Bouncer: Difference between revisions

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[[File:1RB0LB 1LA0RA.png|alt=1RB0LB_1LA0RA|thumb|A close-up of the bouncer {{TM|1RB0LB_1LA0RA}} with 2 states, the smallest number of states for which a bouncer can appear.]]
[[File:1RB0LB 1LA0RA.png|alt=1RB0LB_1LA0RA|thumb|A close-up of the bouncer {{TM|1RB0LB_1LA0RA}} with 2 states, the smallest number of states for which a bouncer can appear.]]{{Stub}}
A '''bouncer''' is a [[non-halting Turing machine]] whose tape head, roughly speaking, alternates back and forth between the two edges of the tape in a linear fashion, growing the tape along one or both edges with each iteration.
A '''bouncer''' is a [[non-halting Turing machine]] whose tape head, roughly speaking, alternates back and forth between the two edges of the tape in a linear fashion, growing the tape along one or both edges with each iteration.


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=== Analysis ===
=== Analysis ===
'''Transcript''': (A0 (B0 A1)<sup>2n+1</sup> B0 (A0 B1)<sup>2n+2</sup>)<sup>n≥0</sup>.
'''[[Transcript]]''': (A0 (B0 A1)<sup>2n+1</sup> B0 (A0 B1)<sup>2n+2</sup>)<sup>n≥0</sup>.


The derivation of this transcript is as follows.
The derivation of this transcript is as follows.
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[[Category:Zoology]]
[[Category:Zoology]]
[[Category:Stub]]

Latest revision as of 22:29, 10 August 2025

1RB0LB_1LA0RA
A close-up of the bouncer 1RB0LB_1LA0RA (bbch) with 2 states, the smallest number of states for which a bouncer can appear.

A bouncer is a non-halting Turing machine whose tape head, roughly speaking, alternates back and forth between the two edges of the tape in a linear fashion, growing the tape along one or both edges with each iteration.

Example

1RB0LB_1LA0RA (bbch) is an example of a bouncer, and its spacetime diagram is shown in the picture on the right.

Analysis

Transcript: (A0 (B0 A1)2n+1 B0 (A0 B1)2n+2)n≥0.

The derivation of this transcript is as follows.

Shift rules:

  • B0 A1: B 10< →[2] B <01,
  • A0 B1: A >10 →[2] A 01>.

Bounce rule:

  • Bn := A0 (B0 A1)2n+1 B0 (A0 B1)2n+2: A 00(102n)>00 →[8n+8] A (102(n+1))>

In particular, we have

Bn: A 0(102n)>0 →[8n+8] 0(102(n+1))>0,

which by induction proves that the transcript of 1RB0LB_1LA0RA on the all-zeros tape is (Bn)n≥0. Moreover, we have shown that there are 8n + 8 steps between the nth and the (n+1)st bounce.

See also

  • Section 7 of bbchallenge's deciders write-up.