They drink coffee in the green house. *) t = EliminatePoss[t, "Pall Mall", "Birds"]; (* 7. The man who smokes Pall Mall has birds.*) t = EliminatePoss[t, "Yellow", "Dunhill"]; (* 8. In the yellow house they smoke Dunhill. *) t = EliminatePoss[t, "Blue Master", "Beer"]; (*13. The man who smokes Blue Master drinks beer. *) t = EliminatePoss[t, "German", "Prince"]; (* 14. The German smokes Prince. *) (* 9. In the middle house they drink milk. *) poss = Position[t, "Milk"]; delpos = Select[poss, #2 != 3 &]; t = ReplacePart[t, delpos -> Null];
(* 10. The Norwegian lives in the first house. *) poss = Position[t, "Norwegian"]; delpos = Select[poss, #2 != 1 &]; t = ReplacePart[t, delpos -> Null]; (* 15. The Norwegian lives next to the blue house.*) poss1 = Position[t, "Norwegian"]; poss2 = Position[t, "Blue"]; poss = Tuples[{poss1, poss2}]; poss = Select[poss, #1, 2 + 1 == #2, 2 \[Or] #1, 2 - 1 == #2, 2 &]\[Transpose]; delpos = Complement[poss1, poss1]; t = ReplacePart[t, delpos -> Null]; delpos = Complement[poss2, poss2]; t = ReplacePart[t, delpos -> Null]; (* 5. The green house is immediately to the left of the white house. *) poss1 = Position[t, "Green"]; poss2 = Position[t, "White"]; poss = Tuples[{poss1, poss2}]; poss = Select[poss, #1, 2 + 1 == #2, 2 &]\[Transpose]; delpos = Complement[poss1, poss1]; t = ReplacePart[t, delpos -> Null]; delpos = Complement[poss2, poss2]; t = ReplacePart[t, delpos -> Null]; (*11. The man who smokes Blend lives in the house next to the house with cats.*) poss1 = Position[t, "Blend"]; poss2 = Position[t, "Cats"]; poss = Tuples[{poss1, poss2}]; poss = Select[poss, #1, 2 + 1 == #2, 2 \[Or] #1, 2 - 1 == #2, 2 &]\[Transpose]; delpos = Complement[poss1, poss1]; t = ReplacePart[t, delpos -> Null]; delpos = Complement[poss2, poss2]; t = ReplacePart[t, delpos -> Null]; (* 12. In a house next to the house where they have a horse, they smoke Dunhill. *) poss1 = Position[t, "Horse"]; poss2 = Position[t, "Dunhill"]; poss = Tuples[{poss1, poss2}]; poss = Select[poss, #1, 2 + 1 == #2, 2 \[Or] #1, 2 - 1 == #2, 2 &]\[Transpose]; delpos = Complement[poss1, poss1]; t = ReplacePart[t, delpos -> Null]; delpos = Complement[poss2, poss2]; t = ReplacePart[t, delpos -> Null]; (* 16. They drink water in a house next to the house where they smoke Blend. *) poss1 = Position[t, "Water"]; poss2 = Position[t, "Blend"]; poss = Tuples[{poss1, poss2}]; poss = Select[poss, #1, 2 + 1 == #2, 2 \[Or] #1, 2 - 1 == #2, 2 &]\[Transpose]; delpos = Complement[poss1, poss1]; t = ReplacePart[t, delpos -> Null]; delpos = Complement[poss2, poss2]; t = ReplacePart[t, delpos -> Null]; (*General rule 1 in a line => cross out vertical and horizontal lines*) (* 1 in a row*) vals = Select[t, Count[#, Null] == 4 &]; vals = DeleteCases[Flatten[vals], Null]; poss = Flatten[Position[t, #] & /@ vals, 1]; delpos = With[{r = First[#], c = Last[#]}, {#, c} & /@ (Range[-4, 0] + Ceiling[r, 5])] & /@ poss; (*delete in columns*) delpos = Flatten[MapThread[DeleteCases, {delpos, poss}], 1]; t = ReplacePart[t, delpos -> Null]; (* 1 in a column*) sets = Flatten[Table[{i + k*5, j}, {k, 0, 4}, {j, 1, 5}, {i, 1, 5}],1]; sets = {#, Extract[t, #]} & /@ sets; sets = Select[sets, Count[#2, Null] == 4 &]; sets = Flatten[Transpose /@ sets, 1]; sets = DeleteCases[sets, {{_, _}, Null}]; delpos = setsAll, 1;(*delete in rows*) delpos = With[{r = First[#], c = Last[#]}, {r, #} & /@ (DeleteCases[Range[5], c])] & /@ delpos; delpos = Flatten[delpos, 1]; t = ReplacePart[t, delpos -> Null]; t
] colors = {"Blue", "Green", "Red", "White", "Yellow"}; nationality = {"Dane", "English", "German", "Norwegian", "Swede"}; beverage = {"Beer", "Coffee", "Milk", "Tea", "Water"}; animal = {"Birds", "Cats", "Dog", "Horse", "Zebra"}; smoke = {"Blend", "Blue Master", "Dunhill", "Pall Mall", "Prince"}; vals = {colors, nationality, beverage, animal, smoke}; bigtable = Join @@ (ConstantArray[#, 5]\[Transpose] & /@ vals);
bigtable = FixedPoint[FilterPuzzle, bigtable]; TableForm[DeleteCases[bigtable\[Transpose], Null, \[Infinity]], TableHeadings -> {Range[5], None}]</lang> Using the command FixedPoint, we iteratively filter out these candidates, until we are (hopefully) left with 1 candidate per kind per house.
- Output:
1 Yellow Norwegian Water Cats Dunhill 2 Blue Dane Tea Horse Blend 3 Red English Milk Birds Pall Mall 4 Green German Coffee Zebra Prince 5 White Swede Beer Dog Blue Master
MiniZinc
<lang MiniZinc> %Solve Zebra Puzzle. Nigel Galloway, August 27th., 2019 include "alldifferent.mzn"; enum N={English,Swedish,Danish,German,Norwegian}; enum I={Tea,Coffee,Milk,Beer,Water}; enum G={Dog,Birds,Cats,Horse,Zebra}; enum E={Red,Green,White,Blue,Yellow}; enum L={PallMall,Dunhill,BlueMaster,Prince,Blend}; array[1..5] of var N: Nz; constraint alldifferent(Nz); constraint Nz[1]=Norwegian; %The Norwegian lives in the first house. array[1..5] of var I: Iz; constraint alldifferent(Iz); constraint Iz[3]=Milk; %In the middle house they drink milk. array[1..5] of var G: Gz; constraint alldifferent(Gz); array[1..5] of var E: Ez; constraint alldifferent(Ez); array[1..5] of var L: Lz; constraint alldifferent(Lz); constraint exists(n in 1..5)(Nz[n]=English /\ Ez[n]=Red); %The English man lives in the red house constraint exists(n in 1..5)(Nz[n]=Swedish /\ Gz[n]=Dog); %The Swede has a dog. constraint exists(n in 1..5)(Nz[n]=Danish /\ Iz[n]=Tea); %The Dane drinks tea. constraint exists(n in 1..4)(Ez[n]=Green /\ Ez[n+1]=White); %The green house is immediately to the left of the white house. constraint exists(n in 1..5)(Ez[n]=Green /\ Iz[n]=Coffee); %They drink coffee in the green house. constraint exists(n in 1..5)(Lz[n]=PallMall /\ Gz[n]=Birds); %The man who smokes Pall Mall has birds constraint exists(n in 1..5)(Ez[n]=Yellow /\ Lz[n]=Dunhill); %In the yellow house they smoke Dunhill. constraint exists(n in 1..4)((Lz[n]=Blend /\ Gz[n+1]=Cats) \/ (Lz[n+1]=Blend /\ Gz[n]=Cats)); %The man who smokes Blend lives in the house next to the house with cats. constraint exists(n in 1..4)((Gz[n]=Horse /\ Lz[n+1]=Dunhill) \/ (Gz[n+1]=Horse /\ Lz[n]=Dunhill)); %In a house next to the house where they have a horse, they smoke Dunhill. constraint exists(n in 1..5)(Lz[n]=BlueMaster /\ Iz[n]=Beer); %The man who smokes Blue Master drinks beer. constraint exists(n in 1..5)(Nz[n]=German /\ Lz[n]=Prince); %The German smokes Prince. constraint exists(n in 1..4)((Nz[n]=Norwegian /\ Ez[n+1]=Blue) \/ (Nz[n+1]=Norwegian /\ Ez[n]=Blue));%The Norwegian lives next to the blue house. constraint exists(n in 1..4)((Lz[n]=Blend /\ Iz[n+1]=Water) \/ (Lz[n+1]=Blend /\ Iz[n]=Water)); %They drink water in a house next to the house where they smoke Blend. var 1..5: n; constraint Gz[n]=Zebra; solve satisfy; output ["The "++show(Nz[n])++" owns the zebra"++"\n\n"++show(Nz)++"\n"++show(Iz)++"\n"++show(Gz)++"\n"++show(Ez)++"\n"++show(Lz)++"\n"]; </lang>
- Output:
The German owns the zebra [Norwegian, Danish, English, German, Swedish] [Water, Tea, Milk, Coffee, Beer] [Cats, Horse, Birds, Zebra, Dog] [Yellow, Blue, Red, Green, White] [Dunhill, Blend, PallMall, Prince, BlueMaster] ----------
Nial
I put a backslash ( \ ) at the end of many lines so that the code can be pasted into the REPL.
<lang Nial> remove is op x xs {filter (not (x =)) xs}
append_map is transformer func op seq { \
reduce (op x xs { (func x) link xs}) (seq append []) }
permutations is op seq { \
if empty seq then [[]] else \
(append_map \
(op head {each (op tail {head hitch tail}) \
(permutations (remove head seq))}) \
seq) \
endif}
f is find tokenize is op str{string_split ' ' str} mk is tr pred op str {filter pred permutations tokenize str} eq is op x xs y ys{f x xs = f y ys} adj is op x xs y ys{1 = abs(f x xs - f y ys)}
run is { \
men := mk (op xs {0 = f 'norwegian' xs}) \
'danish english german norwegian swedish'; \
colors := mk (op xs {1 = ((f 'white' xs) - (f 'green' xs))}) \
'blue green red white yellow'; \
drinks := mk (op xs {2 = f 'milk' xs}) 'beer coffee milk tea water'; \
pets := mk (op xs {l}) 'birds cats dog horse zebra'; \
smokes := mk (op xs {l}) 'blend blue-master dunhill pall-mall prince'; \
for m with men do \
for c with colors do \
if (eq 'english' m 'red' c) and \
(adj 'norwegian' m 'blue' c) then \
for d with drinks do \
if (eq 'danish' m 'tea' d) and \
(eq 'coffee' d 'green' c) then \
for s with smokes do \
if (eq 'yellow' c 'dunhill' s) and \
(eq 'blue-master' s 'beer' d) and \
(eq 'german' m 'prince' s) then \
for p with pets do \
if (eq 'birds' p 'pall-mall' s) and \
(eq 'swedish' m 'dog' p) and \
(adj 'blend' s 'cats' p) and \
(adj 'horse' p 'dunhill' s) then \
write (0 blend (p m c d s)) \
endif \
endfor \
endif \
endfor \
endif \
endfor \
endif \
endfor \
endfor }
abs(time - (run; time)) </lang>
- Output:
+-----------------------------------------+ |cats |norwegian|yellow|water |dunhill | +-----+---------+------+------+-----------| |horse|danish |blue |tea |blend | +-----+---------+------+------+-----------| |birds|english |red |milk |pall-mall | +-----+---------+------+------+-----------| |zebra|german |green |coffee|prince | +-----+---------+------+------+-----------| |dog |swedish |white |beer |blue-master| +-----------------------------------------+ 0.03
Pari/Gp
<lang pari/gp> perm(arr) = { n=#arr;i=n-1; while(i > -1,if (arr[i] < arr[i+1],break);i--); j=n; while(arr[j]<= arr[i],j -=1); tmp = arr[i] ;arr[i]=arr[j];arr[j]=tmp; i +=1; j = n; while(i < j ,tmp = arr[i] ;arr[i]=arr[j];arr[j]=tmp; i +=1; j -=1); return(arr); } perms(arr)={ n=#arr; result = List(); listput(result,arr); for(i=1,n!-1,arr=perm(arr);listput(result,arr)); return(result); }
adj(x,xs,y,ys)={
abs(select(z->z==x,xs,1)[1] - select(z->z==y,ys,1)[1])==1;
} eq(x,xs,y,ys)={
select(z->z==x,xs,1) == select(z->z==y,ys,1);
}
colors =Vec(perms( ["Blue", "Green", "Red", "White", "Yellow"]));; drinks =Vec(perms( ["Beer", "Coffee", "Milk", "Tea", "Water"]));; nations =Vec(perms( ["Denmark", "England", "Germany", "Norway", "Sweden"]));; smokes =Vec(perms( ["Blend", "BlueMaster", "Dunhill", "PallMall", "Prince"]));; pets =Vec(perms( ["Birds", "Cats", "Dog", "Horse", "Zebra"]));;; colors= select(x->select(z->z=="White",x,1)[1] - select(z->z=="Green",x,1)[1]==1,colors); drinks=select(x->x[3]=="Milk",drinks); nations=select(x->x[1]=="Norway",nations);
for(n=1,#nations,for(c=1,#colors,\ if(eq("Red",colors[c],"England",nations[n]) && adj("Norway",nations[n],"Blue",colors[c]),\ for(d=1,#drinks,\ if(eq("Denmark",nations[n],"Tea",drinks[d])&& eq("Coffee",drinks[d],"Green",colors[c]),\ for(s=1,#smokes,\ if(eq("Yellow",colors[c],"Dunhill",smokes[s]) &&\ eq("BlueMaster",smokes[s],"Beer",drinks[d]) &&\ eq("Germany",nations[n],"Prince",smokes[s]),\ for(p=1,#pets,\ if(eq("Birds",pets[p],"PallMall",smokes[s]) &&\ eq("Sweden",nations[n],"Dog",pets[p]) &&\ adj("Blend",smokes[s],"Cats",pets[p]) &&\ adj("Horse",pets[p],"Dunhill",smokes[s]),\ print("Zebra is owned by ",nations[n][select(z->z=="Zebra",pets[p],1)[1]]);print();\ for(i=1,5,printf("House:%s %6s %10s %10s %10s %10s\n",i,colors[c][i],nations[n][i],pets[p][i],drinks[d][i],smokes[s][i]));\ ))))))))); </lang>
- Output:
Zebra is owned by Germany House:1 Yellow Norway Cats Water Dunhill House:2 Blue Denmark Horse Tea Blend House:3 Red England Birds Milk PallMall House:4 Green Germany Zebra Coffee Prince House:5 White Sweden Dog Beer BlueMaster
Perl
Basically the same idea as C, though of course it's much easier to have Perl generate Perl code. <lang perl>#!/usr/bin/perl
use utf8; use strict; binmode STDOUT, ":utf8";
my (@tgt, %names); sub setprops { my %h = @_; my @p = keys %h; for my $p (@p) { my @v = @{ $h{$p} }; @tgt = map(+{idx=>$_-1, map{ ($_, undef) } @p}, 1 .. @v) unless @tgt; $names{$_} = $p for @v; } }
my $solve = sub { for my $i (@tgt) { printf("%12s", ucfirst($i->{$_} // "¿Qué?")) for reverse sort keys %$i; print "\n"; } "there is only one" # <--- change this to a false value to find all solutions (if any) };
sub pair { my ($a, $b, @v) = @_; if ($a =~ /^(\d+)$/) { $tgt[$1]{ $names{$b} } = $b; return; }
@v = (0) unless @v; my %allowed; $allowed{$_} = 1 for @v;
my ($p1, $p2) = ($names{$a}, $names{$b});
my $e = $solve; $solve = sub { # <--- sorta like how TeX \let...\def macro my ($x, $y);
($x) = grep { $_->{$p1} eq $a } @tgt; ($y) = grep { $_->{$p2} eq $b } @tgt;
$x and $y and return $allowed{ $x->{idx} - $y->{idx} } && $e->();
my $try_stuff = sub { my ($this, $p, $v, $sign) = @_; for (@v) { my $i = $this->{idx} + $sign * $_; next unless $i >= 0 && $i < @tgt && !$tgt[$i]{$p}; local $tgt[$i]{$p} = $v; $e->() and return 1; } return };
$x and return $try_stuff->($x, $p2, $b, 1); $y and return $try_stuff->($y, $p1, $a, -1);
for $x (@tgt) { next if $x->{$p1}; local $x->{$p1} = $a; $try_stuff->($x, $p2, $b, 1) and return 1; } }; }
- ---- above should be generic for all similar puzzles ---- #
- ---- below: per puzzle setup ---- #
- property names and values
setprops ( # Svensk n. a Swede, not a swede (kålrot). # AEnglisk (from middle Viking "Æŋløsåksen") n. a Brit. 'Who' => [ qw(Deutsch Svensk Norske Danske AEnglisk) ], 'Pet' => [ qw(birds dog horse zebra cats) ], 'Drink' => [ qw(water tea milk beer coffee) ], 'Smoke' => [ qw(dunhill blue_master prince blend pall_mall) ], 'Color' => [ qw(red green yellow white blue) ] );
- constraints
pair qw( AEnglisk red ); pair qw( Svensk dog ); pair qw( Danske tea ); pair qw( green white 1 ); # "to the left of" can mean either 1 or -1: ambiguous pair qw( coffee green ); pair qw( pall_mall birds ); pair qw( yellow dunhill ); pair qw( 2 milk ); pair qw( 0 Norske ); pair qw( blend cats -1 1 ); pair qw( horse dunhill -1 1 ); pair qw( blue_master beer ); # Nicht das Deutsche Bier trinken? Huh. pair qw( Deutsch prince ); pair qw( Norske blue -1 1 ); pair qw( water blend -1 1 );
$solve->();</lang>
Incidentally, the same logic can be used to solve the dwelling problem, if somewhat awkwardly: <lang perl>...
