Minizinc array sorting

Question

Lets say I have an array declaration looking like this

array[1..5] of int: temp = [1,0,5,0,3];

Is there a way to initiate a new array looking the same as temp but without the 0's? The result would look like the following

[1,5,3]

or sort the array in such a way that the 0's would be either in the beginning or in the end of the array, which would be

[0,0,1,5,3]

or

[1,5,3,0,0]

Thanks

Answer 1

Even though Axel has answered this, I'll show another approach which - in my book is a little neater.

• Case 1: the array ("temp") is a constant array. Then one can simply write

  array[int] of int: temp2 = [temp[i] | i in index_set(temp) where temp[i] != 0];

MiniZinc 2 (in contrast to version 1.*) don't need the size declaration if it can be calculated; it suffices to just use "array[int]". Also, "index_set" is used to be a little more general, e.g. to handle cases where the indices are from 0..4 (see the commented line).

• Case 2: the array ("s") is an array of decision variables

If the array to handle is decision variables, we don't know (per definition) how many 0's there are and must rely on the alternative variant, namely to sort the array. One can then use the "sort" function, as shown in the model.

 include "globals.mzn";

 % constant
 array[1..5] of int: temp = [1,0,5,0,3];
 % array[0..4] of int: temp = array1d(0..4, [1,0,5,0,3]);
 array[int] of int: temp2 = [temp[i] | i in index_set(temp) where temp[i] != 0];

 % decision variables
 array[1..5] of var int: s;
 array[1..5] of var int: s2 = sort(s); % NOT CORRECT, see model below

 solve satisfy;

 constraint
    s = [1,0,5,0,3]
 ;

 %  show our variables
 output 
 [
    "temp: \(temp)\n",
    "temp2: \(temp2)\n",

    "s: \(s)\n",
    "s2: \(s2)\n",
 ];

Update

For the stable version of decision variables, this works what I can see. It calculating the position where to place this number depending on if "s[i]" is 0 or not. Not very pretty though.

 int: n = 5;
 array[1..n] of var 0..5: s;
 array[1..n] of var lb_array(s)..ub_array(s): s2;

 solve satisfy;

 constraint
   s = [1,0,5,0,3] /\
   forall(i in 1..n) (
      if s[i] != 0 then
        s2[sum([s[j]!=0 | j in 1..i-1])+1] = s[i]
      else 
        s2[sum([s[j]!=0 | j in 1..n]) + sum([s[j]=0 | j in 1..i-1])+1 ] = 0
      endif
   )
;

output 
[ 
  "s: \(s)\n",
  "s2: \(s2)\n",
]
;

The output is

s: [1, 0, 5, 0, 3]
s2: [1, 5, 3, 0, 0]