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Background

For the purposes of this challenge, an n-state cellular automaton is simply a binary function f that takes two numbers from the state set {0, 1, ..., n-1} as inputs, and returns another number from that set as output. It can be applied to a list of numbers L = [x0, x1, x2, ..., xk-1] of length at least 2 by

f(L) = [f(x0, x1), f(x1, x2), f(x2, x3), ..., f(xk-2, xk-1)]

Note that the resulting list has one fewer element than the original. A spacetime diagram of f starting from L is the list of lists obtained by repeatedly applying f to L, and collecting the results in a list. The final list has length 1. We say that the list L is an identifying sequence for f, if every two-element list over the state set is a contiguous sublist of some row of the spacetime diagram starting from L. This is equivalent to the condition that no other n-state CA has that exact spacetime diagram.

Input

Your inputs are an n-by-n integer matrix M, a list of integers L of length at least 2, and optionally the number n. The matrix M defines an n-state CA f by f(a,b) = M[a][b] (using 0-based indexing). It is guaranteed that n > 0, and that M and L only contain elements of the state set {0, 1, ..., n-1}.

Output

Your output shall be a consistent truthy value if L is an identifying sequence for the CA f, and a consistent falsy value otherwise. This means that all "yes"-instances result in the same truthy value, and all "no"-instances result in the same falsy value.

Example

Consider the inputs n = 2, M = [[0,1],[1,0]], and L = [1,0,1,1]. The matrix M defines the binary XOR automaton f(a,b) = a+b mod 2, and the spacetime diagram starting from L is

1 0 1 1
1 1 0
0 1
1

This diagram does not contain 0 0 on any row, so L is not an identifying sequence, and the correct output is False. If we input L = [0,1,0,0] instead, the spacetime diagram is

0 1 0 0
1 1 0
0 1
1

The rows of this diagram contain all pairs drawn from the state set, namely 0 0, 0 1, 1 0 and 1 1, so L is an identifying sequence and the correct output is True.

Rules

You can write a full program or a function. The lowest byte count wins, and standard loopholes are disallowed.

Test Cases

Trivial automaton
[[0]] [0,0] 1 -> True
Binary XOR
[[0,1],[1,0]] [1,0,1,1] 2 -> False
[[0,1],[1,0]] [1,0,1,0] 2 -> True
[[0,1],[1,0]] [0,1,0,0] 2 -> True
Addition mod 3
[[0,1,2],[1,2,0],[2,0,1]] [0,1,1,0,0,0,1,0,0] 3 -> False
[[0,1,2],[1,2,0],[2,0,1]] [0,1,1,0,0,0,1,0,1] 3 -> True
Multiplication mod 3
[[0,0,0],[0,1,2],[0,2,1]] [0,1,1,2,0,0,1,0,1] 3 -> False
[[0,0,0],[0,1,2],[0,2,1]] [0,1,1,2,2,2,1,0,1] 3 -> True
Some 4-state automata
[[3,2,2,1],[0,0,0,1],[2,1,3,1],[0,1,2,3]] [0,0,0,0,1,1,1,1] 4 -> False
[[3,2,2,1],[0,0,0,1],[2,1,3,1],[0,1,2,3]] [0,0,0,1,0,1,1,1] 4 -> False
[[3,2,2,1],[0,0,0,1],[2,1,3,1],[0,1,2,3]] [0,1,2,3,3,1,2,3,0] 4 -> True
[[0,1,2,1],[1,0,2,0],[2,2,1,0],[1,2,0,0]] [0,0,1,1,2,2,0,2,1] 4 -> False
[[0,1,2,1],[1,0,2,0],[2,2,1,0],[1,2,0,0]] [0,3,1,3,2,3,3,0,1] 4 -> False
[[0,1,2,1],[1,0,2,0],[2,2,1,0],[1,2,0,0]] [0,3,1,3,2,3,3,0,1,2] 4 -> True
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CJam, 53 43 42 bytes

l~:M;_,({_[\1>]zW<_{M\{=}/}%}*;](_*\L*_&,=

This is a very straight-forward implementation of the definition (I took some inspiration from Jakube after my first attempt). It expects input in reverse order on STDIN, using CJam-style arrays:

2 [1 0 1 1] [[0 1][1 0]]

Here is a test harness which runs the code against all inputs (converting them to the correct input format first). The results in the input field aren't actually used. Remove them if you don't trust me. ;)

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5
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Python 2: 93 byte

M,L,n=input();a=[]
while L:z=zip(L,L[1:]);a+=z;L=[M[i][j]for i,j in z]
print len(set(a))==n*n

Straightforward implementation: Find all pairs by zipping, memorize them for later and apply M to L. Repeat. Compare the number of unique pairs found.

Input is of the form [[0,1],[1,0]], [0,1,0,0], 2.

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2
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Mathematica, 90 83 82 bytes

f=Length[Union@@Last@Reap[#2//.l_List:>Extract[#,Sow/@Partition[l+1,2,1]]]]==#3^2&

Another straight-forward implementation.

Usage:

f[{{0, 1}, {1, 0}}, {0, 1, 0, 0}, 2]

True

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