Department numbers
You are encouraged to solve this task according to the task description, using any language you may know.
There is a highly organized city that has decided to assign a number to each of their departments:
- police department
- sanitation department
- fire department
Each department can have a number between 1 and 7 (inclusive).
The three department numbers are to be unique (different from each other) and must add up to 12.
The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.
- Task
Write a computer program which outputs all valid combinations.
Possible output (for the 1st and 14th solutions):
--police-- --sanitation-- --fire--
2 3 7
6 5 1
11l
print(‘Police Sanitation Fire’)
print(‘----------------------------------’)
L(police) (2..6).step(2)
L(sanitation) 1..7
L(fire) 1..7
I police!=sanitation & sanitation!=fire & fire!=police & police+fire+sanitation==12
print(police"\t\t"sanitation"\t\t"fire)- Output:
Police Sanitation Fire ---------------------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
8080 Assembly
org 100h
lxi h,obuf ; HL = output buffer
mvi b,2 ; B = police
pol: mvi c,1 ; C = sanitation
san: mvi d,1 ; D = fire
fire: mov a,b ; Fire equal to police?
cmp d
jz next ; If so, invalid combination
mov a,c ; Fire equal to sanitation?
cmp d
jz next ; If so, invalid combination
mov a,b ; Total equal to 12?
add c
add d
cpi 12
jnz next ; If not, invalid combination
mov a,b ; Combination is valid, add to output
call num
mov a,c
call num
mov a,d
call num
mvi m,13 ; Add a newline to the output
inx h
mvi m,10
inx h
next: mvi a,7 ; Load 7 to compare to
inr d ; Next fire number
cmp d ; Reached the end?
jnc fire ; If not, next fire number
inr c ; Otherwise, next sanitation number
cmp c ; Reached the end?
jnc san ; If not, next sanitation number
inr b ; Increment police number twice
inr b ; (twice, because it must be even)
cmp b ; Reached the end?
jnc pol ; If not, next police number
mvi m,'$' ; If so, we're done - add CP/M string terminator
mvi c,9 ; Print the output string
lxi d,ohdr
jmp 5
num: adi '0' ; Add number A and space to the output
mov m,a
inx h
mvi m,' '
inx h
ret
ohdr: db 'P S F',13,10
obuf: equ $ ; Output buffer goes after program- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
8086 Assembly
cpu 8086
bits 16
org 100h
section .text
mov di,obuf ; Output buffer
mov bl,2 ; BL = police
pol: mov cl,1 ; CL = sanitation
san: mov dl,1 ; DL = fire
fire: cmp bl,cl ; Police equal to sanitation?
je next ; Invalid combination
cmp bl,dl ; Police equal to fire?
je next ; Invalid combination
cmp cl,dl ; Sanitation equal to fire?
je next ; Invalid combination
mov al,bl ; Total equal to 12?
add al,cl
add al,dl
cmp al,12
jne next ; If not, invalid combination
mov al,bl ; Combination is valid, write the three numbers
call num
mov al,cl
call num
mov al,dl
call num
mov ax,0A0Dh ; And a newline
stosw
next: mov al,7 ; Load 7 to compare to
inc dx ; Increment fire number
cmp al,dl ; If 7 or less,
jae fire ; next fire number.
inc cx ; Otherwise, ncrement sanitation number
cmp al,cl ; If 7 or less,
jae san ; next sanitation number
inc bx ; Increment police number twice
inc bx ; (it must be even)
cmp al,bl ; If 7 or less,
jae pol ; next police number.
mov byte [di],'$' ; At the end, terminate the string
mov dx,ohdr ; Tell MS-DOS to print it
mov ah,9
int 21h
ret
num: mov ah,' ' ; Space
add al,'0' ; Add number to output
stosw ; Store number and space
ret
section .data
ohdr: db 'P S F',13,10 ; Header
obuf: equ $ ; Place to write output
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
ABC
PUT 7 IN max.department.number
PUT 12 IN department.sum
WRITE "police sanitation fire" /
PUT 2 IN police
WHILE police <= max.department.number:
FOR sanitation IN { 1 .. max.department.number }:
IF sanitation <> police:
PUT ( department.sum - police ) - sanitation IN fire
IF fire > 0 AND fire <= max.department.number AND fire <> sanitation AND fire <> police:
WRITE police>>6, sanitation>>11, fire>>5 /
PUT police + 2 IN police- Output:
police sanitation fire
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Action!
PROC Main()
BYTE p,s,f
PrintE("P S F")
FOR p=2 TO 6 STEP 2
DO
FOR s=1 TO 7
DO
FOR f=1 TO 7
DO
IF p#s AND p#f AND s#f AND p+s+f=12 THEN
PrintF("%B %B %B%E",p,s,f)
FI
OD
OD
OD
RETURN- Output:
Screenshot from Atari 8-bit computer
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Ada
with Ada.Text_IO;
procedure Department_Numbers is
use Ada.Text_IO;
begin
Put_Line (" P S F");
for Police in 2 .. 6 loop
for Sanitation in 1 .. 7 loop
for Fire in 1 .. 7 loop
if
Police mod 2 = 0 and
Police + Sanitation + Fire = 12 and
Sanitation /= Police and
Sanitation /= Fire and
Police /= Fire
then
Put_Line (Police'Image & Sanitation'Image & Fire'Image);
end if;
end loop;
end loop;
end loop;
end Department_Numbers;
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Aime
integer p, s, f;
p = 0;
while ((p += 2) <= 7) {
s = 0;
while ((s += 1) <= 7) {
f = 0;
while ((f += 1) <= 7) {
if (p + s + f == 12 && p != s && p != f && s != f) {
o_form(" ~ ~ ~\n", p, s, f);
}
}
}
}ALGOL 68
As noted in the Fortran sample, once the police and sanitation departments are posited, the fire department value is fixed
BEGIN
# show possible department number allocations for police, sanitation and fire departments #
# the police department number must be even, all department numbers in the range 1 .. 7 #
# the sum of the department numbers must be 12 #
INT max department number = 7;
INT department sum = 12;
print( ( "police sanitation fire", newline ) );
FOR police FROM 2 BY 2 TO max department number DO
FOR sanitation TO max department number DO
IF sanitation /= police THEN
INT fire = ( department sum - police ) - sanitation;
IF fire > 0 AND fire <= max department number
AND fire /= sanitation
AND fire /= police
THEN
print( ( whole( police, -6 )
, whole( sanitation, -11 )
, whole( fire, -5 )
, newline
)
)
FI
FI
OD
OD
END- Output:
police sanitation fire
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
ALGOL W
begin
% show possible department number allocations for police, sanitation and fire departments %
% the police department number must be even, all department numbers in the range 1 .. 7 %
% the sum of the department numbers must be 12 %
integer MAX_DEPARTMENT_NUMBER, DEPARTMENT_SUM;
MAX_DEPARTMENT_NUMBER := 7;
DEPARTMENT_SUM := 12;
write( "police sanitation fire" );
for police := 2 step 2 until MAX_DEPARTMENT_NUMBER do begin
for sanitation := 1 until MAX_DEPARTMENT_NUMBER do begin
IF sanitation not = police then begin
integer fire;
fire := ( DEPARTMENT_SUM - police ) - sanitation;
if fire > 0 and fire <= MAX_DEPARTMENT_NUMBER and fire not = sanitation and fire not = police then begin
write( s_w := 0, i_w := 6, police, i_w := 11, sanitation, i_w := 5, fire )
end if_valid_combination
end if_sanitation_ne_police
end for_sanitation
end for_police
end.- Output:
police sanitation fire
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
APL
'PSF'⍪(⊢(⌿⍨)((∪≡⊢)¨↓∧(0=2|1⌷[2]⊢)∧12=+/))↑,⍳3/7
This prints each triplet of numbers from 1 to 7 for which:
- the elements are unique:
(∪≡⊢)¨↓ - the first element is even:
(0=2|1⌷[2]⊢) - the sum of the elements is 12:
12=+/
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
AppleScript
Briefly, composing a solution from generic functions:
on run
script
on |λ|(x)
script
on |λ|(y)
script
on |λ|(z)
if y ≠ z and 1 ≤ z and z ≤ 7 then
{{x, y, z} as string}
else
{}
end if
end |λ|
end script
concatMap(result, {12 - (x + y)}) --Z
end |λ|
end script
concatMap(result, {1, 2, 3, 4, 5, 6, 7}) --Y
end |λ|
end script
unlines(concatMap(result, {2, 4, 6})) --X
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lst to {}
set lng to length of xs
tell mReturn(f)
repeat with i from 1 to lng
set lst to (lst & |λ|(contents of item i of xs, i, xs))
end repeat
end tell
return lst
end concatMap
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- unlines :: [String] -> String
on unlines(xs)
intercalate(linefeed, xs)
end unlines
- Output:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
Or more generally:
-- NUMBERING CONSTRAINTS ------------------------------------------------------
-- options :: Int -> Int -> Int -> [(Int, Int, Int)]
on options(lo, hi, total)
set ds to enumFromTo(lo, hi)
script Xeven
on |λ|(x)
script Ydistinct
on |λ|(y)
script ZinRange
on |λ|(z)
if y ≠ z and lo ≤ z and z ≤ hi then
{{x, y, z}}
else
{}
end if
end |λ|
end script
concatMap(ZinRange, {total - (x + y)}) -- Z IS IN RANGE
end |λ|
end script
script notX
on |λ|(d)
d ≠ x
end |λ|
end script
concatMap(Ydistinct, filter(notX, ds)) -- Y IS NOT X
end |λ|
end script
concatMap(Xeven, filter(my even, ds)) -- X IS EVEN
end options
-- TEST -----------------------------------------------------------------------
on run
set xs to options(1, 7, 12)
intercalate("\n\n", ¬
{"(Police, Sanitation, Fire)", ¬
unlines(map(show, xs)), ¬
"Number of options: " & |length|(xs)})
end run
-- GENERIC FUNCTIONS ----------------------------------------------------------
-- concatMap :: (a -> [b]) -> [a] -> [b]
on concatMap(f, xs)
set lst to {}
set lng to length of xs
tell mReturn(f)
repeat with i from 1 to lng
set lst to (lst & |λ|(contents of item i of xs, i, xs))
end repeat
end tell
return lst
end concatMap
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if n < m then
set d to -1
else
set d to 1
end if
set lst to {}
repeat with i from m to n by d
set end of lst to i
end repeat
return lst
end enumFromTo
-- even :: Int -> Bool
on even(x)
x mod 2 = 0
end even
-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
tell mReturn(f)
set lst to {}
set lng to length of xs
repeat with i from 1 to lng
set v to item i of xs
if |λ|(v, i, xs) then set end of lst to v
end repeat
return lst
end tell
end filter
-- intercalate :: Text -> [Text] -> Text
on intercalate(strText, lstText)
set {dlm, my text item delimiters} to {my text item delimiters, strText}
set strJoined to lstText as text
set my text item delimiters to dlm
return strJoined
end intercalate
-- length :: [a] -> Int
on |length|(xs)
length of xs
end |length|
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn
-- show :: a -> String
on show(e)
set c to class of e
if c = list then
script serialized
on |λ|(v)
show(v)
end |λ|
end script
"[" & intercalate(", ", map(serialized, e)) & "]"
else if c = record then
script showField
on |λ|(kv)
set {k, ev} to kv
"\"" & k & "\":" & show(ev)
end |λ|
end script
"{" & intercalate(", ", ¬
map(showField, zip(allKeys(e), allValues(e)))) & "}"
else if c = date then
"\"" & iso8601Z(e) & "\""
else if c = text then
"\"" & e & "\""
else if (c = integer or c = real) then
e as text
else if c = class then
"null"
else
try
e as text
on error
("«" & c as text) & "»"
end try
end if
end show
-- unlines :: [String] -> String
on unlines(xs)
intercalate(linefeed, xs)
end unlines
- Output:
(Police, Sanitation, Fire) [2, 3, 7] [2, 4, 6] [2, 6, 4] [2, 7, 3] [4, 1, 7] [4, 2, 6] [4, 3, 5] [4, 5, 3] [4, 6, 2] [4, 7, 1] [6, 1, 5] [6, 2, 4] [6, 4, 2] [6, 5, 1] Number of options: 14
Arturo
loop 1..7 'x [
loop 1..7 'y [
loop 1..7 'z [
if all? @[
even? x
12 = sum @[x y z]
3 = size unique @[x y z]
] -> print [x y z]
]
]
]
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Asymptote
write("--police-- --sanitation-- --fire--");
for(int police = 2; police < 6; police += 2) {
for(int sanitation = 1; sanitation < 7; ++sanitation) {
for(int fire = 1; fire < 7; ++fire) {
if(police != sanitation && sanitation != fire && fire != police && police+fire+sanitation == 12){
write(" ", police, suffix=none);
write(" ", sanitation, suffix=none);
write(" ", fire);
}
}
}
}
AutoHotkey
perm(elements, n, opt:="", Delim:="", str:="", res:="", j:=0, dup:="") {
res := IsObject(res) ? res : [], dup := IsObject(dup) ? dup : []
if (n > j)
Loop, parse, elements, % Delim
res := !(InStr(str, A_LoopField) && !(InStr(opt, "rep"))) ? perm(elements, n, opt, Delim, trim(str Delim A_LoopField, Delim), res, j+1, dup) : res
else if !(dup[x := perm_sort(str, Delim)] && (InStr(opt, "comb")))
dup[x] := 1, res.Insert(str)
return res, j++
}
perm_sort(str, Delim){
Loop, Parse, str, % Delim
res .= A_LoopField "`n"
Sort, res, D`n
return StrReplace(res, "`n", Delim)
}
Example:
elements := "1234567", n := 3
for k, v in perm(elements, n)
if (SubStr(v, 1, 1) + SubStr(v, 2, 1) + SubStr(v, 3, 1) = 12) && (SubStr(v, 1, 1) / 2 = Floor(SubStr(v, 1, 1)/2))
res4 .= v "`n"
MsgBox, 262144, , % res4
return
Outputs:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
AWK
# syntax: GAWK -f DEPARTMENT_NUMBERS.AWK
BEGIN {
print(" # FD PD SD")
for (fire=1; fire<=7; fire++) {
for (police=1; police<=7; police++) {
for (sanitation=1; sanitation<=7; sanitation++) {
if (rules() ~ /^1+$/) {
printf("%2d %2d %2d %2d\n",++count,fire,police,sanitation)
}
}
}
}
exit(0)
}
function rules( stmt1,stmt2,stmt3) {
stmt1 = fire != police && fire != sanitation && police != sanitation
stmt2 = fire + police + sanitation == 12
stmt3 = police % 2 == 0
return(stmt1 stmt2 stmt3)
}
- Output:
# FD PD SD 1 1 4 7 2 1 6 5 3 2 4 6 4 2 6 4 5 3 2 7 6 3 4 5 7 4 2 6 8 4 6 2 9 5 4 3 10 5 6 1 11 6 2 4 12 6 4 2 13 7 2 3 14 7 4 1
BCPL
get "libhdr"
let start() be
for p=2 to 6 by 2
for s=1 to 7
for f=1 to 7
if p~=s & s~=f & p~=f & p+s+f=12 then
writef("%N %N %N*N", p, s, f)- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
BASIC
Applesoft BASIC
The Sinclair ZX81 BASIC solution works without any changes.
