Lucy's Moonlight: Difference between revisions

Revision as of 18:21, 2 March 2025

1RB0RD_0RC1RE_1RD0LA_1LE1LC_1RF0LD_---0RA (bbch) is a probviously halting tetrational BB(6) Cryptid found by Racheline on 1 Mar 2025 (Discord link).

Analysis by Racheline

https://discord.com/channels/960643023006490684/1345810396136865822/1345820781363597312

A(x,y) := 0^inf (1011)^x 10 <A (01)^y 0^inf
B(x) := 0^inf 1^x B> 0^inf

A(x+1,3y) -> A(x,8y+3)
A(x+2,3y+1) -> A(x,8y+11)
A(x+2,3y+2) -> A(x,8y+12)
A(0,y) -> B(2y+1)
A(1,3y+1) -> A(4y+4,4)
A(1,3y+2) -> halt
B(3y) -> B(8y-2)
B(3y+1) -> A(2y,4)
B(3y+2) -> B(8y+6)

a is the sequence such that A(x,a_n) goes to A(x',a_(n+1)) in one step assuming x>=2
b is the sequence such that A(x,a_0) goes to A(x-b_n,a_n) in n rules (without using the A(0,y) or A(1,y) rules) assuming x>=b_n
c is the sequence such that A(c_n,a_0) goes to A(c_(n+1),a_0) after only one application of the A(0,y) or A(1,y) rules

f(3n) = 8n+3
f(3n+1) = 8n+11
f(3n+2) = 8n+12
a_0 = 4
a_(n+1) = f(a_n)
b_0 = 0
b_(n+1) = b_n+(1 if 3|a_n else 2)
c_0 = 14
c_(n+1) = idk i'll finish this later but it's somewhere around a_(largest i such that b_i<=c_n)

@@ Line 2: / Line 2: @@
 {{TM|1RB0RD_0RC1RE_1RD0LA_1LE1LC_1RF0LD_---0RA}} is a [[probviously]] halting tetrational [[BB(6)]] [[Cryptid]] found by Racheline on 1 Mar 2025 ([https://discord.com/channels/960643023006490684/1239205785913790465/1345551751016878272 Discord link]).
-<pre>
+== Analysis by Racheline ==
+https://discord.com/channels/960643023006490684/1345810396136865822/1345820781363597312<pre>
+A(x,y) := 0^inf (1011)^x 10 <A (01)^y 0^inf
+B(x) := 0^inf 1^x B> 0^inf
+A(x+1,3y) -> A(x,8y+3)
+A(x+2,3y+1) -> A(x,8y+11)
+A(x+2,3y+2) -> A(x,8y+12)
+A(0,y) -> B(2y+1)
+A(1,3y+1) -> A(4y+4,4)
+A(1,3y+2) -> halt
+B(3y) -> B(8y-2)
+B(3y+1) -> A(2y,4)
+B(3y+2) -> B(8y+6)
+a is the sequence such that A(x,a_n) goes to A(x',a_(n+1)) in one step assuming x>=2
+b is the sequence such that A(x,a_0) goes to A(x-b_n,a_n) in n rules (without using the A(0,y) or A(1,y) rules) assuming x>=b_n
+c is the sequence such that A(c_n,a_0) goes to A(c_(n+1),a_0) after only one application of the A(0,y) or A(1,y) rules
 f(3n) = 8n+3
 f(3n+1) = 8n+11
@@ Line 10: / Line 28: @@
 b_0 = 0
 b_(n+1) = b_n+(1 if 3|a_n else 2)
-c_0 = 14(?)
+c_0 = 14
-c_(n+1) is somewhere around the smallest g^k(a_i) that is (something) mod 3, where i is maximal such that b_i <= c_n
+c_(n+1) = idk i'll finish this later but it's somewhere around a_(largest i such that b_i<=c_n)
-where g is similar to f, just with probably different constants and it only has two defined cases
-so c_(n+1) is around (8/3)^(3/5*c_n)
-but it actually depends
-if b_i = c_n then c_(n+1) should have exactly that form (*some constant ±some constant) always
-if b_i = c_n-1 (which is the only other option) then it depends on a_i mod 3
-for one value of a_i mod 3 it halts, for another you directly get c_(n+1) without iterating g, and for the remaining one you do iterate g but the input is probably slightly different
-so
-the probability of b_i = c_n-1 is 2/5 since the average b_(m+1)-b_m is 5/3
-and if that happens, the probability of halting is 1/3
-so probability of halting is 2/15 at each step
-which means probability of beating the champion is (13/15)^(something around 15)
-wait thats actually a pretty good chance
-we should be able to compute c_2 (and check if it exists or if the TM halts before that)
-if it doesnt halt, i think c_2 should be 10^(something in the hundreds/thousands), so with the information that it doesnt halt there, we will know the TM runs for >10^^3 steps, which means (depending on whether the top of the tower is smaller or larger than that of the current champion) we need either 12 or 13 more elements of the c sequence
-in the case of 13, thats a ~15.56% chance of beating the champion, and in the case of 12, its ~17.95%
 </pre>
 [[Category:Cryptids]]

Lucy's Moonlight: Difference between revisions

Revision as of 18:21, 2 March 2025

Analysis by Racheline

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