$title Peacefully Coexisting Armies of Queens - tight (COEXX,SEQ=218)

$onText
This is a tighter formulation than the original COEX problem.
We have set the size of the board to 5 in order to find
solutions quickly. In addition we fix the position of one queen.

Two armies of queens (black and white) peacefully coexist on a
chessboard when they are placed on the board in such a way that
no two queens from opposing armies can attack each other. The
problem is to find the maximum two equal-sized armies.

Bosch, R, Mind Sharpener. OPTIMA MPS Newsletter (2000).

Keywords: mixed integer linear programming, mathematical games, combinatorial optimization,
          peaceably coexisting armies of queens
$offText

$eolCom //

Set
   i 'size of chess board' / 1*5 /    // use 8 for chess (13)
   s 'diagonal offsets'    / 1*7 /;   // 2i-3 diagonals

Scalar idiags 'correct size of s';
idiags = 2*card(i) - 3;
abort$(card(s) <> idiags) 's has incorrect size', idiags;

Alias (i,j);

Parameter
   sh(s)    'shift values for diagonals'
   rev(s,i) 'reverse shift order';

sh(s)    = ord(s) - card(i) + 1 ;
rev(s,i) = card(i) + 1 - 2*ord(i) + sh(s);

Binary Variable
   xw(i,j) 'has a white queen'
   xb(i,j) 'has a black queen'
   wa(i)   'white in row i'
   wb(i)   'white in column j'
   wc(s)   'white in diagonal s'
   wd(s)   'white in backward diagonal s';

Variable tot;

Equation
   aw(i,j) 'white in row i'
   bw(j,i) 'white in column j'
   cw(s,i) 'white in diagonal s'
   dw(s,i) 'white in backward diagonal s'
   ew      'total white'
   ab(i,j) 'black in row i'
   bb(j,i) 'black in column j'
   cb(s,i) 'black in diagonal s'
   db(s,i) 'black in backward diagonal s'
   eb      'total black';

aw(i,j)..     wa(i) =g= xw(i,j);

bw(j,i)..     wb(j) =g= xw(i,j);

cw(s,i)..     wc(s) =g= xw(i, i + sh(s));

dw(s,i)..     wd(s) =g= xw(i, i + rev(s,i));

ab(i,j).. 1 - wa(i) =g= xb(i,j);

bb(j,i).. 1 - wb(j) =g= xb(i,j);

cb(s,i).. 1 - wc(s) =g= xb(i, i + sh(s));

db(s,i).. 1 - wd(s) =g= xb(i, i + rev(s,i));

eb..           tot  =e= sum((i,j), xb(i,j));

ew..           tot  =e= sum((i,j), xw(i,j));

Model army / all /;

option limCol = 0, limRow = 0;

xb.fx('1','1') = 1; // fix one position in the NW corner

solve army maximizing tot using mip;