- property names and values
setprops 'Who' => [ qw(baker cooper fletcher miller smith) ], 'Level' => [ qw(one two three four five) ];
- constraints
pair qw(0 one); pair qw(1 two); pair qw(2 three); pair qw(3 four); pair qw(4 five); pair qw(baker five -4 -3 -2 -1 1 2 3 4); pair qw(cooper one -4 -3 -2 -1 1 2 3 4); pair qw(fletcher one -4 -3 -2 -1 1 2 3 4); pair qw(fletcher five -4 -3 -2 -1 1 2 3 4); pair qw(miller cooper -1 -2 -3 -4); pair qw(smith fletcher 4 3 2 -2 -3 -4); pair qw(cooper fletcher 4 3 2 -2 -3 -4);
$solve->();</lang>
Phix
<lang Phix>enum Colour, Nationality, Drink, Smoke, Pet constant Colours = {"red","white","green","yellow","blue"},
Nationalities = {"English","Swede","Dane","Norwegian","German"},
Drinks = {"tea","coffee","milk","beer","water"},
Smokes = {"Pall Mall","Dunhill","Blend","Blue Master","Prince"},
Pets = {"dog","birds","cats","horse","zebra"},
Sets = {Colours,Nationalities,Drinks,Smokes,Pets}
constant tagset5 = tagset(5) -- {1,2,3,4,5}, oft-permuted sequence perm = repeat(tagset5,5) -- perm[1] is Colour of each house, etc
function house(integer i, string name)
return find(find(name,Sets[i]),perm[i])
end function
function left_of(integer h1, integer h2)
return (h1-h2)==1
end function
function next_to(integer h1, integer h2)
return abs(h1-h2)==1
end function
procedure print_house(integer i)
printf(1,"%d:%s,%s,%s,%s,%s\n",{i,
Colours[perm[Colour][i]],
Nationalities[perm[Nationality][i]],
Drinks[perm[Drink][i]],
Smokes[perm[Smoke][i]],
Pets[perm[Pet][i]]})
end procedure
integer solutions = 0 sequence solperms = {} atom t0 = time() constant factorial5 = factorial(5) for C=1 to factorial5 do
perm[Colour] = permute(C,tagset5)
if left_of(house(Colour,"green"),house(Colour,"white")) then
for N=1 to factorial5 do
perm[Nationality] = permute(N,tagset5)
if house(Nationality,"Norwegian")==1
and house(Nationality,"English")==house(Colour,"red")
and next_to(house(Nationality,"Norwegian"),house(Colour,"blue")) then
for D=1 to factorial5 do
perm[Drink] = permute(D,tagset5)
if house(Nationality,"Dane")==house(Drink,"tea")
and house(Drink,"coffee")==house(Colour,"green")
and house(Drink,"milk")==3 then
for S=1 to factorial5 do
perm[Smoke] = permute(S,tagset5)
if house(Colour,"yellow")==house(Smoke,"Dunhill")
and house(Nationality,"German")==house(Smoke,"Prince")
and house(Smoke,"Blue Master")==house(Drink,"beer")
and next_to(house(Drink,"water"),house(Smoke,"Blend")) then
for P=1 to factorial5 do
perm[Pet] = permute(P,tagset5)
if house(Nationality,"Swede")==house(Pet,"dog")
and house(Smoke,"Pall Mall")==house(Pet,"birds")
and next_to(house(Smoke,"Blend"),house(Pet,"cats"))
and next_to(house(Pet,"horse"),house(Smoke,"Dunhill")) then
for i=1 to 5 do
print_house(i)
end for
solutions += 1
solperms = append(solperms,perm)
end if
end for
end if
end for
end if
end for
end if
end for
end if
end for printf(1,"%d solution%s found (%3.3fs).\n",{solutions,iff(solutions>1,"s",""),time()-t0}) for i=1 to length(solperms) do
perm = solperms[i]
printf(1,"The %s owns the Zebra\n",{Nationalities[house(Pet,"zebra")]})
end for</lang>
- Output:
1:yellow,Norwegian,water,Dunhill,cats 2:blue,Dane,tea,Blend,horse 3:red,English,milk,Pall Mall,birds 4:green,German,coffee,Prince,zebra 5:white,Swede,beer,Blue Master,dog 1 solution found (0.016s). The German owns the Zebra
PicoLisp
<lang PicoLisp>(be match (@House @Person @Drink @Pet @Cigarettes)
(permute (red blue green yellow white) @House) (left-of @House white @House green)
(permute (Norwegian English Swede German Dane) @Person) (has @Person English @House red) (equal @Person (Norwegian . @)) (next-to @Person Norwegian @House blue)
(permute (tea coffee milk beer water) @Drink) (has @Drink tea @Person Dane) (has @Drink coffee @House green) (equal @Drink (@ @ milk . @))
(permute (dog birds cats horse zebra) @Pet) (has @Pet dog @Person Swede)
(permute (Pall-Mall Dunhill Blend Blue-Master Prince) @Cigarettes) (has @Cigarettes Pall-Mall @Pet birds) (has @Cigarettes Dunhill @House yellow) (next-to @Cigarettes Blend @Pet cats) (next-to @Cigarettes Dunhill @Pet horse) (has @Cigarettes Blue-Master @Drink beer) (has @Cigarettes Prince @Person German)
(next-to @Drink water @Cigarettes Blend) )
(be has ((@A . @X) @A (@B . @Y) @B)) (be has ((@ . @X) @A (@ . @Y) @B)
(has @X @A @Y @B) )
(be right-of ((@A . @X) @A (@ @B . @Y) @B)) (be right-of ((@ . @X) @A (@ . @Y) @B)
(right-of @X @A @Y @B) )
(be left-of ((@ @A . @X) @A (@B . @Y) @B)) (be left-of ((@ . @X) @A (@ . @Y) @B)
(left-of @X @A @Y @B) )
(be next-to (@X @A @Y @B) (right-of @X @A @Y @B)) (be next-to (@X @A @Y @B) (left-of @X @A @Y @B))</lang> Test: <lang PicoLisp>(pilog '((match @House @Person @Drink @Pet @Cigarettes))
(let Fmt (-8 -11 -8 -7 -11)
(tab Fmt "HOUSE" "PERSON" "DRINKS" "HAS" "SMOKES")
(mapc '(@ (pass tab Fmt))
@House @Person @Drink @Pet @Cigarettes ) ) )</lang>
Output:
HOUSE PERSON DRINKS HAS SMOKES yellow Norwegian water cats Dunhill blue Dane tea horse Blend red English milk birds Pall-Mall green German coffee zebra Prince white Swede beer dog Blue-Master
Prolog
In Prolog we can specify the domain by selecting elements from it, making mutually exclusive choices for efficiency:
<lang Prolog>select([A|As],S):- select(A,S,S1),select(As,S1). select([],_).
next_to(A,B,C):- left_of(A,B,C) ; left_of(B,A,C). left_of(A,B,C):- append(_,[A,B|_],C).
zebra(Owns, HS):- % color,nation,pet,drink,smokes
HS = [ h(_,norwegian,_,_,_), _, h(_,_,_,milk,_), _, _],
select( [ h(red,englishman,_,_,_), h(_,swede,dog,_,_),
h(_,dane,_,tea,_), h(_,german,_,_,prince) ], HS),
select( [ h(_,_,birds,_,pallmall), h(yellow,_,_,_,dunhill),
h(_,_,_,beer,bluemaster) ], HS),
left_of( h(green,_,_,coffee,_), h(white,_,_,_,_), HS),
next_to( h(_,_,_,_,dunhill), h(_,_,horse,_,_), HS),
next_to( h(_,_,_,_,blend), h(_,_,cats, _,_), HS),
next_to( h(_,_,_,_,blend), h(_,_,_,water,_), HS),
next_to( h(_,norwegian,_,_,_), h(blue,_,_,_,_), HS),
member( h(_,Owns,zebra,_,_), HS).
- - ?- time(( zebra(Who, HS), maplist(writeln,HS), nl, write(Who), nl, nl, fail
; write('No more solutions.') )).</lang>
Output:
h(yellow,norwegian, cats, water, dunhill) h(blue, dane, horse, tea, blend) h(red, englishman,birds, milk, pallmall) h(green, german, zebra, coffee,prince) h(white, swede, dog, beer, bluemaster) german No more solutions. % 5,959 inferences, 0.000 CPU in 0.060 seconds (0% CPU, Infinite Lips) true.
Works with SWI-Prolog. More verbose translation of the specification works as well.
Direct rule by rule translation
Using extensible records, letting the houses' attributes be discovered from rules, not predetermined by a programmer (as any "a house has five attributes" solution is doing).
<lang Prolog>% attribute store is 'Name - Value' pairs with unique names attrs( H, [N-V | R]) :- !, memberchk( N-X, H), X = V, (R = [], ! ; attrs( H, R)). attrs( HS, AttrsL) :- maplist( attrs, HS, AttrsL).
in( HS, Attrs) :- in( member, HS, Attrs). in( G, HS, Attrs) :- call( G, A, HS), attrs( A, Attrs).
left_of( [A,B], HS) :- append( _, [A,B | _], HS). next_to( [A,B], HS) :- left_of( [A,B], HS) ; left_of( [B,A], HS).
zebra( Owner, Houses):-
Houses = [A,_,C,_,_], % 1
maplist( in(Houses), [ [ nation-englishman, color-red ] % 2
, [ nation-swede, owns -dog ] % 3
, [ nation-dane, drink-tea ] % 4
, [ drink -coffee, color-green ] % 6
, [ smoke -'Pall Mall', owns -birds ] % 7
, [ color -yellow, smoke-'Dunhill' ] % 8
, [ drink -beer, smoke-'Blue Master'] % 13
, [ nation-german, smoke-'Prince' ] % 14
] ),
in( left_of, Houses, [[color -green ], [color -white ]]), % 5
in( left_of, [C,A], [[drink -milk ], [nation-norwegian]]), % 9, 10
maplist( in( next_to, Houses),
[ [[smoke -'Blend' ], [owns -cats ]] % 11
, [[owns -horse ], [smoke-'Dunhill' ]] % 12
, [[nation-norwegian], [color-blue ]] % 15
, [[drink -water ], [smoke-'Blend' ]] % 16
] ),
in( Houses, [owns-zebra, nation-Owner]).</lang>
Output: <lang Prolog>?- time(( zebra(Z,HS), (maplist(length,HS,_) -> maplist(sort,HS,S),
maplist(writeln,S),nl,writeln(Z)), false ; writeln('No More Solutions'))).
