Python: changing element of a list at a random index

Question

I'm working on a simple minesweeper game in python, and am trying to create a function that inserts mines into an empty board.

Note that the board is in the form of a list of lists, as so:

[[col, col, col...],[col,col,col...],[...],[...]...]

my function looks like this:

def insert_mines(NUM_MINES):
for mine in range(NUM_MINES):
    rand_row = randint(0, NUM_ROWS - 1)
    rand_col = randint(0, NUM_COLS - 1)
    if BOARD[rand_row][rand_col] == 'O':
        BOARD[rand_row][rand_col] = 'X'
        NUM_MINES = NUM_MINES - 1
    else:
        BOARD[rand_row][rand_col] = 'X'

print NUM_MINES

when I run it with num_mines = 96, it never prints zero, which is what it should print if all the mines have been inserted.

What could be the cause of this?

note that 'X' represents a mine and 'O' is just a board space with no mine.

By the way, the most uniform way to distribute mines is to put the required number of mines at the front of your grid, then shuffle the whole lot. — C. K. Young
– C. K. Young, Commented Oct 17, 2014 at 4:02
It's not guaranteed that after 96 iterations all empty space will be selected and replaced with mine, since sampling is done uniformly over the grid; i.e. the sampled indices might be repeated. — YS-L
– YS-L, Commented Oct 17, 2014 at 4:06

John La Rooy · Accepted Answer · 2014-10-17 04:26:58Z

1

The only way to have zero printed is if

BOARD[rand_row][rand_col] == 'O'

for every iteration. Depending on the NUM_ROWS and NUM_COLS this may be surprisingly unlikely due to the "birthday paradox"

An easier way to make your board as suggested by Chris

cells = ['O'] * NUM_ROWS * NUM_COLS
cells[:NUM_MINES] = ['X'] * NUM_MINES
random.shuffle(cells)
BOARD = [cells[i::NUM_ROWS] for i in range(NUM_ROWS)]

edited Oct 17, 2014 at 4:26

answered Oct 17, 2014 at 4:18

John La Rooy

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1 Comment

JavascriptLoser Over a year ago

a much more elegant answer than mine. Thank you

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