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 | /* Langford's number problem in SICStus Prolog. Langford's number problem (CSP lib problem 24) """ Arrange 2 sets of positive integers 1..k to a sequence, such that, following the first occurence of an integer i, each subsequent occurrence of i, appears i+1 indices later than the last. For example, for k=4, a solution would be 41312432 """ * John E. Miller: Langford's Problem http://www.lclark.edu/~miller/langford.html * Encyclopedia of Integer Sequences for the number of solutions for each k Also, see the following models: * MiniZinc: http://www.hakank.org/minizinc/langford2.mzn * Comet : http://www.hakank.org/comet/langford.co * Gecode/R: http://www.hakank.org/gecode_r/langford.rb * ECLiPSe : http://www.hakank.org/eclipse/langford.ecl Model created by Hakan Kjellerstrand, hakank@gmail.com*/% Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/:-use_module(library(clpfd)).:-use_module(library(lists)).go :- K in 2..8, indomain(K), write(K),nl, findall([K,Solution,Position], langford(K,Solution,Position), L), length(L,Len), write(L),nl, write(len:Len),nl, nl, fd_statistics, fail.langford(K, Solution, Position) :- K2 is 2*K, length(Position, K2), domain(Position,1,K2), length(Solution,K2), domain(Solution,1,K), all_different(Position), % symmetry breaking nth1(1,Solution,Solution1), nth1(K2,Solution,SolutionK2), Solution1 #< SolutionK2, ( for(I,1,K), param(Position,Solution,K) do IK is I+K, element(IK, Position, PositionIK), element(I, Position, PositionI), I1 is I+1, PositionIK #= PositionI + I1, element(PositionI,Solution,SolutionPositionI), SolutionPositionI #= I, element(PositionIK,Solution,SolutionPositionIK), SolutionPositionIK #= I ), append(Solution,Position, Vars), labeling([ffc,bisect,down], Vars). |