Dinesman's multiple-dwelling problem
From Rosetta Code
You are encouraged to solve this task according to the task description, using any language you may know.
The task is to solve Dinesman's multiple dwelling problem but in a way that most naturally follows the problem statement given below. Solutions are allowed (but not required) to parse and interpret the problem text, but should remain flexible and should state what changes to the problem text are allowed. Flexibility and ease of expression are valued.
Examples may be be split into "setup", "problem statement", and "output" sections where the ease and naturalness of stating the problem and getting an answer, as well as the ease and flexibility of modifying the problem are the primary concerns.
Example output should be shown here, as well as any comments on the examples flexibility.
- The problem
- Baker, Cooper, Fletcher, Miller, and Smith live on different floors of an apartment house that contains only five floors. Baker does not live on the top floor. Cooper does not live on the bottom floor. Fletcher does not live on either the top or the bottom floor. Miller lives on a higher floor than does Cooper. Smith does not live on a floor adjacent to Fletcher's. Fletcher does not live on a floor adjacent to Cooper's. Where does everyone live?
Contents |
[edit] Ada
Uses an enum type People to attempt to be naturally reading. Problem is easily changed by altering subtype Floor, type people and the somewhat naturally reading constraints in the Constrained function. If for example you change the floor range to 1..6 and add Superman to people, all possible solutions will be printed.
with Ada.Text_IO; use Ada.Text_IO;
procedure Dinesman is
subtype Floor is Positive range 1 .. 5;
type People is (Baker, Cooper, Fletcher, Miller, Smith);
type Floors is array (People'Range) of Floor;
type PtFloors is access all Floors;
function Constrained (f : PtFloors) return Boolean is begin
if f (Baker) /= Floor'Last and
f (Cooper) /= Floor'First and
Floor'First < f (Fletcher) and f (Fletcher) < Floor'Last and
f (Miller) > f (Cooper) and
abs (f (Smith) - f (Fletcher)) /= 1 and
abs (f (Fletcher) - f (Cooper)) /= 1
then return True; end if;
return False;
end Constrained;
procedure Solve (list : PtFloors; n : Natural) is
procedure Swap (I : People; J : Natural) is
temp : constant Floor := list (People'Val (J));
begin list (People'Val (J)) := list (I); list (I) := temp;
end Swap;
begin
if n = 1 then
if Constrained (list) then
for p in People'Range loop
Put_Line (p'Img & " on floor " & list (p)'Img);
end loop;
end if;
return;
end if;
for i in People'First .. People'Val (n - 1) loop
Solve (list, n - 1);
if n mod 2 = 1 then Swap (People'First, n - 1);
else Swap (i, n - 1); end if;
end loop;
end Solve;
thefloors : aliased Floors;
begin
for person in People'Range loop
thefloors (person) := People'Pos (person) + Floor'First;
end loop;
Solve (thefloors'Access, Floors'Length);
end Dinesman;
- Output:
BAKER on floor 3 COOPER on floor 2 FLETCHER on floor 4 MILLER on floor 5 SMITH on floor 1
[edit] Bracmat
( Baker Cooper Fletcher Miller Smith:?people
& ( constraints
=
. !arg
: ~(? Baker)
: ~(Cooper ?)
: ~(Fletcher ?|? Fletcher)
: ? Cooper ? Miller ?
: ~(? Smith Fletcher ?|? Fletcher Smith ?)
: ~(? Cooper Fletcher ?|? Fletcher Cooper ?)
)
& ( solution
= floors persons A Z person
. !arg:(?floors.?persons)
& ( !persons:
& constraints$!floors
& out$("Inhabitants, from bottom to top:" !floors)
| !persons
: ?A
%?`person
(?Z&solution$(!floors !person.!A !Z))
)
)
& solution$(.!people)
&
);
Inhabitants, from bottom to top: Smith Cooper Baker Fletcher Miller
[edit] C
#include <stdio.h>Output
#include <stdlib.h>
int verbose = 0;
#define COND(a, b) int a(int *s) { return (b); }
typedef int(*condition)(int *);
/* BEGIN problem specific setup */
#define N_FLOORS 5
#define TOP (N_FLOORS - 1)
int solution[N_FLOORS] = { 0 };
int occupied[N_FLOORS] = { 0 };
enum tenants {
baker = 0,
cooper,
fletcher,
miller,
smith,
phantom_of_the_opera,
};
char *names[] = {
"baker",
"cooper",
"fletcher",
"miller",
"smith",
};
COND(c0, s[baker] != TOP);
COND(c1, s[cooper] != 0);
COND(c2, s[fletcher] != 0 && s[fletcher] != TOP);
COND(c3, s[miller] > s[cooper]);
COND(c4, abs(s[smith] - s[fletcher]) != 1);
COND(c5, abs(s[cooper] - s[fletcher]) != 1);
#define N_CONDITIONS 6
condition cond[] = { c0, c1, c2, c3, c4, c5 };
/* END of problem specific setup */
int solve(int person)
{
int i, j;
if (person == phantom_of_the_opera) {
/* check condition */
for (i = 0; i < N_CONDITIONS; i++) {
if (cond[i](solution)) continue;
if (verbose) {
for (j = 0; j < N_FLOORS; j++)
printf("%d %s\n", solution[j], names[j]);
printf("cond %d bad\n\n", i);
}
return 0;
}
printf("Found arrangement:\n");
for (i = 0; i < N_FLOORS; i++)
printf("%d %s\n", solution[i], names[i]);
return 1;
}
for (i = 0; i < N_FLOORS; i++) {
if (occupied[i]) continue;
solution[person] = i;
occupied[i] = 1;
if (solve(person + 1)) return 1;
occupied[i] = 0;
}
return 0;
}
int main()
{
verbose = 0;
if (!solve(0)) printf("Nobody lives anywhere\n");
return 0;
}
Found arrangement:C, being its compiled self, is not terribly flexible in dynamically changing runtime code content. Parsing some external problem specification would be one way, but for a small problem, it might as well just recompile with conditions hard coded. For this program, to change conditions, one needs to edit content between BEGIN and END of problem specific setup. Those could even be setup in an external file and gets
2 baker
1 cooper
3 fletcher
4 miller
0 smith
#included if need be.
