1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 | /* Set partition problem in B-Prolog. Problem formulation from """ This is a partition problem. Given the set S = {1, 2, ..., n}, it consists in finding two sets A and B such that: <ul> <li>A U B = S,</li> <li>|A| = |B|,</li> <li>sum(A) = sum(B),</li> <li>sum_squares(A) = sum_squares(B).</li> </ul> """ This model just implementa a two set version. Model created by Hakan Kjellerstrand, hakank@gmail.com*/% Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/% find all (7) solutions for N = 16.go :- findall(Sets,set_partition(16, Sets),L), length(L,Len), format("It was ~d solutions.\n", [Len]).go2 :- N :: 2..50, indomain(N), once(set_partition(N, _Sets)), nl,fail.set_partition(N,Sets) :- writeln('\nN':N), (N mod 2 > 0 -> format("~d is not a multiple of 2.\n", [N]), fail ; true ), (N mod 4 > 0 -> format("~d is not a multiple of 4.\n", [N]), fail ; true ), N2 is N //2, [A,B] :: {}..{1..N}, Sets = [A,B], % same cardinality #A #= N2, #A #= #B, % A and B are disjoint A #<> B, % Symmetry breaking: 1 is in the A set 1 #<- A, % All numbers 1..N must be in some set foreach(I in 1..N, (I #<- A ; I #<- B)), % It seems that we must have the labeling % already here... indomain(A), indomain(B), set_to_list(A,AList), set_to_list(B,BList), % sum of the sets must be equal ASum #= sum(AList), BSum #= sum(BList), ASum #= BSum, % sum of the squares must be equal ASumSquared #= sum([ T**2 : I in 1..AList^length, [T], T @= AList[I]]), BSumSquared #= sum([ T**2 : I in 1..BList^length, [T], T @= BList[I]]), ASumSquared #= BSumSquared, % labeling([ASum,BSum,ASumSquared,BSumSquared]), writeln(A), writeln(B), writeln(sums:[ASum,BSum]), writeln(sumSquares:[ASumSquared,BSumSquared]), nl. |