Department numbers

There is a highly organized city that has decided to assign a number to each of their departments. Police, Sanitation and Fire Department.

Each Department can have a number between 1 and 7.
The 3 numbers have to be different and have to add up to the number 12.
For some reason the Chief of the Police Department doesn't like odd numbers and wants to have an even number for his department.

Write a programm which outputs all valid combinations.

Possible Output:

1 2 9
5 3 4

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>

REXX

<lang rexx>/*REXX program finds/displays all possible variants of (3) department numbering puzzle.*/ parse arg high sum . if high== | high=="," then high= 7 if sum== | sum=="," then sum=12

=0; @pd='police'; @fd="fire"; @sd='sanitation'

          Lpd=length(@pd);  Lfd=length(@fd);  Lsd=length(@sd)

 do PD=2  by 2  to high
    do FD=1   for  high
    if FD==PD       then iterate
    if FD+PD>sum-1  then iterate PD
    $2=PD+FD
       do SD=1  for  high
       if SD==PD    then iterate
       if SD==FD    then iterate
       $3=$2+SD
       if $3>  sum  then iterate FD
       if $3\==sum  then iterate
       #=# + 1
       if #==1      then say           @pd                 @fd                 @sd
       if #==1      then say center(,Lpd,"═")  center(,Lfd,"═")  center(,Lsd,"═")
                         say center(PD,Lpd)      center(FD,Lfd)      center(SD,Lsd)
       end   /*SD*/
    end      /*FD*/
 end         /*PD*/

say if #==0 then #='no' say # "solutions found."</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

14 solutions found.

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) .filter("apply","isEven") // at least one number is even .println();</lang>

Output:

L(L(1,4,7),L(1,5,6),L(2,3,7),L(2,4,6),L(3,4,5))

For a table with repeated solutions: <lang zkl>ns:=Utils.Helpers.pickNFrom(3,[1..7].walk()) // 35 combos

 .filter(fcn(numbers){ numbers.sum(0)==12 })  // which all sum to 12 (==5)
 .filter("apply","isEven")		// at least one number is even
 .pump(List,Utils.Helpers.permute)	// expand 5 results --> list of lists
 .flatten()				// ( (),()..) --> ()
 .filter(fcn([(p,_,_)]){ p.isEven });	// with even first number

println("Police Fire Sanitation"); foreach pfs in (ns){ "%d\t%d\t%d".fmt(pfs.xplode()).println() }</lang>

Output:

Police  Fire  Sanitation
4	7	1
4	1	7
6	1	5
6	5	1
2	3	7
2	7	3
2	4	6
2	6	4
6	2	4
6	4	2
4	6	2
4	2	6
4	5	3
4	3	5