[edit] D
This code uses second lazy permutations function of Permutations#Lazy_version.
As for flexibility: the solve code works with an arbitrary number of people and predicates.
import std.stdio, std.math, std.algorithm, std.traits;
import permutations2: permutations; // from rosettacode
void main() {
enum Names { Baker, Cooper, Fletcher, Miller, Smith }
immutable(bool function(in Names[]) pure nothrow)[] predicates = [
s => s[Names.Baker] != s.length-1,
s => s[Names.Cooper] != 0,
s => s[Names.Fletcher] != 0 && s[Names.Fletcher] != s.length-1,
s => s[Names.Miller] > s[Names.Cooper],
s => abs(cast(int)(s[Names.Smith] - s[Names.Fletcher])) != 1,
s => abs(cast(int)(s[Names.Cooper] - s[Names.Fletcher])) != 1];
foreach (sol; permutations([EnumMembers!Names]))
//if (predicates.all!(p => p(sol))())
if (!predicates.canFind!(p => !p(sol))())
writeln(sol);
}
- Output:
[Fletcher, Cooper, Miller, Smith, Baker]
[edit] Haskell
The List monad is perfect for this kind of problem. One can express the problem statements in a very natural and concise way:
import Data.List (permutations)
import Control.Monad (guard)
dinesman :: [(Int,Int,Int,Int,Int)]
dinesman = do
-- baker, cooper, fletcher, miller, smith are integers representing
-- the floor that each person lives on, from 1 to 5
-- Baker, Cooper, Fletcher, Miller, and Smith live on different floors
-- of an apartment house that contains only five floors.
[baker, cooper, fletcher, miller, smith] <- permutations [1..5]
-- Baker does not live on the top floor.
guard $ baker /= 5
-- Cooper does not live on the bottom floor.
guard $ cooper /= 1
-- Fletcher does not live on either the top or the bottom floor.
guard $ fletcher /= 5 && fletcher /= 1
-- Miller lives on a higher floor than does Cooper.
guard $ miller > cooper
-- Smith does not live on a floor adjacent to Fletcher's.
guard $ abs (smith - fletcher) /= 1
-- Fletcher does not live on a floor adjacent to Cooper's.
guard $ abs (fletcher - cooper) /= 1
-- Where does everyone live?
return (baker, cooper, fletcher, miller, smith)
main :: IO ()
main = do
print $ head dinesman -- print first solution: (3,2,4,5,1)
print dinesman -- print all solutions (only one): [(3,2,4,5,1)]
Or as a list comprehension:
import Data.List (permutations)
main = print [ (b,c,f,m,s) | [b,c,f,m,s] <- permutations [1..5], b/=5,c/=1,f/=1,f/=5,m>c,abs(s-f)>1,abs(c-f)>1]
[edit] Icon and Unicon
This solution uses string invocation to call operators and the fact the Icon/Unicon procedures are first class values. The procedure names could also be given as strings and it would be fairly simple to read the names and all the rules directly from a file. Each name and rule recurses and relies on the inherent backtracking in the language to achieve the goal.
The rules explicitly call stop() after showing the solution. Removing the stop would cause the solver to try all possible cases and report all possible solutions (if there were multiple ones).
invocable allOutput:
global nameL, nameT, rules
procedure main() # Dinesman
nameT := table()
nameL := ["Baker", "Cooper", "Fletcher", "Miller", "Smith"]
rules := [ [ distinct ],
[ "~=", "Baker", top() ],
[ "~=", "Cooper", bottom() ],
[ "~=", "Fletcher", top() ],
[ "~=", "Fletcher", bottom() ],
[ ">", "Miller", "Cooper" ],
[ notadjacent, "Smith", "Fletcher" ],
[ notadjacent, "Fletcher", "Cooper" ],
[ showsolution ],
[ stop ] ]
if not solve(1) then
write("No solution found.")