BASIC256
print "--police-- --sanitation-- --fire--"
for police = 2 to 7 step 2
for fire = 1 to 7
if fire = police then continue for
sanitation = 12 - police - fire
if sanitation = fire or sanitation = police then continue for
if sanitation >= 1 and sanitation <= 7 then
print rjust(police, 6); rjust(fire, 13); rjust(sanitation, 12)
end if
next fire
next police- Output:
--police-- --sanitation-- --fire--
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Chipmunk Basic
10 CLS
20 PRINT "--police-- --sanitation-- --fire--"
30 FOR police = 2 TO 7 STEP 2
40 FOR fire = 1 TO 7
50 sanitation = 12-police-fire
60 IF sanitation >= 1 AND sanitation <= 7 THEN
70 PRINT TAB (5)police TAB (18)fire TAB (30)sanitation
80 endif
90 NEXT fire
100 NEXT police
- Output:
Same as FreeBASIC entry.
Craft Basic
print "P S F"
for p = 2 to 7 step 2
for s = 1 to 7
if s <> p then
let f = (12 - p) - s
if f > 0 and f <= 7 and f <> s and f <> p then
print p, " ", s, " ", f
endif
endif
next s
next p
end
- Output:
P S F2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2
6 5 1
Run BASIC
print "police fire sanitation"
for police = 2 to 7 step 2
for fire = 1 to 7
if fire = police then [cont]
sanitation = (12-police)-fire
if sanitation <= 0 or sanitation > 7 or sanitation = fire or sanitation = police then [cont]
print " "; police; chr$(9); fire; chr$(9); sanitation
[cont]
next fire
next policePureBasic
OpenConsole()
PrintN("--police-- --sanitation-- --fire--")
For police = 2 To 7 Step 2
For fire = 1 To 7
If fire = police:
Continue
EndIf
sanitation = 12 - police - fire
If sanitation = fire Or sanitation = police: Continue : EndIf
If sanitation >= 1 And sanitation <= 7:
PrintN(" " + Str(police) + #TAB$ + #TAB$ + Str(fire) + #TAB$ + #TAB$ + Str(sanitation))
EndIf
Next fire
Next police
Input()
CloseConsole()
- Output:
Same as BASIC256 entry.
Yabasic
print "--police-- --sanitation-- --fire--"
for police = 2 to 7 step 2
for fire = 1 to 7
if fire = police continue
sanitation = 12 - police - fire
if sanitation = fire or sanitation = police continue
if sanitation >= 1 and sanitation <= 7 print police using "######", fire using "############", sanitation using "###########"
next fire
next police
- Output:
Same as BASIC256 entry.
IS-BASIC
100 PRINT "Police","San.","Fire"
110 FOR P=2 TO 7 STEP 2
120 FOR S=1 TO 7
130 IF S<>P THEN
131 LET F=(12-P)-S
140 IF F>0 AND F<=7 AND F<>S AND F<>P THEN PRINT P,S,F
141 END IF
150 NEXT
160 NEXTMinimal BASIC
10 REM Department numbers
20 PRINT "POLICE SANITATION FIRE"
30 FOR P = 2 TO 7 STEP 2
40 FOR S = 1 TO 7
50 IF S = P THEN 120
60 LET F = (12-P)-S
70 IF F <= 0 THEN 120
80 IF F > 7 THEN 120
90 IF F = S THEN 120
100 IF F = P THEN 120
110 PRINT TAB(3); P; TAB(11); S; TAB(19); F
120 NEXT S
130 NEXT P
140 END
Quite BASIC
10 REM Department numbers
20 PRINT "POLICE SANITATION FIRE"
30 FOR P = 2 TO 7 STEP 2
40 FOR S = 1 TO 7
50 IF S = P THEN 80
60 LET F = (12-P)-S
70 IF F > 0 AND F <= 7 AND F <> S AND F <> P THEN PRINT " ";P;" ";S;" ";F
80 NEXT S
90 NEXT P
100 END
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Sinclair ZX81 BASIC
Works with 1k of RAM. This program ought not to need more than minimal changes to be compatible with any old-style BASIC dialect.
10 PRINT "POLICE SANITATION FIRE"
20 FOR P=2 TO 7 STEP 2
30 FOR S=1 TO 7
40 IF S=P THEN NEXT S
50 LET F=(12-P)-S
60 IF F>0 AND F<=7 AND F<>S AND F<>P THEN PRINT " ";P;" ";S;" ";F
70 NEXT S
80 NEXT P
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
BBC BASIC
REM >deptnums
max_dept_num% = 7
dept_sum% = 12
PRINT "police sanitation fire"
FOR police% = 2 TO max_dept_num% STEP 2
FOR sanitation% = 1 TO max_dept_num%
IF sanitation% <> police% THEN
fire% = (dept_sum% - police%) - sanitation%
IF fire% > 0 AND fire% <= max_dept_num% AND fire% <> sanitation% AND fire% <> police% THEN PRINT " "; police%; " "; sanitation%; " "; fire%
ENDIF
NEXT
NEXT
END
- Output:
police sanitation fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
QBasic
PRINT "--police-- --sanitation-- --fire--"
FOR police = 2 TO 7 STEP 2
FOR fire = 1 TO 7
IF fire = police THEN GOTO cont
sanitation = 12 - police - fire
IF sanitation = fire OR sanitation = police THEN GOTO cont
IF sanitation >= 1 AND sanitation <= 7 THEN
PRINT USING " # # #"; police; fire; sanitation
END IF
cont:
NEXT fire
NEXT police
XBasic
PROGRAM "Depar/num"
DECLARE FUNCTION Entry ()
FUNCTION Entry ()
PRINT "police sanitation fire"
FOR police = 2 TO 7 STEP 2
FOR fire = 1 TO 7
IF fire = police THEN GOTO cont
sanitation = 12 - police - fire
IF sanitation = fire OR sanitation = police THEN GOTO cont
IF sanitation >= 1 AND sanitation <= 7 THEN
PRINT TAB(3); police; TAB(13); fire; TAB(22); sanitation
END IF
cont:
NEXT fire
NEXT police
END FUNCTION
END PROGRAM
C
Weird that such a simple task was still not implemented in C, would be great to see some really creative ( obfuscated ) solutions for this one.
#include<stdio.h>
int main()
{
int police,sanitation,fire;
printf("Police Sanitation Fire\n");
printf("----------------------------------");
for(police=2;police<=6;police+=2){
for(sanitation=1;sanitation<=7;sanitation++){
for(fire=1;fire<=7;fire++){
if(police!=sanitation && sanitation!=fire && fire!=police && police+fire+sanitation==12){
printf("\n%d\t\t%d\t\t%d",police,sanitation,fire);
}
}
}
}
return 0;
}
Output:
Police Sanitation Fire ---------------------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
C#
using System;
public class Program
{
public static void Main() {
for (int p = 2; p <= 7; p+=2) {
for (int s = 1; s <= 7; s++) {
int f = 12 - p - s;
if (s >= f) break;
if (f > 7) continue;
if (s == p || f == p) continue; //not even necessary
Console.WriteLine($"Police:{p}, Sanitation:{s}, Fire:{f}");
Console.WriteLine($"Police:{p}, Sanitation:{f}, Fire:{s}");
}
}
}
}
- Output:
Police:2, Sanitation:3, Fire:7 Police:2, Sanitation:7, Fire:3 Police:2, Sanitation:4, Fire:6 Police:2, Sanitation:6, Fire:4 Police:4, Sanitation:1, Fire:7 Police:4, Sanitation:7, Fire:1 Police:4, Sanitation:2, Fire:6 Police:4, Sanitation:6, Fire:2 Police:4, Sanitation:3, Fire:5 Police:4, Sanitation:5, Fire:3 Police:6, Sanitation:1, Fire:5 Police:6, Sanitation:5, Fire:1 Police:6, Sanitation:2, Fire:4 Police:6, Sanitation:4, Fire:2
C++
#include <iostream>
#include <iomanip>
int main( int argc, char* argv[] ) {
int sol = 1;
std::cout << "\t\tFIRE\t\tPOLICE\t\tSANITATION\n";
for( int f = 1; f < 8; f++ ) {
for( int p = 1; p < 8; p++ ) {
for( int s = 1; s < 8; s++ ) {
if( f != p && f != s && p != s && !( p & 1 ) && ( f + s + p == 12 ) ) {
std::cout << "SOLUTION #" << std::setw( 2 ) << sol++ << std::setw( 2 )
<< ":\t" << std::setw( 2 ) << f << "\t\t " << std::setw( 3 ) << p
<< "\t\t" << std::setw( 6 ) << s << "\n";
}
}
}
}
return 0;
}
- Output:
FIRE POLICE SANITATION SOLUTION # 1: 1 4 7 SOLUTION # 2: 1 6 5 SOLUTION # 3: 2 4 6 SOLUTION # 4: 2 6 4 SOLUTION # 5: 3 2 7 SOLUTION # 6: 3 4 5 SOLUTION # 7: 4 2 6 SOLUTION # 8: 4 6 2 SOLUTION # 9: 5 4 3 SOLUTION #10: 5 6 1 SOLUTION #11: 6 2 4 SOLUTION #12: 6 4 2 SOLUTION #13: 7 2 3 SOLUTION #14: 7 4 1
Clojure
(let [n (range 1 8)]
(for [police n
sanitation n
fire n
:when (distinct? police sanitation fire)
:when (even? police)
:when (= 12 (+ police sanitation fire))]
(println police sanitation fire)))
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
CLU
start_up = proc ()
po: stream := stream$primary_output()
stream$putl(po, "P S F\n- - -")
for police: int in int$from_to_by(2,7,2) do
for sanitation: int in int$from_to(1,7) do
for fire: int in int$from_to(1,7) do
if police~=sanitation
& sanitation~=fire
& police~=fire
& police+sanitation+fire = 12
then
stream$putl(po, int$unparse(police) || " " ||
int$unparse(sanitation) || " " ||
int$unparse(fire))
end
end
end
end
end start_up- Output:
P S F - - - 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
COBOL
IDENTIFICATION DIVISION.
PROGRAM-ID. DEPARTMENT-NUMBERS.
DATA DIVISION.
WORKING-STORAGE SECTION.
01 BANNER PIC X(24) VALUE "POLICE SANITATION FIRE".
01 COMBINATION.
03 FILLER PIC X(5) VALUE SPACES.
03 POLICE PIC 9.
03 FILLER PIC X(11) VALUE SPACES.
03 SANITATION PIC 9.
03 FILLER PIC X(5) VALUE SPACES.
03 FIRE PIC 9.
01 TOTAL PIC 99.
PROCEDURE DIVISION.
BEGIN.
DISPLAY BANNER.
PERFORM POLICE-LOOP VARYING POLICE FROM 2 BY 2
UNTIL POLICE IS GREATER THAN 6.
STOP RUN.
POLICE-LOOP.
PERFORM SANITATION-LOOP VARYING SANITATION FROM 1 BY 1
UNTIL SANITATION IS GREATER THAN 7.
SANITATION-LOOP.
PERFORM FIRE-LOOP VARYING FIRE FROM 1 BY 1
UNTIL FIRE IS GREATER THAN 7.
