My cousin recently bought a Bomb Trap puzzle, he was crazy for solving that puzzle and tried a couple of times and often came to me for rechecking his solution, and me being lazy, I thought of making a simple python script to check his solutions and save me from the brain drain. The Code bomb_trap.py Code: #!/bin/env/python # Checker for a 8x8 bomb trap # Correct Solution game = [ [0, 0, 0, 0, 0, 0, 1, 0], [0, 0, 0, 0, 1, 0, 0, 0], [0, 0, 1, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 0, 0, 0, 1], [0, 1, 0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0, 0, 0] ] def check_ones(array) : ones = 0 for h in array: if isinstance(h, list) : ones = ones + check_ones(h) elif h == 1: ones = ones + 1 return ones def diagonal_check(game): errors = '' for i in range(0, 7): for j in range(0, 7): if game[i][j] != 1: continue # diagonal 1 diag1 = [] h, k = i, j # addition loop while h <= 7 and h >= 0 and k <= 7 and k >= 0: diag1.append(game[h][k]) h = h + 1 k = k + 1 # subtration loop h, k = i-1, j-1 while h <= 7 and h >= 0 and k <= 7 and k >= 0: diag1.append(game[h][k]) h = h - 1 k = k - 1 # diagonal 2 diag2 = [] # add k loop h, k = i, j while h <= 7 and h >= 0 and k <= 7 and k >= 0: diag2.append(game[h][k]) h = h - 1 k = k + 1 # add h loop h, k = i + 1, j - 1 while h <= 7 and h >= 0 and k <= 7 and k >= 0: diag2.append(game[h][k]) h = h + 1 k = k - 1 # at this point we have 2 diagonal arrays and we can simply check # for multiple 1's, if there are multiple 1's in any diag list # it means we have 2 points in a diagonal i.e check failed if check_ones(diag1) > 1: errors += "Diagonal 1 check @ point [%d, %d] evaluated to FALSE\n" % (i+1, j+1) if check_ones(diag2) > 1: errors += "Diagonal 2 check @ point [%d, %d] evaluated to FALSE\n" % (i+1, j+1) if errors != '': print errors return False return True check1 = True # Let's be +ve ;) errors = '' # let the game begin if check_ones(game) != 8: print "Please fill up exactly 8 places." exit() for i in range(0, 7): horizontal = [] vertical = [] for j in range(0, 7): horizontal.append(game[i][j]) vertical.append(game[j][i]) if check_ones(horizontal) > 1: errors += "Multiple mines in Horizontal, @ Row [%d]\n" % (i + 1) if check_ones(vertical) > 1: errors += "Multiple mines in Vertical, @ Column [%d]\n" % (i + 1) if errors != '': print errors check1 = False # Lets do some diagonal checking now check2 = diagonal_check(game) if check1 and check2 : print "Correct :)" After some hours of trying i was able to get the right solution. I think only 4 solutions are possible but feel free to hunt for more.

Hey dude i think there are 92 solutions to it....... And it is a sort of 8queens problem would like to check out this link....