How to resolve the algorithm Department numbers step by step in the Picat programming language

Published on 12 May 2024 09:40 PM
#Picat

How to resolve the algorithm Department numbers step by step in the Picat programming language

Table of Contents

Problem Statement

There is a highly organized city that has decided to assign a number to each of their departments:

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):

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Department numbers step by step in the Picat programming language

Source code in the picat programming language

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]).

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.

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")).

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.

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