In reply to dave_59:
Hi Dave ,
Have a question related to unrolling of queen_b.sum(b) :
foreach(queen_b[i]) queen_b[i].sum() with (int'(item)) == 1;
Here item refers to each element of queen_b[i] i.e queen_b[i][0] to queen_b[i][7]
As i iterates from 0 to 7 , the resultant expression will be sum i.e addition (+) of 8 resultant with_expression i.e :
For queen_b[0].sum() i.e Row 0 :: int’( queen_b[0][0] ) + int’( queen_b[0][1] ) + … + int’( queen_b[0][7] ) == 1 ;
Similarly for queen_b[7].sum() i.e Row 7 :: int’( queen_b[7][0] ) + int’( queen_b[7][1] ) + … + int’( queen_b[7][7] ) == 1 ;
For queen_b.sum(b) , will the resultant expression comprise of sum of 8x8 i.e 64 with expressions ? ( with some elements being 0 due to false condition )