[color-yellow,drink-water, nation-norwegian, owns-cats, smoke-Dunhill ] [color-blue, drink-tea, nation-dane, owns-horse, smoke-Blend ] [color-red, drink-milk, nation-englishman,owns-birds, smoke-Pall Mall ] [color-green, drink-coffee,nation-german, owns-zebra, smoke-Prince ] [color-white, drink-beer, nation-swede, owns-dog, smoke-Blue Master]
german No More Solutions % 263,486 inferences, 0.047 CPU in 0.063 seconds (74% CPU, 5630007 Lips)</lang>
Alternative version
<lang prolog>:- initialization(main).
zebra(X) :-
houses(Hs), member(h(_,X,zebra,_,_), Hs)
, findall(_, (member(H,Hs), write(H), nl), _), nl
, write('the one who keeps zebra: '), write(X), nl
.
houses(Hs) :-
Hs = [_,_,_,_,_] % 1 , H3 = h(_,_,_,milk,_), Hs = [_,_,H3,_,_] % 9 , H1 = h(_,nvg,_,_,_ ), Hs = [H1|_] % 10
, maplist( flip(member,Hs),
[ h(red,eng,_,_,_) % 2
, h(_,swe,dog,_,_) % 3
, h(_,dan,_,tea,_) % 4
, h(green,_,_,coffe,_) % 6
, h(_,_,birds,_,pm) % 7
, h(yellow,_,_,_,dh) % 8
, h(_,_,_,beer,bm) % 13
, h(_,ger,_,_,pri) % 14
])
, infix([ h(green,_,_,_,_)
, h(white,_,_,_,_) ], Hs) % 5
, maplist( flip(nextto,Hs),
[ [h(_,_,_,_,bl ), h(_,_,cats,_,_)] % 11
, [h(_,_,horse,_,_), h(_,_,_,_,dh )] % 12
, [h(_,nvg,_,_,_ ), h(blue,_,_,_,_)] % 15
, [h(_,_,_,water,_), h(_,_,_,_,bl )] % 16
])
.
flip(F,X,Y) :- call(F,Y,X).
infix(Xs,Ys) :- append(Xs,_,Zs) , append(_,Zs,Ys). nextto(P,Xs) :- permutation(P,R), infix(R,Xs).
main :- findall(_, (zebra(_), nl), _), halt.
</lang>
- Output:
h(yellow,nvg,cats,water,dh) h(blue,dan,horse,tea,bl) h(red,eng,birds,milk,pm) h(green,ger,zebra,coffe,pri) h(white,swe,dog,beer,bm) the one who keeps zebra: ger
Constraint Programming version
Original source code is 'ECLiPSe ( http://eclipseclp.org/ )' example: http://eclipseclp.org/examples/zebra.ecl.txt <lang prolog>:- use_module(library(clpfd)).
zebra :-
Nation = [Englishman, Spaniard, Japanese, Ukrainian, Norwegian ], Color = [Red, Green, White, Yellow, Blue ], Smoke = [Oldgold, Kools, Chesterfield, Luckystrike, Parliament], Pet = [Dog, Snails, Fox, Horse, Zebra ], Drink = [Tea, Coffee, Milk, Orangejuice, Water ],
% house numbers 1 to 5 Nation ins 1..5, Color ins 1..5, Smoke ins 1..5, Pet ins 1..5, Drink ins 1..5, % the values in each list are exclusive all_different(Nation), all_different(Color), all_different(Smoke), all_different(Pet), all_different(Drink), % actual constraints Englishman #= Red, Spaniard #= Dog, Green #= Coffee, Ukrainian #= Tea, Green #= White + 1, Oldgold #= Snails, Yellow #= Kools, Milk #= 3, Norwegian #= 1, (Chesterfield #= Fox - 1 #\/ Chesterfield #= Fox + 1), (Kools #= Horse - 1 #\/ Kools #= Horse + 1), Luckystrike #= Orangejuice, Japanese #= Parliament, (Norwegian #= Blue - 1 #\/ Norwegian #= Blue + 1),
% get solution flatten([Nation, Color, Smoke, Pet, Drink], List), label(List), % print the answers sort([Englishman-englishman, Spaniard-spaniard, Japanese-japanese, Ukrainian-ukrainian, Norwegian-norwegian], NationNames), sort([Red-red, Green-green, White-white, Yellow-yellow, Blue-blue], ColorNames), sort([Oldgold-oldgold, Kools-kools, Chesterfield-chesterfield, Luckystrike-luckystrike, Parliament-parliament], SmokeNames), sort([Dog-dog, Snails-snails, Fox-fox, Horse-horse, Zebra-zebra], PetNames), sort([Tea-tea, Coffee-coffee, Milk-milk, Orangejuice-orangejuice, Water-water], DrinkNames), Fmt = '~w~16|~w~32|~w~48|~w~64|~w~n', format(Fmt, NationNames), format(Fmt, ColorNames), format(Fmt, SmokeNames), format(Fmt, PetNames), format(Fmt, DrinkNames).
</lang>
- Output:
1-norwegian 2-ukrainian 3-englishman 4-spaniard 5-japanese 1-yellow 2-blue 3-red 4-white 5-green 1-kools 2-chesterfield 3-oldgold 4-luckystrike 5-parliament 1-fox 2-horse 3-snails 4-dog 5-zebra 1-water 2-tea 3-milk 4-orangejuice 5-coffee
Python
Using 'logpy': https://github.com/logpy/logpy <lang python> from logpy import * from logpy.core import lall import time
def lefto(q, p, list): # give me q such that q is left of p in list # zip(list, list[1:]) gives a list of 2-tuples of neighboring combinations # which can then be pattern-matched against the query return membero((q,p), zip(list, list[1:]))
def nexto(q, p, list): # give me q such that q is next to p in list # match lefto(q, p) OR lefto(p, q) # requirement of vector args instead of tuples doesn't seem to be documented return conde([lefto(q, p, list)], [lefto(p, q, list)])
houses = var()
zebraRules = lall( # there are 5 houses (eq, (var(), var(), var(), var(), var()), houses), # the Englishman's house is red (membero, ('Englishman', var(), var(), var(), 'red'), houses), # the Swede has a dog (membero, ('Swede', var(), var(), 'dog', var()), houses), # the Dane drinks tea (membero, ('Dane', var(), 'tea', var(), var()), houses), # the Green house is left of the White house (lefto, (var(), var(), var(), var(), 'green'), (var(), var(), var(), var(), 'white'), houses), # coffee is the drink of the green house (membero, (var(), var(), 'coffee', var(), 'green'), houses), # the Pall Mall smoker has birds (membero, (var(), 'Pall Mall', var(), 'birds', var()), houses), # the yellow house smokes Dunhills (membero, (var(), 'Dunhill', var(), var(), 'yellow'), houses), # the middle house drinks milk (eq, (var(), var(), (var(), var(), 'milk', var(), var()), var(), var()), houses), # the Norwegian is the first house (eq, (('Norwegian', var(), var(), var(), var()), var(), var(), var(), var()), houses), # the Blend smoker is in the house next to the house with cats (nexto, (var(), 'Blend', var(), var(), var()), (var(), var(), var(), 'cats', var()), houses), # the Dunhill smoker is next to the house where they have a horse (nexto, (var(), 'Dunhill', var(), var(), var()), (var(), var(), var(), 'horse', var()), houses), # the Blue Master smoker drinks beer (membero, (var(), 'Blue Master', 'beer', var(), var()), houses), # the German smokes Prince (membero, ('German', 'Prince', var(), var(), var()), houses), # the Norwegian is next to the blue house (nexto, ('Norwegian', var(), var(), var(), var()), (var(), var(), var(), var(), 'blue'), houses), # the house next to the Blend smoker drinks water (nexto, (var(), 'Blend', var(), var(), var()), (var(), var(), 'water', var(), var()), houses), # one of the houses has a zebra--but whose? (membero, (var(), var(), var(), 'zebra', var()), houses) )
t0 = time.time() solutions = run(0, houses, zebraRules) t1 = time.time() dur = t1-t0
count = len(solutions) zebraOwner = [house for house in solutions[0] if 'zebra' in house][0][0]
print "%i solutions in %.2f seconds" % (count, dur) print "The %s is the owner of the zebra" % zebraOwner print "Here are all the houses:" for line in solutions[0]: print str(line) </lang>
Alternative Version
<lang python>import psyco; psyco.full()
class Content: elems= """Beer Coffee Milk Tea Water
Danish English German Norwegian Swedish
Blue Green Red White Yellow
Blend BlueMaster Dunhill PallMall Prince
Bird Cat Dog Horse Zebra""".split()
class Test: elems= "Drink Person Color Smoke Pet".split() class House: elems= "One Two Three Four Five".split()
for c in (Content, Test, House):
c.values = range(len(c.elems)) for i, e in enumerate(c.elems): exec "%s.%s = %d" % (c.__name__, e, i)
def finalChecks(M):
def diff(a, b, ca, cb):
for h1 in House.values:
for h2 in House.values:
if M[ca][h1] == a and M[cb][h2] == b:
return h1 - h2
assert False
return abs(diff(Content.Norwegian, Content.Blue,
Test.Person, Test.Color)) == 1 and \
diff(Content.Green, Content.White,
Test.Color, Test.Color) == -1 and \
abs(diff(Content.Horse, Content.Dunhill,
Test.Pet, Test.Smoke)) == 1 and \
abs(diff(Content.Water, Content.Blend,
Test.Drink, Test.Smoke)) == 1 and \
abs(diff(Content.Blend, Content.Cat,
Test.Smoke, Test.Pet)) == 1
def constrained(M, atest):
if atest == Test.Drink:
return M[Test.Drink][House.Three] == Content.Milk
elif atest == Test.Person:
for h in House.values:
if ((M[Test.Person][h] == Content.Norwegian and
h != House.One) or
(M[Test.Person][h] == Content.Danish and
M[Test.Drink][h] != Content.Tea)):
return False
return True
elif atest == Test.Color:
for h in House.values:
if ((M[Test.Person][h] == Content.English and
M[Test.Color][h] != Content.Red) or
(M[Test.Drink][h] == Content.Coffee and
M[Test.Color][h] != Content.Green)):
return False
return True
elif atest == Test.Smoke:
for h in House.values:
if ((M[Test.Color][h] == Content.Yellow and
M[Test.Smoke][h] != Content.Dunhill) or
(M[Test.Smoke][h] == Content.BlueMaster and
M[Test.Drink][h] != Content.Beer) or
(M[Test.Person][h] == Content.German and
M[Test.Smoke][h] != Content.Prince)):
return False
return True
elif atest == Test.Pet:
for h in House.values:
if ((M[Test.Person][h] == Content.Swedish and
M[Test.Pet][h] != Content.Dog) or
(M[Test.Smoke][h] == Content.PallMall and
M[Test.Pet][h] != Content.Bird)):
return False
return finalChecks(M)
def show(M):
for h in House.values:
print "%5s:" % House.elems[h],
for t in Test.values:
print "%10s" % Content.elems[M[t][h]],
print
def solve(M, t, n):
if n == 1 and constrained(M, t):
if t < 4:
solve(M, Test.values[t + 1], 5)
else:
show(M)
return
for i in xrange(n):
solve(M, t, n - 1)
M[t][0 if n % 2 else i], M[t][n - 1] = \
M[t][n - 1], M[t][0 if n % 2 else i]
def main():
M = [[None] * len(Test.elems) for _ in xrange(len(House.elems))]
for t in Test.values:
for h in House.values:
M[t][h] = Content.values[t * 5 + h]
solve(M, Test.Drink, 5)
main()</lang>
- Output:
One: Water Norwegian Yellow Dunhill Cat Two: Tea Danish Blue Blend Horse Three: Milk English Red PallMall Bird Four: Coffee German Green Prince Zebra Five: Beer Swedish White BlueMaster Dog
Runtime about 0.18 seconds.