end
procedure dontstop() # use if you want to search for all solutions
end
procedure showsolution() # show the soluton
write("The solution is:")
every write(" ",n := !nameL, " lives in ", nameT[n])
return
end
procedure eval(n) # evaluate a rule
r := copy(rules[n-top()])
every r[i := 2 to *r] := rv(r[i])
if get(r)!r then suspend
end
procedure rv(x) # return referenced value if it exists
return \nameT[x] | x
end
procedure solve(n) # recursive solver
if n > top() then { # apply rules
if n <= top() + *rules then
( eval(n) & solve(n+1) ) | fail
}
else # setup locations
(( nameT[nameL[n]] := bottom() to top() ) & solve(n + 1)) | fail
return
end
procedure distinct(a,b) # ensure each name is distinct
if nameT[n := !nameL] = nameT[n ~== key(nameT)] then fail
suspend
end
procedure notadjacent(n1,n2) # ensure n1,2 are not adjacent
if not adjacent(n1,n2) then suspend
end
procedure adjacent(n1,n2) # ensure n1,2 are adjacent
if abs(n1 - n2) = 1 then suspend
end
procedure bottom() # return bottom
return if *nameL > 0 then 1 else 0
end
procedure top() # return top
return *nameL
end
The solution is: Baker lives in 3 Cooper lives in 2 Fletcher lives in 4 Miller lives in 5 Smith lives in 1
[edit] J
This problem asks us to pick from one of several possibilities. We can represent these possibilities as permutations of the residents' initials, arranged in order from lowest floor to top floor:
possible=: ((i.!5) A. i.5) { 'BCFMS'
Additionally, we are given a variety of constraints which eliminate some possibilities:
possible=: (#~ 'B' ~: {:"1) possible NB. Baker not on top floor
possible=: (#~ 'C' ~: {."1) possible NB. Cooper not on bottom floor
possible=: (#~ 'F' ~: {:"1) possible NB. Fletcher not on top floor
possible=: (#~ 'F' ~: {."1) possible NB. Fletcher not on bottom floor
possible=: (#~ </@i."1&'CM') possible NB. Miller on higher floor than Cooper
possible=: (#~ 0 = +/@E."1~&'SF') possible NB. Smith not immediately below Fletcher
possible=: (#~ 0 = +/@E."1~&'FS') possible NB. Fletcher not immediately below Smith
possible=: (#~ 0 = +/@E."1~&'CF') possible NB. Cooper not immediately below Fletcher
possible=: (#~ 0 = +/@E."1~&'FC') possible NB. Fletcher not immediately below Cooper
The answer is thus:
possible
SCBFM
(bottom floor) Smith, Cooper, Baker, Fletcher, Miller (top floor)
[edit] Mathematica
floor[x_,y_]:=Flatten[Position[y,x]][[1]]
Select[Permutations[{"Baker","Cooper","Fletcher","Miller","Smith"}],
( floor["Baker",#] < 5 )
&&( Abs[floor["Fletcher",#] - floor["Cooper",#]] > 1 )
&&( Abs[floor["Fletcher",#] - floor["Smith",#]] > 1 )
&&( 1 < floor["Cooper",#] < floor["Miller",#] )
&&( 1 < floor["Fletcher",#] < 5 )
&] [[1]] //Reverse //Column
->
Miller
Fletcher
Baker
Cooper
Smith
[edit] Perl 6
We use permutations because "different floors" are specified. The next_perm subroutine is a variant of the one from the Permutations task.
sub next_perm ( @a is copy ) {Output:
my $j = @a.end - 1;
return Nil if --$j < 0 while [>] @a[ $j, $j+1 ];
my $aj = @a[$j];
my $k = @a.end;
$k-- while [>] $aj, @a[$k];
@a[ $j, $k ] .= reverse;
my $r = @a.end;
my $s = $j + 1;
@a[ $r--, $s++ ] .= reverse while $r > $s;
return @a;
}
# Contains only five floors. 5! = 120 permutations.
for [1..5], &next_perm ...^ !* -> [ $b, $c, $f, $m, $s ] {
say "Baker=$b Cooper=$c Fletcher=$f Miller=$m Smith=$s"
if $b != 5 # Baker !live on top floor.
and $c != 1 # Cooper !live on bottom floor.
and $f != 1|5 # Fletcher !live on top or the bottom floor.
and $m > $c # Miller lives on a higher floor than Cooper.
and $s != $f-1|$f+1 # Smith !live adjacent to Fletcher
and $f != $c-1|$c+1 # Fletcher !live adjacent to Cooper
;
}
Baker=3 Cooper=2 Fletcher=4 Miller=5 Smith=1
[edit] PicoLisp
Using Pilog (PicoLisp Prolog). The problem can be modified by changing just the 'dwelling' rule (the "Problem statement"). This might involve the names and number of dwellers (the list in the first line), and statements about who does (or does not) live on the top floor (using the 'topFloor' predicate), the bottom floor (using the 'bottomFloor' predicate), on a higher floor (using the 'higherFloor' predicate) or on an adjecent floor (using the 'adjacentFloor' predicate). The logic follows an implied AND, and statements may be arbitrarily combined using OR and NOT (using the 'or' and 'not' predicates), or any other Pilog (Prolog) built-in predicates. If the problem statement has several solutions, they will be all generated.
