1RB1RZ 0RC0RE 1LD1LA 1LC0LG 0RF1LF 0RD1LF 1LB0LE: Difference between revisions
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m (Fix math tag) |
(Add note about luck and define shorthand notations.) |
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A(a, b) = 0^inf <F 10 1^a 00 1^b 0^inf | A(a, b) = 0^inf <F 10 1^a 00 1^b 0^inf | ||
B(a, b, c, d, e) = 0^inf <F 1^2a+1 00 1^2b 0^2c+1 1^d 00 1^e 0^inf | B(a, b, c, d, e) = 0^inf <F 1^2a+1 00 1^2b 0^2c+1 1^d 00 1^e 0^inf | ||
Shorthand: | |||
A(a) = A(a, 0) | |||
B(b, d, e) = B(0, b, 0, d, e) | |||
f1(x) = 2x+2 | f1(x) = 2x+2 | ||
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Phase A: | Phase A: | ||
Start --(13)--> A(1 | Start --(13)--> A(1) | ||
A(3k | A(3k) -> B(4, 4k-3, 4) | ||
A(3k+1 | A(3k+1) -> A(4k+4) | ||
A(3k+2 | A(3k+2) -> A(4k+6) | ||
Phase B: | Phase B: | ||
B( | B(b, 3k, e) -> B(f1(f2^k(b)), 4k+e, 0) | ||
B( | B(b, 3k+1, e) -> B(f2^{k+1}(b), 4k+e+4, 0) | ||
B( | B(b, 3k+2, e) -> Halt(2 f2^k(b) + 4k + e + 12) | ||
</pre> | </pre> | ||
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If we ignore the <code>f1</code> calls and note that <math>f_2(x) > 2^x</math> then the sigma score at halt is larger than <math>e > f_2^{519}(4) > (2 \uparrow)^{519} 4 > 10 \uparrow\uparrow 519</math>. | If we ignore the <code>f1</code> calls and note that <math>f_2(x) > 2^x</math> then the sigma score at halt is larger than <math>e > f_2^{519}(4) > (2 \uparrow)^{519} 4 > 10 \uparrow\uparrow 519</math>. | ||
It gets very lucky in Phase A: resetting 9 times before hitting a multiple of 3 and a little bit lucky in Phase B: resetting 4 times before hitting a remainder 2 factor. | |||
Revision as of 05:28, 7 June 2025
1RB1RZ_0RC0RE_1LD1LA_1LC0LG_0RF1LF_0RD1LF_1LB0LE (bbch) is a tetrational halting BB(7) TM with sigma score over . It was found by Andrew Ducharme on 6 Jun 2025 (Discord link).
Analysis by Shawn Ligocki
This TM goes through 2 phases: Phase A and Phase B.
A(a, b) = 0^inf <F 10 1^a 00 1^b 0^inf
B(a, b, c, d, e) = 0^inf <F 1^2a+1 00 1^2b 0^2c+1 1^d 00 1^e 0^inf
Shorthand:
A(a) = A(a, 0)
B(b, d, e) = B(0, b, 0, d, e)
f1(x) = 2x+2
f2(x) = f1^x(4) = 6 2^b - 2
Phase A:
Start --(13)--> A(1)
A(3k) -> B(4, 4k-3, 4)
A(3k+1) -> A(4k+4)
A(3k+2) -> A(4k+6)
Phase B:
B(b, 3k, e) -> B(f1(f2^k(b)), 4k+e, 0)
B(b, 3k+1, e) -> B(f2^{k+1}(b), 4k+e+4, 0)
B(b, 3k+2, e) -> Halt(2 f2^k(b) + 4k + e + 12)
The trajectory is:
Phase A:
1 4 8 14 22 32 46 64 88 120
Phase B:
B(4, 157, 4)
B(a, 216, 0) a = f2^{53}(4)
B(b, 288, 0) b = f1(f2^{72}(a)) = f1( f2^{125}(4) )
B(c, 384, 0) c = f1(f2^{96}(b))
B(d, 512, 0) d = f1(f2^{128}(b))
Halt(2e + 692) e = f2^{170}(d)
If we ignore the f1 calls and note that then the sigma score at halt is larger than .
It gets very lucky in Phase A: resetting 9 times before hitting a multiple of 3 and a little bit lucky in Phase B: resetting 4 times before hitting a remainder 2 factor.