Department numbers: Difference between revisions
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Line 8:
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
The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.
Line 16:
;Task:
Write a computer program which outputs <u>all</u> valid combinations.
Possible output (for the 1<sup>st</sup> and 14<sup>th</sup> solutions):
--police-- --sanitation-- --fire--
2 3 7
6 5 1
<br><br>
=={{header|11l}}==
{{trans|C}}
<syntaxhighlight lang="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)</syntaxhighlight>
{{Output}}
<pre>
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
</pre>
=={{header|8080 Assembly}}==
<syntaxhighlight lang="8080asm"> 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</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|8086 Assembly}}==
<syntaxhighlight lang="asm"> 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</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|ABC}}==
{{Trans|ALGOL 68}}
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Action!}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
[https://gitlab.com/amarok8bit/action-rosetta-code/-/raw/master/images/Department_numbers.png Screenshot from Atari 8-bit computer]
<pre>
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
</pre>
=={{header|Ada}}==
<syntaxhighlight lang="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;</syntaxhighlight>
{{out}}
<pre> 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
</pre>
=={{header|Aime}}==
<syntaxhighlight lang="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);
}
}
}
}</syntaxhighlight>
=={{header|ALGOL 68}}==
As noted in the Fortran sample, once the police and sanitation departments are posited, the fire department value is fixed
<
# 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 #
Line 55 ⟶ 356:
OD
OD
END</
{{out}}
<pre>
Line 74 ⟶ 375:
6 5 1
</pre>
=={{header|ALGOL W}}==
{{Trans|ALGOL 68}}
<syntaxhighlight lang="algolw">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.</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|APL}}==
<syntaxhighlight lang="apl">'PSF'⍪(⊢(⌿⍨)((∪≡⊢)¨↓∧(0=2|1⌷[2]⊢)∧12=+/))↑,⍳3/7</syntaxhighlight>
This prints each triplet of numbers from 1 to 7 for which:
# the elements are unique: <code>(∪≡⊢)¨↓</code>
# the first element is even: <code>(0=2|1⌷[2]⊢)</code>
# the sum of the elements is 12: <code>12=+/</code>
{{out}}
<pre>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</pre>
=={{header|AppleScript}}==
Briefly, composing a solution from generic functions:
<
script
on |λ|(x)
Line 142 ⟶ 512:
on unlines(xs)
intercalate(linefeed, xs)
end unlines</
{{Out}}
<pre>237
Line 162 ⟶ 532:
{{Trans|JavaScript}}
{{Trans|Haskell}}
<
-- options :: Int -> Int -> Int -> [(Int, Int, Int)]
Line 335 ⟶ 705:
on unlines(xs)
intercalate(linefeed, xs)
end unlines</
{{Out}}
<pre>(Police, Sanitation, Fire)
Line 355 ⟶ 725:
Number of options: 14</pre>
=={{header|Arturo}}==
<syntaxhighlight lang="rebol">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]
]
]
]</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Asymptote}}==
<syntaxhighlight lang="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);
}
}
}
}</syntaxhighlight>
=={{header|AutoHotkey}}==
<syntaxhighlight lang="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)
}</syntaxhighlight>
Example:<syntaxhighlight lang="autohotkey">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</syntaxhighlight>
Outputs:<pre>237
246
264
273
417
426
435
453
462
471
615
624
642
651</pre>
=={{header|AWK}}==
<syntaxhighlight lang="awk">
# syntax: GAWK -f DEPARTMENT_NUMBERS.AWK
BEGIN {
Line 377 ⟶ 833:
return(stmt1 stmt2 stmt3)
}
</syntaxhighlight>
{{out}}
<pre>
Line 396 ⟶ 852:
14 7 4 1
</pre>
=={{header|BCPL}}==
<syntaxhighlight lang="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)</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|BASIC}}==
==={{header|Applesoft BASIC}}===
The [[#Sinclair_ZX81_BASIC|Sinclair ZX81 BASIC]] solution works without any changes.
