Department Numbers

From Rosetta Code
Task
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 the number 12.

The Chief of the Police doesn't like odd numbers and wants to have an even number for his department.


Task

Write a program which outputs all valid combinations.


Possible output:

1 2 9
5 3 4

ALGOL 68[edit]

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

AppleScript[edit]

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
Translation of: Haskell
-- 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

AWK[edit]

 
# 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

C++[edit]

 
#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[edit]

(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

D[edit]

Translation of: C++
 
import std.stdio, std.range;
 
void main() {
int sol = 1;
writeln("\t\tFIRE\t\tPOLICE\t\tSANITATION");
foreach( f; iota(7) ) {
foreach( p; iota(7) ) {
foreach( s; iota(7) ) {
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                6                  5    
SOLUTION # 2:    2                4                  6    
SOLUTION # 3:    2                6                  4    
SOLUTION # 4:    3                4                  5    
SOLUTION # 5:    4                2                  6    
SOLUTION # 6:    4                6                  2    
SOLUTION # 7:    5                4                  3    
SOLUTION # 8:    5                6                  1    
SOLUTION # 9:    6                2                  4    
SOLUTION #10:    6                4                  2    

Haskell[edit]

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 =
mapM_ print $
[1 .. 7] >>=
\x ->
[1 .. 7] >>=
\y ->
[1 .. 7] >>=
\z ->
if even x && length (nub [x, y, z]) == 3 && (sum [x, y, z] == 12)
then [(x, y, z)]
else []
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:

options :: Int -> Int -> Int -> [(Int, Int, Int)]
options lo hi total =
(\ds ->
filter even ds >>=
\x ->
filter (/= x) ds >>=
\y ->
[total - (x + y)] >>=
\z ->
case y /= z && lo <= z && z <= hi of
True -> [(x, y, z)]
_ -> [])
[lo .. hi]
 
-- TEST -----------------------------------------------------------------------
main :: IO ()
main = do
let xs = options 1 7 12
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

Fortran[edit]

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[edit]

' 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 

Gambas[edit]

Click this link to run this code

Public Sub Main()
Dim siC0, siC1, siC2 As Short
Dim sOut As New String[]
Dim sTemp As String
 
For siC0 = 2 To 6 Step 2
For siC1 = 1 To 7
For siC2 = 1 To 7
If sic0 + siC1 + siC2 = 12 Then
If siC0 <> siC1 And siC1 <> siC2 And siC0 <> siC2 Then sOut.Add(Str(siC0) & Str(siC1) & Str(siC2))
End If
Next
Next
Next
 
Print "\tPolice\tFire\tSanitation"
siC0 = 0
 
For Each sTemp In sOut
Inc sic0
Print "[" & Format(Str(siC0), "00") & "]\t" & Left(sTemp, 1) & "\t" & Mid(sTemp, 2, 1) & "\t" & Right(sTemp, 1)
Next
 
End

Output:

        Police  Fire    Sanitation
[01]    2       3       7
[02]    2       4       6
[03]    2       6       4
[04]    2       7       3
[05]    4       1       7
[06]    4       2       6
[07]    4       3       5
[08]    4       5       3
[09]    4       6       2
[10]    4       7       1
[11]    6       1       5
[12]    6       2       4
[13]    6       4       2
[14]    6       5       1

J[edit]

Solution:

require 'stats'
permfrom=: ,/@([email protected][ {"_ 1 comb) NB. get permutations of length x from y possible items
 
alluniq=: # = #@~. NB. check items are unique
addto12=: 12 = +/ NB. check items add to 12
iseven=: [email protected](2&|) NB. check items are even
policeeven=: {[email protected] 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 approach[edit]

   (#~ 12=+/"1) 1+3 comb 7 [ load'stats'
1 4 7
1 5 6
2 3 7
2 4 6
3 4 5

Note that we are only showing the distinct combinations here, not all permutations of those combinations.