Alternative Version
<lang python>from itertools import permutations
class Number:elems= "One Two Three Four Five".split() class Color: elems= "Red Green Blue White Yellow".split() class Drink: elems= "Milk Coffee Water Beer Tea".split() class Smoke: elems= "PallMall Dunhill Blend BlueMaster Prince".split() class Pet: elems= "Dog Cat Zebra Horse Bird".split() class Nation:elems= "British Swedish Danish Norvegian German".split()
for c in (Number, Color, Drink, Smoke, Pet, Nation):
for i, e in enumerate(c.elems): exec "%s.%s = %d" % (c.__name__, e, i)
def show_row(t, data):
print "%6s: %12s%12s%12s%12s%12s" % ( t.__name__, t.elems[data[0]], t.elems[data[1]], t.elems[data[2]], t.elems[data[3]], t.elems[data[4]])
def main():
perms = list(permutations(range(5)))
for number in perms:
if number[Nation.Norvegian] == Number.One: # Constraint 10
for color in perms:
if color[Nation.British] == Color.Red: # Constraint 2
if number[color.index(Color.Blue)] == Number.Two: # Constraint 15+10
if number[color.index(Color.White)] - number[color.index(Color.Green)] == 1: # Constraint 5
for drink in perms:
if drink[Nation.Danish] == Drink.Tea: # Constraint 4
if drink[color.index(Color.Green)] == Drink.Coffee: # Constraint 6
if drink[number.index(Number.Three)] == Drink.Milk: # Constraint 9
for smoke in perms:
if smoke[Nation.German] == Smoke.Prince: # Constraint 14
if drink[smoke.index(Smoke.BlueMaster)] == Drink.Beer: # Constraint 13
if smoke[color.index(Color.Yellow)] == Smoke.Dunhill: # Constraint 8
if number[smoke.index(Smoke.Blend)] - number[drink.index(Drink.Water)] in (1, -1): # Constraint 16
for pet in perms:
if pet[Nation.Swedish] == Pet.Dog: # Constraint 3
if pet[smoke.index(Smoke.PallMall)] == Pet.Bird: # Constraint 7
if number[pet.index(Pet.Horse)] - number[smoke.index(Smoke.Dunhill)] in (1, -1): # Constraint 12
if number[smoke.index(Smoke.Blend)] - number[pet.index(Pet.Cat)] in (1, -1): # Constraint 11
print "Found a solution:"
show_row(Nation, range(5))
show_row(Number, number)
show_row(Color, color)
show_row(Drink, drink)
show_row(Smoke, smoke)
show_row(Pet, pet)
print
main()</lang> Output:
Found a solution: Nation: British Swedish Danish Norvegian German Number: Three Five Two One Four Color: Red White Blue Yellow Green Drink: Milk Beer Tea Water Coffee Smoke: PallMall BlueMaster Blend Dunhill Prince Pet: Bird Dog Horse Cat Zebra
Runtime about 0.0013 seconds
Constraint Programming Version
Using 'python-constraint': http://labix.org/python-constraint, Original source code is 'ECLiPSe ( http://eclipseclp.org/ )' example: http://eclipseclp.org/examples/zebra.ecl.txt <lang python>from constraint import *
problem = Problem()
Nation = ["Englishman", "Spaniard", "Japanese", "Ukrainian", "Norwegian" ] Color = ["Red", "Green", "White", "Yellow", "Blue" ] Smoke = ["Oldgold", "Kools", "Chesterfield", "Luckystrike", "Parliament"] Pet = ["Dog", "Snails", "Fox", "Horse", "Zebra" ] Drink = ["Tea", "Coffee", "Milk", "Orangejuice", "Water" ]
- add variables: house numbers 1 to 5
problem.addVariables(Nation, range(1,5+1)) problem.addVariables(Color, range(1,5+1)) problem.addVariables(Smoke, range(1,5+1)) problem.addVariables(Pet, range(1,5+1)) problem.addVariables(Drink, range(1,5+1))
- add constraint: the values in each list are exclusive
problem.addConstraint(AllDifferentConstraint(), Nation) problem.addConstraint(AllDifferentConstraint(), Color) problem.addConstraint(AllDifferentConstraint(), Smoke) problem.addConstraint(AllDifferentConstraint(), Pet) problem.addConstraint(AllDifferentConstraint(), Drink)
- add constraint: actual constraints
problem.addConstraint(lambda a, b: a == b, ["Englishman", "Red" ]) problem.addConstraint(lambda a, b: a == b, ["Spaniard", "Dog" ]) problem.addConstraint(lambda a, b: a == b, ["Green", "Coffee" ]) problem.addConstraint(lambda a, b: a == b, ["Ukrainian", "Tea" ]) problem.addConstraint(lambda a, b: a == b + 1, ["Green", "White" ]) problem.addConstraint(lambda a, b: a == b, ["Oldgold", "Snails" ]) problem.addConstraint(lambda a, b: a == b, ["Yellow", "Kools" ]) problem.addConstraint(lambda a: a == 3, ["Milk" ]) problem.addConstraint(lambda a: a == 1, ["Norwegian" ]) problem.addConstraint(lambda a, b: a == b - 1 or a == b + 1, ["Chesterfield", "Fox" ]) problem.addConstraint(lambda a, b: a == b - 1 or a == b + 1, ["Kools", "Horse" ]) problem.addConstraint(lambda a, b: a == b, ["Luckystrike", "Orangejuice"]) problem.addConstraint(lambda a, b: a == b, ["Japanese", "Parliament" ]) problem.addConstraint(lambda a, b: a == b - 1 or a == b + 1, ["Norwegian", "Blue" ])
- get solution
sol = problem.getSolution()
- print the answers
nation = ["Nation" if i == 0 else "" for i in range(6)] color = ["Color" if i == 0 else "" for i in range(6)] smoke = ["Smoke" if i == 0 else "" for i in range(6)] pet = ["Pet" if i == 0 else "" for i in range(6)] drink = ["Drink" if i == 0 else "" for i in range(6)] for n in Nation:
nation[sol[n]] = n
for n in Color:
color[sol[n]] = n
for n in Smoke:
smoke[sol[n]] = n
for n in Pet:
pet[sol[n]] = n
for n in Drink:
drink[sol[n]] = n
for d in [nation, color, smoke, pet, drink]:
print("%6s: %14s%14s%14s%14s%14s" % (d[0], d[1], d[2], d[3], d[4], d[5]))
</lang> Output:
Nation: Norwegian Ukrainian Englishman Spaniard Japanese Color: Yellow Blue Red White Green Smoke: Kools Chesterfield Oldgold Luckystrike Parliament Pet: Fox Horse Snails Dog Zebra Drink: Water Tea Milk Orangejuice Coffee
R
<lang R>
library(combinat)
col <- factor(c("Red","Green","White","Yellow","Blue")) own <- factor(c("English","Swedish","Danish","German","Norwegian")) pet <- factor(c("Dog","Birds","Cats","Horse","Zebra")) drink <- factor(c("Coffee","Tea","Milk","Beer","Water")) smoke <- factor(c("PallMall", "Blend", "Dunhill", "BlueMaster", "Prince"))
col_p <- permn(levels(col)) own_p <- permn(levels(own)) pet_p <- permn(levels(pet)) drink_p <- permn(levels(drink)) smoke_p <- permn(levels(smoke))
imright <- function(h1,h2){
return(h1-h2==1)
}
nextto <- function(h1,h2){
return(abs(h1-h2)==1)
}
house_with <- function(f,val){
return(which(levels(f)==val))
}
for (i in seq(length(col_p))){
col <- factor(col, levels=col_pi) if (imright(house_with(col,"Green"),house_with(col,"White"))) { for (j in seq(length(own_p))){ own <- factor(own, levels=own_pj) if(house_with(own,"English") == house_with(col,"Red")){ if(house_with(own,"Norwegian") == 1){ if(nextto(house_with(own,"Norwegian"),house_with(col,"Blue"))){ for(k in seq(length(drink_p))){ drink <- factor(drink, levels=drink_pk) if(house_with(drink,"Coffee") == house_with(col,"Green")){ if(house_with(own,"Danish") == house_with(drink,"Tea")){ if(house_with(drink,"Milk") == 3){ for(l in seq(length(smoke_p))){ smoke <- factor(smoke, levels=smoke_pl) if(house_with(smoke,"Dunhill") == house_with(col,"Yellow")){ if(house_with(smoke,"BlueMaster") == house_with(drink,"Beer")){ if(house_with(own,"German") == house_with(smoke,"Prince")){ if(nextto(house_with(smoke,"Blend"),house_with(drink,"Water"))){ for(m in seq(length(pet_p))){ pet <- factor(pet, levels=pet_pm) if(house_with(own,"Swedish") == house_with(pet,"Dog")){ if(house_with(smoke,"PallMall") == house_with(pet,"Birds")){ if(nextto(house_with(smoke,"Blend"),house_with(pet,"Cats"))){ if(nextto(house_with(smoke,"Dunhill"),house_with(pet,"Horse"))){ res <- sapply(list(own,col,pet,smoke,drink),levels) colnames(res) <- c("Nationality","Colour","Pet","Drink","Smoke") print(res) } } } } } } } } } } } } } } } } } } }
}</lang>
- Output:
Nationality Colour Pet Drink Smoke [1,] "Norwegian" "Yellow" "Cats" "Dunhill" "Water" [2,] "Danish" "Blue" "Horse" "Blend" "Tea" [3,] "English" "Red" "Birds" "PallMall" "Milk" [4,] "Swedish" "White" "Dog" "BlueMaster" "Beer" [5,] "German" "Green" "Zebra" "Prince" "Coffee"
Racket
<lang racket>#lang racket
(require racklog)
(define %select
(%rel (x xs S S1)
[(x (cons x xs) xs)]
[(x (cons S xs) (cons S S1)) (%select x xs S1)]
[((cons x xs) S)
(%select x S S1)
(%select xs S1)]
[('() (_))]))
(define %next-to
(%rel (A B C)
[(A B C)
(%or (%left-of A B C)
(%left-of B A C))]))
(define %left-of
(%rel (A B C) [(A B C) (%append (_) (cons A (cons B (_))) C)]))
(define %zebra
(%rel (Owns HS)
[(Owns HS)
(%is HS (list (list (_) 'norwegian (_) (_) (_))
(_)
(list (_) (_) (_) 'milk (_))
(_) (_)))
(%select (list (list 'red 'englishman (_) (_) (_))
(list (_) 'swede 'dog (_) (_))
(list (_) 'dane (_) 'tea (_))
(list (_) 'german (_) (_) 'prince))
HS)
(%select (list (list (_) (_) 'birds (_) 'pallmall)
(list 'yellow (_) (_) (_) 'dunhill)
(list (_) (_) (_) 'beer 'bluemaster))
HS)
(%left-of (list 'green (_) (_) 'coffee (_))
(list 'white (_) (_) (_) (_))
HS)
(%next-to (list (_) (_) (_) (_) 'dunhill)
(list (_) (_) 'horse (_) (_))
HS)
(%next-to (list (_) (_) (_) (_) 'blend)
(list (_) (_) 'cats (_) (_))
HS)
(%next-to (list (_) (_) (_) (_) 'blend)
(list (_) (_) (_) 'water (_))
HS)
(%next-to (list (_) 'norwegian (_) (_) (_))
(list 'blue (_) (_) (_) (_))
HS)
(%member (list (_) Owns 'zebra (_) (_)) HS)]))
(%which (Who HS) (%zebra Who HS))</lang>
Output:
'((Who . german) (HS (yellow norwegian cats water dunhill) (blue dane horse tea blend) (red englishman birds milk pallmall) (green german zebra coffee prince) (white swede dog beer bluemaster)))
Raku
(formerly Perl 6) A rule driven approach: <lang perl6>my Hash @houses = (1 .. 5).map: { %(:num($_)) }; # 1 there are five houses
my @facts = (
{ :nat<English>, :color<red> }, # 2 The English man lives in the red house.
{ :nat<Swede>, :pet<dog> }, # 3 The Swede has a dog.
{ :nat<Dane>, :drink<tea> }, # 4 The Dane drinks tea.
{ :color<green>, :Left-Of(:color<white>) }, # 5 the green house is immediately to the left of the white house
{ :drink<coffee>, :color<green> }, # 6 They drink coffee in the green house.
{ :smoke<Pall-Mall>, :pet<birds> }, # 7 The man who smokes Pall Mall has birds.
{ :color<yellow>, :smoke<Dunhill> }, # 8 In the yellow house they smoke Dunhill.
{ :num(3), :drink<milk> }, # 9 In the middle house they drink milk.
{ :num(1), :nat<Norwegian> }, # 10 The Norwegian lives in the first house.
{ :smoke<Blend>, :Next-To(:pet<cats>) }, # 11 The man who smokes Blend lives in the house next to the house with cats.
{ :pet<horse>, :Next-To(:smoke<Dunhill>) }, # 12 In a house next to the house where they have a horse, they smoke Dunhill.
{ :smoke<Blue-Master>, :drink<beer> }, # 13 The man who smokes Blue Master drinks beer.
{ :nat<German>, :smoke<Prince> }, # 14 The German smokes Prince.
{ :nat<Norwegian>, :Next-To(:color<blue>) }, # 15 The Norwegian lives next to the blue house.
{ :drink<water>, :Next-To(:smoke<Blend>) }, # 16 They drink water in a house next to the house where they smoke Blend.
{ :pet<zebra> }, # who owns this?
);
sub MAIN {
for gather solve(@houses, @facts) {
#-- output
say .[0].pairs.sort.map(*.key.uc.fmt("%-9s")).join(' | ');
say .pairs.sort.map(*.value.fmt("%-9s")).join(' | ')
for .list;
last; # stop after first solution
}
}
- | found a solution that fits all the facts
multi sub solve(@solution, @facts [ ]) {
take @solution;
}
- | extend a scenario to cover the next fact
multi sub solve(@scenario, [ $fact, *@facts ] is copy) {
for gather choices(@scenario, |$fact) {
solve(@$_, @facts)
}
}
- | find all possible solutions for pairs of houses with
- | %a attributes, left of a house with %b attributes
multi sub choices(@houses, :Left-Of(%b)!, *%a) {
my @scenarios;
for @houses {
my $idx = .<num> - 1;
if $idx > 0 && plausible(@houses[$idx-1], %a) && plausible(@houses[$idx], %b) {
my @scenario = @houses.clone;
@scenario[$idx-1] = %( %(@houses[$idx-1]), %a );
@scenario[$idx] = %( %(@houses[$idx]), %b );
take @scenario;
}
}
}
- | find all possible pairs of houses with %a attributes, either side
- ! of a house with %b attributes
multi sub choices(@houses, :Next-To(%b)!, *%a ) {
choices(@houses, |%a, :Left-Of(%b) ); choices(@houses, |%b, :Left-Of(%a) );
}
- | find all possible houses that match the given attributes
multi sub choices(@houses, *%fact) {
for @houses.grep({plausible($_, %fact)}) -> $house {
my @scenario = @houses.clone;
my $idx = $house<num> - 1;
@scenario[$idx] = %( %$house, %fact );
take @scenario;
}
}
- | plausible if doesn't conflict with anything
sub plausible(%house, %atts) {
all %atts.keys.map: { (%house{$_}:!exists) || %house{$_} eq %atts{$_} };
} </lang>
- Output:
COLOR | DRINK | NAT | NUM | PET | SMOKE yellow | water | Norwegian | 1 | cats | Dunhill blue | tea | Dane | 2 | horse | Blend red | milk | English | 3 | birds | Pall-Mall green | coffee | German | 4 | zebra | Prince white | beer | Swede | 5 | dog | Blue-Master
Note: Facts can be shuffled by changing line 3 to my @facts = pick *, (. It seems to reliably find solutions, although execution times will vary (from under 1 sec up to about 10sec).