# Problem statement
(be dwelling (@Tenants)
(permute (Baker Cooper Fletcher Miller Smith) @Tenants)
(not (topFloor Baker @Tenants))
(not (bottomFloor Cooper @Tenants))
(not (or ((topFloor Fletcher @Tenants)) ((bottomFloor Fletcher @Tenants))))
(higherFloor Miller Cooper @Tenants)
(not (adjacentFloor Smith Fletcher @Tenants))
(not (adjacentFloor Fletcher Cooper @Tenants)) )
# Utility rules
(be topFloor (@Tenant @Lst)
(equal (@ @ @ @ @Tenant) @Lst) )
(be bottomFloor (@Tenant @Lst)
(equal (@Tenant @ @ @ @) @Lst) )
(be higherFloor (@Tenant1 @Tenant2 @Lst)
(append @ @Rest @Lst)
(equal (@Tenant2 . @Higher) @Rest)
(member @Tenant1 @Higher) )
(be adjacentFloor (@Tenant1 @Tenant2 @Lst)
(append @ @Rest @Lst)
(or
((equal (@Tenant1 @Tenant2 . @) @Rest))
((equal (@Tenant2 @Tenant1 . @) @Rest)) ) )
Output:
: (? (dwelling @Result)) @Result=(Smith Cooper Baker Fletcher Miller) # Only one solution -> NIL
[edit] Prolog
Works with SWI-Prolog and library(clpfd) written by Markus Triska.
:- use_module(library(clpfd)).
:- dynamic top/1, bottom/1.
% Baker does not live on the top floor
rule1(L) :-
member((baker, F), L),
top(Top),
F #\= Top.
% Cooper does not live on the bottom floor.
rule2(L) :-
member((cooper, F), L),
bottom(Bottom),
F #\= Bottom.
% Fletcher does not live on either the top or the bottom floor.
rule3(L) :-
member((fletcher, F), L),
top(Top),
bottom(Bottom),
F #\= Top,
F #\= Bottom.
% Miller lives on a higher floor than does Cooper.
rule4(L) :-
member((miller, Fm), L),
member((cooper, Fc), L),
Fm #> Fc.
% Smith does not live on a floor adjacent to Fletcher's.
rule5(L) :-
member((smith, Fs), L),
member((fletcher, Ff), L),
abs(Fs-Ff) #> 1.
% Fletcher does not live on a floor adjacent to Cooper's.
rule6(L) :-
member((cooper, Fc), L),
member((fletcher, Ff), L),
abs(Fc-Ff) #> 1.
init(L) :-
% we need to define top and bottom
assert(bottom(1)),
length(L, Top),
assert(top(Top)),
% we say that they are all in differents floors
bagof(F, X^member((X, F), L), LF),
LF ins 1..Top,
all_different(LF),
% Baker does not live on the top floor
rule1(L),
% Cooper does not live on the bottom floor.
rule2(L),
% Fletcher does not live on either the top or the bottom floor.
rule3(L),
% Miller lives on a higher floor than does Cooper.
rule4(L),
% Smith does not live on a floor adjacent to Fletcher's.
rule5(L),
% Fletcher does not live on a floor adjacent to Cooper's.
rule6(L).
solve(L) :-
bagof(F, X^member((X, F), L), LF),
label(LF).
dinners :-
retractall(top(_)), retractall(bottom(_)),
L = [(baker, _Fb), (cooper, _Fc), (fletcher, _Ff), (miller, _Fm), (smith, _Fs)],
init(L),
solve(L),
maplist(writeln, L).
Output :
?- dinners. baker,3 cooper,2 fletcher,4 miller,5 smith,1 true ; false.
true ==> predicate succeeded.
false ==> no other solution.
About flexibility : each name is associated with a floor, (contiguous floors differs from 1).
Bottom is always 1 but Top is defined from the number of names.
Each statement of the problem is translated in a Prolog rule, (a constraint on the floors), we can add so much of rules that we want, and a modification of one statement only modified one rule.
To solve the problem, library clpfd does the job.