==={{header|BASIC256}}===
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>--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</pre>
==={{header|Chipmunk Basic}}===
{{works with|Chipmunk Basic|3.6.4}}
<syntaxhighlight lang="qbasic">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</syntaxhighlight>
{{out}}
<pre>Same as FreeBASIC entry.</pre>
==={{header|Craft Basic}}===
<syntaxhighlight lang="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</syntaxhighlight>
{{out| Output}}<pre>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</pre>
==={{header|Run BASIC}}===
<syntaxhighlight lang="runbasic">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 police</syntaxhighlight>
==={{header|PureBasic}}===
<syntaxhighlight lang="purebasic">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()</syntaxhighlight>
{{out}}
<pre>Same as BASIC256 entry.</pre>
==={{header|Yabasic}}===
<syntaxhighlight lang="freebasic">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</syntaxhighlight>
{{out}}
<pre>Same as BASIC256 entry.</pre>
==={{header|IS-BASIC}}===
<syntaxhighlight lang="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 NEXT</syntaxhighlight>
==={{header|Minimal BASIC}}===
{{trans|Sinclair ZX81 BASIC}}
<syntaxhighlight lang="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
</syntaxhighlight>
==={{header|Quite BASIC}}===
{{trans|Sinclair ZX81 BASIC}}
<syntaxhighlight lang="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
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
==={{header|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.
<
20 FOR P=2 TO 7 STEP 2
30 FOR S=1 TO 7
Line 407 ⟶ 1,088:
60 IF F>0 AND F<=7 AND F<>S AND F<>P THEN PRINT " ";P;" ";S;" ";F
70 NEXT S
80 NEXT P</
{{out}}
<pre>POLICE SANITATION FIRE
Line 425 ⟶ 1,106:
6 5 1</pre>
==={{header|BBC BASIC}}===
{{trans|ALGOL 68}}
<
max_dept_num% = 7
dept_sum% = 12
Line 439 ⟶ 1,120:
NEXT
NEXT
END</
{{out}}
<pre>police sanitation fire
Line 456 ⟶ 1,137:
6 4 2
6 5 1</pre>
==={{header|QBasic}}===
<syntaxhighlight lang="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</syntaxhighlight>
==={{header|XBasic}}===
{{works with|Windows XBasic}}
<syntaxhighlight lang="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 </syntaxhighlight>
=={{header|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.
<syntaxhighlight lang="c">
#include<stdio.h>
Line 483 ⟶ 1,200:
return 0;
}
</syntaxhighlight>
Output:
<pre>
Line 503 ⟶ 1,220:
6 5 1
</pre>
=={{header|C sharp|C#}}==
<syntaxhighlight lang="csharp">using System;
public class Program
{
Line 519 ⟶ 1,237:
}
}
}</
{{out}}
<pre>
Line 539 ⟶ 1,257:
=={{header|C++}}==
<
#include <iostream>
#include <iomanip>
Line 558 ⟶ 1,276:
}
return 0;
}</
{{out}}
<pre>
Line 579 ⟶ 1,297:
=={{header|Clojure}}==
<
(for [police n
sanitation n
Line 586 ⟶ 1,304:
:when (even? police)
:when (= 12 (+ police sanitation fire))]
(println police sanitation fire)))</
{{Out}}
<pre>
Line 604 ⟶ 1,322:
6 5 1
</pre>
=={{header|CLU}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|COBOL}}==
<syntaxhighlight lang="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.</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Cowgol}}==
<syntaxhighlight lang="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;</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|D}}==
{{Trans|C++}}
<syntaxhighlight lang="d">
import std.stdio, std.range;
Line 624 ⟶ 1,486:
}
}
</syntaxhighlight>
Output:
<pre>
Line 642 ⟶ 1,504:
SOLUTION #13: 7 2 3
SOLUTION #14: 7 4 1
</pre>
=={{header|Delphi}}==
{{libheader| System.SysUtils}}
{{Trans|Go}}
<syntaxhighlight lang="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.</syntaxhighlight>
=={{header|Draco}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|EasyLang}}==
{{trans|C}}
<syntaxhighlight lang="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
.
.
.
.
.
