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Program to find next state of next cell matrix state in Python?
Suppose we have a 2D binary matrix where a 1 means a live cell and a 0 means a dead cell. A cell's neighbors are its immediate horizontal, vertical and diagonal cells. We have to find the next state of the matrix using these rules
Any living cell with two or three living neighbors lives.
Any dead cell with three living neighbors becomes a live cell.
All other cells die.
So, if the input is like
1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 |
0 | 1 | 0 | 1 |
1 | 1 | 0 | 1 |
then the output will be
1 | 1 | 0 | 0 |
0 | 1 | 0 | 0 |
0 | 1 | 0 | 0 |
1 | 1 | 0 | 0 |
To solve this, we will follow these steps:
n := row size of matrix, m := column size of matrix
res := a matrix of size n x m, and fill with 0
for i in range 0 to n, do
for j in range 0 to m, do
s := 0
if matrix[i, j] is same as 0, then
for k in range i - 1 to i + 1, do
or h in range j - 1 to j + 1, do
if 0 <= k < n and 0 <= h < m, then
s := s + matrix[k, h]
res[i, j] := [0, 1, true when s is same as 3]
otherwise,
for k in range i - 1 to i + 1, do
for h in range j - 1 to j + 1, do
if 0 <= k < n and 0 <= h < m, then
s := s + matrix[k, h]
if s is either 3 or 4, then
res[i, j] := 1
return res
Let us see the following implementation to get better understanding:
Example
class Solution: def solve(self, matrix): n, m = len(matrix), len(matrix[0]) res = [[0 for j in range(m)] for i in range(n)] for i in range(n): for j in range(m): s = 0 if matrix[i][j] == 0: for k in range(i - 1, i + 2): for h in range(j - 1, j + 2): if 0 <= k < n and 0 <= h < m: s += matrix[k][h] res[i][j] = [0, 1][s == 3] else: for k in range(i - 1, i + 2): for h in range(j - 1, j + 2): if 0 <= k < n and 0 <= h < m: s += matrix[k][h] if s in [3, 4]: res[i][j] = 1 return res ob = Solution() matrix = [ [1, 1, 0, 0], [0, 1, 0, 0], [0, 1, 0, 1], [1, 1, 0, 1] ] print(ob.solve(matrix))
Input
[[1, 1, 0, 0], [0, 1, 0, 0], [0, 1, 0, 1], [1, 1, 0, 1] ]
Output
[[1, 1, 0, 0], [0, 1, 0, 0], [0, 1, 0, 0], [1, 1, 0, 0]]