[edit] PureBasic
Prototype cond(Array t(1))
Enumeration #Null
#Baker
#Cooper
#Fletcher
#Miller
#Smith
EndEnumeration
Procedure checkTenands(Array tenants(1), Array Condions.cond(1))
Protected i, j
Protected.cond *f
j=ArraySize(Condions())
For i=0 To j
*f=Condions(i) ; load the function pointer to the current condition
If *f(tenants()) = #False
ProcedureReturn #False
EndIf
Next
ProcedureReturn #True
EndProcedure
Procedure C1(Array t(1))
If Int(Abs(t(#Fletcher)-t(#Cooper)))<>1
ProcedureReturn #True
EndIf
EndProcedure
Procedure C2(Array t(1))
If t(#Baker)<>5
ProcedureReturn #True
EndIf
EndProcedure
Procedure C3(Array t(1))
If t(#Cooper)<>1
ProcedureReturn #True
EndIf
EndProcedure
Procedure C4(Array t(1))
If t(#Miller) >= t(#Cooper)
ProcedureReturn #True
EndIf
EndProcedure
Procedure C5(Array t(1))
If t(#Fletcher)<>1 And t(#Fletcher)<>5
ProcedureReturn #True
EndIf
EndProcedure
Procedure C6(Array t(1))
If Int(Abs(t(#Smith)-t(#Fletcher)))<>1
ProcedureReturn #True
EndIf
EndProcedure
If OpenConsole()
Dim People(4)
Dim Conditions(5)
Define a, b, c, d, e, i
;
;- Load all conditions
Conditions(i)=@C1(): i+1
Conditions(i)=@C2(): i+1
Conditions(i)=@C3(): i+1
Conditions(i)=@C4(): i+1
Conditions(i)=@C5(): i+1
Conditions(i)=@C6()
;
; generate and the all legal combinations
For a=1 To 5
For b=1 To 5
If a=b: Continue: EndIf
For c=1 To 5
If a=c Or b=c: Continue: EndIf
For d=1 To 5
If d=a Or d=b Or d=c : Continue: EndIf
For e=1 To 5
If e=a Or e=b Or e=c Or e=d: Continue: EndIf
People(#Baker)=a
People(#Cooper)=b
People(#Fletcher)=c
People(#Miller)=d
People(#Smith)=e
If checkTenands(People(), Conditions())
PrintN("Solution found;")
PrintN("Baker="+Str(a)+#CRLF$+"Cooper="+Str(b)+#CRLF$+"Fletcher="+Str(c))
PrintN("Miller="+Str(d)+#CRLF$+"Smith="+Str(e)+#CRLF$)
EndIf
Next
Next
Next
Next
Next
Print("Press ENTER to exit"): Input()
EndIf
Solution found; Baker=3 Cooper=2 Fletcher=4 Miller=5 Smith=1
[edit] Python
[edit] By parsing the problem statement
This example parses the statement of the problem as given and allows some variability such as the number of people, floors and constraints can be varied although the type of constraints allowed and the sentence structure is limited
- Setup
Parsing is done with the aid of the multi-line regular expression at the head of the program.
import re
from itertools import product
problem_re = re.compile(r"""(?msx)(?:
# Multiple names of form n1, n2, n3, ... , and nK
(?P<namelist> [a-zA-Z]+ (?: , \s+ [a-zA-Z]+)* (?: ,? \s+ and) \s+ [a-zA-Z]+ )
# Flexible floor count (2 to 10 floors)
| (?: .* house \s+ that \s+ contains \s+ only \s+
(?P<floorcount> two|three|four|five|six|seven|eight|nine|ten ) \s+ floors \s* \.)
# Constraint: "does not live on the n'th floor"
|(?: (?P<not_live> \b [a-zA-Z]+ \s+ does \s+ not \s+ live \s+ on \s+ the \s+
(?: top|bottom|first|second|third|fourth|fifth|sixth|seventh|eighth|ninth|tenth) \s+ floor \s* \. ))
# Constraint: "does not live on either the I'th or the J'th [ or the K'th ...] floor
|(?P<not_either> \b [a-zA-Z]+ \s+ does \s+ not \s+ live \s+ on \s+ either
(?: \s+ (?: or \s+)? the \s+
(?: top|bottom|first|second|third|fourth|fifth|sixth|seventh|eighth|ninth|tenth))+ \s+ floor \s* \. )
# Constraint: "P1 lives on a higher/lower floor than P2 does"
|(?P<hi_lower> \b [a-zA-Z]+ \s+ lives \s+ on \s+ a \s (?: higher|lower)
\s+ floor \s+ than (?: \s+ does) \s+ [a-zA-Z]+ \s* \. )
# Constraint: "P1 does/does not live on a floor adjacent to P2's"
|(?P<adjacency> \b [a-zA-Z]+ \s+ does (?:\s+ not)? \s+ live \s+ on \s+ a \s+
floor \s+ adjacent \s+ to \s+ [a-zA-Z]+ (?: 's )? \s* \. )
# Ask for the solution
|(?P<question> Where \s+ does \s+ everyone \s+ live \s* \?)
)
""")
names, lennames = None, None
floors = None
constraint_expr = 'len(set(alloc)) == lennames' # Start with all people on different floors
def do_namelist(txt):
" E.g. 'Baker, Cooper, Fletcher, Miller, and Smith'"
global names, lennames
names = txt.replace(' and ', ' ').split(', ')
lennames = len(names)
def do_floorcount(txt):
" E.g. 'five'"
global floors
floors = '||two|three|four|five|six|seven|eight|nine|ten'.split('|').index(txt)
def do_not_live(txt):
" E.g. 'Baker does not live on the top floor.'"
