# 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.

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

Translation of: C
```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?
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?
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
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
```

## 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:
```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
```

```with Ada.Text_IO;

procedure Department_Numbers is
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

Translation of: ALGOL 68
```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⌷⊢)∧12=+/))↑,⍳3/7
```

This prints each triplet of numbers from 1 to 7 for which:

1. the elements are unique: `(∪≡⊢)¨↓`
2. the first element is even: `(0=2|1⌷⊢)`
3. 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:

Translation of: JavaScript
```-- 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

Works with: Chipmunk Basic version 3.6.4
```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 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```

### 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 police```

### 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()```
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 NEXT```

### Minimal BASIC

Translation of: Sinclair ZX81 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

Translation of: Sinclair ZX81 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

Translation of: ALGOL 68
```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

Works with: Windows 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 := p + '0';
out := s + '0';
out := f + '0';
if p != s and p != f and f != s and p+s+f == 12 then
print(&out);
end if;
end sub;

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

Translation of: C++
```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

Translation of: Go
```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');
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

Translation of: C
```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:

```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:

 =departmentNumbers fx 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

```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
    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
```

## Go

Translation of: Kotlin
```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

Translation of: Java
```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```

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
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

Translation of: Kotlin
```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 . % 2 == 0)
| {fire: ., police: ., sanitation: .}```

## 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
```

```           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, {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}}```

## Modula-2

```MODULE DepartmentNumbers;
FROM Conversions IMPORT IntToStr;

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;

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

Translation of: C++
```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

Translation of: Prolog
```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))
(<> 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

Translation of: ALGOL W
Works with: 8080 PL/M Compiler
... 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 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;

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:

Works with: Python version 3
```'''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):
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:

Works with: Python version 3.7
```'''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 ```

```# 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:

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

Translation of: Forth
```  [ 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(*. %%  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
next```

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
```

## 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

Translation of: C
```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

Translation of: Raku
```@(1..7)->combinations(3, {|*a|
a.sum == 12 || next
a.permutations {|*b|
b.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 10```

## Transd

```#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

Translation of: Sinclair ZX81 BASIC
```// 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

Works with: Bourne Again SHell
Works with: Korn Shell
Works with: Z 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

Translation of: C#
```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)

Translation of: 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")
}
```
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

Translation of: Kotlin
```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

Translation of: Sinclair ZX81 BASIC
Works with: EXPL-32
```\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

Works with: Zig version 0.11.0dev
```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

Works with: Zig version 0.11.0

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) {
if (ov == 1) {
self.u.together = 0;
return null; // overflowed, completed
} else {
self.u.together = ov;
// 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
```