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Analysis of the Oct 2000 Marxen-Buntrock BB6 Record-Setter 'q'
References:
Heiner Marxen, table of record setters
Heiner Marxen, table of BB(6) candidates found by a search in 2000
Robert Munafo, Detailed Analysis of the Oct 2000 Marxen-Buntrock BB(6) Record-Setter 'q'.
Ligocki and Kropitz 2022 BB(6) Candidates
In May 2022 a series of BB(6) candidates were announced, raising the record four times in succession, to 6.0×1039456, 1.7×10646,456,993, 1010101.6926214×1020823, and 10↑↑15.
Shawn Ligocki's BB(6) Candidate "pt4-21k" 3pt3.57331185×1020823
The latter of the 2022 BB(6) machines from Shawn Ligocki has a canonical state looking like this:
canonical(n,m) := ...000[E](0)n100(01)m000...
in which the machine is in state E with the head having just overwritten a 1 with a 0, which has n-1 0's to the right followed by 100 and then m copies of 01. The rules for getting to the next canonical state, given by Ligocki on the 27th May 2022, are:
canonical(3k+0,m) -→ canonical((19×4k-13)/3,m+1)
canonical(3k+1,m) -→ ...000[h]11(011)k00(01)(m+1)000...
canonical(3k+2,m) -→ canonical((28×4k-13)/3,m+1)
To determine when this machine halts, one must iterate a process in which at each step given a k whose remainder modulo 0 is known, one much then compute a value of 4k modulo 9. This becomes difficult once the number of digits is too big to fit in memory. It is necessary ti use methods such as those described in my page on sequence A92188. Based on this type of computation Ligocki determined that the machine terminates with S() and S() bounded by
444434587 < S() < S() < 444434589
This calculation was confirmed by Wietze Koops on the 30th May 2022.
To calculate the all important modulo values for each step (+0, +1, or +2 in the "canonical" expressions above) Koops defined the values n0, n1, n2, etc. and corresponding k0, k1, k2, etc.:
|
Before the conversation had completed on the verification of pt4-21k, Pavel Kropitz announced the machine that came to be called "t15", discussed next.
Pavel Kropitz's BB(6) Candidate "t15" 10↑↑15
On the 30th May 2022, Pavel Kropitz announced to the busy-beaver-discuss group that this machine:
1,_: 2,1,> 1,1: 4,_,<
2,_: 3,1,> 2,1: 6,_,>
3,_: 3,1,< 3,1: 1,1,<
4,_: 5,_,< 4,1: H,1,>
5,_: 6,1,< 5,1: 2,_,>
6,_: 3,_,> 6,1: 5,_,>
halts with exactly this many consecutive 1's on the tape:
((81×(3↑( (27×(3↑( (27×(3↑( (27×(3↑( (27×(3↑(
(27×(3↑( (27×(3↑( (27×(3↑( (27×(3↑( (27×(3↑(
(27×(3↑( (27×(3↑( (27×(3↑( (27×(3↑( (27×(3↑(
(27×(3↑(1)) - 17)/8)
) - 3)/8)) - 17)/8)) - 1)/8)) - 17)/8)) - 3)/8)
) - 17)/8)) - 1)/8)) - 17)/8)) - 1)/8)) - 17)/8)
) - 3)/8)) - 17)/8)) - 1)/8)) - 19)/8)) - 13)/2)
which is about (10↑)×13(1.26720185×1010565), in which the "(10↑)×13" means 13 repetitions of "10↑". Hereafter I call this pt14-10k signifying a power tower with fourteen 10's and "10k" (about 10,000) on top.
Less than 12 hours later, Shawn Ligocki reported that his "preliminary analysis" shows that this machine's M (number of "marks", i.e. 1's left on the tape) is greater than (3↑)×12(22144) ≈ (10↑)×10(1.12640164×1010565). He notes that this is a shorter power-tower than what Kropitz had found, but shows his analysis of the rules that can be used to infer the progression of the machine through canonical states, allowing computation with suitable modulo arithmetic algorithms. Pascal Michel and Shawn Ligocki worked out a few days later that the larger result is correct.
Further Developments
As Allan Wechsler stated to the SeqFan list early in 2025, There has been quite a lot of movement in aschalocastorology over the past four or five years, and nobody can be blamed for missing an announcement. Much work has been performed in this area, particularly in the use of finite field arithmetic to determine the number of iterations before halting for penrational-growth candidates such as pt4-21k and pt14-10k above.
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