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queens.gms : Maximum Queens Chess Problem

This model finds all possible ways to arrange eight queens on a chess
board in such a way that no two queens check against any other queen.
The solution proceeds in two steps. In the first step, we find any
solution. In the second step we find all solutions by adding cuts to
to the original problem set. Finally, a reporting step is added
to print all solutions found. This problem has a long history. In 1850
it was investigated by C.F. Gauss, but he was unable to solve it
completely. There are 92 possible solutions.
References:
Large Model of Types: MIP gams
$Title Maximum Queens Chess Problem (QUEENS,SEQ=103) $Eolcom ! $Ontext This model finds all possible ways to arrange eight queens on a chess board in such a way that no two queens check against any other queen. The solution proceeds in two steps. In the first step, we find any solution. In the second step we find all solutions by adding cuts to to the original problem set. Finally, a reporting step is added to print all solutions found. This problem has a long history. In 1850 it was investigated by C.F. Gauss, but he was unable to solve it completely. There are 92 possible solutions. Dudeney, H E, Amusements in Mathematics. Dover, New York, 1970. Beauvais, J, Solving the Maximum Queens Chess Problem with OSL. IBM Kingston, EKKNEWS 2 (1991). $Offtext Sets i size of chess board / 1*8 / s diagonal offsets / 1*13 / ! 2i-3 diagonals Alias (i,j) Parameter sh(s) shift values for diagonals rev(i) reverse order; sh(s) = ord(s) - card(i) + 1 ; rev(i) = card(i) + 1 - 2*ord(i); Display sh, rev; Binary Variable x(i,j) square occupied by Queen Variable tot total squares occupied by queens Equations a(i) no to queens may be in the same rank b(j) no to queens may be in the same file c(s) no to queens may be in the same diagonal (forward) d(s) no to queens may be in the same diagonal (backward) obj objective definition ; a(i).. sum(j, x(i,j)) =e= 1; b(j).. sum(i, x(i,j)) =e= 1; c(s).. sum(i, x(i,i+sh(s))) =l= 1; d(s).. sum(i, x(i,i+(rev(i)+sh(s)))) =l= 1; obj.. tot =e= sum((i,j), x(i,j)); Model queen1 first model for queens / all /; Option optcr = 0; Solve queen1 maximizing tot using mip; $Stitle Find all remaining solutions Sets nn max number of solutions groups / 1*20 / n(nn) dynamic set for solution groups t multiple solutions via rotations and reflections / found original solution rot-90 original solution rotated 90 degrees rot-180 original solution rotated 180 degrees rot-270 original solution rotated 270 degrees ref-h original solution reflected horizontally ref-v original solution reflected vertically ref-r original solution reflected diagonally main ref-l original solution reflected diagonally back / Scalar saverow, coloff; Parameter cutval all possible solutions for cut generation; Equation cut(nn,t) known solutions to be eliminated; cut(n,t).. sum((i,j), cutval(n,t,i,j)*x(i,j)) =l= card(i)-1; Model queens queens model with cuts / all /; Option limrow=0, limcol=0, solprint=off; n(nn) = no; ! clear set of cuts queens.solvestat = %solvestat.NormalCompletion%; queens.modelstat = %modelstat.Optimal%; Loop(nn$(queens.solvestat=%solvestat.NormalCompletion% and queens.modelstat=%modelstat.Optimal%), n(nn) = yes; ! add element to set cutval(nn,'found' ,i,j) = x.l(i ,j ); cutval(nn,'rot-90' ,i,j) = x.l(j+rev(j),i ); cutval(nn,'rot-180' ,i,j) = x.l(i+rev(i),j+rev(j)); cutval(nn,'rot-270' ,i,j) = x.l(j ,i+rev(i)); cutval(nn,'ref-h' ,i,j) = x.l(i+rev(i),j ); cutval(nn,'ref-v' ,i,j) = x.l(i ,j+rev(j)); cutval(nn,'ref-r' ,i,j) = x.l(j ,i ); cutval(nn,'ref-l' ,i,j) = x.l(j+rev(j),i+rev(i)); Solve queens maximizing tot using mip; ) Option cutval:0:3:1; display cutval; $Stitle Write solution report to PUT file File queen report file for solution groups ; put queen; queen.pc=3; Scalar saverow row position to start second report column coloff column offset for board displays; Puttl 'Queens Solution Summary ' system.date ' ' system.time Loop(n(nn), Putpage; Put @12 // 'Solution Number ' ord(nn):3:0 ' of ' card(n):<3:0 ; saverow = file.cr; coloff = 1; Loop(t, if(ord(t)=5, file.cr = saverow; coloff = 31); Put // @coloff 'Type = ' t.tl // @(coloff+2) ; Loop(j, Put j.tl:>2); Loop(i, Put / @coloff i.tl:>2; Loop(j, If(cutval(n,t,i,j), Put ' x' Else Put ' ' ) ) ) ) );