Bomb Trap (Puzzle) Checker in Python

Discussion in 'Python' started by lionaneesh, Mar 11, 2012.

1. lionaneeshActive Member

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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
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 = []
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

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.

2. lionaneeshActive Member

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Edit: The problem has 12 unique solutions and 92 distinct solutions

Joined:
Sep 10, 2010
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2