JavaScript[edit]

ES5[edit]

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:

Translation of: Haskell
(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[edit]

Briefly:

(() => {
 
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
 
return '(Police, Sanitation, Fire)\n' +
concatMap(x =>
concatMap(y =>
concatMap(z =>
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, by composition of generic functions:

Translation of: Haskell
(() => {
'use strict';
 
// NUMBERING CONSTRAINTS --------------------------------------------------
 
// options :: Int -> Int -> Int -> [(Int, Int, Int)]
const options = (lo, hi, total) => {
const
bind = flip(concatMap),
ds = enumFromTo(lo, hi);
 
return bind(filter(even, ds),
x => bind(filter(d => d !== x, ds),
y => bind([total - (x + y)],
z => (z !== y && lo <= z && z <= hi) ? [
[x, y, z]
] : []
)
)
)
};
 
// GENERIC FUNCTIONS ------------------------------------------------------
 
// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = (f, xs) => [].concat.apply([], xs.map(f));
 
// enumFromTo :: Int -> Int -> [Int]
const enumFromTo = (m, n) =>
Array.from({
length: Math.floor(n - m) + 1
}, (_, i) => m + i);
 
// even :: Integral a => a -> Bool
const even = n => n % 2 === 0;
 
// filter :: (a -> Bool) -> [a] -> [a]
const filter = (f, xs) => xs.filter(f);
 
// flip :: (a -> b -> c) -> b -> a -> c
const flip = f => (a, b) => f.apply(null, [b, a]);
 
// length :: [a] -> Int
const length = xs => xs.length;
 
// map :: (a -> b) -> [a] -> [b]
const map = (f, xs) => xs.map(f);
 
// show :: a -> String
const show = x => JSON.stringify(x) //, null, 2);
 
// unlines :: [String] -> String
const unlines = xs => xs.join('\n');
 
// TEST -------------------------------------------------------------------
const 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

jq[edit]

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 first can easily be made more efficient by adding some pruning.

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: $fire, police: $police, sanitation: $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 )

Pruning

range(1;8) as $fire
| (range(1;8) | select(. != $fire)) as $police
| (range(1;8) | select(. != $fire and . != $police)) as $sanitation
| {fire: $fire, police: $police, sanitation: $sanitation}
| select( ([.[]] | add) == 12)

Kotlin[edit]

// 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[edit]

 
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

PARI/GP[edit]

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[edit]

 
#!/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
 

Perl 6[edit]

for (1..7).combinations(3).grep(*.sum == 12) {
for .permutations\ .grep(*.[0] %% 2) {
say <police fire sanitation> Z=> .list;
}
}
 
Output:
(police => 4 fire => 1 sanitation => 7)
(police => 4 fire => 7 sanitation => 1)
(police => 6 fire => 1 sanitation => 5)
(police => 6 fire => 5 sanitation => 1)
(police => 2 fire => 3 sanitation => 7)
(police => 2 fire => 7 sanitation => 3)
(police => 2 fire => 4 sanitation => 6)
(police => 2 fire => 6 sanitation => 4)
(police => 4 fire => 2 sanitation => 6)
(police => 4 fire => 6 sanitation => 2)
(police => 6 fire => 2 sanitation => 4)
(police => 6 fire => 4 sanitation => 2)
(police => 4 fire => 3 sanitation => 5)
(police => 4 fire => 5 sanitation => 3)

Python[edit]

 
from sys import stdout
 
 
def create():
stdout.write("\t\t Police\t\tFire\tSanitation\n")
c = 1
for p in range(1, 8):
for f in range(1, 8):
for s in range(1, 8):
if p != s and p != f and s != f and p + s + f == 12 and p % 2 == 0:
stdout.write("{0}:\t\t\t{1}\t\t {2}\t\t\t{3}\n".format(c, p, f, s))
c += 1
 
 
# entry point
create()
 
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

Racket[edit]

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))

REXX[edit]

bare bones[edit]

/*REXX program finds/displays all possible variants of (3) department numbering  puzzle.*/
say 'police fire sanitation' /*display a crude title for the output.*/
do p=2 to 7 by 2 /*try numbers for the police department*/
do f=1 for 7 /* " " " " fire " */
do s=1 for 7; $=p+f+s /* " " " " sanitation " */
if f\==p & s\==p & s\==f & $==12 then say center(p,6) center(f,5) center(s,10)
end /*s*/
end /*f*/
end /*p*/ /*stick a fork in it, we're all done. */
output   when using the default inputs:
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

options and optimizing[edit]

A little extra code was added to allow the specification for the high department number as well as the sum.

