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'''
The algorithm works as follows:
It simply checks all positions, with positions being marked with None (unknown) x (part of a ship) or o (no ship)
Because the description says that ships can't touch (including diagonally) you know for sure that if a position
has the part of a ship that the positions diagonal to it don't, like so:
o.o
.x.
o.o
This is the only tactic that my program uses, although it does check the inner square first
(with a border of 1 position around it) to increase the chance that positions marked as
being empty don't fall outside the board
'''
import sys
board = []
TOTAL_SHIPPARTS = 4+2*3+3*2+4*1
hits = 0
LETTERS = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j']
for x in xrange(10):
board.append([])
for y in xrange(10):
board[x].append(None)
def printBoard():
print "!"
for y in xrange(10):
for x in xrange(10):
if board[x][y] is "x":
sys.stdout.write("x")
else:
sys.stdout.write("o")
if y < 9:
sys.stdout.write("\n")
sys.exit(0)
def processSquare(x, y):
global hits
if board[x][y] is not None:
return
sys.stdout.write(LETTERS[x] + str(9-y)+"\n")
sys.stdout.flush()
answer = raw_input()
if answer[0] is "x":
board[x][y] = "x"
if not (x-1 < 0):
if not (y-1 < 0):
board[x-1][y-1] = "o"
if not (y+1 > 9):
board[x-1][y+1] = "o"
if not (x+1 > 9):
if not (y-1 < 0):
board[x+1][y-1] = "o"
if not (y+1 > 9):
board[x+1][y+1] = "o"
hits = hits + 1
if hits >= TOTAL_SHIPPARTS:
printBoard()
else:
board[x][y] = "o"
for y in xrange(1,9):
for x in xrange(1,9):
processSquare(x, y)
for y in xrange(10):
for x in xrange(10):
processSquare(x, y)
printBoard()
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