Skelet 33: Difference between revisions
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Added see also section |
Added missing definition of b(n) and see also section |
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<pre> | <pre> | ||
b(n) is the number we have to add to n to reach the “next” power of 2 (strictly greater than n) or the smallest number which when added to n “expands” the binary representation to a larger length. | |||
L(2k) = L(k) 0000 | L(2k) = L(k) 0000 | ||
L(2k+1) = L(k) 0001 | L(2k+1) = L(k) 0001 | ||
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== See also == | == See also == | ||
* [[Skelet 26]] | * [[Skelet 26]] | ||
* [https://www.sligocki.com/2023/02/02/skelet-34.html Skelet #34 is Infinite] | |||
== References == | == References == | ||
[[Category:BB(5)]] | [[Category:BB(5)]] | ||
Latest revision as of 19:18, 7 March 2026
1LC1RD_1RE---_0LD0LC_1RB0RA_1RA1LE (bbch), called Skelet #33, was one of Skelet's 43 holdouts and one of the last holdouts in BB(5). It is a Shift overflow counter and has an individual proof of non-halting in Coq-BB5.[1] It is adjacent to Skelet 34.[2]
Skelet 33 was formally proven to be non-halting in December 2023 by int-y1[3] and meithecatte.[4]
Analysis by Shawn Ligocki[2]
b(n) is the number we have to add to n to reach the “next” power of 2 (strictly greater than n) or the smallest number which when added to n “expands” the binary representation to a larger length.
L(2k) = L(k) 0000
L(2k+1) = L(k) 0001
G(n, m) = L(n) <A 010 R(m)
D(n, m) = L(n) 000 <A 010 R(m)
G(n, m) -> G(n+1, m+1) (if b(m) > 1)
L(n) <A -> L(n+1) B>
0001 <A -> <A 1111
0000 <A -> 0001 B>
E> R(n) -> <C R(n+1)
E> 11 -> 11 E>
E> 10 -> <C 11
11 <C -> <C 10
B> 010 -> 111 E>
111 <C -> <A 010
G(2n, 2^k - 1) -> D(n, 2^k)
0 <A 010 11^k 0 -> <A 010 10^k 11
D(n, m) -> D(n', 0, 2 m' + 1) (if b(m) > 2n) (n' < n)
1000 0000^k <A 010 R(m) -> 1001 0001^k <A 010 R(m + 2^{k+1} - 1)
-> <A 010 11 10 R(m + 2^{k+1} - 1)
Start -> G(0, 0, 0, 13) @ Step 83