global constraint_expr
t = txt.strip().split()
who, floor = t[0], t[-2]
w, f = (names.index(who),
('|first|second|third|fourth|fifth|sixth|' +
'seventh|eighth|ninth|tenth|top|bottom|').split('|').index(floor)
)
if f == 11: f = floors
if f == 12: f = 1
constraint_expr += ' and alloc[%i] != %i' % (w, f)
def do_not_either(txt):
" E.g. 'Fletcher does not live on either the top or the bottom floor.'"
global constraint_expr
t = txt.replace(' or ', ' ').replace(' the ', ' ').strip().split()
who, floor = t[0], t[6:-1]
w, fl = (names.index(who),
[('|first|second|third|fourth|fifth|sixth|' +
'seventh|eighth|ninth|tenth|top|bottom|').split('|').index(f)
for f in floor]
)
for f in fl:
if f == 11: f = floors
if f == 12: f = 1
constraint_expr += ' and alloc[%i] != %i' % (w, f)
def do_hi_lower(txt):
" E.g. 'Miller lives on a higher floor than does Cooper.'"
global constraint_expr
t = txt.replace('.', '').strip().split()
name_indices = [names.index(who) for who in (t[0], t[-1])]
if 'lower' in t:
name_indices = name_indices[::-1]
constraint_expr += ' and alloc[%i] > alloc[%i]' % tuple(name_indices)
def do_adjacency(txt):
''' E.g. "Smith does not live on a floor adjacent to Fletcher's."'''
global constraint_expr
t = txt.replace('.', '').replace("'s", '').strip().split()
name_indices = [names.index(who) for who in (t[0], t[-1])]
constraint_expr += ' and abs(alloc[%i] - alloc[%i]) > 1' % tuple(name_indices)
def do_question(txt):
global constraint_expr, names, lennames
exec_txt = '''
for alloc in product(range(1,floors+1), repeat=len(names)):
if %s:
break
else:
alloc = None
''' % constraint_expr
exec(exec_txt, globals(), locals())
a = locals()['alloc']
if a:
output= ['Floors are numbered from 1 to %i inclusive.' % floors]
for a2n in zip(a, names):
output += [' Floor %i is occupied by %s' % a2n]
output.sort(reverse=True)
print('\n'.join(output))
else:
print('No solution found.')
print()
handler = {
'namelist': do_namelist,
'floorcount': do_floorcount,
'not_live': do_not_live,
'not_either': do_not_either,
'hi_lower': do_hi_lower,
'adjacency': do_adjacency,
'question': do_question,
}
def parse_and_solve(problem):
p = re.sub(r'\s+', ' ', problem).strip()
for x in problem_re.finditer(p):
groupname, txt = [(k,v) for k,v in x.groupdict().items() if v][0]
#print ("%r, %r" % (groupname, txt))
handler[groupname](txt)
- Problem statement
This is not much more than calling a function on the text of the problem!
if __name__ == '__main__':
parse_and_solve("""
Baker, Cooper, Fletcher, Miller, and Smith
live on different floors of an apartment house that contains
only five floors. Baker does not live on the top floor. Cooper
does not live on the bottom floor. Fletcher does not live on
either the top or the bottom floor. Miller lives on a higher
floor than does Cooper. Smith does not live on a floor
adjacent to Fletcher's. Fletcher does not live on a floor
adjacent to Cooper's. Where does everyone live?""")
print('# Add another person with more constraints and more floors:')
parse_and_solve("""
Baker, Cooper, Fletcher, Miller, Guinan, and Smith
live on different floors of an apartment house that contains
only seven floors. Guinan does not live on either the top or the third or the fourth floor.
Baker does not live on the top floor. Cooper
does not live on the bottom floor. Fletcher does not live on
either the top or the bottom floor. Miller lives on a higher
floor than does Cooper. Smith does not live on a floor
adjacent to Fletcher's. Fletcher does not live on a floor
adjacent to Cooper's. Where does everyone live?""")
- Output
This shows the output from the original problem and then for another, slightly different problem to cover some of the variability asked for in the task.
Floors are numbered from 1 to 5 inclusive. Floor 5 is occupied by Miller Floor 4 is occupied by Fletcher Floor 3 is occupied by Baker Floor 2 is occupied by Cooper Floor 1 is occupied by Smith # Add another person with more constraints and more floors: Floors are numbered from 1 to 7 inclusive. Floor 7 is occupied by Smith Floor 6 is occupied by Guinan Floor 4 is occupied by Fletcher Floor 3 is occupied by Miller Floor 2 is occupied by Cooper Floor 1 is occupied by Baker
[edit] By using the Amb operator
In this example, the problem needs to be turned into valid Python code for use with the Amb operator. Setup is just to import Amb.