</syntaxhighlight>
=={{header|Elixir}}==
<syntaxhighlight lang="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)
</syntaxhighlight>
<syntaxhighlight lang="elixir">
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
</syntaxhighlight>
=={{header|Excel}}==
===LAMBDA===
Binding the name '''departmentNumbers''' to the following expression in the Name Manager of the Excel WorkBook:
(See [https://www.microsoft.com/en-us/research/blog/lambda-the-ultimatae-excel-worksheet-function/ LAMBDA: The ultimate Excel worksheet function])
{{Works with|Office 365 betas 2021}}
<syntaxhighlight lang="lisp">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)
)
)
)</syntaxhighlight>
and also assuming the following generic bindings in the Name Manager for the WorkBook:
<syntaxhighlight lang="lisp">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)
)
)
)
)</syntaxhighlight>
{{Out}}
The formula in cell B2 below defines a two-dimensional array which populates the range '''B2:D15''':
{| class="wikitable"
|-
|||style="text-align:right; font-family:serif; font-style:italic; font-size:120%;"|fx
! colspan="4" style="text-align:left; vertical-align: bottom; font-family:Arial, Helvetica, sans-serif !important;"|=departmentNumbers
|- style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff;"
|
| A
| B
| C
| D
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 1
| colspan="4" style="font-weight:bold" | Valid department numbers
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 2
|
| style="background-color:#cbcefb" | 2
| 3
| 7
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 3
|
| 2
| 4
| 6
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 4
|
| 2
| 6
| 4
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 5
|
| 2
| 7
| 3
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 6
|
| 4
| 1
| 7
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 7
|
| 4
| 2
| 6
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 8
|
| 4
| 3
| 5
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 9
|
| 4
| 5
| 3
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 10
|
| 4
| 6
| 2
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 11
|
| 4
| 7
| 1
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 12
|
| 6
| 1
| 5
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 13
|
| 6
| 2
| 4
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 14
|
| 6
| 4
| 2
|-
| style="text-align:center; font-family:Arial, Helvetica, sans-serif !important; background-color:#000000; color:#ffffff" | 15
|
| 6
| 5
| 1
|}
=={{header|F_Sharp|F#}}==
<syntaxhighlight lang="fsharp">
// 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")
</syntaxhighlight>
{{out}}
<pre>
{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;}
</pre>
=={{header|Factor}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>
{ 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 }
</pre>
=={{header|Fermat}}==
<syntaxhighlight lang="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;</syntaxhighlight>
{{out}}<pre>
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
</pre>
=={{header|FOCAL}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>= 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</pre>
=={{header|Forth}}==
<syntaxhighlight lang="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 ;</syntaxhighlight>
{{out}}
<pre>
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
</pre>
Line 651 ⟶ 1,996:
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. <
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.
Line 661 ⟶ 2,006:
8 END DO SS !Next S
9 END DO PP !Next P.
END !Well, that was straightforward.</
Output:
<pre>
Line 679 ⟶ 2,024:
6 5 1
</pre>
=={{header|FreeBASIC}}==
<
' compile with: fbc -s console
Line 703 ⟶ 2,049:
Print : Print "hit any key to end program"
Sleep
End</
{{out}}
<pre>police fire sanitation
Line 721 ⟶ 2,067:
6 4 2
6 5 1 </pre>
=={{header|FutureBasic}}==
<syntaxhighlight lang="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
</syntaxhighlight>
{{output}}
<pre>
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
</pre>
=={{header|Gambas}}==
'''[https://gambas-playground.proko.eu/?gist=cdfeadc26aaeb0e23f4523626b6fe7c9 Click this link to run this code]'''