FIRE-LOOP.
ADD POLICE, SANITATION, FIRE GIVING TOTAL.
IF POLICE IS NOT EQUAL TO SANITATION
AND POLICE IS NOT EQUAL TO FIRE
AND SANITATION IS NOT EQUAL TO FIRE
AND TOTAL IS EQUAL TO 12,
DISPLAY COMBINATION.
- Output:
POLICE SANITATION FIRE
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Cowgol
include "cowgol.coh";
typedef Dpt is int(1, 7);
# print combination if valid
sub print_comb(p: Dpt, s: Dpt, f: Dpt) is
var out: uint8[] := {'*',' ','*',' ','*','\n',0};
out[0] := p + '0';
out[2] := s + '0';
out[4] := f + '0';
if p != s and p != f and f != s and p+s+f == 12 then
print(&out[0]);
end if;
end sub;
print("P S F\n"); # header
var pol: Dpt := 2;
while pol <= 7 loop
var san: Dpt := 1;
while san <= 7 loop
var fire: Dpt := 1;
while fire <= 7 loop
print_comb(pol, san, fire);
fire := fire + 1;
end loop;
san := san + 1;
end loop;
pol := pol + 2;
end loop;- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
D
import std.stdio, std.range;
void main() {
int sol = 1;
writeln("\t\tFIRE\t\tPOLICE\t\tSANITATION");
foreach( f; iota(1,8) ) {
foreach( p; iota(1,8) ) {
foreach( s; iota(1,8) ) {
if( f != p && f != s && p != s && !( p & 1 ) && ( f + s + p == 12 ) ) {
writefln("SOLUTION #%2d:\t%2d\t\t%3d\t\t%6d", sol++, f, p, s);
}
}
}
}
}
Output:
FIRE POLICE SANITATION SOLUTION # 1: 1 4 7 SOLUTION # 2: 1 6 5 SOLUTION # 3: 2 4 6 SOLUTION # 4: 2 6 4 SOLUTION # 5: 3 2 7 SOLUTION # 6: 3 4 5 SOLUTION # 7: 4 2 6 SOLUTION # 8: 4 6 2 SOLUTION # 9: 5 4 3 SOLUTION #10: 5 6 1 SOLUTION #11: 6 2 4 SOLUTION #12: 6 4 2 SOLUTION #13: 7 2 3 SOLUTION #14: 7 4 1
Delphi
program Department_numbers;
{$APPTYPE CONSOLE}
uses
System.SysUtils;
var
i, j, k, count: Integer;
begin
writeln('Police Sanitation Fire');
writeln('------ ---------- ----');
count := 0;
i := 2;
while i < 7 do
begin
for j := 1 to 7 do
begin
if j = i then
Continue;
for k := 1 to 7 do
begin
if (k = i) or (k = j) then
Continue;
if i + j + k <> 12 then
Continue;
writeln(format(' %d %d %d', [i, j, k]));
inc(count);
end;
end;
inc(i, 2);
end;
writeln(#10, count, ' valid combinations');
readln;
end.
Draco
proc main() void:
byte police, sanitation, fire;
writeln("Police Sanitation Fire");
for police from 2 by 2 upto 7 do
for sanitation from 1 upto 7 do
for fire from 1 upto 7 do
if police /= sanitation
and police /= fire
and sanitation /= fire
and police + sanitation + fire = 12 then
writeln(police:6, " ", sanitation:10, " ", fire:4)
fi
od
od
od
corp- Output:
Police Sanitation Fire
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
EasyLang
numfmt 0 3
for pol = 2 step 2 to 6
for san = 1 to 7
for fire = 1 to 7
if pol <> san and san <> fire and fire <> pol
if pol + fire + san = 12
print pol & san & fire
.
.
.
.
.Elixir
IO.puts("P - F - S")
for p <- [2,4,6],
f <- 1..7,
s <- 1..7,
p != f and p != s and f != s and p + f + s == 12 do
"#{p} - #{f} - #{s}"
end
|> Enum.each(&IO.puts/1)
P - F - S
2 - 3 - 7
2 - 4 - 6
2 - 6 - 4
2 - 7 - 3
4 - 1 - 7
4 - 2 - 6
4 - 3 - 5
4 - 5 - 3
4 - 6 - 2
4 - 7 - 1
6 - 1 - 5
6 - 2 - 4
6 - 4 - 2
6 - 5 - 1
Excel
LAMBDA
Binding the name departmentNumbers to the following expression in the Name Manager of the Excel WorkBook:
(See LAMBDA: The ultimate Excel worksheet function)
departmentNumbers
=validRows(
LAMBDA(ab,
LET(
x, INDEX(ab, 0, 1),
y, INDEX(ab, 0, 2),
z, 12 - (x + y),
IF(y <> z,
IF(1 <= z,
IF(7 >= z,
CHOOSE({1, 2, 3}, x, y, z),
NA()
),
NA()
),
NA()
)
)
)(
cartesianProduct({2;4;6})(
SEQUENCE(7, 1, 1, 1)
)
)
)
and also assuming the following generic bindings in the Name Manager for the WorkBook:
cartesianProduct
=LAMBDA(xs,
LAMBDA(ys,
LET(
ny, ROWS(ys),
ixs, SEQUENCE(ROWS(xs) * ny, 2, 1, 1),
IF(0 <> MOD(ixs, 2),
INDEX(xs,
1 + QUOTIENT(ixs, ny * 2)
),
LET(
r, MOD(QUOTIENT(ixs, 2), ny),
INDEX(ys, IF(0 = r, ny, r))
)
)
)
)
)
validRows
=LAMBDA(xs,
LET(
ixs, SEQUENCE(ROWS(xs), 1, 1, 1),
valids, FILTER(
ixs,
LAMBDA(i,
NOT(ISNA(INDEX(xs, i)))
)(ixs)
),
INDEX(
xs,
valids,
SEQUENCE(
1,
COLUMNS(xs)
)
)
)
)
- Output:
The formula in cell B2 below defines a two-dimensional array which populates the range B2:D15:
| fx | =departmentNumbers | ||||
|---|---|---|---|---|---|
| A | B | C | D | ||
| 1 | Valid department numbers | ||||
| 2 | 2 | 3 | 7 | ||
| 3 | 2 | 4 | 6 | ||
| 4 | 2 | 6 | 4 | ||
| 5 | 2 | 7 | 3 | ||
| 6 | 4 | 1 | 7 | ||
| 7 | 4 | 2 | 6 | ||
| 8 | 4 | 3 | 5 | ||
| 9 | 4 | 5 | 3 | ||
| 10 | 4 | 6 | 2 | ||
| 11 | 4 | 7 | 1 | ||
| 12 | 6 | 1 | 5 | ||
| 13 | 6 | 2 | 4 | ||
| 14 | 6 | 4 | 2 | ||
| 15 | 6 | 5 | 1 | ||
F#
// A function to generate department numbers. Nigel Galloway: May 2nd., 2018
type dNum = {Police:int; Fire:int; Sanitation:int}
let fN n=n.Police%2=0&&n.Police+n.Fire+n.Sanitation=12&&n.Police<>n.Fire&&n.Police<>n.Sanitation&&n.Fire<>n.Sanitation
List.init (7*7*7) (fun n->{Police=n%7+1;Fire=(n/7)%7+1;Sanitation=(n/49)+1})|>List.filter fN|>List.iter(printfn "%A")
- Output:
{Police = 6;
Fire = 5;
Sanitation = 1;}
{Police = 4;
Fire = 7;
Sanitation = 1;}
{Police = 6;
Fire = 4;
Sanitation = 2;}
{Police = 4;
Fire = 6;
Sanitation = 2;}
{Police = 4;
Fire = 5;
Sanitation = 3;}
{Police = 2;
Fire = 7;
Sanitation = 3;}
{Police = 6;
Fire = 2;
Sanitation = 4;}
{Police = 2;
Fire = 6;
Sanitation = 4;}
{Police = 6;
Fire = 1;
Sanitation = 5;}
{Police = 4;
Fire = 3;
Sanitation = 5;}
{Police = 4;
Fire = 2;
Sanitation = 6;}
{Police = 2;
Fire = 4;
Sanitation = 6;}
{Police = 4;
Fire = 1;
Sanitation = 7;}
{Police = 2;
Fire = 3;
Sanitation = 7;}
Factor
USING: formatting io kernel math math.combinatorics math.ranges
sequences sets ;
IN: rosetta-code.department-numbers
7 [1,b] 3 <k-permutations>
[ [ first even? ] [ sum 12 = ] bi and ] filter
"{ Police, Sanitation, Fire }" print nl
[ "%[%d, %]\n" printf ] each
- Output:
{ Police, Sanitation, Fire }
{ 2, 3, 7 }
{ 2, 4, 6 }
{ 2, 6, 4 }
{ 2, 7, 3 }
{ 4, 1, 7 }
{ 4, 2, 6 }
{ 4, 3, 5 }
{ 4, 5, 3 }
{ 4, 6, 2 }
{ 4, 7, 1 }
{ 6, 1, 5 }
{ 6, 2, 4 }
{ 6, 4, 2 }
{ 6, 5, 1 }
Fermat
!!'Police Sanitation Fire';
!!'------|----------|----';
for p = 2 to 6 by 2 do
for s = 1 to 7 do
for f = 1 to 7 do
if p+f+s=12 and f<>p and f<>s and s<>p then !!(' ',p,' ',s,' ',f);
fi od od od;- Output:
Police Sanitation Fire ------|----------|----
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
FOCAL
01.10 F P=2,2,6;F S=1,7;F G=1,7;D 2
01.20 Q
02.10 I (P-S)2.2,2.6,2.2
02.20 I (P-G)2.3,2.6,2.3
02.30 I (S-G)2.4,2.6,2.4
02.40 I (P+S+G-12)2.6,2.5,2.6
02.50 T %1,P,S,G,!
02.60 R- Output:
= 2= 3= 7 = 2= 4= 6 = 2= 6= 4 = 2= 7= 3 = 4= 1= 7 = 4= 2= 6 = 4= 3= 5 = 4= 5= 3 = 4= 6= 2 = 4= 7= 1 = 6= 1= 5 = 6= 2= 4 = 6= 4= 2 = 6= 5= 1
Forth
\ if department numbers are valid, print them on a single line
: fire ( pol san fir -- )
2dup = if 2drop drop exit then
2 pick over = if 2drop drop exit then
rot . swap . . cr ;
\ tries to assign numbers with given policeno and sanitationno
\ and fire = 12 - policeno - sanitationno
: sanitation ( pol san -- )
2dup = if 2drop exit then \ no repeated numbers
12 over - 2 pick - \ calculate fireno
dup 1 < if 2drop drop exit then \ cannot be less than 1
dup 7 > if 2drop drop exit then \ cannot be more than 7
fire ;
\ tries to assign numbers with given policeno
\ and sanitation = 1, 2, 3, ..., or 7
: police ( pol -- )
8 1 do dup i sanitation loop drop ;
\ tries to assign numbers with police = 2, 4, or 6
: departments cr \ leave input line
8 2 do i police 2 +loop ;
- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 ok
Fortran
This uses the ability standardised in F90 of labelling a DO-loop so that its start and end are linked by usage of the same name, with this checked by the compiler. Further, in avoiding the use of the dreaded GO TO statement, the CYCLE statement can be employed instead with the same effect, and it too can bear the same name so that it is clear which loop is involved. These names prefix the DO-loop, and so, force some additional indentation. They are not statement labels and must be unique themselves. Notably, they cannot be the same text as the name of the index variable for their DO-loop, unlike the lead given by BASIC with its FOR I ... NEXT I arrangement.
The method is just to generate all the possibilities, discarding those that fail the specified tests. However, the requirement that the codes add up to twelve means that after the first two are chosen the third is determined, and blandly looping through all the possibilities is too much brute force and ignorance, though other collections of rules could make that bearable.
Since the modernisers of Fortran made a point of specifying that it does not specify the manner of evaluation of compound boolean expressions, specifically, that there is to be no reliance on Short-circuit_evaluation, both parts of the compound expression of the line labelled 5 "may" be evaluated even though the first may have determined the result. Prior to the introduction of LOGICAL variables with F66, one employed integer arithmetic as is demonstrated in the arithmetic-IF test of the line labelled 6. On the B6700, this usage ran faster than the corresponding boolean expression - possibly because there was no test for short-circuiting the expression when the first part of a multiply was zero...
Note that the syntax enables two classes of labels: the old-style numerical label in columns one to five, and the special label-like prefix of a DO-loop that is not in columns one to five. And yes, a line can have both.
INTEGER P,S,F !Department codes for Police, Sanitation, and Fire. Values 1 to 7 only.
1 PP:DO P = 2,7,2 !The police demand an even number. They're special and use violence.
2 SS:DO S = 1,7 !The sanitation department accepts any value.
3 IF (P.EQ.S) CYCLE SS !But it must differ from the others.
4 F = 12 - (P + S) !The fire department accepts any number, but the sum must be twelve.
5 IF (F.LE.0 .OR. F.GT.7) CYCLE SS !Ensure that the only option is within range.
6 IF ((F - S)*(F - P)) 7,8,7 !And F is to differ from S and from P
7 WRITE (6,"(3I2)") P,S,F !If we get here, we have a possible set.
8 END DO SS !Next S
9 END DO PP !Next P.
END !Well, that was straightforward.
Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
FreeBASIC
' version 15-08-2017
' compile with: fbc -s console
Dim As Integer fire, police, sanitation
Print "police fire sanitation"
Print "----------------------"
For police = 2 To 7 Step 2
For fire = 1 To 7
If fire = police Then Continue For
sanitation = 12 - police - fire
If sanitation = fire Or sanitation = police Then Continue For
If sanitation >= 1 And sanitation <= 7 Then
Print Using " # # # "; police; fire; sanitation
End If
Next
Next
' empty keyboard buffer
While Inkey <> "" : Wend
Print : Print "hit any key to end program"
Sleep
End
- Output:
police fire sanitation ---------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
FutureBasic
include "NSLog.incl"
void local fn DepartmentNumbers
long police, sanitation, fire
printf @"Police Sanitation Fire"
printf @"-------------------------------"
for police = 2 to 7 step 2
for fire = 1 to 7
if ( fire = police ) then continue
sanitation = 12 - police - fire
if ( sanitation == fire ) or ( sanitation == police ) then continue
if ( sanitation >= 1 ) and ( sanitation <= 7 )
printf @"%4d%12d%13d", police, fire, sanitation
end if
next
next
end fn
window 1
fn DepartmentNumbers
HandleEvents- Output:
Police Sanitation Fire ------------------------------- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Gambas
Click this link to run this code
Public Sub Main()
Dim siC0, siC1, siC2 As Short
Dim sOut As New String[]
Dim sTemp As String
For siC0 = 2 To 6 Step 2
For siC1 = 1 To 7
For siC2 = 1 To 7
If sic0 + siC1 + siC2 = 12 Then
If siC0 <> siC1 And siC1 <> siC2 And siC0 <> siC2 Then sOut.Add(Str(siC0) & Str(siC1) & Str(siC2))
End If
Next
Next
Next
Print "\tPolice\tFire\tSanitation"
siC0 = 0
For Each sTemp In sOut
Inc sic0
Print "[" & Format(Str(siC0), "00") & "]\t" & Left(sTemp, 1) & "\t" & Mid(sTemp, 2, 1) & "\t" & Right(sTemp, 1)
Next
End
Output:
Police Fire Sanitation [01] 2 3 7 [02] 2 4 6 [03] 2 6 4 [04] 2 7 3 [05] 4 1 7 [06] 4 2 6 [07] 4 3 5 [08] 4 5 3 [09] 4 6 2 [10] 4 7 1 [11] 6 1 5 [12] 6 2 4 [13] 6 4 2 [14] 6 5 1
Go
package main
import "fmt"
func main() {
fmt.Println("Police Sanitation Fire")
fmt.Println("------ ---------- ----")
count := 0
for i := 2; i < 7; i += 2 {
for j := 1; j < 8; j++ {
if j == i { continue }
for k := 1; k < 8; k++ {
if k == i || k == j { continue }
if i + j + k != 12 { continue }
fmt.Printf(" %d %d %d\n", i, j, k)
count++
}
}
}
fmt.Printf("\n%d valid combinations\n", count)
}
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
Groovy
class DepartmentNumbers {
static void main(String[] args) {
println("Police Sanitation Fire")
println("------ ---------- ----")
int count = 0
for (int i = 2; i <= 6; i += 2) {
for (int j = 1; j <= 7; ++j) {
if (j == i) continue
for (int k = 1; k <= 7; ++k) {
if (k == i || k == j) continue
if (i + j + k != 12) continue
println(" $i $j $k")
count++
}
}
}
println()
println("$count valid combinations")
}
}
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
GW-BASIC
10 PRINT "Police Sanitation Fire"
20 PRINT "------|----------|----"
30 FOR P = 2 TO 7 STEP 2
40 FOR S = 1 TO 7
50 IF S = P THEN GOTO 100
60 FOR F = 1 TO 7
70 IF S = F OR F = P THEN GOTO 90
80 IF P+S+F = 12 THEN PRINT USING " # # #";P;F;S
90 NEXT F
100 NEXT S
110 NEXT P
Haskell
Bare minimum:
main :: IO ()
main =
mapM_ print $
[2, 4, 6] >>=
\x ->
[1 .. 7] >>=
\y ->
[12 - (x + y)] >>=
\z ->
case y /= z && 1 <= z && z <= 7 of
True -> [(x, y, z)]
_ -> []
or, resugaring this into list comprehension format:
main :: IO ()
main =
mapM_
print
[ (x, y, z)
| x <- [2, 4, 6]
, y <- [1 .. 7]
, z <- [12 - (x + y)]
, y /= z && 1 <= z && z <= 7 ]
Do notation:
main :: IO ()
main =
mapM_ print $
do x <- [2, 4, 6]
y <- [1 .. 7]
z <- [12 - (x + y)]
if y /= z && 1 <= z && z <= 7
then [(x, y, z)]
else []
Unadorned brute force – more than enough at this small scale:
import Data.List (nub)
main :: IO ()
main =
let xs = [1 .. 7]
in mapM_ print $
xs >>=
\x ->
xs >>=
\y ->
xs >>=
\z ->
[ (x, y, z)
| even x && 3 == length (nub [x, y, z]) && 12 == sum [x, y, z] ]
- Output:
(2,3,7) (2,4,6) (2,6,4) (2,7,3) (4,1,7) (4,2,6) (4,3,5) (4,5,3) (4,6,2) (4,7,1) (6,1,5) (6,2,4) (6,4,2) (6,5,1)
Or, more generally:
-------------------- DEPARTMENT NUMBERS ------------------
options :: Int -> Int -> Int -> [(Int, Int, Int)]
options lo hi total =
( \ds ->
filter even ds
>>= \x ->
filter (/= x) ds
>>= \y ->
[total - (x + y)]
>>= \z ->
[ (x, y, z)
| y /= z && lo <= z && z <= hi
]
)
[lo .. hi]
--------------------------- TEST -------------------------
main :: IO ()
main =
let xs = options 1 7 12
in putStrLn "(Police, Sanitation, Fire)\n"
>> mapM_ print xs
>> mapM_
putStrLn
[ "\nNumber of options: ",
show (length xs)
]
Reaching again for a little more syntactic sugar, the options function above could also be re-written either as a list comprehension,
options :: Int -> Int -> Int -> [(Int, Int, Int)]
options lo hi total =
let ds = [lo .. hi]
in [ (x, y, z)
| x <- filter even ds
, y <- filter (/= x) ds
, let z = total - (x + y)
, y /= z && lo <= z && z <= hi ]
or in Do notation:
import Control.Monad (guard)
options :: Int -> Int -> Int -> [(Int, Int, Int)]
options lo hi total =
let ds = [lo .. hi]
in do x <- filter even ds
y <- filter (/= x) ds
let z = total - (x + y)
guard $ y /= z && lo <= z && z <= hi
return (x, y, z)
- Output:
(Police, Sanitation, Fire) (2,3,7) (2,4,6) (2,6,4) (2,7,3) (4,1,7) (4,2,6) (4,3,5) (4,5,3) (4,6,2) (4,7,1) (6,1,5) (6,2,4) (6,4,2) (6,5,1) Number of options: 14
J
Solution:
require 'stats'
permfrom=: ,/@(perm@[ {"_ 1 comb) NB. get permutations of length x from y possible items
alluniq=: # = #@~. NB. check items are unique
addto12=: 12 = +/ NB. check items add to 12
iseven=: -.@(2&|) NB. check items are even
policeeven=: {.@iseven NB. check first item is even
conditions=: policeeven *. addto12 *. alluniq
Validnums=: >: i.7 NB. valid Department numbers
getDeptNums=: [: (#~ conditions"1) Validnums {~ permfrom
Example usage:
3 getDeptNums 7
4 1 7
4 7 1
6 1 5
6 5 1
2 3 7
2 7 3
2 4 6
2 6 4
4 2 6
4 6 2
6 2 4
6 4 2
4 3 5
4 5 3
Alternate approaches
(/:"#. 2|]) (#~ 12=+/"1) 1+3 comb 7 [ load'stats'
4 1 7
6 1 5
2 3 7
2 4 6
4 3 5
Note that we are only showing the distinct valid combinations here, not all valid permutations of those combinations. (Valid permutations would be: swapping the last two values in a combination, and all permutations of 2 4 6.)
Another variation would be more constraint based, blindly implementing the rules of the task and yielding all valid permutations. (Shown here with the number of possibilities at each step):
NB. 3 departments, 1..7 in each
#rule1=. >,{3#<1+i.7
343
NB. total must be 12, numbers must be unique
#rule2=. (#~ ((3=#@~.) * 12=+/)"1) rule1
30
NB. no odd numbers in police department (first department)
#rule3=. (#~ 0=2|{."1) rule2
14
rule3
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Java
public class DepartmentNumbers {
public static void main(String[] args) {
System.out.println("Police Sanitation Fire");
System.out.println("------ ---------- ----");
int count = 0;
for (int i = 2; i <= 6; i += 2) {
for (int j = 1; j <= 7; ++j) {
if (j == i) continue;
for (int k = 1; k <= 7; ++k) {
if (k == i || k == j) continue;
if (i + j + k != 12) continue;
System.out.printf(" %d %d %d\n", i, j, k);
count++;
}
}
}
System.out.printf("\n%d valid combinations", count);
}
}
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
JavaScript
ES5
Briefly:
(function () {
'use strict';
// concatMap :: (a -> [b]) -> [a] -> [b]
function concatMap(f, xs) {
return [].concat.apply([], xs.map(f));
};
return '(Police, Sanitation, Fire)\n' +
concatMap(function (x) {
return concatMap(function (y) {
return concatMap(function (z) {
return z !== y && 1 <= z && z <= 7 ? [
[x, y, z]
] : [];
}, [12 - (x + y)]);
}, [1, 2, 3, 4, 5, 6, 7]);
}, [2, 4, 6])
.map(JSON.stringify)
.join('\n');
})();
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1]
Or, more generally:
(function () {
'use strict';
// NUMBERING CONSTRAINTS --------------------------------------------------
// options :: Int -> Int -> Int -> [(Int, Int, Int)]
function options(lo, hi, total) {
var bind = flip(concatMap),
ds = enumFromTo(lo, hi);
return bind(filter(even, ds),
function (x) { // X is even,
return bind(filter(function (d) { return d !== x; }, ds),
function (y) { // Y is distinct from X,
return bind([total - (x + y)],
function (z) { // Z sums with x and y to total, and is in ds.
return z !== y && lo <= z && z <= hi ? [
[x, y, z]
] : [];
})})})};
// GENERIC FUNCTIONS ------------------------------------------------------
// concatMap :: (a -> [b]) -> [a] -> [b]
function concatMap(f, xs) {
return [].concat.apply([], xs.map(f));
};
// enumFromTo :: Int -> Int -> [Int]
function enumFromTo(m, n) {
return Array.from({
length: Math.floor(n - m) + 1
}, function (_, i) {
return m + i;
});
};
// even :: Integral a => a -> Bool
function even(n) {
return n % 2 === 0;
};
// filter :: (a -> Bool) -> [a] -> [a]
function filter(f, xs) {
return xs.filter(f);
};
// flip :: (a -> b -> c) -> b -> a -> c
function flip(f) {
return function (a, b) {
return f.apply(null, [b, a]);
};
};
// length :: [a] -> Int
function length(xs) {
return xs.length;
};
// map :: (a -> b) -> [a] -> [b]
function map(f, xs) {
return xs.map(f);
};
// show :: a -> String
function show(x) {
return JSON.stringify(x);
}; //, null, 2);
// unlines :: [String] -> String
function unlines(xs) {
return xs.join('\n');
};
// TEST -------------------------------------------------------------------
var xs = options(1, 7, 12);
return '(Police, Sanitation, Fire)\n\n' +
unlines(map(show, xs)) + '\n\nNumber of options: ' + length(xs);
})();
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1] Number of options: 14
ES6
Briefly:
(() => {
"use strict";
const
label = "(Police, Sanitation, Fire)",
solutions = [2, 4, 6]
.flatMap(
x => [1, 2, 3, 4, 5, 6, 7]
.flatMap(
y => [12 - (x + y)]
.flatMap(
z => z !== y && 1 <= z && z <= 7 ? [
[x, y, z]
] : []
)
)
)
.map(JSON.stringify)
.join("\n");
return `${label}\n${solutions}`;
})();
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1]
Or, more generally, by composition of generic functions:
(() => {
"use strict";
// -------------- NUMBERING CONSTRAINTS --------------
// options :: Int -> Int -> Int -> [(Int, Int, Int)]
const options = lo => hi => total => {
const
bind = xs => f => xs.flatMap(f),
ds = enumFromTo(lo)(hi);
return bind(ds.filter(even))(
x => bind(ds.filter(d => d !== x))(
y => bind([total - (x + y)])(
z => (z !== y && lo <= z && z <= hi) ? [
[x, y, z]
] : []
)
)
);
};
// ---------------------- TEST -----------------------
const main = () => {
const
label = "(Police, Sanitation, Fire)",
solutions = options(1)(7)(12),
n = solutions.length,
list = solutions
.map(JSON.stringify)
.join("\n");
return (
`${label}\n\n${list}\n\nNumber of options: ${n}`
);
};
// ---------------- GENERIC FUNCTIONS ----------------
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = m =>
n => Array.from({
length: 1 + n - m
}, (_, i) => m + i);
// even :: Integral a => a -> Bool
const even = n => n % 2 === 0;
// MAIN ---
return main();
})();
- Output:
(Police, Sanitation, Fire) [2,3,7] [2,4,6] [2,6,4] [2,7,3] [4,1,7] [4,2,6] [4,3,5] [4,5,3] [4,6,2] [4,7,1] [6,1,5] [6,2,4] [6,4,2] [6,5,1] Number of options: 14
jq
In this section, we present three solutions.