REXX
Permutations algorithm taken from REXX <lang rexx>/* REXX ---------------------------------------------------------------
- Solve the Zebra Puzzle
- --------------------------------------------------------------------*/
Call mk_perm /* compute all permutations */
Call encode /* encode the elements of the specifications */
/* ex2 .. eg16 the formalized specifications */
solutions=0
Call time 'R'
Do nation_i = 1 TO 120
Nations = perm.nation_i
IF ex10() Then Do
Do color_i = 1 TO 120
Colors = perm.color_i
IF ex5() & ex2() & ex15() Then Do
Do drink_i = 1 TO 120
Drinks = perm.drink_i
IF ex9() & ex4() & ex6() Then Do
Do smoke_i = 1 TO 120
Smokes = perm.smoke_i
IF ex14() & ex13() & ex16() & ex8() Then Do
Do animal_i = 1 TO 120
Animals = perm.animal_i
IF ex3() & ex7() & ex11() & ex12() Then Do
/* Call out 'Drinks =' Drinks 54321 Wat Tea Mil Cof Bee */
/* Call out 'Nations=' Nations 41235 Nor Den Eng Ger Swe */
/* Call out 'Colors =' Colors 51324 Yel Blu Red Gre Whi */
/* Call out 'Smokes =' Smokes 31452 Dun Ble Pal Pri Blu */
/* Call out 'Animals=' Animals 24153 Cat Hor Bir Zeb Dog */
Call out 'House Drink Nation Colour'||,
' Smoke Animal'
Do i=1 To 5
di=substr(drinks,i,1)
ni=substr(nations,i,1)
ci=substr(colors,i,1)
si=substr(smokes,i,1)
ai=substr(animals,i,1)
ol.i=right(i,3)' '||left(drink.di,11),
||left(nation.ni,11),
||left(color.ci,11),
||left(smoke.si,11),
||left(animal.ai,11)
Call out ol.i
End
solutions=solutions+1
End
End /* animal_i */
End
End /* smoke_i */
End
End /* drink_i */
End
End /* color_i */
End
End /* nation_i */
Say 'Number of solutions =' solutions
Say 'Solved in' time('E') 'seconds'
Exit
/*------------------------------------------------------------------------------
#There are five houses.
ex2: #The English man lives in the red house. ex3: #The Swede has a dog. ex4: #The Dane drinks tea. ex5: #The green house is immediately to the left of the white house. ex6: #They drink coffee in the green house. ex7: #The man who smokes Pall Mall has birds. ex8: #In the yellow house they smoke Dunhill. ex9: #In the middle house they drink milk. ex10: #The Norwegian lives in the first house. ex11: #The man who smokes Blend lives in the house next to the house with cats. ex12: #In a house next to the house where they have a horse, they smoke Dunhill. ex13: #The man who smokes Blue Master drinks beer. ex14: #The German smokes Prince. ex15: #The Norwegian lives next to the blue house. ex16: #They drink water in a house next to the house where they smoke Blend.
*/
ex2: Return pos(England,Nations)=pos(Red,Colors) ex3: Return pos(Sweden,Nations)=pos(Dog,Animals) ex4: Return pos(Denmark,Nations)=pos(Tea,Drinks) ex5: Return pos(Green,Colors)=pos(White,Colors)-1 ex6: Return pos(Coffee,Drinks)=pos(Green,Colors) ex7: Return pos(PallMall,Smokes)=pos(Birds,Animals) ex8: Return pos(Dunhill,Smokes)=pos(Yellow,Colors) ex9: Return substr(Drinks,3,1)=Milk ex10: Return left(Nations,1)=Norway ex11: Return abs(pos(Blend,Smokes)-pos(Cats,Animals))=1 ex12: Return abs(pos(Dunhill,Smokes)-pos(Horse,Animals))=1 ex13: Return pos(BlueMaster,Smokes)=pos(Beer,Drinks) ex14: Return pos(Germany,Nations)=pos(Prince,Smokes) ex15: Return abs(pos(Norway,Nations)-pos(Blue,Colors))=1 ex16: Return abs(pos(Blend,Smokes)-pos(Water,Drinks))=1
mk_perm: Procedure Expose perm. /*---------------------------------------------------------------------
- Make all permutations of 12345 in perm.*
- --------------------------------------------------------------------*/
perm.=0 n=5 Do pop=1 For n
p.pop=pop End
Call store Do While nextperm(n,0)
Call store End
Return
nextperm: Procedure Expose p. perm.
Parse Arg n,i
nm=n-1
Do k=nm By-1 For nm
kp=k+1
If p.k<p.kp Then Do
i=k
Leave
End
End
Do j=i+1 While j<n
Parse Value p.j p.n With p.n p.j
n=n-1
End
If i>0 Then Do
Do j=i+1 While p.j<p.i
End
Parse Value p.j p.i With p.i p.j
End
Return i>0
store: Procedure Expose p. perm.
z=perm.0+1
_=
Do j=1 To 5
_=_||p.j
End
perm.z=_
perm.0=z
Return
encode:
Beer=1 ; Drink.1='Beer' Coffee=2 ; Drink.2='Coffee' Milk=3 ; Drink.3='Milk' Tea=4 ; Drink.4='Tea' Water=5 ; Drink.5='Water' Denmark=1 ; Nation.1='Denmark' England=2 ; Nation.2='England' Germany=3 ; Nation.3='Germany' Norway=4 ; Nation.4='Norway' Sweden=5 ; Nation.5='Sweden' Blue=1 ; Color.1='Blue' Green=2 ; Color.2='Green' Red=3 ; Color.3='Red' White=4 ; Color.4='White' Yellow=5 ; Color.5='Yellow' Blend=1 ; Smoke.1='Blend' BlueMaster=2 ; Smoke.2='BlueMaster' Dunhill=3 ; Smoke.3='Dunhill' PallMall=4 ; Smoke.4='PallMall' Prince=5 ; Smoke.5='Prince' Birds=1 ; Animal.1='Birds' Cats=2 ; Animal.2='Cats' Dog=3 ; Animal.3='Dog' Horse=4 ; Animal.4='Horse' Zebra=5 ; Animal.5='Zebra' Return
out:
Say arg(1) Return</lang>
- Output:
House Drink Nation Colour Smoke Animal 1 Water Norway Yellow Dunhill Cats 2 Tea Denmark Blue Blend Horse 3 Milk England Red PallMall Birds 4 Coffee Germany Green Prince Zebra 5 Beer Sweden White BlueMaster Dog Number of solutions = 1 Solved in 0.063000 seconds
Ruby
<lang ruby>CONTENT = { House: ,
Nationality: %i[English Swedish Danish Norwegian German],
Colour: %i[Red Green White Blue Yellow],
Pet: %i[Dog Birds Cats Horse Zebra],
Drink: %i[Tea Coffee Milk Beer Water],
Smoke: %i[PallMall Dunhill BlueMaster Prince Blend] }
def adjacent? (n,i,g,e)
(0..3).any?{|x| (n[x]==i and g[x+1]==e) or (n[x+1]==i and g[x]==e)}
end
def leftof? (n,i,g,e)
(0..3).any?{|x| n[x]==i and g[x+1]==e}
end
def coincident? (n,i,g,e)
n.each_index.any?{|x| n[x]==i and g[x]==e}
end
def solve_zebra_puzzle
CONTENT[:Nationality].permutation{|nation|
next unless nation.first == :Norwegian # 10
CONTENT[:Colour].permutation{|colour|
next unless leftof?(colour, :Green, colour, :White) # 5
next unless coincident?(nation, :English, colour, :Red) # 2
next unless adjacent?(nation, :Norwegian, colour, :Blue) # 15
CONTENT[:Pet].permutation{|pet|
next unless coincident?(nation, :Swedish, pet, :Dog) # 3
CONTENT[:Drink].permutation{|drink|
next unless drink[2] == :Milk # 9
next unless coincident?(nation, :Danish, drink, :Tea) # 4
next unless coincident?(colour, :Green, drink, :Coffee) # 6
CONTENT[:Smoke].permutation{|smoke|
next unless coincident?(smoke, :PallMall, pet, :Birds) # 7
next unless coincident?(smoke, :Dunhill, colour, :Yellow) # 8
next unless coincident?(smoke, :BlueMaster, drink, :Beer) # 13
next unless coincident?(smoke, :Prince, nation, :German) # 14
next unless adjacent?(smoke, :Blend, pet, :Cats) # 11
next unless adjacent?(smoke, :Blend, drink, :Water) # 16
next unless adjacent?(smoke, :Dunhill,pet, :Horse) # 12
print_out(nation, colour, pet, drink, smoke)
} } } } }
end
def print_out (nation, colour, pet, drink, smoke)
width = CONTENT.map{|x| x.flatten.map{|y|y.size}.max}
fmt = width.map{|w| "%-#{w}s"}.join(" ")
national = nation[ pet.find_index(:Zebra) ]
puts "The Zebra is owned by the man who is #{national}",""
puts fmt % CONTENT.keys, fmt % width.map{|w| "-"*w}
[nation,colour,pet,drink,smoke].transpose.each.with_index(1){|x,n| puts fmt % [n,*x]}
end
solve_zebra_puzzle</lang>
- Output:
The Zebra is owned by the man who is German House Nationality Colour Pet Drink Smoke ----- ----------- ------ ----- ------ ---------- 1 Norwegian Yellow Cats Water Dunhill 2 Danish Blue Horse Tea Blend 3 English Red Birds Milk PallMall 4 German Green Zebra Coffee Prince 5 Swedish White Dog Beer BlueMaster
Another approach
<lang ruby> class String; def brk; split(/(?=[A-Z])/); end; end men,drinks,colors,pets,smokes = "NorwegianGermanDaneSwedeEnglish
MilkTeaBeerWaterCoffeeGreenWhiteRedYellowBlueZebraDogCatsHorseBirds
PallmallDunhillBlendBluemasterPrince".delete(" \n").
brk.each_slice(5).map{|e| e.permutation.to_a};
men.select!{|x| "Norwegian"==x[0]}; drinks.select!{|x| "Milk"==x[2]}; colors.select!{|x| x.join[/GreenWhite/]};
dis = proc{|s,*a| s.brk.map{|w| a.map{|p| p.index(w)}.
compact[0]}.each_slice(2).map{|a,b| (a-b).abs}}
men.each{|m| colors.each{|c|
next unless dis["RedEnglishBlueNorwegian",c,m]==[0,1]
drinks.each{|d| next unless dis["DaneTeaCoffeeGreen",m,d,c]==[0,0]
smokes.each{|s|
next unless dis["YellowDunhillBluemasterBeerGermanPrince",
c,s,d,m]==[0,0,0]
pets.each{|p|
next unless dis["SwedeDogBirdsPallmallCatsBlendHorseDunhill",
m,p,s]==[0,0,1,1]
x = [p,m,c,d,s].transpose
puts "The #{x.find{|y|y[0]=="Zebra"}[1]} owns the zebra.",
x.map{|y| y.map{|z| z.ljust(11)}.join}}}}}}
</lang> Output:
The German owns the zebra. Cats Norwegian Yellow Water Dunhill Horse Dane Blue Tea Blend Birds English Red Milk Pallmall Zebra German Green Coffee Prince Dog Swede White Beer Bluemaster
Scala
Idiomatic (for comprehension)
<lang Scala>/* Note to the rules:
* * It can further concluded that: * 5a: The green house cannot be at the h1 position * 5b: The white house cannot be at the h5 position * * 16: This rule is redundant. */
object Einstein extends App {
val possibleMembers = for { // pair clues results in 78 members
nationality <- List("Norwegian", "German", "Dane", "Englishman", "Swede")
color <- List("Red", "Green", "Yellow", "White", "Blue")
beverage <- List("Milk", "Coffee", "Tea", "Beer", "Water")
animal <- List("Dog", "Horse", "Birds", "Cats", "Zebra")
brand <- List("Blend", "Pall Mall", "Prince", "Blue Master", "Dunhill")
if (color == "Red") == (nationality == "Englishman") // #2
if (nationality == "Swede") == (animal == "Dog") // #3
if (nationality == "Dane") == (beverage == "Tea") // #4
if (color == "Green") == (beverage == "Coffee") // #6
if (brand == "Pall Mall") == (animal == "Birds") // #7
if (brand == "Dunhill") == (color == "Yellow") // #8
if (brand == "Blue Master") == (beverage == "Beer") // #13
if (brand == "Prince") == (nationality == "German") // #14
} yield new House(nationality, color, beverage, animal, brand)
val members = for { // Neighborhood clues
h1 <- housesLeftOver().filter(p => (p.nationality == "Norwegian" /* #10 */) && (p.color != "Green") /* #5a */) // 28
h3 <- housesLeftOver(h1).filter(p => p.beverage == "Milk") // #9 // 24
h2 <- housesLeftOver(h1, h3).filter(_.color == "Blue") // #15
if matchMiddleBrandAnimal(h1, h2, h3, "Blend", "Cats") // #11
if matchCornerBrandAnimal(h1, h2, "Horse", "Dunhill") // #12
h4 <- housesLeftOver(h1, h2, h3).filter(_.checkAdjacentWhite(h3) /* #5 */)
h5 <- housesLeftOver(h1, h2, h3, h4)
// Redundant tests if h2.checkAdjacentWhite(h1) if h3.checkAdjacentWhite(h2) if matchCornerBrandAnimal(h5, h4, "Horse", "Dunhill") if matchMiddleBrandAnimal(h2, h3, h4, "Blend", "Cats") if matchMiddleBrandAnimal(h3, h4, h5, "Blend", "Cats") } yield Seq(h1, h2, h3, h4, h5)
def matchMiddleBrandAnimal(home1: House, home2: House, home3: House, brand: String, animal: String) =
(home1.animal == animal || home2.brand != brand || home3.animal == animal) &&
(home1.brand == brand || home2.animal != animal || home3.brand == brand)
def matchCornerBrandAnimal(corner: House, inner: House, animal: String, brand: String) = (corner.brand != brand || inner.animal == animal) && (corner.animal == animal || inner.brand != brand)
def housesLeftOver(pickedHouses: House*): List[House] = {
possibleMembers.filter(house => pickedHouses.forall(_.totalUnEqual(house)))
}
class House(val nationality: String, val color: String, val beverage: String, val animal: String, val brand: String) {
override def toString = {
f"$nationality%10s, ${color + ", "}%-8s$beverage,\t$animal,\t$brand."
