Pebble Automaton: Difference between revisions

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|1
|1
|<code><nowiki>1 --> 0 | 1 | 0</nowiki></code>
|<code><nowiki>1 --> 0 | 1 | 0</nowiki></code>
<code>2 --> 0 | 1 | 1</code>
<code>2 --> 0 | 1 | 1</code>
|-
|-
Line 33: Line 34:
|3
|3
|<code><nowiki>1 --> 0 | 1 | 0</nowiki></code>
|<code><nowiki>1 --> 0 | 1 | 0</nowiki></code>
<code>2 --> 1 | 1 | 0</code>
<code>2 --> 1 | 1 | 0</code>
<code>3 --> 0 | 1 | 2</code>
<code>3 --> 0 | 1 | 2</code>
|-
|4
|≥ 6
|<code><nowiki>1 --> 0 | 1 | 0</nowiki></code>
<code>2 --> 0 | 1 | 1</code>
<code>3 --> 2 | 1 | 0</code>
<code>4 --> 0 | 1 | 3</code>
|}
|}


== Sources ==
== Sources ==
Discord thread: https://discord.com/channels/960643023006490684/1535695107448381591
Discord thread: https://discord.com/channels/960643023006490684/1535695107448381591

Revision as of 08:02, 11 August 2026

A pebble automaton is defined by two integer functions L(n) and R(n), defined over positive integers n, under the constraints:

  • L(n) >= 0
  • R(n) >= 0
  • L(n) + R(n) <= n for all n

p(x, t) represents the number of pebbles at position x and time t.

  • p(0, 0) = n
  • p(x, 0) = 0 for x != 0
  • p(x, t+1) = R(p(x-1, t)) + L(p(x+1, t)) + (p(x,t) - L(p(x,t)) - R(p(x,t)))

Informally, when a cell has n pebbles, it pushes L(n) to its left neighbor, and R(n) to its right neighbor.

Busy Beaver function

Let define peBBle(n) as the maximum number of steps it takes for any pebble automaton to stabilize, starting from an initial state of n pebbles in a single cell.

n Value Champion
1 0 1 --> 0 | 1 | 0
2 1 1 --> 0 | 1 | 0

2 --> 0 | 1 | 1

3 3 1 --> 0 | 1 | 0

2 --> 1 | 1 | 0

3 --> 0 | 1 | 2

4 ≥ 6 1 --> 0 | 1 | 0

2 --> 0 | 1 | 1

3 --> 2 | 1 | 0

4 --> 0 | 1 | 3

Sources

Discord thread: https://discord.com/channels/960643023006490684/1535695107448381591