Problem 1970. Kaggle: Reverse Game of Life - Periods of Oscillators

Created by Richard Zapor

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<a href = "http://www.kaggle.com/c/conway-s-reverse-game-of-life">Kaggle's Conway's Reverse Game of Life</a> contest inspires this Period of Life challenge. The kaggle contest runs from Oct-14-2013 thru Mar-02-2014. References: <a href = "http://mathworld.wolfram.com/GameofLife.html">Game of Life at Wolfram</a>. <a href = "http://en.wikipedia.org/wiki/Conway%27s_Game_of_Life">Wiki Life</a>.<pre class="language-matlab">1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by overcrowding.
4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.
5. No wrap around. Beyond edge is zero. Eight Neighbors.
</pre>Determine the period of life for a given binary matrix. A stable configuration, Still Life, has a cycle of 1.Input: M , an [m,n] arrayOutput: N, Period of Life cycle (Period &lt;11)Examples: A few matrices of varying periods<pre class="language-matlab">N=1 N=2 N=3 Caterer
0000 00000 0000000000
0110 00000 0001000000
0110 01110 0100011110
0000 00000 0100010000
 00000 0100000000
 0000100000
 0011000000
 0000000000
</pre>Additional References: <a href = "http://www.conwaylife.com/wiki/Oscillator">Oscillators</a>, <a href = "http://www.conwaylife.com/wiki/Still_life">Still Life</a>

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Last Solution submitted on Oct 02, 2020

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Rafael S.T. Vieira on 4 Jul 2020

I recommend seeing this: https://en.wikipedia.org/wiki/Oscillator_(cellular_automaton)
In this problem, all he wants is how many steps are necessary for the automaton to repeat itself.

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Problem 1970. Kaggle: Reverse Game of Life - Periods of Oscillators

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