N-queens problem: Difference between revisions

=={{header|MiniScript}}==

This GUI implementation is for use with [http://miniscript.org/MiniMicro Mini Micro]. It displays a chess board with animation of the possibilities. At the end, after all of the solutions have been calculated, you can scroll through them with the left/right cursor keys.

<syntaxhighlight lang="miniscript">

clear

N = 8

SOLUTIONCOUNT = 0

getTileDisplay = function

gfx.clear

queen = file.loadImage("/sys/pics/gamePieces/blackQueen.png")

gfx.color = color.white

gfx.fillRect 0, 0, 80, 80

gfx.fillRect 160, 0, 80, 80

gfx.color = color.brown

gfx.fillRect 80, 0, 80, 80

gfx.fillRect 240, 0, 80, 80

gfx.drawImage queen, 172, 14

gfx.drawImage queen, 252, 14

tiles = gfx.getImage(0,0, 320, 80)

gfx.clear

display(4).mode = displayMode.tile

td = display(4)

td.cellSize = 640 / N

td.extent = [N, N]

td.overlap = 0

td.tileSet = tiles

td.tileSetTileSize = 80

td.scrollX = -160

td.clear

return td

end function

updateBoard = function(td, arr)

for y in range(0, N - 1)

ix = y % 2

for x in range(0, N - 1)

td.setCell x, y, ix

ix += 1

ix %= 2

end for

end for

y = 0

for x in arr

td.setCell x, y, td.cell(x, y) + 2

y += 1

end for

yield

end function

list.has = function(n)

return self.indexOf(n) != null

end function

queens = function(n, i, a, b, c, td)

solutions = []

updateBoard(td, a)

if i < n then

for j in range(0, n - 1)

if not a.has(j) and not b.has(i + j) and not c.has(i - j) then

solution = queens(n, i + 1, a + [j], b + [i + j], c + [i - j], td)

if solution != null then solutions += solution

end if

end for

else

globals.SOLUTIONCOUNT += 1

text.row = 25

text.print "SOLUTIONS"

text.print globals.SOLUTIONCOUNT

solutions.push(a)

end if

return solutions

end function

td = getTileDisplay

solutions = queens(N, 0, [], [], [], td)

ix = 0

while true

text.row = 25

text.print "SOLUTION # "

text.print (ix + 1) + (" " * 10)

text.print

text.print char(17) + "/" + char(18) + " keys"

updateBoard(td, solutions[ix])

k = key.get

kcode = code(k)

if kcode == 27 then break

ix = ix - (kcode == 17) + (kcode == 18) + solutions.len

ix %= solutions.len

end while

</syntaxhighlight>