Two optimizing statements were added (for speed),   but for this simple puzzle they aren't needed.

Also, extra code was added to nicely format a title (header) for the output, as well as displaying the number of solutions found.

/*REXX program finds/displays all possible variants of (3) department numbering  puzzle.*/
parse arg high sum . /*obtain optional arguments from the CL*/
if high=='' | high=="," then high= 7 /*Not specified? Then use the default.*/
if sum=='' | sum=="," then sum=12 /* " " " " " " */
@pd= ' police '; @fd= " fire "  ; @sd= ' sanitation ' /*define names of departments.*/
@dept= ' department '; L=length(@dept) /*literal; and also its length*/
#=0 /*initialize the number of solutions. */
do PD=2 by 2 to high /*try numbers for the police department*/
do FD=1 for high /* " " " " fire " */
if FD==PD then iterate /*Same FD# & PD#? They must be unique.*/
if FD+PD>sum-1 then iterate PD /*Is sum too large? Try another PD#. */ /* ◄■■■■■■ optimizing code*/
do SD=1 for high /*try numbers for the sanitation dept. */
if SD==PD | SD==FD then iterate /*Is SD# ¬unique? They must be unique,*/
$=PD+FD+SD /*compute sum of department numbers. */
if $> sum then iterate FD /*Is the sum too high? Try another FD#*/ /* ◄■■■■■■ optimizing code*/
if $\==sum then iterate /*Is the sum ¬correct? " " SD#*/
#=# + 1 /*bump the number of solutions (so far)*/
if #==1 then do /*Is this the 1st solution? Show hdr.*/
say center(@pd, L) center(@fd, L) center(@sd, L)
say copies(center( @dept, L)' ', 3)
say copies(center('number', L)' ', 3)
say center('', L, "═") center('', L, "═") center('', L, "═")
end
say center(PD, L) center(FD, L) center(SD, L) /*display a solution.*/
end /*SD*/
end /*FD*/
end /*PD*/
say /*display a blank line before the #sols*/
if #==0 then #= 'no' /*use a better word for bupkis. */
say # "solutions found." /*stick a fork in it, we're all done. */
output   when using the default inputs:
   police        fire      sanitation
 department   department   department
   number       number       number
════════════ ════════════ ════════════
     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[edit]

 
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

Scala[edit]

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[edit]

Translation of: Perl 6
@(1..7)->combinations(3, {|*a|
a.sum == 12 || next
a.permutations {|*b|
b[0].is_even || next
say (%w(police fire sanitation) ~Z b -> join(" "))
}
})
Output:
["police", 4] ["fire", 1] ["sanitation", 7]
["police", 4] ["fire", 7] ["sanitation", 1]
["police", 6] ["fire", 1] ["sanitation", 5]
["police", 6] ["fire", 5] ["sanitation", 1]
["police", 2] ["fire", 3] ["sanitation", 7]
["police", 2] ["fire", 7] ["sanitation", 3]
["police", 2] ["fire", 4] ["sanitation", 6]
["police", 2] ["fire", 6] ["sanitation", 4]
["police", 4] ["fire", 2] ["sanitation", 6]
["police", 4] ["fire", 6] ["sanitation", 2]
["police", 6] ["fire", 2] ["sanitation", 4]
["police", 6] ["fire", 4] ["sanitation", 2]
["police", 4] ["fire", 3] ["sanitation", 5]
["police", 4] ["fire", 5] ["sanitation", 3]

zkl[edit]

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