The first illustrates how a straightforward generate-and-test algorithm using familiar for-loops can be translated into jq.
The second illustrates how essentially the same algorithm can be written in a more economical way, without sacrificing comprehensibility.
The third illustrates how the built-in function `combinations/1' can be used to achieve greater efficiency.
The solutions in all cases are presented as a stream of JSON objects such as:
{"fire":1,"police":4,"sanitation":7}
as these are self-explanatory, though it would be trivial to present them in another format. For brevity, the solutions are omitted here.
Nested for-loop
def check(fire; police; sanitation):
(fire != police) and (fire != sanitation) and (police != sanitation)
and (fire + police + sanitation == 12)
and (police % 2 == 0);
range(1;8) as $fire
| range(1;8) as $police
| range(1;8) as $sanitation
| select( check($fire; $police; $sanitation) )
| {$fire, $police, $sanitation}In Brief
{fire: range(1;8), police: range(1;8), sanitation: range(1;8)}
| select( .fire != .police and .fire != .sanitation and .police != .sanitation
and .fire + .police + .sanitation == 12
and .police % 2 == 0 )combinations
[range(1;8)]
| combinations(3)
| select( add == 12 and .[1] % 2 == 0)
| {fire: .[0], police: .[1], sanitation: .[2]}Julia
using Printf
function findsolution(rng=1:7)
rst = Matrix{Int}(0, 3)
for p in rng, f in rng, s in rng
if p != s != f != p && p + s + f == 12 && iseven(p)
rst = [rst; p s f]
end
end
return rst
end
function printsolutions(sol::Matrix{Int})
println(" Pol. Fire San.")
println(" ---- ---- ----")
for row in 1:size(sol, 1)
@printf("%2i | %4i%7i%7i\n", row, sol[row, :]...)
end
end
printsolutions(findsolution())
- Output:
Pol. Fire San.
---- ---- ----
1 | 2 7 3
2 | 2 6 4
3 | 2 4 6
4 | 2 3 7
5 | 4 7 1
6 | 4 6 2
7 | 4 5 3
8 | 4 3 5
9 | 4 2 6
10 | 4 1 7
11 | 6 5 1
12 | 6 4 2
13 | 6 2 4
14 | 6 1 5
Kotlin
// version 1.1.2
fun main(args: Array<String>) {
println("Police Sanitation Fire")
println("------ ---------- ----")
var count = 0
for (i in 2..6 step 2) {
for (j in 1..7) {
if (j == i) continue
for (k in 1..7) {
if (k == i || k == j) continue
if (i + j + k != 12) continue
println(" $i $j $k")
count++
}
}
}
println("\n$count valid combinations")
}
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
Lua
print( "Fire", "Police", "Sanitation" )
sol = 0
for f = 1, 7 do
for p = 1, 7 do
for s = 1, 7 do
if s + p + f == 12 and p % 2 == 0 and f ~= p and f ~= s and p ~= s then
print( f, p, s ); sol = sol + 1
end
end
end
end
print( string.format( "\n%d solutions found", sol ) )
- Output:
Fire Police Sanitation 1 4 7 1 6 5 2 4 6 2 6 4 3 2 7 3 4 5 4 2 6 4 6 2 5 4 3 5 6 1 6 2 4 6 4 2 7 2 3 7 4 1 14 solutions found
MAD
NORMAL MODE IS INTEGER
PRINT COMMENT $ POLICE SANITATION FIRE$
THROUGH LOOP, FOR P=2, 2, P.G.7
THROUGH LOOP, FOR S=1, 1, S.G.7
THROUGH LOOP, FOR F=1, 1, F.G.7
WHENEVER P.E.S .OR. P.E.F .OR. S.E.F, TRANSFER TO LOOP
WHENEVER P+S+F .E. 12, PRINT FORMAT OCC, P, S, F
LOOP CONTINUE
VECTOR VALUES OCC = $I6,S2,I10,S2,I4*$
END OF PROGRAM- Output:
POLICE SANITATION FIRE
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Maple
#determines if i, j, k are exclusive numbers
exclusive_numbers := proc(i, j, k)
if (i = j) or (i = k) or (j = k) then
return false;
end if;
return true;
end proc;
#outputs all possible combinations of numbers that statisfy given conditions
department_numbers := proc()
local i, j, k;
printf("Police Sanitation Fire\n");
for i to 7 do
for j to 7 do
k := 12 - i - j;
if (k <= 7) and (k >= 1) and (i mod 2 = 0) and exclusive_numbers(i,j,k) then
printf("%d %d %d\n", i, j, k);
end if;
end do;
end do;
end proc;
department_numbers();- Output:
Police Sanitation Fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Mathematica /Wolfram Language
Select[Permutations[Range[7], {3}], Total[#] == 12 && EvenQ[First[#]] &]
- Output:
{{2, 3, 7}, {2, 4, 6}, {2, 6, 4}, {2, 7, 3}, {4, 1, 7}, {4, 2, 6}, {4, 3, 5}, {4, 5, 3},
{4, 6, 2}, {4, 7, 1}, {6, 1, 5}, {6, 2, 4}, {6, 4, 2}, {6, 5, 1}}
MATLAB
% Execute the functions
clear all;close all;clc;
sol = findsolution();
disp(table(sol(:, 1), sol(:, 2), sol(:, 3), 'VariableNames',{'Pol.','Fire','San.'}))
function sol = findsolution()
rng = 1:7;
sol = [];
for p = rng
for f = rng
for s = rng
if p ~= s && s ~= f && f ~= p && p + s + f == 12 && mod(p, 2) == 0
sol = [sol; p s f];
end
end
end
end
end
- Output:
Pol. Fire San.
____ ____ ____
2 7 3
2 6 4
2 4 6
2 3 7
4 7 1
4 6 2
4 5 3
4 3 5
4 2 6
4 1 7
6 5 1
6 4 2
6 2 4
6 1 5
MiniScript
print "Police Sanitation Fire"
print "------ ---------- ----"
count = 0
for h in range(1, 3)
i = h * 2
for j in range(1, 7)
if j != i then
for k in range(1, 7)
if k != i and k != j and i + j + k == 12 then
print " " + i + " " + j + " " + k
count += 1
end if
end for
end if
end for
end for
print char(10) + count + " valid combinations"
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
Modula-2
MODULE DepartmentNumbers;
FROM Conversions IMPORT IntToStr;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
PROCEDURE WriteInt(num : INTEGER);
VAR str : ARRAY[0..16] OF CHAR;
BEGIN
IntToStr(num,str);
WriteString(str);
END WriteInt;
VAR i,j,k,count : INTEGER;
BEGIN
count:=0;
WriteString("Police Sanitation Fire");
WriteLn;
WriteString("------ ---------- ----");
WriteLn;
FOR i:=2 TO 6 BY 2 DO
FOR j:=1 TO 7 DO
IF j=i THEN CONTINUE; END;
FOR k:=1 TO 7 DO
IF (k=i) OR (k=j) THEN CONTINUE; END;
IF i+j+k # 12 THEN CONTINUE; END;
WriteString(" ");
WriteInt(i);
WriteString(" ");
WriteInt(j);
WriteString(" ");
WriteInt(k);
WriteLn;
INC(count);
END;
END;
END;
WriteLn;
WriteInt(count);
WriteString(" valid combinations");
WriteLn;
ReadChar;
END DepartmentNumbers.
Mercury
:- module department_numbers.
:- interface.
:- import_module io.
:- pred main(io::di, io::uo) is cc_multi.
:- implementation.
:- import_module int, list, solutions, string.
main(!IO) :-
io.print_line("P S F", !IO),
unsorted_aggregate(department_number, print_solution, !IO).
:- pred print_solution({int, int, int}::in, io::di, io::uo) is det.
print_solution({P, S, F}, !IO) :-
io.format("%d %d %d\n", [i(P), i(S), i(F)], !IO).
:- pred department_number({int, int, int}::out) is nondet.
department_number({Police, Sanitation, Fire}) :-
list.member(Police, [2, 4, 6]),
list.member(Sanitation, 1 .. 7),
list.member(Fire, 1 .. 7),
Police \= Sanitation,
Police \= Fire,
Sanitation \= Fire,
Police + Sanitation + Fire = 12.- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Nim
type Solution = tuple[p, s, f: int]
iterator solutions(max, total: Positive): Solution =
for p in countup(2, max, 2):
for s in 1..max:
if s == p: continue
let f = total - p - s
if f notin [p, s] and f in 1..max:
yield (p, s, f)
echo "P S F"
for sol in solutions(7, 12):
echo sol.p, " ", sol.s, " ", sol.f
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Objeck
class Program {
function : Main(args : String[]) ~ Nil {
sol := 1;
"\t\tFIRE\tPOLICE\tSANITATION"->PrintLine();
for( f := 1; f < 8; f+=1; ) {
for( p := 1; p < 8; p+=1; ) {
for( s:= 1; s < 8; s+=1; ) {
if( f <> p & f <> s & p <> s & ( p and 1 ) = 0 & ( f + s + p = 12 ) ) {
"SOLUTION #{$sol}: \t{$f}\t{$p}\t{$s}"->PrintLine();
sol += 1;
};
};
};
};
}
}Output:
FIRE POLICE SANITATION SOLUTION #1: 1 4 7 SOLUTION #2: 1 6 5 SOLUTION #3: 2 4 6 SOLUTION #4: 2 6 4 SOLUTION #5: 3 2 7 SOLUTION #6: 3 4 5 SOLUTION #7: 4 2 6 SOLUTION #8: 4 6 2 SOLUTION #9: 5 4 3 SOLUTION #10: 5 6 1 SOLUTION #11: 6 2 4 SOLUTION #12: 6 4 2 SOLUTION #13: 7 2 3 SOLUTION #14: 7 4 1
OCaml
(*
* Caution: This is my first Ocaml program and anyone with Ocaml experience probably thinks it's horrible
* So please don't use this as an example for "good ocaml code" see it more as
* "this is what my first lines of ocaml might look like"
*
* The only reason im publishing this is that nobody has yet submitted an example in ocaml
*)
(* sfp is just a convenience to put a combination if sanitation (s) fire (f) and police (p) department in one record*)
type sfp = {s : int; f : int; p : int}
(* Convenience Function to print a single sfp Record *)
let print_sfp e =
Printf.printf "%d %d %d\n" e.s e.f e.p
(* Convenience Function to print a list of sfp Records*)
let print_sfp_list l =
l |> List.iter print_sfp
(* Computes sum of list l *)
let sum l = List.fold_left (+) 0 l
(* checks if element e is in list l *)
let element_in_list e l =
l |> List.find_map (fun x -> if x == e then Some(e) else None) <> None
(* returns a list with only the unique elements of list l *)
let uniq l =
let rec uniq_helper acc l =
match l with
| [] -> acc
| h::t -> if element_in_list h t then uniq_helper acc t else uniq_helper (h::acc) t in
uniq_helper [] l |> List.rev
(* checks wheter or not list l only contains unique elements *)
let is_uniq l = uniq l = l
(* computes all combinations for a given list of sanitation, fire & police departments
im not very proud of this function...maybe someone with some experience can clean it up? ;)
*)
let department_numbers sl fl pl =
sl |> List.fold_left (fun aa s ->
fl |> List.fold_left (fun fa f ->
pl |> List.fold_left (fun pa p ->
if
sum [s;f;p] == 12 &&
is_uniq [s;f;p] then
{s = s; f = f; p = p} :: pa
else
pa) []
|> List.append fa) []
|> List.append aa) []
(* "main" function *)
let _ =
let s = [1;2;3;4;5;6;7] in
let f = [1;2;3;4;5;6;7] in
let p = [2;4;6] in
let result = department_numbers s f p in
print_endline "S F P";
print_sfp_list result;
- Output:
S F P 1 5 6 1 7 4 2 4 6 2 6 4 3 5 4 3 7 2 4 2 6 4 6 2 5 1 6 5 3 4 6 2 4 6 4 2 7 1 4 7 3 2
PARI/GP
forstep(p=2,6,2, for(f=1,7, s=12-p-f; if(p!=f && p!=s && f!=s && s>0 && s<8, print(p" "f" "s))))- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Perl
#!/usr/bin/perl
my @even_numbers;
for (1..7)
{
if ( $_ % 2 == 0)
{
push @even_numbers, $_;
}
}
print "Police\tFire\tSanitation\n";
foreach my $police_number (@even_numbers)
{
for my $fire_number (1..7)
{
for my $sanitation_number (1..7)
{
if ( $police_number + $fire_number + $sanitation_number == 12 &&
$police_number != $fire_number &&
$fire_number != $sanitation_number &&
$sanitation_number != $police_number)
{
print "$police_number\t$fire_number\t$sanitation_number\n";
}
}
}
}
Above Code cleaned up and shortened
#!/usr/bin/perl
use strict; # Not necessary but considered good perl style
use warnings; # this one too
print "Police\t-\tFire\t-\tSanitation\n";
for my $p ( 1..7 ) # Police Department
{
for my $f ( 1..7) # Fire Department
{
for my $s ( 1..7 ) # Sanitation Department
{
if ( $p % 2 == 0 && $p + $f + $s == 12 && $p != $f && $f != $s && $s != $p && $f != $s) # Check if the combination of numbers is valid
{
print "$p\t-\t$f\t-\t$s\n";
}
}
}
}
Output:
Police - Fire - Sanitation
2 - 3 - 7
2 - 4 - 6
2 - 6 - 4
2 - 7 - 3
4 - 1 - 7
4 - 2 - 6
4 - 3 - 5
4 - 5 - 3
4 - 6 - 2
4 - 7 - 1
6 - 1 - 5
6 - 2 - 4
6 - 4 - 2
6 - 5 - 1
Alternate with Regex
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Department_numbers
use warnings;
print "P S F\n\n";
'246 1234567 1234567' =~
/(.).* \s .*?(?!\1)(.).* \s .*(?!\1)(?!\2)(.)
