Question

*Note i refer to thing as a matrix but it is not, it is just a collection of 1’s and zeros

Suppose you have a matrix that is always square (n x n). Is it possible to determine if there exists a single column/row/diagonal such that each item is a 1.

Take the matrix below for instance (True):

1 0 0
1 1 0
0 0 1

Another example (True):

1 1 1
0 0 0
0 1 0

And finally one without a solution (False):

0 1 1
1 0 0
0 0 0

Notice how there is a diagonal filled with 1’s. The rule is there is either there is a solution or there is no solution. There can be any number of 1’s or zeros within the matrix. All i really need to do is, if you have (n x n) then there should be a row/column/diagonal with n elements the same.

If this is not possible with recursions, please let me know what is the best and most efficient method. Thanks a lot, i have been stuck on this for hours so any help is appreciated (if you could post samples that would be great).

EDIT
This is one solution that i came up with but it gets really complex after a while.

Take the first example i gave and string all the rows together so you get:

1 0 0, 1 1 0, 0 0 1
Then add zeros between the rows to get:
1 0 0 0, 1 1 0 0, 0 0 1 0

Now if you look closely, you will see that the distances between the 1’s that form a solution are equal. I dont know how this can be implemented though.

Answer 1

In search for an elegant solution I came up with this:

class LineOfOnesChecker(object):

    _DIAG_INDICES = (lambda i: i, lambda i: -i - 1)

    def __init__(self, matrix):
        self._matrix = matrix
        self._len_range = range(len(self._matrix))

    def has_any(self):
        return self.has_row() or self.has_col() or self.has_diag()

    def has_row(self):
        return any(all(elem == 1 for elem in row)
                   for row in self._matrix)

    def has_col(self):
        return any(all(self._matrix[i][j] == 1 for i in self._len_range)
                   for j in self._len_range)

    def has_diag(self):
        return any(all(self._matrix[transf(i)][i] == 1 for i in self._len_range)
                   for transf in self._DIAG_INDICES)

Usage:

print LineOfOnesChecker(matrix).has_any()