}
def totalUnEqual(home2: House) =
this.animal != home2.animal &&
this.beverage != home2.beverage &&
this.brand != home2.brand &&
this.color != home2.color &&
this.nationality != home2.nationality
//** Checks if the this green house is next to the other white house*/ def checkAdjacentWhite(home2: House) = (this.color == "Green") == (home2.color == "White") // #5 }
{ // Main program
val beest = "Zebra"
members.flatMap(p => p.filter(p => p.animal == beest)).
foreach(s => println(s"The ${s.nationality} is the owner of the ${beest.toLowerCase}."))
println(s"The ${members.size} solution(s) are:")
members.foreach(solution => solution.zipWithIndex.foreach(h => println(s"House ${h._2 + 1} ${h._1}")))
}
} // loc 58</lang>
- Output:
See it in running in your browser by ScalaFiddle (JavaScript executed in browser) or by Scastie (remote JVM).
The German is the owner of the zebra. The 1 solution(s) are: House 1 Norwegian, Yellow, Water, Cats, Dunhill. House 2 Dane, Blue, Tea, Horse, Blend. House 3 Englishman, Red, Milk, Birds, Pall Mall. House 4 Swede, White, Beer, Dog, Blue Master. House 5 German, Green, Coffee, Zebra, Prince.
Scala Alternate Version (Over-engineered)
<lang scala>import scala.util.Try
object Einstein extends App {
// The strategy here is to mount a brute-force attack on the solution space, pruning very aggressively. // The scala standard `permutations` method is extremely helpful here. It turns out that by pruning // quickly and smartly we can solve this very quickly (45ms on my machine) compared to days or weeks // required to fully enumerate the solution space.
// We set up a for comprehension with an enumerator for each of the 5 variables, with if clauses to // prune. The hard part is the pruning logic, which is basically just translating the rules to code // and the data model. The data model is basically Seq[Seq[String]]
// Rules are encoded as for comprehension filters. There is a natural cascade of rules from // depending on more or less criteria. The rules about smokes are the most complex and depend // on the most other factors
// 4. The green house is just to the left of the white one.
def colorRules(colors: Seq[String]) = Try(colors(colors.indexOf("White") - 1) == "Green").getOrElse(false)
// 1. The Englishman lives in the red house.
// 9. The Norwegian lives in the first house.
// 14. The Norwegian lives next to the blue house.
def natRules(colors: Seq[String], nats: Seq[String]) =
nats.head == "Norwegian" && colors(nats.indexOf("Brit")) == "Red" &&
(Try(colors(nats.indexOf("Norwegian") - 1) == "Blue").getOrElse(false) ||
Try(colors(nats.indexOf("Norwegian") + 1) == "Blue").getOrElse(false))
// 3. The Dane drinks tea.
// 5. The owner of the green house drinks coffee.
// 8. The man in the center house drinks milk.
def drinkRules(colors: Seq[String], nats: Seq[String], drinks: Seq[String]) =
drinks(nats.indexOf("Dane")) == "Tea" &&
drinks(colors.indexOf("Green")) == "Coffee" &&
drinks(2) == "Milk"
// 2. The Swede keeps dogs.
def petRules(nats: Seq[String], pets: Seq[String]) = pets(nats.indexOf("Swede")) == "Dogs"
// 6. The Pall Mall smoker keeps birds.
// 7. The owner of the yellow house smokes Dunhills.
// 10. The Blend smoker has a neighbor who keeps cats.
// 11. The man who smokes Blue Masters drinks bier.
// 12. The man who keeps horses lives next to the Dunhill smoker.
// 13. The German smokes Prince.
// 15. The Blend smoker has a neighbor who drinks water.
def smokeRules(colors: Seq[String], nats: Seq[String], drinks: Seq[String], pets: Seq[String], smokes: Seq[String]) =
pets(smokes.indexOf("Pall Mall")) == "Birds" &&
smokes(colors.indexOf("Yellow")) == "Dunhill" &&
(Try(pets(smokes.indexOf("Blend") - 1) == "Cats").getOrElse(false) ||
Try(pets(smokes.indexOf("Blend") + 1) == "Cats").getOrElse(false)) &&
drinks(smokes.indexOf("BlueMaster")) == "Beer" &&
(Try(smokes(pets.indexOf("Horses") - 1) == "Dunhill").getOrElse(false) ||
Try(pets(pets.indexOf("Horses") + 1) == "Dunhill").getOrElse(false)) &&
smokes(nats.indexOf("German")) == "Prince" &&
(Try(drinks(smokes.indexOf("Blend") - 1) == "Water").getOrElse(false) ||
Try(drinks(smokes.indexOf("Blend") + 1) == "Water").getOrElse(false))
// once the rules are created it, the actual solution is simple: iterate brute force, pruning early.
val solutions = for {
colors <- Seq("Red", "Blue", "White", "Green", "Yellow").permutations if colorRules(colors)
nats <- Seq("Brit", "Swede", "Dane", "Norwegian", "German").permutations if natRules(colors, nats)
drinks <- Seq("Tea", "Coffee", "Milk", "Beer", "Water").permutations if drinkRules(colors, nats, drinks)
pets <- Seq("Dogs", "Birds", "Cats", "Horses", "Fish").permutations if petRules(nats, pets)
smokes <- Seq("BlueMaster", "Blend", "Pall Mall", "Dunhill", "Prince").permutations if smokeRules(colors, nats, drinks, pets, smokes)
} yield Seq(colors, nats, drinks, pets, smokes)
// There *should* be just one solution...
solutions.foreach { solution =>
// so we can pretty-print, find out the maximum string length of all cells
val maxLen = solution.flatten.map(_.length).max
def pretty(str: String): String = str + (" " * (maxLen - str.length + 1))
// a labels column
val labels = ("" +: Seq("Color", "Nation", "Drink", "Pet", "Smoke").map(_ + ":")).toIterator
// print each row including a column header
((1 to 5).map(n => s"House $n") +: solution).map(_.map(pretty)).map(x => (pretty(labels.next) +: x).mkString(" ")).foreach(println)
println(s"\nThe ${solution(1)(solution(3).indexOf("Fish"))} owns the Fish")
}
}// loc 38</lang>
- Output:
Experience running it in your browser by ScalaFiddle (JavaScript executed in browser) or by Scastie (remote JVM).
- Output:
House 1 House 2 House 3 House 4 House 5 Color: Yellow Blue Red Green White Nation: Norwegian Dane Brit German Swede Drink: Water Tea Milk Coffee Beer Pet: Cats Horses Birds Fish Dogs Smoke: Dunhill Blend Pall Mall Prince BlueMaster The German owns the Fish
Sidef
<lang ruby>var CONTENT = Hash(
:House => nil,
:Nationality => [:English, :Swedish, :Danish, :Norwegian, :German],
:Colour => [:Red, :Green, :White, :Blue, :Yellow],
:Pet => [:Dog, :Birds, :Cats, :Horse, :Zebra],
:Drink => [:Tea, :Coffee, :Milk, :Beer, :Water],
:Smoke => [:PallMall, :Dunhill, :BlueMaster, :Prince, :Blend]
)
func adjacent(n,i,g,e) {
(0..3).any {|x| (n[x]==i && g[x+1]==e) || (n[x+1]==i && g[x]==e) }
}
func leftof(n,i,g,e) {
(0..3).any {|x| n[x]==i && g[x+1]==e }
}
func coincident(n,i,g,e) {
n.indices.any {|x| n[x]==i && g[x]==e }
}
func solve {
CONTENT{:Nationality}.permutations{|*nation|
nation.first == :Norwegian ->
&& CONTENT{:Colour}.permutations {|*colour|
leftof(colour,:Green,colour,:White) ->
&& coincident(nation,:English,colour,:Red) ->
&& adjacent(nation,:Norwegian,colour,:Blue) ->
&& CONTENT{:Pet}.permutations {|*pet|
coincident(nation,:Swedish,pet,:Dog) ->
&& CONTENT{:Drink}.permutations {|*drink|
drink[2] == :Milk ->
&& coincident(nation,:Danish,drink,:Tea) ->
&& coincident(colour,:Green,drink,:Coffee) ->
&& CONTENT{:Smoke}.permutations {|*smoke|
coincident(smoke,:PallMall,pet,:Birds) ->
&& coincident(smoke,:Dunhill,colour,:Yellow) ->
&& coincident(smoke,:BlueMaster,drink,:Beer) ->
&& coincident(smoke,:Prince,nation,:German) ->
&& adjacent(smoke,:Blend,pet,:Cats) ->
&& adjacent(smoke,:Blend,drink,:Water) ->
&& adjacent(smoke,:Dunhill,pet,:Horse) ->
&& return [nation,colour,pet,drink,smoke]
} } } } } }
var res = solve(); var keys = [:House, :Nationality, :Colour, :Pet, :Drink, :Smoke] var width = keys.map{ .len } var fmt = width.map{|w| "%-#{w+2}s" }.join(" ") say "The Zebra is owned by the man who is #{res[0][res[2].first_index(:Zebra)]}\n" say (fmt % keys..., "\n", fmt % width.map{|w| "-"*w }...) res[0].indices.map{|i| res.map{|a| a[i] }}.each_kv {|k,v| say fmt%(k,v...) }</lang>
- Output:
The Zebra is owned by the man who is German House Nationality Colour Pet Drink Smoke ----- ----------- ------ --- ----- ----- 0 Norwegian Yellow Cats Water Dunhill 1 Danish Blue Horse Tea Blend 2 English Red Birds Milk PallMall 3 German Green Zebra Coffee Prince 4 Swedish White Dog Beer BlueMaster
Standard ML
(This implementation uses the search algorithm of the C++ implementation, but the rules check algorithm is different.)