<
Dim siC0, siC1, siC2 As Short
Dim sOut As New String[]
Line 747 ⟶ 2,145:
Next
End</
Output:
<pre>
Line 769 ⟶ 2,167:
=={{header|Go}}==
{{trans|Kotlin}}
<
import "fmt"
Line 789 ⟶ 2,187:
}
fmt.Printf("\n%d valid combinations\n", count)
}</
{{out}}
Line 812 ⟶ 2,210:
14 valid combinations
</pre>
=={{header|Groovy}}==
{{trans|Java}}
<syntaxhighlight lang="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")
}
}</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|GW-BASIC}}==
<syntaxhighlight lang="gwbasic">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</syntaxhighlight>
=={{header|Haskell}}==
Bare minimum:
<
main =
mapM_ print $
Line 826 ⟶ 2,279:
case y /= z && 1 <= z && z <= 7 of
True -> [(x, y, z)]
_ -> []</
or, resugaring this into list comprehension format:
<
main =
mapM_
Line 836 ⟶ 2,289:
, y <- [1 .. 7]
, z <- [12 - (x + y)]
, y /= z && 1 <= z && z <= 7 ]</
Do notation:
<
main =
mapM_ print $
Line 846 ⟶ 2,299:
if y /= z && 1 <= z && z <= 7
then [(x, y, z)]
else []</
Unadorned brute force – more than enough at this small scale:
<
main :: IO ()
main =
let xs = [1 .. 7]
\z ->
| even x && 3 == length (nub [x, y, z]) && 12 == sum [x, y, z] ]</syntaxhighlight>
{{Out}}
<pre>(2,3,7)
Line 879 ⟶ 2,332:
Or, more generally:
<syntaxhighlight lang="haskell">-------------------- 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 ->
)
[lo .. hi]
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)
]</syntaxhighlight>
Reaching again for a little more syntactic sugar, the options function above could also be re-written either as a list comprehension,
<
options lo hi total =
let ds = [lo .. hi]
Line 909 ⟶ 2,369:
, y <- filter (/= x) ds
, let z = total - (x + y)
, y /= z && lo <= z && z <= hi ]</
or in Do notation:
<
options :: Int -> Int -> Int -> [(Int, Int, Int)]
Line 921 ⟶ 2,381:
let z = total - (x + y)
guard $ y /= z && lo <= z && z <= hi
return (x, y, z)</
{{Out}}
<pre>(Police, Sanitation, Fire)
Line 945 ⟶ 2,405:
=={{header|J}}==
'''Solution:'''
<
permfrom=: ,/@(perm@[ {"_ 1 comb) NB. get permutations of length x from y possible items
Line 956 ⟶ 2,416:
Validnums=: >: i.7 NB. valid Department numbers
getDeptNums=: [: (#~ conditions"1) Validnums {~ permfrom</
'''Example usage:'''
<
4 1 7
4 7 1
Line 972 ⟶ 2,432:
6 4 2
4 3 5
4 5 3</
===Alternate
<
6 1 5
2 3 7
2 4 6
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):
<syntaxhighlight lang=J> 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</syntaxhighlight>
=={{header|Java}}==
{{trans|Kotlin}}
<
public static void main(String[] args) {
System.out.println("Police Sanitation Fire");
Line 1,005 ⟶ 2,492:
System.out.printf("\n%d valid combinations", count);
}
}</
{{out}}
<pre>Police Sanitation Fire
Line 1,027 ⟶ 2,514:
===ES5===
Briefly:
<
'use strict';
Line 1,047 ⟶ 2,534:
.map(JSON.stringify)
.join('\n');
})();</
{{Out}}
<pre>(Police, Sanitation, Fire)
Line 1,067 ⟶ 2,554:
Or, more generally:
{{Trans|Haskell}}
<
'use strict';
Line 1,145 ⟶ 2,632:
return '(Police, Sanitation, Fire)\n\n' +
unlines(map(show, xs)) + '\n\nNumber of options: ' + length(xs);
})();</
{{Out}}
<pre>(Police, Sanitation, Fire)
Line 1,168 ⟶ 2,655:
===ES6===
Briefly:
<
"use strict";
const
label = "(Police, Sanitation, Fire)",
solutions = [2, 4, 6]
.flatMap(
z => z !== y && 1 <= z && z <= 7 ? [
[x, y, z]
] : [
)
)
)
.map(JSON.stringify)
.join(
return `${label}\n${solutions}`;
})();</syntaxhighlight>
{{Out}}
<pre>(Police, Sanitation, Fire)
Line 1,205 ⟶ 2,696:
Or, more generally, by composition of generic functions:
{{Trans|Haskell}}
<
//
// options :: Int -> Int -> Int -> [(Int, Int, Int)]
const options =
const
bind =
ds = enumFromTo(lo
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]
Line 1,224 ⟶ 2,715:
)
)
);
};
//
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 =
n => Array.from({
length:
}, (_, i) => m + i);
Line 1,241 ⟶ 2,744:
const even = n => n % 2 === 0;
//
return main();
})();</syntaxhighlight>
{{Out}}
<pre>(Police, Sanitation, Fire)
Line 1,292 ⟶ 2,774:
The second illustrates how essentially the same algorithm can be written in a more economical way, without sacrificing comprehensibility.