(??{$1+$2+$3!=12})
(?{ print "@{^CAPTURE}\n" })(*FAIL)/x;
- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Alternate with Glob
#!/usr/bin/perl
use strict; # https://rosettacode.org/wiki/Department_numbers
use warnings;
print "P S F\n\n";
print tr/+/ /r, "\n" for
grep !/(\d).*\1/ && 12 == eval,
glob '{2,4,6}' . '+{1,2,3,4,5,6,7}' x 2;
Output same as with Regex
Phix
printf(1,"Police Sanitation Fire\n") printf(1,"------ ---------- ----\n") integer solutions = 0 for police=2 to 7 by 2 do for sanitation=1 to 7 do if sanitation!=police then integer fire = 12-(police+sanitation) if fire>=1 and fire<=7 and fire!=police and fire!=sanitation then printf(1," %d %d %d\n", {police,sanitation,fire}) solutions += 1 end if end if end for end for printf(1,"\n%d solutions found\n", solutions)
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 solutions found
PHP
<?php
$valid = 0;
for ($police = 2 ; $police <= 6 ; $police += 2) {
for ($sanitation = 1 ; $sanitation <= 7 ; $sanitation++) {
$fire = 12 - $police - $sanitation;
if ((1 <= $fire) and ($fire <= 7) and ($police != $sanitation) and ($sanitation != $fire)) {
echo 'Police: ', $police, ', Sanitation: ', $sanitation, ', Fire: ', $fire, PHP_EOL;
$valid++;
}
}
}
echo $valid, ' valid combinations found.', PHP_EOL;
- Output:
Police: 2, Sanitation: 3, Fire: 7 Police: 2, Sanitation: 4, Fire: 6 Police: 2, Sanitation: 6, Fire: 4 Police: 2, Sanitation: 7, Fire: 3 Police: 4, Sanitation: 1, Fire: 7 Police: 4, Sanitation: 2, Fire: 6 Police: 4, Sanitation: 3, Fire: 5 Police: 4, Sanitation: 5, Fire: 3 Police: 4, Sanitation: 6, Fire: 2 Police: 4, Sanitation: 7, Fire: 1 Police: 6, Sanitation: 1, Fire: 5 Police: 6, Sanitation: 2, Fire: 4 Police: 6, Sanitation: 4, Fire: 2 Police: 6, Sanitation: 5, Fire: 1 14 valid combinations found.
Picat
Constraint model
import cp.
go ?=>
N = 7,
Sols = findall([P,S,F], department_numbers(N, P,S,F)),
println(" P S F"),
foreach([P,S,F] in Sols)
printf("%2d %2d %2d\n",P,S,F)
end,
nl,
printf("Number of solutions: %d\n", Sols.len),
nl.
go => true.
department_numbers(N, Police,Sanitation,Fire) =>
Police :: 1..N,
Sanitation :: 1..N,
Fire :: 1..N,
all_different([Police,Sanitation,Fire]),
Police + Sanitation + Fire #= 12,
Police mod 2 #= 0,
solve([Police,Sanitation,Fire]).- Output:
P S F 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 Number of solutions: 14
Loop
go2 => department_numbers2(N) =>
println(" P S F"),
foreach(P in 1..N, P mod 2 == 0)
foreach(S in 1..N, P != S)
foreach(F in 1..N, F != P, F != S, P + S + F == 12)
printf("%2d %2d %2d\n",P,S,F)
end
end
end.List comprehension
import util.
department_numbers3(N) =>
println("P S F"),
L = [[P.to_string,S.to_string,F.to_string] : P in 1..N, P mod 2 == 0,
S in 1..N, P != S,
F in 1..N,
F != P, F != S, P + S + F == 12],
println(map(L,join).join("\n")).Prolog style
go :-
println("P F S"),
assign(Police, Fire, Sanitation),
printf("%w %w %w\n", Police, Fire, Sanitation),
fail,
nl.
dept(X) :- between(1, 7, X).
police(X) :- member(X, [2, 4, 6]).
fire(X) :- dept(X).
san(X) :- dept(X).
assign(A, B, C) :-
police(A), fire(B), san(C),
A != B, A != C, B != C,
12 is A + B + C.
PicoLisp
(de numbers NIL
(co 'numbers
(let N 7
(for P N
(for S N
(for F N
(yield (list P S F)) ) ) ) ) ) )
(de departments NIL
(use (L)
(while (setq L (numbers))
(or
(bit? 1 (car L))
(= (car L) (cadr L))
(= (car L) (caddr L))
(= (cadr L) (caddr L))
(<> 12 (apply + L))
(println L) ) ) ) )
(departments)- Output:
(2 3 7) (2 4 6) (2 6 4) (2 7 3) (4 1 7) (4 2 6) (4 3 5) (4 5 3) (4 6 2) (4 7 1) (6 1 5) (6 2 4) (6 4 2) (6 5 1)
Pilog
(be departments (@Pol @Fire @San)
(member @Pol (2 4 6))
(for @Fire 1 7)
(for @San 1 7)
(different @Pol @Fire)
(different @Pol @San)
(different @Fire @San)
(^ @
(= 12
(+ (-> @Pol) (-> @Fire) (-> @San)) ) ) )- Output:
: (? (departments @Police @Fire @Sanitation)) @Police=2 @Fire=3 @Sanitation=7 @Police=2 @Fire=4 @Sanitation=6 @Police=2 @Fire=6 @Sanitation=4 @Police=2 @Fire=7 @Sanitation=3 @Police=4 @Fire=1 @Sanitation=7 @Police=4 @Fire=2 @Sanitation=6 @Police=4 @Fire=3 @Sanitation=5 @Police=4 @Fire=5 @Sanitation=3 @Police=4 @Fire=6 @Sanitation=2 @Police=4 @Fire=7 @Sanitation=1 @Police=6 @Fire=1 @Sanitation=5 @Police=6 @Fire=2 @Sanitation=4 @Police=6 @Fire=4 @Sanitation=2 @Police=6 @Fire=5 @Sanitation=1 -> NIL
Prolog
dept(X) :- between(1, 7, X).
police(X) :- member(X, [2, 4, 6]).
fire(X) :- dept(X).
san(X) :- dept(X).
assign(A, B, C) :-
police(A), fire(B), san(C),
A =\= B, A =\= C, B =\= C,
12 is A + B + C.
main :-
write("P F S"), nl,
forall(assign(Police, Fire, Sanitation), format("~w ~w ~w~n", [Police, Fire, Sanitation])),
halt.
?- main.
- Output:
P F S 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
PL/M
... under CP/M (or an emulator)
100H: /* SHOW POSSIBLE DEPARTMENT NUMBERS FOR POLICE, SANITATION AND FIRE */
/* THE POLICE DEPARTMENT NUMBER MUST BE EVEN, ALL DEPARTMENT NUMBERS */
/* MUST BE IN THE RANGE 1 .. 7 AND THE NUMBERS MUST SUM TO 12 */
/* CP/M SYSTEM CALL AND I/O ROUTINES */
BDOS: PROCEDURE( FN, ARG ); DECLARE FN BYTE, ARG ADDRESS; GOTO 5; END;
PR$CHAR: PROCEDURE( C ); DECLARE C BYTE; CALL BDOS( 2, C ); END;
PR$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;
PR$NL: PROCEDURE; CALL PR$CHAR( 0DH ); CALL PR$CHAR( 0AH ); END;
PR$NUMBER: PROCEDURE( N ); /* PRINTS A NUMBER IN THE MINIMUN FIELD WIDTH */
DECLARE N ADDRESS;
DECLARE V ADDRESS, N$STR ( 6 )BYTE, W BYTE;
V = N;
W = LAST( N$STR );
N$STR( W ) = '$';
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
DO WHILE( ( V := V / 10 ) > 0 );
N$STR( W := W - 1 ) = '0' + ( V MOD 10 );
END;
CALL PR$STRING( .N$STR( W ) );
END PR$NUMBER;
/* TASK */
DECLARE MAX$DEPARTMENT$NUMBER LITERALLY '7';
DECLARE DEPARTMENT$SUM LITERALLY '12';
DECLARE ( POLICE, SANITATION, FIRE ) BYTE;
CALL PR$STRING( .'POLICE SANITATION FIRE$' );
CALL PR$NL;
DO POLICE = 2 TO MAX$DEPARTMENT$NUMBER BY 2;
DO SANITATION = 1 TO MAX$DEPARTMENT$NUMBER;
IF SANITATION <> POLICE THEN DO;
FIRE = ( DEPARTMENT$SUM - POLICE ) - SANITATION;
IF FIRE <= MAX$DEPARTMENT$NUMBER
AND FIRE <> SANITATION
AND FIRE <> POLICE
THEN DO;
CALL PR$STRING( .' $' ); CALL PR$NUMBER( POLICE );
CALL PR$STRING( .' $' ); CALL PR$NUMBER( SANITATION );
CALL PR$STRING( .' $' ); CALL PR$NUMBER( FIRE );
CALL PR$NL;
END;
END;
END;
END;
EOF- Output:
POLICE SANITATION FIRE
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Python
Procedural
from itertools import permutations
def solve():
c, p, f, s = "\\,Police,Fire,Sanitation".split(',')
print(f"{c:>3} {p:^6} {f:^4} {s:^10}")
c = 1
for p, f, s in permutations(range(1, 8), r=3):
if p + s + f == 12 and p % 2 == 0:
print(f"{c:>3}: {p:^6} {f:^4} {s:^10}")
c += 1
if __name__ == '__main__':
solve()
- Output:
\ Police Fire Sanitation 1: 2 3 7 2: 2 4 6 3: 2 6 4 4: 2 7 3 5: 4 1 7 6: 4 2 6 7: 4 3 5 8: 4 5 3 9: 4 6 2 10: 4 7 1 11: 6 1 5 12: 6 2 4 13: 6 4 2 14: 6 5 1
Composition of pure functions
Expressing the options directly and declaratively in terms of a bind operator, without importing permutations:
'''Department numbers'''
from itertools import (chain)
from operator import (ne)
# options :: Int -> Int -> Int -> [(Int, Int, Int)]
def options(lo, hi, total):
'''Eligible integer triples.'''
ds = enumFromTo(lo)(hi)
return bind(filter(even, ds))(
lambda x: bind(filter(curry(ne)(x), ds))(
lambda y: bind([total - (x + y)])(
lambda z: [(x, y, z)] if (
z != y and lo <= z <= hi
) else []
)
)
)
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Test'''
xs = options(1, 7, 12)
print(('Police', 'Sanitation', 'Fire'))
for tpl in xs:
print(tpl)
print('\nNo. of options: ' + str(len(xs)))
# GENERIC ABSTRACTIONS ------------------------------------
# bind (>>=) :: [a] -> (a -> [b]) -> [b]
def bind(xs):
'''List monad injection operator.