<lang sml> (* Attributes and values *) val str_attributes = Vector.fromList ["Color", "Nation", "Drink", "Pet", "Smoke"] val str_colors = Vector.fromList ["Red", "Green", "White", "Yellow", "Blue"] val str_nations = Vector.fromList ["English", "Swede", "Dane", "German", "Norwegian"] val str_drinks = Vector.fromList ["Tea", "Coffee", "Milk", "Beer", "Water"] val str_pets = Vector.fromList ["Dog", "Birds", "Cats", "Horse", "Zebra"] val str_smokes = Vector.fromList ["PallMall", "Dunhill", "Blend", "BlueMaster", "Prince"]
val (Color, Nation, Drink, Pet, Smoke) = (0, 1, 2, 3, 4) (* Attributes *) val (Red, Green, White, Yellow, Blue) = (0, 1, 2, 3, 4) (* Color *) val (English, Swede, Dane, German, Norwegian) = (0, 1, 2, 3, 4) (* Nation *) val (Tea, Coffee, Milk, Beer, Water) = (0, 1, 2, 3, 4) (* Drink *) val (Dog, Birds, Cats, Horse, Zebra) = (0, 1, 2, 3, 4) (* Pet *) val (PallMall, Dunhill, Blend, BlueMaster, Prince) = (0, 1, 2, 3, 4) (* Smoke *)
type attr = int type value = int type houseno = int
(* Rules *) datatype rule = AttrPairRule of (attr * value) * (attr * value) | NextToRule of (attr * value) * (attr * value) | LeftOfRule of (attr * value) * (attr * value)
(* Conditions *) val rules = [ AttrPairRule ((Nation, English), (Color, Red)), (* #02 *) AttrPairRule ((Nation, Swede), (Pet, Dog)), (* #03 *) AttrPairRule ((Nation, Dane), (Drink, Tea)), (* #04 *) LeftOfRule ((Color, Green), (Color, White)), (* #05 *) AttrPairRule ((Color, Green), (Drink, Coffee)), (* #06 *) AttrPairRule ((Smoke, PallMall), (Pet, Birds)), (* #07 *) AttrPairRule ((Smoke, Dunhill), (Color, Yellow)), (* #08 *) NextToRule ((Smoke, Blend), (Pet, Cats)), (* #11 *) NextToRule ((Smoke, Dunhill), (Pet, Horse)), (* #12 *) AttrPairRule ((Smoke, BlueMaster), (Drink, Beer)), (* #13 *) AttrPairRule ((Nation, German), (Smoke, Prince)), (* #14 *) NextToRule ((Nation, Norwegian), (Color, Blue)), (* #15 *) NextToRule ((Smoke, Blend), (Drink, Water))] (* #16 *)
type house = value option * value option * value option * value option * value option
fun houseval ((a, b, c, d, e) : house, 0 : attr) = a
| houseval ((a, b, c, d, e) : house, 1 : attr) = b | houseval ((a, b, c, d, e) : house, 2 : attr) = c | houseval ((a, b, c, d, e) : house, 3 : attr) = d | houseval ((a, b, c, d, e) : house, 4 : attr) = e | houseval _ = raise Domain
fun sethouseval ((a, b, c, d, e) : house, 0 : attr, a2 : value option) = (a2, b, c, d, e )
| sethouseval ((a, b, c, d, e) : house, 1 : attr, b2 : value option) = (a, b2, c, d, e ) | sethouseval ((a, b, c, d, e) : house, 2 : attr, c2 : value option) = (a, b, c2, d, e ) | sethouseval ((a, b, c, d, e) : house, 3 : attr, d2 : value option) = (a, b, c, d2, e ) | sethouseval ((a, b, c, d, e) : house, 4 : attr, e2 : value option) = (a, b, c, d, e2) | sethouseval _ = raise Domain
fun getHouseVal houses (no, attr) = houseval (Array.sub (houses, no), attr) fun setHouseVal houses (no, attr, newval) = Array.update (houses, no, sethouseval (Array.sub (houses, no), attr, newval))
fun match (house, (rule_attr, rule_val)) =
let
val value = houseval (house, rule_attr)
in
isSome value andalso valOf value = rule_val
end
fun matchNo houses (no, rule) = match (Array.sub (houses, no), rule)
fun compare (house1, house2, ((rule_attr1, rule_val1), (rule_attr2, rule_val2))) = let val val1 = houseval (house1, rule_attr1) val val2 = houseval (house2, rule_attr2) in if isSome val1 andalso isSome val2 then (valOf val1 = rule_val1 andalso valOf val2 <> rule_val2) orelse (valOf val1 <> rule_val1 andalso valOf val2 = rule_val2) else false end
fun compareNo houses (no1, no2, rulepair) = compare (Array.sub (houses, no1), Array.sub (houses, no2), rulepair)
fun invalid houses no (AttrPairRule rulepair) =
compareNo houses (no, no, rulepair)
| invalid houses no (NextToRule rulepair) = (if no > 0
then compareNo houses (no, no-1, rulepair) else true) andalso (if no < 4 then compareNo houses (no, no+1, rulepair) else true)
| invalid houses no (LeftOfRule rulepair) = if no > 0
then compareNo houses (no-1, no, rulepair) else matchNo houses (no, #1rulepair)
(*
* val checkRulesForNo : house vector -> houseno -> bool * Check all rules for a house; * Returns true, when one rule was invalid. *)
fun checkRulesForNo (houses : house array) no = let exception RuleError in (map (fn rule => if invalid houses no rule then raise RuleError else ()) rules; false) handle RuleError => true end
(*
* val checkAll : house vector -> bool * Check all rules; * return true if everything is ok. *)
fun checkAll (houses : house array) = let exception RuleError in (map (fn no => if checkRulesForNo houses no then raise RuleError else ()) [0,1,2,3,4]; true) handle RuleError => false end
(*
* * House printing for debugging * *)
fun valToString (0, SOME a) = Vector.sub (str_colors, a)
| valToString (1, SOME b) = Vector.sub (str_nations, b) | valToString (2, SOME c) = Vector.sub (str_drinks, c) | valToString (3, SOME d) = Vector.sub (str_pets, d) | valToString (4, SOME e) = Vector.sub (str_smokes, e) | valToString _ = "-"
(*
* Note: * Format needs SML NJ *)
fun printHouse no ((a, b, c, d, e) : house) = ( print (Format.format "%12d" [Format.LEFT (12, Format.INT no)]); print (Format.format "%12s%12s%12s%12s%12s" (map (fn (x, y) => Format.LEFT (12, Format.STR (valToString (x, y)))) [(0,a), (1,b), (2,c), (3,d), (4,e)])); print ("\n") )
fun printHouses houses = ( print (Format.format "%12s" [Format.LEFT (12, Format.STR "House")]); Vector.map (fn a => print (Format.format "%12s" [Format.LEFT (12, Format.STR a)])) str_attributes; print "\n"; Array.foldli (fn (no, house, _) => printHouse no house) () houses )
(*
* * Solving * *)
exception SolutionFound
fun search (houses : house array, used : bool Array2.array) (no : houseno, attr : attr) = let val i = ref 0 val (nextno, nextattr) = if attr < 4 then (no, attr + 1) else (no + 1, 0) in if isSome (getHouseVal houses (no, attr)) then ( search (houses, used) (nextno, nextattr) ) else ( while (!i < 5) do ( if Array2.sub (used, attr, !i) then () else ( Array2.update (used, attr, !i, true); setHouseVal houses (no, attr, SOME (!i));
if checkAll houses then ( if no = 4 andalso attr = 4 then raise SolutionFound else search (houses, used) (nextno, nextattr) ) else (); Array2.update (used, attr, !i, false) ); (* else *) i := !i + 1 ); (* do *) setHouseVal houses (no, attr, NONE) ) (* else *) end
fun init () = let val unknown : house = (NONE, NONE, NONE, NONE, NONE) val houses = Array.fromList [unknown, unknown, unknown, unknown, unknown] val used = Array2.array (5, 5, false) in (houses, used) end
fun solve () = let val (houses, used) = init() in setHouseVal houses (2, Drink, SOME Milk); (* #09 *) Array2.update (used, Drink, Milk, true); setHouseVal houses (0, Nation, SOME Norwegian); (* #10 *) Array2.update (used, Nation, Norwegian, true); (search (houses, used) (0, 0); NONE) handle SolutionFound => SOME houses end
(*
* * Execution * *)
fun main () = let val solution = solve() in if isSome solution then printHouses (valOf solution) else print "No solution found!\n" end </lang>
- Output:
- main(); House Color Nation Drink Pet Smoke 0 Yellow Norwegian Water Cats Dunhill 1 Blue Dane Tea Horse Blend 2 Red English Milk Birds PallMall 3 Green German Coffee Zebra Prince 4 White Swede Beer Dog BlueMaster val it = () : unit
Tailspin
General solver with relational algebra
A general solver for this type of puzzle, using relational algebra. This solver can also be used for "Dinesman's multiple-dwelling problem" <lang tailspin> processor EinsteinSolver
@: [{|{}|}]; // A list of possible relations, start with one relation with an empty tuple
$ -> \(
when <[](1..)> do
def variableRange: $(1);
@EinsteinSolver: [($@EinsteinSolver(1) join {|{by $variableRange...}|})];
$(2..last) -> #
\) -> !VOID
sink isFact
def fact: $;
def parts: [$... -> {$}];
@EinsteinSolver: [$@EinsteinSolver... -> \(
def new: ($ matching {|$fact|});
@: $;
$parts... -> @: ($@ notMatching {| $ |});
($new union $@) !
\)];
end isFact
operator (a nextTo&{byField:, bMinusA:} b)
@EinsteinSolver: [$@EinsteinSolver... -> \(
def in: $;
def temp: {| $... -> {$, ES_temp__: $(byField)} |};
def numbers: [$temp({ES_temp__:})... -> $.ES_temp__];
$numbers... -> \(
def aNumber: $;
def bNumbers: [$bMinusA... -> $ + $aNumber];
def new: ($temp matching {| {$a, ES_temp__: $aNumber} |});
@: ($new union (($temp notMatching {| $a |}) notMatching {| {ES_temp__: $aNumber} |}));
$numbers... -> \(<~ ?($bNumbers <[<=$>]>)> $! \) -> @: ($@ notMatching {| {$b, ES_temp__: $} |});
($in matching $@) !
\) !
\)];
end nextTo
source solutions&{required:}
templates resolve&{rows:}
when <?($rows <=1>)?($::count <=1>)> do $ !
when <?($::count <$rows..>)> do
def in: $;
def selected: [$...] -> $(1);
($in minus {|$selected|}) -> resolve&{rows: $rows} ! // Alternative solutions
@: $;
$selected... -> {$} -> @: ($@ notMatching {| $ |});
[$@ -> resolve&{rows: $rows-1}] -> \(
$ -> \(
when <~=[]> do
$... -> {| $..., $selected |} !
\) !
\) !
end resolve
[$@EinsteinSolver... -> resolve&{rows: $required}] !
end solutions
end EinsteinSolver
def numbers: [1..5 -> (no: $)]; def nationalities: [['Englishman', 'Swede', 'Dane', 'Norwegian', 'German']... -> (nationality:$)]; def colours: [['red', 'green', 'white', 'yellow', 'blue']... -> (colour:$)]; def pets: [['dog', 'birds', 'cats', 'horse', 'zebra']... -> (pet:$)]; def drinks: [['tea', 'coffee', 'milk', 'beer', 'water']... -> (drink:$)]; def smokes: [['Pall Mall', 'Dunhill', 'Blend', 'Blue Master', 'Prince']... -> (smoke: $)];
def solutions: [$numbers, $nationalities, $colours, $pets, $drinks, $smokes] -> \(
def solver: $ -> EinsteinSolver;
{nationality: 'Englishman', colour: 'red'} -> !solver::isFact
{nationality: 'Swede', pet: 'dog'} -> !solver::isFact
{nationality: 'Dane', drink: 'tea'} -> !solver::isFact
({colour: 'green'} solver::nextTo&{byField: :(no:), bMinusA: [1]} {colour: 'white'}) -> !VOID
{drink: 'coffee', colour: 'green'} -> !solver::isFact
{smoke: 'Pall Mall', pet: 'birds'} -> !solver::isFact
{colour: 'yellow', smoke: 'Dunhill'} -> !solver::isFact
{no: 3, drink: 'milk'} -> !solver::isFact
{nationality: 'Norwegian', no: 1} -> !solver::isFact
({smoke: 'Blend'} solver::nextTo&{byField: :(no:), bMinusA: [-1, 1]} {pet: 'cats'}) -> !VOID
({smoke: 'Dunhill'} solver::nextTo&{byField: :(no:), bMinusA: [-1, 1]} {pet: 'horse'}) -> !VOID
{smoke: 'Blue Master', drink: 'beer'} -> !solver::isFact
{nationality: 'German', smoke: 'Prince'} -> !solver::isFact
({nationality: 'Norwegian'} solver::nextTo&{byField: :(no:), bMinusA: [-1, 1]} {colour: 'blue'}) -> !VOID
({drink: 'water'} solver::nextTo&{byField: :(no:), bMinusA: [-1, 1]} {smoke: 'Blend'}) -> !VOID
$solver::solutions&{required: 5}!
\);
$solutions... -> ($ matching {| {pet: 'zebra'} |}) ... -> 'The $.nationality; owns the zebra.
' -> !OUT::write
$solutions -> \[i]('Solution $i;: $... -> '$; '; '! \)... -> !OUT::write 'No more solutions ' -> !OUT::write </lang>
- Output:
The German owns the zebra.
Solution 1:
{colour=green, drink=coffee, nationality=German, no=4, pet=zebra, smoke=Prince}
{colour=white, drink=beer, nationality=Swede, no=5, pet=dog, smoke=Blue Master}
{colour=blue, drink=tea, nationality=Dane, no=2, pet=horse, smoke=Blend}
{colour=yellow, drink=water, nationality=Norwegian, no=1, pet=cats, smoke=Dunhill}
{colour=red, drink=milk, nationality=Englishman, no=3, pet=birds, smoke=Pall Mall}
No more solutions
Manual "Norvig" solution
The streaming syntax of Tailspin lends itself naturally to this solution <lang tailspin> templates permutations
when <=1> do [1] !
otherwise
def n: $;
templates expand
def p: $;
1..$n -> \(def k: $;
[$p(1..$k-1)..., $n, $p($k..last)...] !\) !
end expand
$n - 1 -> permutations -> expand !
end permutations
def permutationsOf5: [5 -> permutations];
def nationalities: ['Englishman', 'Swede', 'Dane', 'Norwegian', 'German']; def colours: ['red', 'green', 'white', 'yellow', 'blue']; def pets: ['dog', 'birds', 'cats', 'horse', 'zebra']; def drinks: ['tea', 'coffee', 'milk', 'beer', 'water']; def smokes: ['Pall Mall', 'Dunhill', 'Blend', 'Blue Master', 'Prince'];
$permutationsOf5... -> $colours($) -> [$... -> {colour: $}]
-> \(<[(<{colour: <='green'>}>:<{colour: <='white'>}>)]> $! \)
-> \(def current: $;
$permutationsOf5... -> $nationalities($) -> \[i]({$current($i), nationality: $}! \) !
\)
-> \(<?($(1) <{nationality: <='Norwegian'>}>)> $! \)
-> \(<[<{nationality: <='Englishman'>, colour: <='red'>}>]> $! \)
-> \(<[(<{nationality: <='Norwegian'>}>:<{colour: <='blue'>}>)] | [(<{colour: <='blue'>}>:<{nationality: <='Norwegian'>}>)]> $! \)
-> \(def current: $;
$permutationsOf5... -> $drinks($) -> \[i]({$current($i), drink: $}! \) !
\)
-> \(<?($(3) <{drink: <='milk'>}>)> $! \)
-> \(<[<{drink: <='coffee'>, colour: <='green'>}>]> $! \)
-> \(<[<{drink: <='tea'>, nationality: <='Dane'>}>]> $! \)
-> \(def current: $;
$permutationsOf5... -> $pets($) -> \[i]({$current($i), pet: $}! \)!
\)
-> \(<[<{nationality: <='Swede'>, pet: <='dog'>}>]> $! \)
-> \(def current: $;
$permutationsOf5... -> $smokes($) -> \[i]({$current($i), smoke: $}! \)!
\)
-> \(<[<{smoke: <='Pall Mall'>, pet: <='birds'>}>]> $! \)
-> \(<[<{smoke: <='Dunhill'>, colour: <='yellow'>}>]> $! \)
-> \(<[(<{smoke: <='Blend'>}>:<{pet: <='cats'>}>)] | [(<{pet: <='cats'>}>:<{smoke: <='Blend'>}>)]> $! \)
-> \(<[(<{smoke: <='Dunhill'>}>:<{pet: <='horse'>}>)] | [(<{pet: <='horse'>}>:<{smoke: <='Dunhill'>}>)]> $! \)
-> \(<[<{smoke: <='Blue Master'>, drink: <='beer'>}>]> $! \)
-> \(<[<{smoke: <='Prince'>, nationality: <='German'>}>]> $! \)
-> \(<[(<{smoke: <='Blend'>}>:<{drink: <='water'>}>)] | [(<{drink: <='water'>}>:<{smoke: <='Blend'>}>)]> $! \)
-> \[i](when <{pet: <='zebra'>}> do 'The $.nationality; owns the zebra.$#10;' -> !OUT::write $!
otherwise $! \)
-> \[i]('$i;: $;'! \) -> '$... -> '$;$#10;';$#10;'
-> !OUT::write
</lang>
- Output:
The German owns the zebra.