The third illustrates how the
The solutions in all cases are presented as a stream of JSON objects such as:
Line 1,301 ⟶ 2,783:
'''Nested for-loop'''
<
(fire != police) and (fire != sanitation) and (police != sanitation)
and (fire + police + sanitation == 12)
Line 1,310 ⟶ 2,792:
| range(1;8) as $sanitation
| select( check($fire; $police; $sanitation) )
| {
'''In Brief'''
<
| select( .fire != .police and .fire != .sanitation and .police != .sanitation
and .fire + .police + .sanitation == 12
and .police % 2 == 0 )</
'''
<syntaxhighlight lang="jq">
| combinations(3)
| select( add == 12 and .[1] % 2 == 0)
| {fire: .[0], police: .[1], sanitation: .[2]}</syntaxhighlight>
=={{header|Julia}}==
<syntaxhighlight lang="julia">using Printf
rst = Matrix{Int}(0, 3)
for p in rng, f in rng, s in rng
Line 1,347 ⟶ 2,829:
printsolutions(findsolution())
</syntaxhighlight>
{{out}}
Line 1,368 ⟶ 2,850:
=={{header|Kotlin}}==
<
fun main(args: Array<String>) {
Line 1,386 ⟶ 2,868:
}
println("\n$count valid combinations")
}</
{{out}}
Line 1,411 ⟶ 2,893:
=={{header|Lua}}==
<
print( "Fire", "Police", "Sanitation" )
sol = 0
Line 1,424 ⟶ 2,906:
end
print( string.format( "\n%d solutions found", sol ) )
</syntaxhighlight>
{{out}}
<pre>
Line 1,446 ⟶ 2,928:
</pre>
=={{header|
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Maple}}==
<syntaxhighlight lang="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();
</syntaxhighlight>
{{out|Output}}
<pre>
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
</pre>
=={{header|Mathematica}}/{{header|Wolfram Language}}==
<syntaxhighlight lang="mathematica">Select[Permutations[Range[7], {3}], Total[#] == 12 && EvenQ[First[#]] &]</syntaxhighlight>
{{out}}
<pre>{{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}}</pre>
=={{header|MATLAB}}==
<syntaxhighlight lang="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
</syntaxhighlight>
{{out}}
<pre>
Pol. Fire
6 2 4
6 1 5
</pre>
=={{header|MiniScript}}==
{{trans|Kotlin}}
<syntaxhighlight lang="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"</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|Modula-2}}==
<
FROM Conversions IMPORT IntToStr;
FROM Terminal IMPORT WriteString,WriteLn,ReadChar;
Line 1,535 ⟶ 3,135:
ReadChar;
END DepartmentNumbers.</
=={{header|Mercury}}==
<syntaxhighlight lang="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.</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Nim}}==
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Objeck}}==
{{Trans|C++}}
<syntaxhighlight lang="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;
};
};
};
};
}
}</syntaxhighlight>
Output:
<pre>
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
</pre>
=={{header|OCaml}}==
<syntaxhighlight lang="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;
</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|PARI/GP}}==
<
{{out}}
<pre>2 3 7
Line 1,558 ⟶ 3,360:
=={{header|Perl}}==
<syntaxhighlight lang="perl">
#!/usr/bin/perl
Line 1,589 ⟶ 3,391:
}
}
</syntaxhighlight>
Above Code cleaned up and shortened
<syntaxhighlight lang="perl">
#!/usr/bin/perl
Line 1,614 ⟶ 3,416:
}
}
</syntaxhighlight>
Output:
<syntaxhighlight lang="perl">
Police - Fire - Sanitation
2 - 3 - 7
Line 1,633 ⟶ 3,435:
6 - 4 - 2
6 - 5 - 1
</syntaxhighlight>
===Alternate with Regex===
<syntaxhighlight lang="perl">#!/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;</syntaxhighlight>
{{out}}
<pre>
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
</pre>
===Alternate with Glob===
<syntaxhighlight lang="perl">#!/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;</syntaxhighlight>
Output same as with Regex
=={{header|Phix}}==
<!--<syntaxhighlight lang="phix">(phixonline)-->
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"Police Sanitation Fire\n"</span><span style="color: #0000FF;">)</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"------ ---------- ----\n"</span><span style="color: #0000FF;">)</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">solutions</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">0</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">police</span><span style="color: #0000FF;">=</span><span style="color: #000000;">2</span> <span style="color: #008080;">to</span> <span style="color: #000000;">7</span> <span style="color: #008080;">by</span> <span style="color: #000000;">2</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">for</span> <span style="color: #000000;">sanitation</span><span style="color: #0000FF;">=</span><span style="color: #000000;">1</span> <span style="color: #008080;">to</span> <span style="color: #000000;">7</span> <span style="color: #008080;">do</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">sanitation</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">police</span> <span style="color: #008080;">then</span>
<span style="color: #004080;">integer</span> <span style="color: #000000;">fire</span> <span style="color: #0000FF;">=</span> <span style="color: #000000;">12</span><span style="color: #0000FF;">-(</span><span style="color: #000000;">police</span><span style="color: #0000FF;">+</span><span style="color: #000000;">sanitation</span><span style="color: #0000FF;">)</span>