Two computations sequentially composed,
with any value produced by the first
passed as an argument to the second.'''
return lambda f: list(
chain.from_iterable(
map(f, xs)
)
)
# curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
'''A curried function derived
from an uncurried function.'''
return lambda a: lambda b: f(a, b)
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
# even :: Int -> Bool
def even(x):
'''True if x is an integer
multiple of two.'''
return 0 == x % 2
if __name__ == '__main__':
main()
- Output:
('Police', 'Sanitation', 'Fire')
(2, 3, 7)
(2, 4, 6)
(2, 6, 4)
(2, 7, 3)
(4, 1, 7)
(4, 2, 6)
(4, 3, 5)
(4, 5, 3)
(4, 6, 2)
(4, 7, 1)
(6, 1, 5)
(6, 2, 4)
(6, 4, 2)
(6, 5, 1)
No. of options: 14
List comprehension
Nested bind (or concatMap) expressions (like those above) can also be translated into list comprehension notation:
'''Department numbers'''
from operator import ne
# options :: Int -> Int -> Int -> [(Int, Int, Int)]
def options(lo, hi, total):
'''Eligible triples.'''
ds = enumFromTo(lo)(hi)
return [
(x, y, z)
for x in filter(even, ds)
for y in filter(curry(ne)(x), ds)
for z in [total - (x + y)]
if y != z and lo <= z <= hi
]
# Or with less tightly-constrained generation,
# and more winnowing work downstream:
# options2 :: Int -> Int -> Int -> [(Int, Int, Int)]
def options2(lo, hi, total):
'''Eligible triples.'''
ds = enumFromTo(lo)(hi)
return [
(x, y, z)
for x in ds
for y in ds
for z in [total - (x + y)]
if even(x) and y not in [x, z] and lo <= z <= hi
]
# GENERIC -------------------------------------------------
# curry :: ((a, b) -> c) -> a -> b -> c
def curry(f):
'''A curried function derived
from an uncurried function.'''
return lambda a: lambda b: f(a, b)
# enumFromTo :: (Int, Int) -> [Int]
def enumFromTo(m):
'''Integer enumeration from m to n.'''
return lambda n: list(range(m, 1 + n))
# even :: Int -> Bool
def even(x):
'''True if x is an integer
multiple of two.'''
return 0 == x % 2
# unlines :: [String] -> String
def unlines(xs):
'''A single string derived by the intercalation
of a list of strings with the newline character.'''
return '\n'.join(xs)
# TEST ----------------------------------------------------
# main :: IO ()
def main():
'''Test'''
xs = options(1, 7, 12)
print(('Police', 'Sanitation', 'Fire'))
print(unlines(map(str, xs)))
print('\nNo. of options: ' + str(len(xs)))
if __name__ == '__main__':
main()
- Output:
('Police', 'Sanitation', 'Fire')
(2, 3, 7)
(2, 4, 6)
(2, 6, 4)
(2, 7, 3)
(4, 1, 7)
(4, 2, 6)
(4, 3, 5)
(4, 5, 3)
(4, 6, 2)
(4, 7, 1)
(6, 1, 5)
(6, 2, 4)
(6, 4, 2)
(6, 5, 1)
No. of options: 14
Adapted from C# Example
# We start with the Police Department.
# Range is the start, stop, and step. This returns only even numbers.
for p in range(2, 7, 2):
#Next, the Sanitation Department. A simple range.
for s in range(1, 7):
# And now the Fire Department. After determining the Police and Fire
# numbers we just have to subtract those from 12 to get the FD number.
f = 12 - p -s
if s >= f:
break
elif f > 7:
continue
print("Police: ", p, " Sanitation:", s, " Fire: ", f)
print("Police: ", p, " Sanitation:", f, " Fire: ", s)
- Output:
Police: 2 Sanitation: 3 Fire: 7 Police: 2 Sanitation: 7 Fire: 3 Police: 2 Sanitation: 4 Fire: 6 Police: 2 Sanitation: 6 Fire: 4 Police: 4 Sanitation: 1 Fire: 7 Police: 4 Sanitation: 7 Fire: 1 Police: 4 Sanitation: 2 Fire: 6 Police: 4 Sanitation: 6 Fire: 2 Police: 4 Sanitation: 3 Fire: 5 Police: 4 Sanitation: 5 Fire: 3 Police: 6 Sanitation: 1 Fire: 5 Police: 6 Sanitation: 5 Fire: 1 Police: 6 Sanitation: 2 Fire: 4 Police: 6 Sanitation: 4 Fire: 2
Using a constraint solver
A problem this trivial is amenable to brute-force solutions such as the above, but it is a good example of the type of problem for which a constraint solver can be useful. This is how one could solve it using the `python-constraint` library:
import constraint
depts = ( 'police', 'sanitation', 'fire' )
p = constraint.Problem()
for var in depts:
p.addVariable(var, range(1,8))
p.addConstraint(constraint.AllDifferentConstraint())
p.addConstraint(lambda *vars: sum(vars)==12, depts)
p.addConstraint(lambda p: p%2==0, ('police',))
for s in p.getSolutions():
print(s)
- Output:
{'police': 6, 'fire': 5, 'sanitation': 1}
{'police': 6, 'fire': 4, 'sanitation': 2}
{'police': 6, 'fire': 2, 'sanitation': 4}
{'police': 6, 'fire': 1, 'sanitation': 5}
{'police': 4, 'fire': 6, 'sanitation': 2}
{'police': 4, 'fire': 7, 'sanitation': 1}
{'police': 4, 'fire': 5, 'sanitation': 3}
{'police': 4, 'fire': 3, 'sanitation': 5}
{'police': 4, 'fire': 2, 'sanitation': 6}
{'police': 4, 'fire': 1, 'sanitation': 7}
{'police': 2, 'fire': 4, 'sanitation': 6}
{'police': 2, 'fire': 6, 'sanitation': 4}
{'police': 2, 'fire': 7, 'sanitation': 3}
{'police': 2, 'fire': 3, 'sanitation': 7}
Quackery
[ 2dup = iff
[ 2drop drop ] done
dip over swap over = iff
[ 2drop drop ] done
rot echo sp
swap echo sp
echo cr ] is fire ( pol san fir --> )
[ 2dup = iff 2drop done
12 over -
dip over swap -
dup 1 < iff
[ 2drop drop ] done
dup 7 > iff
[ 2drop drop ] done
fire ] is sanitation ( pol san --> )
[ 7 times
[ dup
i^ 1+ sanitation ]
drop ] is police ( pol --> )
[ cr ' [ 2 4 6 ]
witheach police ] is departments ( --> )
departments- Output:
2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
R
We solve this task in two lines. The rest of the code is to make the result look nice.
allPermutations <- setNames(expand.grid(seq(2, 7, by = 2), 1:7, 1:7), c("Police", "Sanitation", "Fire"))
solution <- allPermutations[which(rowSums(allPermutations)==12 & apply(allPermutations, 1, function(x) !any(duplicated(x)))),]
solution <- solution[order(solution$Police, solution$Sanitation),]
row.names(solution) <- paste0("Solution #", seq_len(nrow(solution)), ":")
print(solution)
- Output:
Police Sanitation Fire Solution #1: 2 3 7 Solution #2: 2 4 6 Solution #3: 2 6 4 Solution #4: 2 7 3 Solution #5: 4 1 7 Solution #6: 4 2 6 Solution #7: 4 3 5 Solution #8: 4 5 3 Solution #9: 4 6 2 Solution #10: 4 7 1 Solution #11: 6 1 5 Solution #12: 6 2 4 Solution #13: 6 4 2 Solution #14: 6 5 1
Racket
We filter the Cartesian product of the lists of candidate department numbers.
#lang racket
(cons '(police fire sanitation)
(filter (λ (pfs) (and (not (check-duplicates pfs))
(= 12 (apply + pfs))
pfs))
(cartesian-product (range 2 8 2) (range 1 8) (range 1 8))))
- Output:
'((police fire sanitation) (2 3 7) (2 4 6) (2 6 4) (2 7 3) (4 1 7) (4 2 6) (4 3 5) (4 5 3) (4 6 2) (4 7 1) (6 1 5) (6 2 4) (6 4 2) (6 5 1))
Raku
(formerly Perl 6)
for (1..7).combinations(3).grep(*.sum == 12) {
for .permutations\ .grep(*.[0] %% 2) {
say <police fire sanitation> Z=> .list;
}
}
- Output:
(police => 4 fire => 1 sanitation => 7) (police => 4 fire => 7 sanitation => 1) (police => 6 fire => 1 sanitation => 5) (police => 6 fire => 5 sanitation => 1) (police => 2 fire => 3 sanitation => 7) (police => 2 fire => 7 sanitation => 3) (police => 2 fire => 4 sanitation => 6) (police => 2 fire => 6 sanitation => 4) (police => 4 fire => 2 sanitation => 6) (police => 4 fire => 6 sanitation => 2) (police => 6 fire => 2 sanitation => 4) (police => 6 fire => 4 sanitation => 2) (police => 4 fire => 3 sanitation => 5) (police => 4 fire => 5 sanitation => 3)
REXX
This REXX example essentially uses brute force approach for so simple a puzzle.
/*REXX program finds/displays all possible variants of (3) department numbering puzzle.*/
say 'police sanitation fire' /*display simple title for the output*/
say '══════ ══════════ ════' /* " head separator " " " */
#=0 /*number of solutions found (so far). */
do p=1 for 7; if p//2 then iterate /*try numbers for the police department*/
do s=1 for 7; if s==p then iterate /* " " " " fire " */
do f=1 for 7; if f==s then iterate /* " " " " sanitation " */
if p + s + f \== 12 then iterate /*check if sum of department nums ¬= 12*/
#= # + 1 /*bump count of the number of solutions*/
say center(p,6) center(s,10) center(f,4) /*display one possible solution. */
end /*s*/
end /*f*/
end /*p*/
say '══════ ══════════ ════' /* " head separator " " " */
say /*stick a fork in it, we're all done. */
say # ' solutions found.' /*also, show the # of solutions found. */
- output when using the default inputs:
police sanitation fire ══════ ══════════ ════ 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 ══════ ══════════ ════ 14 solutions found.
Ring
sanitation= 0
see "police fire sanitation" + nl
for police = 2 to 7 step 2
for fire = 1 to 7
if fire = police
loop
ok
sanitation = 12 - police - fire
if sanitation = fire or sanitation = police
loop
ok
if sanitation >= 1 and sanitation <= 7
see " " + police + " " + fire + " " + sanitation + nl
ok
next
nextOutput:
police fire sanitation 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
RPL
≪ { }
2 6 FOR police
1 7 FOR sanitation
IF police sanitation ≠ THEN
12 police - sanitation -
IF DUP 0 >
OVER 7 ≤ AND
OVER police ≠ AND
OVER sanitation ≠ AND
THEN
police sanitation ROT
3 →ARRY +
ELSE DROP END
END
NEXT
2 STEP
DUP SIZE
≫ 'PSF' STO
- Output:
2: { [ 2 3 7 ] [ 2 4 6 ] [ 2 6 4 ] [ 2 7 3 ] [ 4 1 7 ] [ 4 2 6 ] [ 4 3 5 ] [ 4 5 3 ] [ 4 6 2 ] [ 4 7 1 ] [ 6 1 5 ] [ 6 2 4 ] [ 6 4 2 ] [ 6 5 1 ] }
1: 14
Ruby
(1..7).to_a.permutation(3){|p| puts p.join if p.first.even? && p.sum == 12 }
- Output:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
Rust
extern crate num_iter;
fn main() {
println!("Police Sanitation Fire");
println!("----------------------");
for police in num_iter::range_step(2, 7, 2) {
for sanitation in 1..8 {
for fire in 1..8 {
if police != sanitation
&& sanitation != fire
&& fire != police
&& police + fire + sanitation == 12
{
println!("{:6}{:11}{:4}", police, sanitation, fire);
}
}
}
}
}
Scala
val depts = {
(1 to 7).permutations.map{ n => (n(0),n(1),n(2)) }.toList.distinct // All permutations of possible department numbers
.filter{ n => n._1 % 2 == 0 } // Keep only even numbers favored by Police Chief
.filter{ n => n._1 + n._2 + n._3 == 12 } // Keep only numbers that add to 12
}
{
println( "(Police, Sanitation, Fire)")
println( depts.mkString("\n") )
}
- Output:
(Police, Sanitation, Fire) (2,3,7) (2,4,6) (2,6,4) (2,7,3) (4,1,7) (4,2,6) (4,3,5) (4,5,3) (4,6,2) (4,7,1) (6,1,5) (6,2,4) (6,4,2) (6,5,1)
Sidef
@(1..7)->combinations(3, {|*a|
a.sum == 12 || next
a.permutations {|*b|
b[0].is_even || next
say (%w(police fire sanitation) ~Z b -> join(" "))
}
})
- Output:
["police", 4] ["fire", 1] ["sanitation", 7] ["police", 4] ["fire", 7] ["sanitation", 1] ["police", 6] ["fire", 1] ["sanitation", 5] ["police", 6] ["fire", 5] ["sanitation", 1] ["police", 2] ["fire", 3] ["sanitation", 7] ["police", 2] ["fire", 7] ["sanitation", 3] ["police", 2] ["fire", 4] ["sanitation", 6] ["police", 2] ["fire", 6] ["sanitation", 4] ["police", 4] ["fire", 2] ["sanitation", 6] ["police", 4] ["fire", 6] ["sanitation", 2] ["police", 6] ["fire", 2] ["sanitation", 4] ["police", 6] ["fire", 4] ["sanitation", 2] ["police", 4] ["fire", 3] ["sanitation", 5] ["police", 4] ["fire", 5] ["sanitation", 3]
Swift
Functional approach:
let res = [2, 4, 6].map({x in
return (1...7)
.filter({ $0 != x })
.map({y -> (Int, Int, Int)? in
let z = 12 - (x + y)
guard y != z && 1 <= z && z <= 7 else {
return nil
}
return (x, y, z)
}).compactMap({ $0 })
}).flatMap({ $0 })
for result in res {
print(result)
}
Iterative approach:
var res = [(Int, Int, Int)]()
for x in [2, 4, 6] {
for y in 1...7 where x != y {
let z = 12 - (x + y)
guard y != z && 1 <= z && z <= 7 else {
continue
}
res.append((x, y, z))
}
}
for result in res {
print(result)
}
- Output:
(2, 3, 7) (2, 4, 6) (2, 6, 4) (2, 7, 3) (4, 1, 7) (4, 2, 6) (4, 3, 5) (4, 5, 3) (4, 6, 2) (4, 7, 1) (6, 1, 5) (6, 2, 4) (6, 4, 2) (6, 5, 1)
Tcl
Since Tool Command Language is a multi-paradigm language, very different solutions are possible.