1: {colour=yellow, drink=water, nationality=Norwegian, pet=cats, smoke=Dunhill}
2: {colour=blue, drink=tea, nationality=Dane, pet=horse, smoke=Blend}
3: {colour=red, drink=milk, nationality=Englishman, pet=birds, smoke=Pall Mall}
4: {colour=green, drink=coffee, nationality=German, pet=zebra, smoke=Prince}
5: {colour=white, drink=beer, nationality=Swede, pet=dog, smoke=Blue Master}
Tcl
<lang tcl>package require struct::list
- Implements the constants by binding them directly into the named procedures.
- This is much faster than the alternatives!
proc initConstants {args} {
global {}
set remap {}
foreach {class elems} {
Number {One Two Three Four Five} Color {Red Green Blue White Yellow} Drink {Milk Coffee Water Beer Tea} Smoke {PallMall Dunhill Blend BlueMaster Prince} Pet {Dog Cat Horse Bird Zebra} Nation {British Swedish Danish Norwegian German}
} {
set i -1 foreach e $elems {lappend remap "\$${class}($e)" [incr i]} set ($class) $elems
}
foreach procedure $args {
proc $procedure [info args $procedure] \ [string map $remap [info body $procedure]]
}
}
proc isPossible {number color drink smoke pet} {
if {[llength $number] && [lindex $number $Nation(Norwegian)] != $Number(One)} {
return false
} elseif {[llength $color] && [lindex $color $Nation(British)] != $Color(Red)} {
return false
} elseif {[llength $drink] && [lindex $drink $Nation(Danish)] != $Drink(Tea)} {
return false
} elseif {[llength $smoke] && [lindex $smoke $Nation(German)] != $Smoke(Prince)} {
return false
} elseif {[llength $pet] && [lindex $pet $Nation(Swedish)] != $Pet(Dog)} {
return false
}
if {!([llength $number] && [llength $color] && [llength $drink] && [llength $smoke] && [llength $pet])} {
return true
}
for {set i 0} {$i < 5} {incr i} {
if {[lindex $color $i] == $Color(Green) && [lindex $drink $i] != $Drink(Coffee)} { return false } elseif {[lindex $smoke $i] == $Smoke(PallMall) && [lindex $pet $i] != $Pet(Bird)} { return false } elseif {[lindex $color $i] == $Color(Yellow) && [lindex $smoke $i] != $Smoke(Dunhill)} { return false } elseif {[lindex $number $i] == $Number(Three) && [lindex $drink $i] != $Drink(Milk)} { return false } elseif {[lindex $smoke $i] == $Smoke(BlueMaster) && [lindex $drink $i] != $Drink(Beer)} { return false } elseif {[lindex $color $i] == $Color(Blue) && [lindex $number $i] != $Number(Two)} { return false }
for {set j 0} {$j < 5} {incr j} { if {[lindex $color $i] == $Color(Green) && [lindex $color $j] == $Color(White) && [lindex $number $j] - [lindex $number $i] != 1} { return false }
set diff [expr {abs([lindex $number $i] - [lindex $number $j])}] if {[lindex $smoke $i] == $Smoke(Blend) && [lindex $pet $j] == $Pet(Cat) && $diff != 1} { return false } elseif {[lindex $pet $i] == $Pet(Horse) && [lindex $smoke $j] == $Smoke(Dunhill) && $diff != 1} { return false } elseif {[lindex $smoke $i] == $Smoke(Blend) && [lindex $drink $j] == $Drink(Water) && $diff != 1} { return false } }
}
return true
}
proc showRow {t data} {
upvar #0 ($t) elems puts [format "%6s: %12s%12s%12s%12s%12s" $t \
[lindex $elems [lindex $data 0]] \ [lindex $elems [lindex $data 1]] \ [lindex $elems [lindex $data 2]] \ [lindex $elems [lindex $data 3]] \ [lindex $elems [lindex $data 4]]] }
proc main {} {
set perms [struct::list permutations {0 1 2 3 4}]
foreach number $perms {
if {![isPossible $number {} {} {} {}]} continue foreach color $perms { if {![isPossible $number $color {} {} {}]} continue foreach drink $perms { if {![isPossible $number $color $drink {} {}]} continue foreach smoke $perms { if {![isPossible $number $color $drink $smoke {}]} continue foreach pet $perms { if {[isPossible $number $color $drink $smoke $pet]} { puts "Found a solution:" showRow Nation {0 1 2 3 4} showRow Number $number showRow Color $color showRow Drink $drink showRow Smoke $smoke showRow Pet $pet puts "" } } } } }
}
}
initConstants isPossible main</lang>
- Output:
Found a solution:
Nation: British Swedish Danish Norwegian German
Number: Three Five Two One Four
Color: Red White Blue Yellow Green
Drink: Milk Beer Tea Water Coffee
Smoke: PallMall BlueMaster Blend Dunhill Prince
Pet: Bird Dog Horse Cat Zebra
VBA
<lang vb>Option Base 1 Public Enum attr
Colour = 1 Nationality Beverage Smoke Pet
End Enum Public Enum Drinks_
Beer = 1 Coffee Milk Tea Water
End Enum Public Enum nations
Danish = 1 English German Norwegian Swedish
End Enum Public Enum colors
Blue = 1 Green Red White Yellow
End Enum Public Enum tobaccos
Blend = 1 BlueMaster Dunhill PallMall Prince
End Enum Public Enum animals
Bird = 1 Cat Dog Horse Zebra
End Enum Public permutation As New Collection Public perm(5) As Variant Const factorial5 = 120 Public Colours As Variant, Nationalities As Variant, Drinks As Variant, Smokes As Variant, Pets As Variant
Private Sub generate(n As Integer, A As Variant)
If n = 1 Then
permutation.Add A
Else
For i = 1 To n
generate n - 1, A
If n Mod 2 = 0 Then
tmp = A(i)
A(i) = A(n)
A(n) = tmp
Else
tmp = A(1)
A(1) = A(n)
A(n) = tmp
End If
Next i
End If
End Sub
Function house(i As Integer, name As Variant) As Integer
Dim x As Integer
For x = 1 To 5
If perm(i)(x) = name Then
house = x
Exit For
End If
Next x
End Function
Function left_of(h1 As Integer, h2 As Integer) As Boolean
left_of = (h1 - h2) = -1
End Function
Function next_to(h1 As Integer, h2 As Integer) As Boolean
next_to = Abs(h1 - h2) = 1
End Function
Private Sub print_house(i As Integer)
Debug.Print i & ": "; Colours(perm(Colour)(i)), Nationalities(perm(Nationality)(i)), _
Drinks(perm(Beverage)(i)), Smokes(perm(Smoke)(i)), Pets(perm(Pet)(i))
End Sub Public Sub Zebra_puzzle()
Colours = [{"blue","green","red","white","yellow"}]
Nationalities = [{"Dane","English","German","Norwegian","Swede"}]
Drinks = [{"beer","coffee","milk","tea","water"}]
Smokes = [{"Blend","Blue Master","Dunhill","Pall Mall","Prince"}]
Pets = [{"birds","cats","dog","horse","zebra"}]
Dim solperms As New Collection
Dim solutions As Integer
Dim b(5) As Integer, i As Integer
For i = 1 To 5: b(i) = i: Next i
'There are five houses.
generate 5, b
For c = 1 To factorial5
perm(Colour) = permutation(c)
'The green house is immediately to the left of the white house.
If left_of(house(Colour, Green), house(Colour, White)) Then
For n = 1 To factorial5
perm(Nationality) = permutation(n)
'The Norwegian lives in the first house.
'The English man lives in the red house.
'The Norwegian lives next to the blue house.
If house(Nationality, Norwegian) = 1 _
And house(Nationality, English) = house(Colour, Red) _
And next_to(house(Nationality, Norwegian), house(Colour, Blue)) Then
For d = 1 To factorial5
perm(Beverage) = permutation(d)
'The Dane drinks tea.
'They drink coffee in the green house.
'In the middle house they drink milk.
If house(Nationality, Danish) = house(Beverage, Tea) _
And house(Beverage, Coffee) = house(Colour, Green) _
And house(Beverage, Milk) = 3 Then
For s = 1 To factorial5
perm(Smoke) = permutation(s)
'In the yellow house they smoke Dunhill.
'The German smokes Prince.
'The man who smokes Blue Master drinks beer.
'They Drink water in a house next to the house where they smoke Blend.
If house(Colour, Yellow) = house(Smoke, Dunhill) _
And house(Nationality, German) = house(Smoke, Prince) _
And house(Smoke, BlueMaster) = house(Beverage, Beer) _
And next_to(house(Beverage, Water), house(Smoke, Blend)) Then
For p = 1 To factorial5
perm(Pet) = permutation(p)
'The Swede has a dog.
'The man who smokes Pall Mall has birds.
'The man who smokes Blend lives in the house next to the house with cats.
'In a house next to the house where they have a horse, they smoke Dunhill.
If house(Nationality, Swedish) = house(Pet, Dog) _
And house(Smoke, PallMall) = house(Pet, Bird) _
And next_to(house(Smoke, Blend), house(Pet, Cat)) _
And next_to(house(Pet, Horse), house(Smoke, Dunhill)) Then
For i = 1 To 5
print_house i
Next i
Debug.Print
solutions = solutions + 1
solperms.Add perm
End If
Next p
End If
Next s
End If
Next d
End If
Next n
End If
Next c
Debug.Print Format(solutions, "@"); " solution" & IIf(solutions > 1, "s", "") & " found"
For i = 1 To solperms.Count
For j = 1 To 5
perm(j) = solperms(i)(j)
Next j
Debug.Print "The " & Nationalities(perm(Nationality)(house(Pet, Zebra))) & " owns the Zebra"
Next i
End Sub</lang>
- Output:
1: yellow Norwegian water Dunhill cats 2: blue Dane tea Blend horse 3: red English milk Pall Mall birds 4: green German coffee Prince zebra 5: white Swede beer Blue Master dog 1 solution found The German owns the Zebra
zkl
The solution space is too large to brute force search so solutions are built up by solving for one or two variables, then solving that set for a one or two more variables until all the variables & constraints are tested.
Lists are used as associated arrays rather than hash tables. With only five entries, it doesn't really matter. <lang zkl>var people,drinks,houses,smokes,pets; // lists treated as associated arrays fcn c2 { people.find(English)==houses.find(Red) } fcn c3 { people.find(Swede)==pets.find(Dog) } fcn c4 { people.find(Dane)==drinks.find(Tea) } fcn c5 { (houses.find(Green) + 1)==houses.find(White) } fcn c5a{ houses.find(Green)!=4 } // deduced constraint (from c5) fcn c5b{ houses.find(White)!=0 } // deduced constraint (from c5) fcn c6 { drinks.find(Coffee)==houses.find(Green) } fcn c7 { smokes.find(PallMall)==pets.find(Bird) } fcn c8 { houses.find(Yellow)==smokes.find(Dunhill) } fcn c9 { drinks[2]==Milk } // 0,1,2,3,4 fcn c10{ people[0]==Norwegian } fcn c11{ (smokes.find(Blend) - pets.find(Cat)).abs()==1 } fcn c12{ (pets.find(Horse) - smokes.find(Dunhill)).abs()==1 } fcn c13{ smokes.find(BlueMaster)==drinks.find(Beer) } fcn c14{ people.find(German)==smokes.find(Prince) } fcn c15{ (people.find(Norwegian) - houses.find(Blue)).abs()==1 } fcn c16{ (drinks.find(Water) - smokes.find(Blend)).abs()==1 }
- <<<#//////////////////////////////////////////////////////////////////////
Showing a solution to c2,c5,c10,c15: |0 1 2 3 4
+-------------------------------------------
houses: |Yellow Blue Red Green White people: |Norwegian Dane English German Swede
- <<<#//////////////////////////////////////////////////////////////////////
const Blue =0,Green =1,Red =2,White =3,Yellow=4,
Dane =0,English =1,German =2,Norwegian=3,Swede =4,
Beer =0,Coffee =1,Milk =2,Tea =3,Water =4,
Blend=0,BlueMaster=1,Dunhill=2,PallMall =3,Prince=4,
Bird =0,Cat =1,Dog =2,Horse =3,Zebra =4;
perm5:=T(0,1,2,3,4) : Utils.Helpers.permute(_); // 120 sets
constraints:=T(c2,c3,c4,c5,c5a,c5b,c6,c7,c8,c9,c10,c11,c12,c13,c14,c15,c16); constraints1:=T(c2,c5,c10,c15); // houses,people: 12 solutions constraints2:=T(c4,c6,c9); // houses,people,drinks: down to 8 solutions foreach _houses,_people in (perm5,perm5){ houses,people=_houses,_people;
if(not constraints1.runNFilter(False)){ // all constraints are True
foreach _drinks in (perm5){ drinks=_drinks;
if(not constraints2.runNFilter(False)){ foreach _smokes,_pets in (perm5,perm5){ smokes,pets=_smokes,_pets; if(not constraints.runNFilter(False)) printSolution(); }// smokes,pets }
} // drinks } // houses,people
} fcn printSolution{
var titles=T("Houses:","People:","Drinks:","Smokes:","Pets:"),
names=T(
T("Blue", "Green", "Red", "White", "Yellow",),
T("Dane", "English", "German", "Norwegian","Swede",), T("Beer", "Coffee", "Milk", "Tea", "Water",), T("Blend","Blue Master","Dunhill","Pall Mall","Prince",), T("Bird", "Cat", "Dog", "Horse", "Zebra",) ),
;
fmt:=("%-7s " + "%-11s "*5).fmt;
foreach list,title,names in (T(houses,people,drinks,smokes,pets)
.zip(titles,names))
{ println(list.apply(names.get):fmt(title,_.xplode())) }
}</lang>
- Output:
Houses: Yellow Blue Red Green White People: Norwegian Dane English German Swede Drinks: Water Tea Milk Coffee Beer Smokes: Dunhill Blend Pall Mall Prince Blue Master Pets: Cat Horse Bird Zebra Dog