<span style="color: #008080;">if</span> <span style="color: #000000;">fire</span><span style="color: #0000FF;">>=</span><span style="color: #000000;">1</span>
<span style="color: #008080;">and</span> <span style="color: #000000;">fire</span><span style="color: #0000FF;"><=</span><span style="color: #000000;">7</span>
<span style="color: #008080;">and</span> <span style="color: #000000;">fire</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">police</span>
<span style="color: #008080;">and</span> <span style="color: #000000;">fire</span><span style="color: #0000FF;">!=</span><span style="color: #000000;">sanitation</span> <span style="color: #008080;">then</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">" %d %d %d\n"</span><span style="color: #0000FF;">,</span> <span style="color: #0000FF;">{</span><span style="color: #000000;">police</span><span style="color: #0000FF;">,</span><span style="color: #000000;">sanitation</span><span style="color: #0000FF;">,</span><span style="color: #000000;">fire</span><span style="color: #0000FF;">})</span>
<span style="color: #000000;">solutions</span> <span style="color: #0000FF;">+=</span> <span style="color: #000000;">1</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">if</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #008080;">end</span> <span style="color: #008080;">for</span>
<span style="color: #7060A8;">printf</span><span style="color: #0000FF;">(</span><span style="color: #000000;">1</span><span style="color: #0000FF;">,</span><span style="color: #008000;">"\n%d solutions found\n"</span><span style="color: #0000FF;">,</span> <span style="color: #000000;">solutions</span><span style="color: #0000FF;">)</span>
<!--</syntaxhighlight>-->
{{out}}
<pre>
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
</pre>
=={{header|PHP}}==
<syntaxhighlight lang="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;</syntaxhighlight>
{{out}}
<pre>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.</pre>
=={{header|Picat}}==
===Constraint model===
<syntaxhighlight lang="picat">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]).</syntaxhighlight>
{{out}}
<pre> 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</pre>
===Loop===
<syntaxhighlight lang="picat">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.</syntaxhighlight>
===List comprehension===
<syntaxhighlight lang="picat">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")).</syntaxhighlight>
===Prolog style===
{{trans|Prolog}}
<syntaxhighlight lang="picat">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.</syntaxhighlight>
=={{header|PicoLisp}}==
<syntaxhighlight lang="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)</syntaxhighlight>
{{out}}
<pre>
(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)
</pre>
===Pilog===
<syntaxhighlight lang="picolisp">(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)) ) ) )</syntaxhighlight>
{{out}}
<pre>
: (? (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
</pre>
=={{header|Prolog}}==
<syntaxhighlight lang="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.
</syntaxhighlight>
{{Out}}
<pre>
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
</pre>
=={{header|PL/M}}==
{{Trans|ALGOL W}}
{{works with|8080 PL/M Compiler}} ... under CP/M (or an emulator)
<syntaxhighlight lang="plm">
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
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Python}}==
===Procedural===
<syntaxhighlight lang="python">from itertools import permutations
def solve():
Line 1,673 ⟶ 3,838:
if __name__ == '__main__':
solve()</
{{out}}
Line 1,691 ⟶ 3,856:
13: 6 4 2
14: 6 5 1 </pre>
===Composition of pure functions===
Expressing the options directly and declaratively in terms of a ''bind'' operator, without importing ''permutations'':
{{Works with|Python|3}}
<syntaxhighlight lang="python">'''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()</syntaxhighlight>
{{Out}}
<pre>('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</pre>
===List comprehension===
Nested ''bind'' (or ''concatMap'') expressions (like those above) can also be translated into list comprehension notation:
{{Works with|Python|3.7}}
<syntaxhighlight lang="python">'''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()</syntaxhighlight>
{{Out}}
<pre>('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 </pre>
===Adapted from C# Example===
<syntaxhighlight lang="python">
# 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)
</syntaxhighlight>
{{Out}}
<pre>
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
</pre>
===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:
{{libheader|python-constraint}}
<syntaxhighlight lang="python">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)</syntaxhighlight>
{{Out}}
<pre>{'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}</pre>
=={{header|Quackery}}==
{{trans|Forth}}
<syntaxhighlight lang="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</syntaxhighlight>
{{out}}
<pre>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
</pre>
=={{header|R}}==
We solve this task in two lines. The rest of the code is to make the result look nice.
