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Find height, width of the largest rectangle containing all 0's in the matrix

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dp.py
Python
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#!/usr/bin/env python
"""Find height, width of the largest rectangle containing all 0's in the matrix.
 
The algorithm for `max_size()` is suggested by @j_random_hacker [1].
The algorithm for `max_rectangle_size()` is from [2].
The Python implementation [3] is under CC BY-SA 3.0
(if you need other license, e-mail me)
 
[1]: http://stackoverflow.com/questions/2478447/find-largest-rectangle-containing-only-zeros-in-an-nn-binary-matrix#comment5169734_4671342
 
[2]: http://blog.csdn.net/arbuckle/archive/2006/05/06/710988.aspx
 
[3]: http://stackoverflow.com/a/4671342
"""
from collections import namedtuple
from operator import mul
 
try:
reduce = reduce
except NameError:
from functools import reduce # py3k
 
Info = namedtuple('Info', 'start height')
 
def max_size(mat, value=0):
"""Find height, width of the largest rectangle containing all `value`'s.
 
For each row solve "Largest Rectangle in a Histrogram" problem [1]:
 
[1]: http://blog.csdn.net/arbuckle/archive/2006/05/06/710988.aspx
"""
it = iter(mat)
hist = [(el==value) for el in next(it, [])]
max_size = max_rectangle_size(hist)
for row in it:
hist = [(1+h) if el == value else 0 for h, el in zip(hist, row)]
max_size = max(max_size, max_rectangle_size(hist), key=area)
return max_size
 
def max_rectangle_size(histogram):
"""Find height, width of the largest rectangle that fits entirely under
the histogram.
 
>>> f = max_rectangle_size
>>> f([5,3,1])
(3, 2)
>>> f([1,3,5])
(3, 2)
>>> f([3,1,5])
(5, 1)
>>> f([4,8,3,2,0])
(3, 3)
>>> f([4,8,3,1,1,0])
(3, 3)
>>> f([1,2,1])
(1, 3)
 
Algorithm is "Linear search using a stack of incomplete subproblems" [1].
 
[1]: http://blog.csdn.net/arbuckle/archive/2006/05/06/710988.aspx
"""
stack = []
top = lambda: stack[-1]
max_size = (0, 0) # height, width of the largest rectangle
pos = 0 # current position in the histogram
for pos, height in enumerate(histogram):
start = pos # position where rectangle starts
while True:
if not stack or height > top().height:
stack.append(Info(start, height)) # push
elif stack and height < top().height:
max_size = max(max_size, (top().height, (pos - top().start)),
key=area)
start, _ = stack.pop()
continue
break # height == top().height goes here
 
pos += 1
for start, height in stack:
max_size = max(max_size, (height, (pos - start)), key=area)
 
return max_size
 
def area(size):
return reduce(mul, size)
 
import unittest
class TestCase(unittest.TestCase):
def test(self):
self.assertEqual(max_size(self.__s2m("""
0 0 0 0 1 0
0 0 1 0 0 1
0 0 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 1
0 0 1 0 0 0""")), (3, 4))
self.assertEqual(max_size([[1, 1], [0, 0]]), (1, 2))
self.assertEqual(max_size([[0, 0], [1, 1]]), (1, 2))
self.assertEqual(max_size([[1, 0], [1, 0]]), (2, 1))
self.assertEqual(max_size([[0, 1], [0, 1]]), (2, 1))
self.assertEqual(max_size(self.__s2m("""
0 0 0 0 1 0
0 0 1 0 0 1
0 0 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 1
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 0 0""")), (7, 2))
self.assertEqual(max_size([[]]), (0, 0))
self.assertEqual(max_size([]), (0, 0))
self.assertEqual(max_size(self.__s2m("""
0 0 0 0 1 0
0 0 1 0 0 1
0 0 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 0
0 0 1 0 0 1
0 0 0 0 0 0
0 0 0 0 0 0""")), (3, 5))
self.assertEqual(max_size(self.__s2m("""
0 0 0 0 1 0
0 0 0 0 0 0
0 0 1 0 0 1
0 0 0 0 0 0
1 0 0 0 0 0
0 0 0 0 0 0
0 0 1 0 0 1
0 0 0 0 0 0
0 0 0 0 0 1""")), (8, 2))
self.assertEqual(max_size(self.__s2m("""
0 0 0 0 1 1 1
0 0 0 0 0 0 0
0 0 0 1 1 1 1
0 0 1 1 1 1 1
1 0 1 1 1 1 1
1 0 1 1 1 1 1
1 0 1 1 1 1 1
""")), (3, 3))
 
def __s2m(self, s):
return [map(int, line.split())
for line in s.splitlines() if line.strip()]
 
if __name__=="__main__":
import unittest; unittest.main()

I got the rather surprising result with the following test case:

print zheights 
array([[0, 0, 0, 0, 0, 0, 0, 0, 3, 2],
   [0, 4, 0, 2, 4, 0, 0, 1, 0, 0],
   [1, 0, 1, 0, 0, 0, 3, 0, 0, 4],
   [0, 0, 0, 0, 4, 2, 0, 0, 0, 0],
   [0, 0, 0, 2, 0, 0, 0, 0, 1, 0],
   [4, 3, 0, 0, 1, 2, 0, 0, 0, 0],
   [3, 0, 0, 0, 2, 0, 0, 0, 0, 4],
   [0, 0, 0, 1, 0, 3, 2, 4, 3, 2],
   [0, 3, 0, 0, 0, 2, 0, 1, 0, 0],
   [0, 0, 2, 0, 0, 0, 0, 1, 0, 0]])

max_size(zheights, value=0)
(True, 8)

I added this to the bottom of the max_size function:

row_length = np.sum(max_size[0])
column_length = np.sum(max_size[1])
return (row_length, column_length)

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