CCC 2016: Circle of Life

Question

Before I begin, this challenge was not mine originally

Credits to The University of Waterloo. This came from the Canadian Computing Competition 2016, Senior Problem 5. Here is a clickable link to the contest PDF:

http://cemc.uwaterloo.ca/contests/computing/2016/stage%201/seniorEn.pdf

Here is a link to the site:

http://cemc.uwaterloo.ca/contests/past_contests.html

Challenge

Given a wrapping array of two constant values, determine the configuration after n evolutions for positive integer input n. These two values represent a living cell and a dead cell. Evolutions work like this:

Evolution!

After each iteration, a cell is alive if it had exactly one living neighbor in the previous iteration. Any less and it dies of loneliness; any more and it dies of overcrowding. The neighbourhood is exclusive: i.e. each cell has two neighbours, not three.

For example, let's see how 1001011010 would evolve, where 1 is a living cell and 0 is a dead cell.

(0) 1 0 0 1 0 1 1 0 1 0 (1)
    *   $         %

The cell at the * has a dead cell on both sides of it so it dies of lonliness.
The cell at the $ has a living cell on one side of it and a dead cell on the other. It becomes alive.
The cel at the % has a living cell on both sides of it so it stays dead from overcrowding.

Winning Criteria

Shortest code wins.

I/O

Input will be a list of the cell states as two consistent values, and an integer representing the number of inputs, in some reasonable format. Output is to be a list of the cell states after the specified number of iterations.

Test Cases

start, iterations -> end
1001011010, 1000 -> 1100001100
100101011010000, 100 -> 000110101001010
0000000101011000010000010010001111110100110100000100011111111100111101011010100010110000100111111010, 1000 -> 1001111111100010010100000100100100111010010110001011001101010111011011011100110110100000100011011001

Test Case
This test case froze hastebin and exceeded the size limit on pastebin

Answer 1

APL (Dyalog),14 bytes

Prompts for start state as Boolean list and then for number of iterations

(1∘⌽≠¯1∘⌽)⍣⎕⊢⎕

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⎕ numeric prompt (for Boolean list of start state)

⊢ on that, apply

(…)⍣⎕ the following tacit function, numeric-prompt times

 ¯1∘⌽ the argument rotated one step right

 ≠ different from (XOR)

 1∘⌽ the argument rotated one step left

Answer 2

Jelly, 7 bytes

ṙ2^ṙ-µ¡

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Extra Test Case (footer for formatting).

Explanation

ṙ2^ṙ-µ¡
     µ¡  - repeat a number of times equal to input 2:
ṙ2         - previous iteration rotated 2 to the left
  ^        - XOR-ed with:
           - (implicit) previous iteration
   ṙ-      - rotate back (by negative 1 to the left)

Answer 3

05AB1E, 6 bytes

FDÀÀ^Á

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Explanation

F        # input_1 times do
 D       # duplicate last iteration (input_2 the first iteration)
  ÀÀ     # rotate left twice
    ^    # XOR
     Á   # rotate right