<syntaxhighlight lang="rsplus">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)</syntaxhighlight>
{{out}}
<pre> 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</pre>
=={{header|Racket}}==
Line 1,696 ⟶ 4,199:
We filter the Cartesian product of the lists of candidate department numbers.
<
(cons '(police fire sanitation)
(filter (λ (pfs) (and (not (check-duplicates pfs))
Line 1,702 ⟶ 4,205:
pfs))
(cartesian-product (range 2 8 2) (range 1 8) (range 1 8))))
</syntaxhighlight>
{{out}}
Line 1,722 ⟶ 4,225:
(6 5 1))</pre>
=={{header|
(formerly Perl 6)
<syntaxhighlight lang="raku" line>for (1..7).combinations(3).grep(*.sum == 12) {
for .permutations\ .grep(*.[0] %% 2) {
say <police fire sanitation> Z=> .list;
}
}
</syntaxhighlight>
{{out}}
<pre>
(police => 4 fire => 1 sanitation => 7)
(police => 4 fire => 7 sanitation => 1)
(police => 6 fire => 1 sanitation => 5)
(police
(police => 2 fire => 3 sanitation => 7
(police => 2 fire => 7 sanitation => 3)
(police
(police => 2 fire => 6 sanitation => 4)
(police => 4 fire => 2 sanitation => 6)
(police => 4 fire
(police => 6 fire => 2 sanitation => 4)
(police => 6 fire => 4 sanitation => 2)
(police => 4 fire => 3 sanitation => 5)
(police
</pre>
=={{header|REXX}}==
This REXX example essentially uses brute force approach for so simple a puzzle.
<syntaxhighlight lang="rexx">/*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. */
{{out|output|text= when using the default inputs:}}
<pre>
══════ ══════════ ════
2
2 4 6
4
4
6
6
══════ ══════════ ════
14 solutions found.
</pre>
=={{header|Ring}}==
<
sanitation= 0
see "police fire sanitation" + nl
Line 1,830 ⟶ 4,312:
next
next
</syntaxhighlight>
Output:
<pre>
Line 1,850 ⟶ 4,332:
</pre>
=={{header|RPL}}==
≪ { }
<span style="color:red">2 6</span> '''FOR''' police
<span style="color:red">1 7</span> '''FOR''' sanitation
'''IF''' police sanitation ≠ '''THEN'''
<span style="color:red">12</span> police - sanitation -
'''IF''' DUP <span style="color:red">0</span> >
OVER <span style="color:red">7</span> ≤ AND
OVER police ≠ AND
OVER sanitation ≠ AND
'''THEN'''
police sanitation ROT
<span style="color:red">3</span> →ARRY +
'''ELSE''' DROP '''END'''
'''END'''
'''NEXT'''
<span style="color:red">2</span> '''STEP'''
DUP SIZE
≫ '<span style="color:blue">PSF</span>' STO
{{out}}
<pre>
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
</pre>
=={{header|Ruby}}==
<
(1..7).to_a.permutation(3){|p| puts p.join if p.first.even? && p.sum == 12 }
</syntaxhighlight>
{{Output}}
<pre>237
Line 1,870 ⟶ 4,377:
651
</pre>
=={{header|Rust}}==
{{trans|C}}
<syntaxhighlight lang="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);
}
}
}
}
}</syntaxhighlight>
=={{header|Scala}}==
<
(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
Line 1,882 ⟶ 4,411:
println( depts.mkString("\n") )
}
</syntaxhighlight>
{{output}}
<pre>(Police, Sanitation, Fire)
Line 1,902 ⟶ 4,431:
=={{header|Sidef}}==
{{trans|
<
a.sum == 12 || next
a.permutations {|*b|
Line 1,909 ⟶ 4,438:
say (%w(police fire sanitation) ~Z b -> join(" "))
}
})</
{{out}}
<pre>
Line 1,927 ⟶ 4,456:
["police", 4] ["fire", 5] ["sanitation", 3]
</pre>
=={{header|Swift}}==
Functional approach:
<syntaxhighlight lang="swift">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({ $0 })
for result in res {
print(result)
}</syntaxhighlight>
Iterative approach:
<syntaxhighlight lang="swift">var res = [(Int, Int, Int)]()
for x in [2, 4, 6] {
for y in 1...7 where x != y {
let z = 12 - (x + y)
guard y != z && 1 <= z && z <= 7 else {
continue
}
res.append((x, y, z))
}
}
for result in res {
print(result)
}</syntaxhighlight>
{{out}}
<pre>(2, 3, 7)
(2, 4, 6)
(2, 6, 4)
(2, 7, 3)
(4, 1, 7)
(4, 2, 6)
(4, 3, 5)
(4, 5, 3)
(4, 6, 2)
(4, 7, 1)
(6, 1, 5)
(6, 2, 4)
(6, 4, 2)
(6, 5, 1)</pre>
=={{header|Tcl}}==
Since Tool Command Language is a multi-paradigm language, very different solutions are possible.