VERSION A - using procedures and list operations
# Procedure named ".." returns list of integers from 1 to max.
proc .. max {
for {set i 1} {$i <= $max} {incr i} {
lappend l $i
}
return $l
}
# Procedure named "anyEqual" returns true if any elements are equal,
# false otherwise.
proc anyEqual l {
if {[llength [lsort -unique $l]] != [llength $l]} {
return 1
}
return 0
}
# Procedure named "odd" tells whether a value is odd or not.
proc odd n {
expr $n %2 != 0
}
# Procedure named "sum" sums its parameters.
proc sum args {
expr [join $args +]
}
# Create lists of candidate numbers using proc ".."
set sanitation [.. 7]
set fire $sanitation
# Filter even numbers for police stations (remove odd ones).
set police [lmap e $sanitation {
if [odd $e] continue
set e
}]
# Try all combinations and display acceptable ones.
set valid 0
foreach p $police {
foreach s $sanitation {
foreach f $fire {
# Check for equal elements in list.
if [anyEqual [list $p $s $f]] continue
# Check for sum of list elements.
if {[sum $p $s $f] != 12} continue
puts "$p $s $f"
incr valid
}
}
}
puts "$valid valid combinations found."
VERSION B - using simple for loops with number literals
set valid 0
for {set police 2} {$police <= 6} {incr police 2} {
for {set sanitation 1} {$sanitation <= 7} {incr sanitation} {
if {$police == $sanitation} continue
for {set fire 1} {$fire <= 7} {incr fire} {
if {$police == $fire || $sanitation == $fire} continue
if {[expr $police + $sanitation + $fire] != 12} continue
puts "$police $sanitation $fire"
incr valid
}
}
}
puts "$valid valid combinations found."
VERSION C - using simple for loops with number variables
set min 1
set max 7
set valid 0
for {set police $min} {$police <= $max} {incr police} {
if {[expr $police % 2] == 1} continue ;# filter even numbers for police
for {set sanitation $min} {$sanitation <= $max} {incr sanitation} {
if {$police == $sanitation} continue
for {set fire $min} {$fire <= $max} {incr fire} {
if {$police == $fire || $sanitation == $fire} continue
if {[expr $police + $sanitation + $fire] != 12} continue
puts "$police $sanitation $fire"
incr valid
}
}
}
puts "$valid valid combinations found."
VERSION D - using list filter with lambda expressions
# Procedure named ".." returns list of integers from 1 to max.
proc .. max {
for {set i 1} {$i <= $max} {incr i} {
lappend l $i
}
return $l
}
# Procedure named "..." returns list of n lists of integers from 1 to max.
proc ... {max n} {
foreach i [.. $n] {
lappend result [.. $max]
}
return $result
}
# Procedure named "crossProduct" returns cross product of lists
proc crossProduct {listOfLists} {
set result [list [list]]
foreach factor $listOfLists {
set newResult {}
foreach combination $result {
foreach elt $factor {
lappend newResult [linsert $combination end $elt]
}
}
set result $newResult
}
return $result
}
# Procedure named "filter" filters list elements by using a
# condition λ (lambda) expression
proc filter {l condition} {
return [lmap el $l {
if {![apply $condition $el]} continue
set el
}]
}
# Here the fun using lambda expressions begins. The following is the main program.
# Set λ expressions
set λPoliceEven {_ {expr [lindex $_ 0] % 2 == 0}}
set λNoEquals {_ {expr [llength [lsort -unique $_]] == [llength $_]}}
set λSumIs12 {_ {expr [join $_ +] == 12}}
# Create all combinations and filter acceptable ones
set numbersOk [filter [filter [filter [crossProduct [... 7 3]] ${λPoliceEven}] ${λSumIs12}] ${λNoEquals}]
puts [join $numbersOk \n]
puts "[llength $numbersOk] valid combinations found."
- Output:
All four versions (A, B, C and D) produce the same result:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations found.
Tiny BASIC
PRINT "Police Sanitation Fire"
PRINT "------|----------|----"
10 LET P = P + 2
LET S = 0
20 LET S = S + 1
LET F = 0
IF S = P THEN GOTO 20
30 LET F = F + 1
IF S = F THEN GOTO 30
IF F = P THEN GOTO 30
IF P + S + F = 12 THEN PRINT " ",P," ", S," ", F
IF F < 7 THEN GOTO 30
IF S < 7 THEN GOTO 20
IF P < 6 THEN GOTO 10Transd
#lang transd
MainModule : {
_start: (lambda
(lout "Police | Sanit. | Fire")
(for i in Range(1 8) where (not (mod i 2)) do
(for j in Range(1 8) where (neq i j) do
(for k in Range(1 8) where (and (neq i k) (neq j k)) do
(if (eq (+ i j k) 12) (lout i " " j " " k)))))
)
}
- Output:
Police | Sanit. | Fire 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
TypeScript
// Department numbers
console.log(`POLICE SANITATION FIRE`);
let f: number;
for (var p = 2; p <= 7; p += 2) {
for (var s = 1; s <= 7; s++) {
if (s != p) {
f = (12 - p) - s;
if ((f > 0) && (f <= 7) && (f != s) && (f != p))
console.log(` ${p} ${s} ${f}`);
}
}
}
- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
UNIX Shell
function main {
set -- Police Sanitation Fire
typeset -i pw=${#1} sw=${#2} fw=${#3}
printf '%s' "$1"
shift
printf '\t%s' "$@"
printf '\n'
for (( p=2; p<8; p+=2 )); do
for (( s=1; s<8; ++s )); do
if (( s == p )); then
continue
fi
(( f = 12 - p - s ))
if (( f == s || f == p || f < 1 || f > 7 )); then
continue
fi
printf "%${pw}d\t%${sw}d\t%${fw}d\n" "$p" "$s" "$f"
done
done
}
main "$@"
- Output:
Police Sanitation Fire
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
Visual Basic .NET
Module Module1
Sub Main()
For p = 2 To 7 Step 2
For s = 1 To 7
Dim f = 12 - p - s
If s >= f Then
Exit For
End If
If f > 7 Then
Continue For
End If
If s = p OrElse f = p Then
Continue For 'not even necessary
End If
Console.WriteLine($"Police:{p}, Sanitation:{s}, Fire:{f}")
Console.WriteLine($"Police:{p}, Sanitation:{f}, Fire:{s}")
Next
Next
End Sub
End Module
- Output:
Police:2, Sanitation:3, Fire:7 Police:2, Sanitation:7, Fire:3 Police:2, Sanitation:4, Fire:6 Police:2, Sanitation:6, Fire:4 Police:4, Sanitation:1, Fire:7 Police:4, Sanitation:7, Fire:1 Police:4, Sanitation:2, Fire:6 Police:4, Sanitation:6, Fire:2 Police:4, Sanitation:3, Fire:5 Police:4, Sanitation:5, Fire:3 Police:6, Sanitation:1, Fire:5 Police:6, Sanitation:5, Fire:1 Police:6, Sanitation:2, Fire:4 Police:6, Sanitation:4, Fire:2
V (Vlang)
fn main() {
println("Police Sanitation Fire")
println("------ ---------- ----")
mut count := 0
for i := 2; i < 7; i += 2 {
for j in 1..8 {
if j == i { continue }
for k in 1..8 {
if k == i || k == j { continue }
if i + j + k != 12 { continue }
println(" $i $j $k")
count++
}
}
}
println("\n$count valid combinations")
}
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
VTL-2
10 P=2
20 S=1
30 F=1
40 #=0<((P=S)+(P=F)+(S=F)+(P+S+F=12=0))*110
50 ?=P
60 $=32
70 ?=S
80 $=32
90 ?=F
100 ?=""
110 F=F+1
120 #=F<8*40
130 S=S+1
140 #=S<8*30
150 P=P+2
160 #=P<8*20- Output:
2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Wren
System.print("Police Sanitation Fire")
System.print("------ ---------- ----")
var count = 0
for (h in 1..3) {
var i = h * 2
for (j in 1..7) {
if (j != i) {
for (k in 1..7) {
if ((k != i && k != j) && (i + j + k == 12) ) {
System.print(" %(i) %(j) %(k)")
count = count + 1
}
}
}
}
}
System.print("\n%(count) valid combinations")
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1 14 valid combinations
XPL0
\Department numbers
code CrLf=9, IntIn=10, IntOut=11, Text=12;
integer P, S, F;
begin
Text(0, "POLICE SANITATION FIRE");
CrLf(0);
P:= 2;
while P <= 7 do
begin
for S:= 1, 7 do
if S # P then
begin
F:= (12 - P) - S;
if (F > 0) & (F <= 7) & (F # S) & (F # P) then
begin
Text(0, " "); IntOut(0, P);
Text(0, " "); IntOut(0, S);
Text(0, " "); IntOut(0, F);
CrLf(0)
end
end;
P:= P + 2
end;
end- Output:
POLICE SANITATION FIRE 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Zig
const std = @import("std");
pub fn main() !void {
const stdout = std.io.getStdOut().writer();
try stdout.writeAll("Police Sanitation Fire\n");
try stdout.writeAll("------ ---------- ----\n");
var p: usize = 2;
while (p <= 7) : (p += 2)
for (1..7 + 1) |s|
for (1..7 + 1) |f|
if (p != s and s != f and f != p and p + f + s == 12) {
try stdout.print(" {d} {d} {d}\n", .{ p, s, f });
};
}
- Output:
Police Sanitation Fire ------ ---------- ---- 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
Zig using an iterator
Using a Zig struct to create an iterator is a common pattern in Zig.
const std = @import("std");
pub fn main() !void {
const stdout = std.io.getStdOut().writer();
try stdout.writeAll("Police Sanitation Fire\n");
try stdout.writeAll("------ ---------- ----\n");
var it = SolutionIterator.init();
while (it.next()) |solution| {
try stdout.print(
" {d} {d} {d}\n",
.{ solution.police, solution.sanitation, solution.fire },
);
}
}
/// 3 bit unsigned (u3) limits 0 <= department <= 7
const Departments = packed struct {
police: u3,
sanitation: u3,
fire: u3,
};
const DepartmentsUnion = packed union {
departments: Departments,
together: u9,
};
const SolutionIterator = struct {
// police is initialized to one as adding one is the first operation in next()
// with the result .police == 2 (an even number) on the first pass.
u: DepartmentsUnion = .{ .departments = .{ .police = 1, .sanitation = 1, .fire = 1 } },
/// init() returns an initialised structure.
/// Using init() is a common Zig pattern.
fn init() SolutionIterator {
return SolutionIterator{};
}
fn next(self: *SolutionIterator) ?Departments {
if (self.u.together == 0) return null; // already completed
while (true) {
const ov = @addWithOverflow(self.u.together, 1);
if (ov[1] == 1) {
self.u.together = 0;
return null; // overflowed, completed
} else {
self.u.together = ov[0];
// None can be zero
if (self.u.departments.police == 0) self.u.departments.police = 2; // even
if (self.u.departments.sanitation == 0) self.u.departments.sanitation = 1;
if (self.u.departments.fire == 0) self.u.departments.fire = 1;
// Police must be even
if (self.u.departments.police & 1 == 1)
continue;
// No two can be the same
if (self.u.departments.police == self.u.departments.sanitation) continue;
if (self.u.departments.sanitation == self.u.departments.fire) continue;
if (self.u.departments.fire == self.u.departments.police) continue;
// Must total twelve (maximum sum 7 + 7 + 7 = 21 requires 5 bits)
const p = @as(u5, self.u.departments.police);
const s = @as(u5, self.u.departments.sanitation);
const f = @as(u5, self.u.departments.fire);
if (p + s + f != 12)
continue;
return self.u.departments;
}
}
}
};zkl
Utils.Helpers.pickNFrom(3,[1..7].walk()) // 35 combos
.filter(fcn(numbers){ numbers.sum(0)==12 }) // which all sum to 12 (==5)
.println();- Output:
L(L(1,4,7),L(1,5,6),L(2,3,7),L(2,4,6),L(3,4,5))
Note: The sum of three odd numbers is odd, so a+b+c=12 means at least one even nmber (1 even, two odd or 3 even). Futher, 2a+b=12, a,b in (2,4,6) has one solution: a=2,b=4
For a table with repeated solutions using nested loops:
println("Police Fire Sanitation");
foreach p,f,s in ([2..7,2], [1..7], [1..7])
{ if((p!=s!=f) and p+f+s==12) println(p,"\t",f,"\t",s) }- Output:
Police Fire Sanitation 2 3 7 2 4 6 2 6 4 2 7 3 4 1 7 4 2 6 4 3 5 4 5 3 4 6 2 4 7 1 6 1 5 6 2 4 6 4 2 6 5 1
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