Download
%
% 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
% http://www.hakank.org/minizinc/all_interval1.mzn
% http://www.hakank.org/minizinc/all_interval2.mzn
% http://www.hakank.org/minizinc/all_interval3.mzn
% http://www.hakank.org/minizinc/all_interval4.mzn
% http://www.hakank.org/minizinc/all_interval5.mzn
% http://www.hakank.org/minizinc/all_interval6.mzn
%
%
% 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)
];