Day 14: Find 2D pattern in matrix

day14/2d-pattern-in-matrix.py

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+########
+# Code #
+########
+
+# pattern and matrix are both 2D arrays
+# return true as soon as the pattern is found for a first time
+def patternInMatrix(pattern, matrix):
+    rows, cols = len(matrix), len(matrix[0])
+    patternRows, patternCols = len(pattern), len(pattern[0])
+    for row in xrange(rows-patternRows):
+        for col in xrange(cols-patternCols):
+            found = False
+            for pRow in xrange(patternRows):
+                for pCol in xrange(patternCols):
+                    if pattern[pRow][pCol] != matrix[row+pRow][col+pCol]:
+                        found = False
+                        break
+                    else:
+                        found = True
+            if found:
+                return True
+    return False
+
+# return a list of all top-left starting indices
+def indicesOfPatternInMatrix(pattern, matrix):
+    rows, cols = len(matrix), len(matrix[0])
+    patternRows, patternCols = len(pattern), len(pattern[0])
+
+    startingSpots = []
+    for row in xrange(rows-patternRows+1):
+        for col in xrange(cols-patternCols+1):
+            found = False
+            for pRow in xrange(patternRows):
+                for pCol in xrange(patternCols):
+                    if pattern[pRow][pCol] != matrix[row+pRow][col+pCol]:
+                        found = False
+                        break
+                    else:
+                        found = True
+            if found:
+                startingSpots.append((row, col))
+    return startingSpots
+
+#########
+# Tests #
+#########
+
+def testPatterInMatrix():
+    assert patternInMatrix([[1]], [[0,0,0],[0,1,0],[0,0,0]])
+    assert not patternInMatrix([[2]], [[0,0,0],[0,1,0],[0,0,0]])
+
+def testIndicesOfPatternInMatrix():
+    p0 = [[1]]
+
+    m0 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]]
+    assert indicesOfPatternInMatrix(p0, m0) == [(1, 1)]
+
+    p1 = [[2]]
+
+    m1 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]]
+    assert indicesOfPatternInMatrix(p1, m1) == []
+
+    p2 = [[0,1,0]]
+
+    m2 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]]
+    assert indicesOfPatternInMatrix(p2, m2) == [(1, 0)]
+
+    p3 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]] 
+
+    m3 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]] 
+    assert indicesOfPatternInMatrix(p3, m3) == [(0,0)]
+
+    p4 = [[0]]
+
+    m4 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]] 
+    assert indicesOfPatternInMatrix(p4, m4) == [(0,0),(0,1),(0,2),
+                                                (1,0),(1,2),
+                                                (2,0),(2,1),(2,2)]
+
+    p4 = [[0, 0]]
+
+    m4 = [[0,0,0],
+          [0,1,0],
+          [0,0,0]] 
+    assert indicesOfPatternInMatrix(p4, m4) == [(0,0),(0,1),
+                                                (2,0),(2,1)]
+
+def main():
+    testPatterInMatrix()
+    testIndicesOfPatternInMatrix()
+
+if __name__ == "__main__":
+    main()

day14/README.md

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+Question of the day: https://www.careercup.com/question?id=5642834636963840
+
+Given a 2D array of digits, try to find the occurence of a given 2D
+pattern of digits. For example, consider the following 2D matrix:
+
+```python
+7283455864  
+6731158619  
+8988242643  
+3839505324  
+9509505813  
+3843845384  
+6473530293  
+7053106601  
+0834282956  
+4607924137  
+```
+
+Assume we need to look for the following 2D pattern:
+
+```python
+950  
+384  
+353  
+```
+
+Return a list of the top left indices of an occurence of the pattern in
+the 2D matrix. If there are multiple occurences, return all of the
+possible indices.
+
+## Ideas
+
+I'm going to make some assumptions to get myself started. I can explore
+these assumptions more in the follow-up section later on.
+
+1. We don't care about finding rotations of the 2D pattern inside the matrix.
+2. We don't care about finding parts of the 2D patten inside the matrix
+3. The dimensions of the 2D pattern are always less than or equal to the respective dimensions of the matrix.
+
+A brute force solution would be to iterate through all possible locations
+in the matrix and check if the 2D pattern matches starting at that
+location. Let's define `width` and `height` to be the width and height
+of the matrix. It would take `O(width*height)` time just to go through
+all the possible starting locations. Let's define `pWidth` and `pHeight`
+to be the width and height of the 2D pattern. So then if the pattern
+matches, there would be an additional `O(pWidth*pHeight)` per location
+with a possible start point, which excludes parts of the matrix. In the
+worst case, the runtime to this brute force solution is
+`O((width - pWidth + 1) * (height - pHeight + 1) * pWidth * pHeight)`. I do a
+subtraction which is key, because in a matrix like this:
+
+```python
+11111  
+11111  
+11111  
+11111  
+11111  
+```
+
+with a 2D pattern like this:
+
+```python
+1111  
+1111  
+1111  
+1111  
+```
+
+there are only 4 possible starting locations: `(0, 0), (0, 1), (1,0), (1,1)`.
+
+I don't think it's possible to do this problem and faster. So it's not
+a very complex problem, now I just have to write the code.
+
+## Code
+
+[Python](./2D-pattern-in-matrix.py)
+
+## Follow up
+
+@jjwon0 made a great [observation](https://github.com/alberthu16/100-day-coding-challenge/pull/9#discussion_r113083672)
+that the Knuth-Morris-Pratt linear time string matching algorithm may
+be applied to 2D pattern matching to make this more efficient.
+
+Read the [wikipedia](https://en.wikipedia.org/wiki/Knuth%E2%80%93Morris%E2%80%93Pratt_algorithm)
+article on the algorithm, and then read this guy's [blog post](http://jakeboxer.com/blog/2009/12/13/the-knuth-morris-pratt-algorithm-in-my-own-words/)
+on what's going on with the preprocessing step.