Bouncer: Difference between revisions

From BusyBeaverWiki
Jump to navigation Jump to search
(Used Template:Stub)
 
(3 intermediate revisions by 2 users not shown)
Line 1: Line 1:
[[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 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 possible classification of [[Non-halting Turing machine|non-halting Turing machines]].
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 ==
== Example ==
Line 6: Line 6:


=== Analysis ===
=== Analysis ===
'''[[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.
'''Shift rules:'''
'''Shift rules:'''


Line 15: Line 19:
* '''B'''<sub>n</sub> := A0 (B0 A1)<sup>2n+1</sup> B0 (A0 B1)<sup>2n+2</sup>: <code>A 00(10<sup>2n</sup>)>00 →[8n+8] A (10<sup>2(n+1)</sup>)></code>
* '''B'''<sub>n</sub> := A0 (B0 A1)<sup>2n+1</sup> B0 (A0 B1)<sup>2n+2</sup>: <code>A 00(10<sup>2n</sup>)>00 →[8n+8] A (10<sup>2(n+1)</sup>)></code>


In particular, we have
In particular, we have<blockquote>'''B'''<sub>n</sub>: <code>A 0<sup>∞</sup>(10<sup>2n</sup>)>0<sup>∞</sup> →[8n+8] 0<sup>∞</sup>(10<sup>2(n+1)</sup>)>0<sup>∞</sup></code>,</blockquote>which by induction proves that the transcript of <code>1RB0LB_1LA0RA</code> on the all-zeros tape is ('''B'''<sub>n</sub>)<sup>n≥0</sup>. Moreover, we have shown that there are 8''n'' + 8 steps between the ''n''<sup>th</sup> and the (''n''+1)<sup>st</sup> bounce.
 
'''B'''<sub>n</sub>: <code>A 0<sup>∞</sup>(10<sup>2n</sup>)>0<sup>∞</sup> →[8n+8] 0<sup>∞</sup>(10<sup>2(n+1)</sup>)>0<sup>∞</sup></code>,
 
which by induction proves that the transcript of <code>1RB0LB_1LA0RA</code> on the all-zeros tape is ('''B'''<sub>n</sub>)<sup>n≥0</sup>. This shows that <code>1RB0LB_1LA0RA</code> is a bouncer. Moreover, we have shown that there are 8''n'' + 8 steps between the ''n''<sup>th</sup> and the (''n''+1)<sup>st</sup> bounce.


== See also ==
== See also ==
Line 26: Line 26:


[[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.