<b>VERSION A</b> - using procedures and list operations
<syntaxhighlight lang="tcl">
# Procedure named ".." returns list of integers from 1 to max.
proc .. max {
for {set i 1} {$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."
</syntaxhighlight>
<b>VERSION B</b> - using simple for loops with number literals
<syntaxhighlight lang="tcl">
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."
</syntaxhighlight>
<b>VERSION C</b> - using simple for loops with number variables
<syntaxhighlight lang="tcl">
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."
</syntaxhighlight>
<b>VERSION D</b> - using list filter with lambda expressions
<syntaxhighlight lang="tcl">
# 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."
</syntaxhighlight>
{{output}}
All four versions (A, B, C and D) produce the same result:
<pre>
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.
</pre>
=={{header|Tiny BASIC}}==
<syntaxhighlight lang="tinybasic"> 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 10</syntaxhighlight>
=={{header|Transd}}==
<syntaxhighlight lang="scheme">#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)))))
)
}</syntaxhighlight>{{out}}
<pre>
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
</pre>
== {{header|TypeScript}} ==
{{trans|Sinclair ZX81 BASIC}}
<syntaxhighlight lang="javascript">
// 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} ${f}`);
}
}
}
</syntaxhighlight>
{{out}}
<pre>
POLICE SANITATION FIRE
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1
</pre>
=={{header|UNIX Shell}}==
{{works with|Bourne Again SHell}}
{{works with|Korn Shell}}
{{works with|Z Shell}}
<syntaxhighlight lang="bash">function main {
set -- Police Sanitation Fire
typeset -i pw=${#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 "$@"</syntaxhighlight>
{{Out}}
<pre>Police Sanitation Fire
2 3 7
2 4 6
2 6 4
2 7 3
4 1 7
4 2 6
4 3 5
4 5 3
4 6 2
4 7 1
6 1 5
6 2 4
6 4 2
6 5 1</pre>
=={{header|Visual Basic .NET}}==
{{trans|C#}}
<syntaxhighlight lang="vbnet">Module Module1
Sub Main()
For p = 2 To 7 Step 2
For s = 1 To 7
Dim f = 12 - p - s
If s >= f Then
Exit For
End If
If f > 7 Then
Continue For
End If
If s = p OrElse f = p Then
Continue For 'not even necessary
End If
Console.WriteLine($"Police:{p}, Sanitation:{s}, Fire:{f}")
Console.WriteLine($"Police:{p}, Sanitation:{f}, Fire:{s}")
Next
Next
End Sub
End Module</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|V (Vlang)}}==
{{trans|Go}}
<syntaxhighlight lang="go">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")
}</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|VTL-2}}==
<syntaxhighlight lang="vtl2">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</syntaxhighlight>
{{out}}
<pre>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</pre>
=={{header|Wren}}==
{{trans|Kotlin}}
<syntaxhighlight lang="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")</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|XPL0}}==
{{trans|Sinclair ZX81 BASIC}}
{{works with|EXPL-32}}
<syntaxhighlight lang="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
</syntaxhighlight>
{{out}}
<pre>
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
</pre>
=={{header|Zig}}==
{{works with|Zig|0.11.0dev}}
<syntaxhighlight lang="zig">const std = @import("std");</syntaxhighlight>
<syntaxhighlight lang="zig">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 });
};
}</syntaxhighlight>
{{out}}
<pre>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</pre>
===Zig using an iterator===
{{works with|Zig|0.11.0}}
Using a Zig ''struct'' to create an iterator is a common pattern in Zig.
<syntaxhighlight lang="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;
}
}
}
};
</syntaxhighlight>
=={{header|zkl}}==
<
.filter(fcn(numbers){ numbers.sum(0)==12 }) // which all sum to 12 (==5)
.println();</
{{output}}
<pre>
Line 1,941 ⟶ 5,149:
For a table with repeated solutions using nested loops:
<
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) }</
{{out}}
<pre>
| |||