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%
% All interval problem in MiniZinc.
%
% Different approaches inspired by
% http://www.dis.uniroma1.it/~tmancini/index.php?currItem=research.publications.webappendices.csplib2x.problemDetails&problemid=007
%
% Also see
%
 
%
% Model created by Hakan Kjellerstrand, hakank@gmail.com
% See also my MiniZinc page: http://www.hakank.org/minizinc
 
% Licenced under CC-BY-4.0 : http://creativecommons.org/licenses/by/4.0/
 
int: n= 12;
set of int: classes = 0..n-1;
set of int: differ = 1..n-1;
 
 
% Search space: The set of permutations of integer range [0..n-1]
array[classes] of var classes: series;
array[0..n-2] of var differ: differences;
 
% solve satisfy;
solve :: int_search(series, occurrence, indomain_min, complete) satisfy;
 
constraint
   % C1: Each pitch class occurs exactly once
   forall(i,j in classes where i != j) (
     series[i] != series[j]
   )
   /\
   % C2: Differences between neighbouring notes are all different
   % AUX: Addition of auxiliary predicates
   % Auxiliary predicate stores the interval between pairs of neighbouring notes
   forall(i in 0..n-2) (
      differences[i]=abs(series[i+1] - series[i])
   )
   /\
   forall(i,j in 0..n-2 where i != j) (
      differences[i] != differences[j]
   )
   /\
   % SBSO: Symmetry-breaking by selective ordering
   % The first note is less than last one
   series[0] < series[n-1]
;
 
output [
  show(series)
];