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 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 | /* All interval problem in Comet. CSPLib problem number 7 http://www.cs.st-andrews.ac.uk/~ianm/CSPLib/prob/prob007/index.html """ Given the twelve standard pitch-classes (c, c , d, ...), represented by numbers 0,1,...,11, find a series in which each pitch-class occurs exactly once and in which the musical intervals between neighbouring notes cover the full set of intervals from the minor second (1 semitone) to the major seventh (11 semitones). That is, for each of the intervals, there is a pair of neigbhouring pitch-classes in the series, between which this interval appears. The problem of finding such a series can be easily formulated as an instance of a more general arithmetic problem on Z_n, the set of integer residues modulo n. Given n in N, find a vector s = (s_1, ..., s_n), such that (i) s is a permutation of Z_n = {0,1,...,n-1}; and (ii) the interval vector v = (|s_2-s_1|, |s_3-s_2|, ... |s_n-s_{n-1}|) is a permutation of Z_n-{0} = {1,2,...,n-1}. A vector v satisfying these conditions is called an all-interval series of size n; the problem of finding such a series is the all-interval series problem of size n. We may also be interested in finding all possible series of a given size. """ Compare with * MiniZinc model: http://www.hakank.org/minizinc/all_interval.mzn * Gecode/R model: http://www.hakank.org/gecode_r/all_interval.rb This Comet model was created by Hakan Kjellerstrand (hakank@gmail.com) Also, see my Comet page: http://www.hakank.org/comet */ // Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/ /* Note: For n = 12, all 1328 solution (no symmetry breaking) took about 16 seconds. With symmetry breaking: 463 solutions in 5.5 seconds. */ import cotfd; int t0 = System.getCPUTime(); int n = 12; int sum_distinct = ((n+1)*n) / 2; Solver<CP> m(); var<CP>{int} x[1..n](m, 1..n); var<CP>{int} diffs[1..n-1](m, 1..n-1); Integer num_solutions(0); exploreall<m> { forall(k in 1..n-1) m.post(diffs[k] == abs(x[k+1] - x[k]), onValues); m.post(alldifferent(x), onValues); m.post(alldifferent(diffs), onValues); // symmetry breaking m.post(x[1] < x[n-1], onValues); m.post(diffs[1] < diffs[2], onValues); } using { /* forall(i in 1..n : !x[i].bound()) by (-x[i].getSize()) { // tryall<m>(v in 1..n) // m.label(x[i], v); label(x[i]); } forall(i in 1..n-1 : !diffs[i].bound()) by (diffs[i].getSize()) { // tryall<m>(v in 1..n) // m.label(x[i], v); label(diffs[i]); } */ //label(m); label(x); // label(diffs); num_solutions := num_solutions + 1; cout << x << " " << sum_distinct << " " << diffs << endl; cout << flush; } // cout << x << endl; cout << "\nnum_solutions: " << num_solutions << endl; int t1 = System.getCPUTime(); cout << "time: " << (t1-t0) << endl; cout << "#choices = " << m.getNChoice() << endl; cout << "#fail = " << m.getNFail() << endl; cout << "#propag = " << m.getNPropag() << endl; |