The second set of results corresponds to this modification to the problem statement:
Baker, Cooper, Fletcher, Miller, Guinan, and Smith live on different floors of an apartment house that contains only seven floors. Guinan does not live on either the top or the third or the fourth floor. Baker does not live on the top floor. Cooper does not live on the bottom floor. Fletcher does not live on either the top or the bottom floor. Miller lives on a higher floor than does Cooper. Smith does not live on a floor adjacent to Fletcher's. Fletcher does not live on a floor adjacent to Cooper's. Where does everyone live
from amb import Amb
if __name__ == '__main__':
amb = Amb()
maxfloors = 5
floors = range(1, maxfloors+1)
# Possible floors for each person
Baker, Cooper, Fletcher, Miller, Smith = (amb(floors) for i in range(5))
for _dummy in amb( lambda Baker, Cooper, Fletcher, Miller, Smith: (
len(set([Baker, Cooper, Fletcher, Miller, Smith])) == 5 # each to a separate floor
and Baker != maxfloors
and Cooper != 1
and Fletcher not in (maxfloors, 1)
and Miller > Cooper
and (Smith - Fletcher) not in (1, -1) # Not adjacent
and (Fletcher - Cooper) not in (1, -1) # Not adjacent
) ):
print 'Floors are numbered from 1 to %i inclusive.' % maxfloors
print '\n'.join(sorted(' Floor %i is occupied by %s'
% (globals()[name], name)
for name in 'Baker, Cooper, Fletcher, Miller, Smith'.split(', ')))
break
else:
print 'No solution found.'
print '# Add another person with more constraints and more floors:'
# The order that Guinan is added to any list of people must stay consistant
amb = Amb()
maxfloors = 7
floors = range(1, maxfloors+1)
# Possible floors for each person
Baker, Cooper, Fletcher, Miller, Guinan, Smith = (amb(floors) for i in range(6))
for _dummy in amb( lambda Baker, Cooper, Fletcher, Miller, Guinan, Smith: (
len(set([Baker, Cooper, Fletcher, Miller, Guinan, Smith])) == 6 # each to a separate floor
and Guinan not in (maxfloors, 3, 4)
and Baker != maxfloors
and Cooper != 1
and Fletcher not in (maxfloors, 1)
and Miller > Cooper
and (Smith - Fletcher) not in (1, -1) # Not adjacent
and (Fletcher - Cooper) not in (1, -1) # Not adjacent
) ):
print 'Floors are numbered from 1 to %i inclusive.' % maxfloors
print '\n'.join(sorted(' Floor %i is occupied by %s'
% (globals()[name], name)
for name in 'Baker, Cooper, Fletcher, Miller, Guinan, Smith'.split(', ')))
break
else:
print 'No solution found.'
- Output
Floors are numbered from 1 to 5 inclusive. Floor 1 is occupied by Smith Floor 2 is occupied by Cooper Floor 3 is occupied by Baker Floor 4 is occupied by Fletcher Floor 5 is occupied by Miller # Add another person with more constraints and more floors: Floors are numbered from 1 to 7 inclusive. Floor 1 is occupied by Baker Floor 2 is occupied by Cooper Floor 3 is occupied by Miller Floor 4 is occupied by Fletcher Floor 5 is occupied by Guinan Floor 6 is occupied by Smith
[edit] Simple Solution
from itertools import permutations
class Names:
Baker, Cooper, Fletcher, Miller, Smith = range(5)
seq = [Baker, Cooper, Fletcher, Miller, Smith]
strings = "Baker Cooper Fletcher Miller Smith".split()
predicates = [
lambda s: s[Names.Baker] != len(s)-1,
lambda s: s[Names.Cooper] != 0,
lambda s: s[Names.Fletcher] != 0 and s[Names.Fletcher] != len(s)-1,
lambda s: s[Names.Miller] > s[Names.Cooper],
lambda s: abs(s[Names.Smith] - s[Names.Fletcher]) != 1,
lambda s: abs(s[Names.Cooper] - s[Names.Fletcher]) != 1];
for sol in permutations(Names.seq):
if all(p(sol) for p in predicates):
print " ".join(Names.strings[s] for s in sol)
- Output:
Fletcher Cooper Miller Smith Baker
[edit] Ruby
def dinesman(floors, names, criteria)
# the "bindVars" method returns a context where the "name" variables are bound to values
eval "
def bindVars(#{names.map {|n| n.downcase}.join ','})
return binding
end
"
expression = criteria.map {|c| "(#{c.downcase})"}.join " and "
floors.permutation.each do |perm|
b = bindVars *perm
return b if b.eval(expression)
end
nil
end
floors = (1..5).to_a
names = %w(Baker Cooper Fletcher Miller Smith)
criteria = [
"Baker != 5",
"Cooper != 1",
"Fletcher != 1",
"Fletcher != 5",
"Miller > Cooper",
"(Smith - Fletcher).abs != 1",
"(Fletcher - Cooper).abs != 1",
]
b = dinesman(floors, names, criteria)
if b.nil?
