Linear Cellular Automata and the Garden-of-Eden

Abstract

Suppose each of the squares of an n x n chessboard is equipped with an indicator light and a button. If the button of a square is pressed, the light of that square will change from off to on and vice versa; the same happens to the lights of all the edge-adjacent squares. Initially all lights are off. Now, consider the following question: is it possible to press a sequence of buttons in such a way that in the end all lights are on? We will refer to this problem as the All-Ones Problem. A moment 's reflection will show that pressing a button twice has the same effect as not pressing it at all. Thus a solution to our problem can be described by a subset of all squares (namely a set of squares whose buttons when pressed in an arbitrary order will render all lights on) rather than a sequence. In fact a set X of squares is a solution to the All-Ones Problem if and only if for every square s the number of squares in X adjacent to or equal to s is odd. Consequently, we will call such a set an odd-parity cover. Trial and error in conjunction with a pad of graph paper will readily produce solutions for n ~ 4. A little more experimentation shows that an odd-parity cover shou ld one exist--is difficult to construct even for n = 5 o r 6 . The brute-force approach to the problem, namely exhaustive search over all subsets of {1 . . . . n} x {1 . . . . n}, presents 2 n2 candidates, and the search becomes infeasible for moderate values of n even with the help of a computer. A less brute-force method would be to try to solve the system terpreted as a matrix over GF(2)) and 1 is the vector with all components equal to 1. This method, which involves n 2 equations, again becomes unwieldy for small values of n. For a similar approach to a game related to the All-Ones Problem, see [3]. In any case, Figure 1 shows odd-parity covers for n = 4, 5, 8. Several questions come to mind. For which n does a solution to the All-Ones Problem exist? More generally, how many odd-parity covers are there for an n x n board? What happens if the adjacency condition is changed--say , to an octal array (where a cell in the center has eight neighbors)? Can one replace an n x n rectangular grid by some other arrangement of sites and still obtain a solution? To answer some of these questions, we first rephrase the problem in terms of cellular automata.

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@inproceedings{Sutner2009LinearCA, title={Linear Cellular Automata and the Garden-of-Eden}, author={Klaus Sutner}, year={2009} }