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
<lang algol68>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; [ 1 : max department number ]BOOL number used; FOR d TO UPB number used DO number used[ d ] := FALSE OD; print( ( "police sanitation fire", newline ) ); FOR police FROM 2 BY 2 TO max department number DO number used[ police ] := TRUE; FOR sanitation TO max department number DO IF NOT number used[ sanitation ] THEN number used[ sanitation ] := TRUE; FOR fire TO max department number DO IF NOT number used[ fire ] THEN IF police + sanitation + fire = 12 THEN print( ( whole( police, -6 ) , whole( sanitation, -11 ) , whole( fire, -5 ) , newline ) ) FI FI OD; number used[ sanitation ] := FALSE FI OD; number used[ police ] := FALSE OD
END</lang>
- 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
Briefly, composing a solution from generic functions:
<lang AppleScript>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</lang>
- Output:
237 246 264 273 417 426 435 453 462 471 615 624 642 651
Or more generally:
<lang AppleScript>-- 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 Template: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</lang>
- 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
C++
<lang cpp>
- 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;
}</lang>
- 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
Haskell
Bare minimum: <lang haskell>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)] _ -> []</lang>
or, resugaring this into list comprehension format: <lang Haskell>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 ]</lang>
or Do notation: <lang Haskell>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 []</lang>
- 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: <lang Haskell>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)]</lang>
Reaching again for a little more syntactic sugar, the options function above could also be re-written either as a list comprehension, <lang Haskell>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 ]</lang>
or in Do notation: <lang haskell>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)</lang>
- 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
Gambas
<lang 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</lang> 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
Solution: <lang j>require 'stats' permfrom=: ,/@(perm@[ {"_ 1 comb) NB. get permutations of length x from y possible items
alluniq=: # = #@~. NB. check items are unique addto12=: 12 = +/ NB. check items add to 12 iseven=: -.@(2&|) NB. check items are even policeeven=: {.@iseven NB. check first item is even conditions=: policeeven *. addto12 *. alluniq
Validnums=: >: i.7 NB. valid Department numbers
getDeptNums=: [: (#~ conditions"1) Validnums {~ permfrom</lang> Example usage: <lang j> 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</lang>
JavaScript
ES5
Briefly: <lang JavaScript>(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');
})();</lang>
- 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:
<lang JavaScript>(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);
})();</lang>
- 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: <lang JavaScript>(() => {
// 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');
})();</lang>
- 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:
<lang JavaScript>(() => {
'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);
})();</lang>
- 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
Kotlin
<lang scala>// 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")
}</lang>
- 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
<lang 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 ) ) </lang>
- 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
Perl
<lang 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"; } } }
} </lang>
Perl 6
<lang perl6>for (1..7).combinations(3).grep(*.sum == 12) {
for .permutations\ .grep(*.[0] %% 2) { say <police fire sanitation> Z=> .list; }
} </lang>
- 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
bare bones
<lang rexx>/*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. */</lang>
- 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
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. <lang rexx>/*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. */</lang>
- 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.
Scala
<lang 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") ) } </lang>
- 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)
zkl
<lang zkl>Utils.Helpers.pickNFrom(3,[1..7].walk()) // 35 combos .filter(fcn(numbers){ numbers.sum(0)==12 }) // which all sum to 12 (==5) .println();</lang>
- 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: <lang zkl>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) }</lang>
- 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