puts "no solution"
else
puts "Found a solution:"
len = names.map {|n| n.length}.max
residents = names.inject({}) {|r, n| r[b.eval(n.downcase)] = n; r}
floors.each {|f| puts " Floor #{f}: #{residents[f]}"}
end
output
Found a solution: Floor 1: Smith Floor 2: Cooper Floor 3: Baker Floor 4: Fletcher Floor 5: Miller
[edit] Run BASIC
people$ = "Baler,Cooper,Fletcher,Miller,Smith"
for baler = 1 to 4 ' can not be in room 5
for cooper = 2 to 5 ' can not be in room 1
for fletcher = 2 to 4 ' can not be in room 1 or 5
for miller = 1 to 5 ' can be in any room
for smith = 1 to 5 ' can be in any room
if miller > cooper and abs(smith - fletcher) > 1 and abs(fletcher - cooper) > 1 then
if baler + cooper + fletcher + miller + smith = 15 then ' that is 1 + 2 + 3 + 4 + 5
rooms$ = baler;cooper;fletcher;miller;smith
bad = 0
for i = 1 to 5 ' make sure each room is unique
rm$ = chr$(i + 48)
r1 = instr(rooms$,rm$)
r2 = instr(rooms$,rm$,r1+1)
if r2 <> 0 then bad = 1
next i
if bad = 0 then goto [roomAssgn] ' if it is not bad it is a good assignment
end if
end if
next smith
next miller
next fletcher
next cooper
next baler
print "Cam't assign rooms" ' print this if it can not find a solution
wait
[roomAssgn]
Print "Room Assignment"
for i = 1 to 5
print mid$(rooms$,i,1);" ";word$(people$,i,",");" "; ' print the room assignments
next i
Room Assignment 3 Baler 2 Cooper 4 Fletcher 5 Miller 1 Smith
[edit] Scala
object Dinesman extends App {
val tenants = List("Baker", "Cooper", "Fletcher", "Miller", "Smith")
val floors = (1 to tenants.size).toList
// define the predicates
import scala.math.abs
val predicates =
List((perm: Map[String, Int]) => !(perm("Baker")==floors.size)
,(perm: Map[String, Int]) => !(perm("Cooper")==1)
,(perm: Map[String, Int]) => !(perm("Fletcher")==1 || perm("Fletcher")==floors.size)
,(perm: Map[String, Int]) => !(perm("Miller")<=perm("Cooper"))
,(perm: Map[String, Int]) => !(abs(perm("Smith")-perm("Fletcher"))==1)
,(perm: Map[String, Int]) => !(abs(perm("Fletcher")-perm("Cooper"))==1)
)
val p: Seq[(String, Int)] => Boolean = perm => !predicates.map(_(perm.toMap)).contains(false)
tenants.permutations.map(_ zip floors).toList
.map(perm=>Pair(perm,p(perm))).filter(_._2==true).map(p=>p._1.toList)
match {
case Nil => println("no solution")
case xss => { println("solutions: "+xss.size)
xss.foreach{l=>
println("possible solution:")
l.foreach(p=>println(" "+p._1+ " lives on floor number "+p._2))
}
}
}
}
Output:
solutions: 1 possible solution: Smith lives on floor number 1 Cooper lives on floor number 2 Baker lives on floor number 3 Fletcher lives on floor number 4 Miller lives on floor number 5
We can extend this problem by adding a tenant resp. adding conditions:
...
val tenants = List("Baker", "Cooper", "Fletcher", "Miller", "Smith", "Rollo")
...
val predicates =
List((perm: Map[String, Int]) => !(perm("Baker")==floors.size)
...
,(perm: Map[String, Int]) => !(perm("Rollo")==floors.size || perm("Rollo")==3 || perm("Rollo")==4)
,(perm: Map[String, Int]) => !(perm("Rollo")>perm("Smith"))
,(perm: Map[String, Int]) => !(perm("Rollo")<perm("Fletcher"))
)
...
Output:
solutions: 1 possible solution: Baker lives on floor number 1 Cooper lives on floor number 2 Miller lives on floor number 3 Fletcher lives on floor number 4 Rollo lives on floor number 5 Smith lives on floor number 6
[edit] Tcl
It's trivial to extend this problem to deal with more floors and people and more constraints; the main internally-generated constraint is that the names of people should begin with an upper case character so that they are distinct from internal variables. This code also relies on the caller encoding the conditions as expressions that produce a value that is/can be interpreted as a boolean.
package require Tcl 8.5
package require struct::list
proc dinesmanSolve {floors people constraints} {
# Search for a possible assignment that satisfies the constraints
struct::list foreachperm p $floors {
lassign $p {*}$people
set found 1
foreach c $constraints {
if {![expr $c]} {
set found 0
break
}
}
if {$found} break
}
# Found something, or exhausted possibilities
if {!$found} {
error "no solution possible"
}
# Generate in "nice" order
foreach f $floors {
foreach person $people {
if {[set $person] == $f} {
lappend result $f $person
break
}
}
}
return $result
}
Solve the particular problem:
set soln [dinesmanSolve {1 2 3 4 5} {Baker Cooper Fletcher Miller Smith} {
{$Baker != 5}
{$Cooper != 1}
{$Fletcher != 1 && $Fletcher != 5}
{$Miller > $Cooper}
{abs($Smith-$Fletcher) != 1}
{abs($Fletcher-$Cooper) != 1}
}]
puts "Solution found:"
foreach {where who} $soln {puts " Floor ${where}: $who"}
Output:
Solution found: Floor 1: Smith Floor 2: Cooper Floor 3: Baker Floor 4: Fletcher Floor 5: Miller
