1RB2LC1RC 2LC---2RB 2LA0LB0RA: Difference between revisions

Revision as of 03:49, 24 June 2024

https://bbchallenge.org/1RB2LC1RC_2LC---2RB_2LA0LB0RA

This is a BB(3, 3) holdout under active exploration. It simulates a complex set of Collatz-like rules with two decreasing parameters and seems as if it may be a new Cryptid (and perhaps even one that "probviously" halts! But this is really speculation at this point.)

This is holdout #758 on Justin's 3x3 mugshots. And if you start in state C it is a permutation of #153: 1RB0LB0RC_2LC2LA1RA_1RA1LC---.

NOTE: These rules are under active development and may have mistakes or typos.

dyuan01's Rules

https://discord.com/channels/960643023006490684/1084047886494470185/1224457633176486041

A_1(a, b, c) = 0^inf 1 2^a <C (22)^b (20)^c 0^inf
A_2(a, b, c) = 0^inf 1 2^a <A2 (22)^b (20)^c 0^inf
B(a, b) = 0^inf 1 2^a <B0 (20)^b 0^inf

See Discord link for rules ...

savask's Rules

https://discord.com/channels/960643023006490684/1084047886494470185/1254085725138190336

Let (m, b, n) = A2(m, b, n) = 0^inf 1 2^m <A2 (22)^b (20)^n 0^inf

(0, b, n) -> (2b+2, 1, n) if n is even
          -> (2b, 1, n+3) if n is odd

(1, b, 0) -> Halt

(2, b, 0) -> (0, b+3, 0) if b is even
          -> (0, 1, b+5) if b is odd

(m, b, 0) -> (m-2, b+3, 0) if b is even
          -> (m-3, 1, b+3) if b is odd

(1, b, n) -> (0, 1, n+b+2) if n is even
          -> Halt if n is odd

(2, b, 1) -> Halt if b is even
          -> (0, 1, b+5) if b is odd

(m, b, 1) -> (m-3, 1, b+3)

(m, b, n) -> (m-2, b+3, n-2)

https://discord.com/channels/960643023006490684/1084047886494470185/1254306301786198116

step (A2 0 b n) | even n = A2 (2*b+2) 1 n
                | otherwise = A2 (2*b) 1 (n+3)
-- From now on m > 0
step (A2 1 b 0) = error $ "Halt A2 1 " ++ show b ++ " 0"
step (A2 2 b 0) | even b = A2 0 (b+3) 0
                | otherwise = A2 0 1 (b+5)
step (A2 m b 0) | even b = A2 (m-2) (b+3) 0
                | otherwise = A2 (m-3) 1 (b+3)
-- From now on n > 0
step (A2 1 b n) | even n = A2 0 1 (n+b+2)
                | otherwise = error $ "Halt A2 1 " ++ show b ++ " " ++ show n
step (A2 2 b 1) | even b = error $ "Halt A2 2 " ++ show b ++ " 1"
                | otherwise = A2 0 1 (b+5)
step (A2 m b 1) = A2 (m-3) 1 (b+3)
-- Here m > 1, n > 1
step (A2 m b n) = let d2 = (min m n) `div` 2 in A2 (m - 2*d2) (b + 3*d2) (n - 2*d2) -- Accelerated

Shawn's Rules

https://discord.com/channels/960643023006490684/1084047886494470185/1254307091863048264

We can reduce the set of rules from savask's list a bit by noticing that we can evaluate so that all rules end with c even:

  (0, b, 2c)    -> (2b+2, 1, 2c)

  (1, b, 0) -> Halt
  (1, 2b,   2c)  -> (0, 1, 2(b+c+1))
  (1, 2b+1, 2c)  -> (2, 1, 2(b+c+3))

  (a, 2b,   0)  -> (a-2, 2b+3, 0)
  (2, 2b+1, 0)  -> (0, 1, 2b+6)
  (a, 2b+1, 0)  -> (a-3, 1, 2b+4)

  (a, b, c) -> (a-2, b+3, c-2)

Phases

We can think of this going through two different phases. "Even Phase" (where a is even) and "Odd Phase" (where a is odd).

Even Phase: a,c even:
  (0, b, 2c) -> (2b+2, 1, 2c)
  (2a+2, 2b, 0) -> (2a, 2b+3, 0)
  (2, 2b+1, 0) -> (0, 1, 2(b+3))

  To Odd Phase:
    (2a+4, 2b+1, 0) -> (2a+1, 1, 2b+4)
 
Odd Phase: a odd, c even
  To Halt:
    (1, b, 0) -> Halt
    (3, 2b, 0) -> (1, 2b+3, 0) -> Halt

  To Even Phase:
    (1, 2b, 2c+2) -> (0, 1, 2(b+c+2))
    (1, 2b+1, 2c+2) -> (0, 1, 2b+2c+5) -> (2, 1, 2(b+c+4))
    
    (2a+5, 2b, 0) -> (2a+3, 2b+3, 0) -> (2a, 1, 2b+6)
    (2a+3, 2b+1, 0)  -> (2a, 1, 2b+4)

So the only way for this to halt is if it is in "Even Phase" and hits (2k+8, 2k+1, 0) or (4k+12, 4k+3, 0) (which will lead to (1, b, 0) or (3, 2b, 0) eventually). If a is bigger or smaller, then "Odd Phase" will end going back to "Even Phase" again.

Repeated (0, b, 2c)

Let $f(n)=3n+4$ , then

(0,b,2c)\to (0,f(b),2(c-b-1))

Let

h(n)=f^{n}(1)+1=3^{n+1}-1

g(n)=\sum _{k=0}^{n-1}h(k)={\frac {3}{2}}(3^{n}-1)-n

Then if $c>g(n)$ :

(0,1,2c)\to (0,f^{n}(1),2(c-g(n)))\to (2h(n),1,2(c-g(n)))

Repeated (0, 1, 2c)

https://discord.com/channels/960643023006490684/1084047886494470185/1254635277020954705

Let $C(n)=(0,1,2n)$ = 0^inf 1 <A2 22 (20)^2n 0^inf

C(g(n)+8k+1)\to C(g(n)+8k+1+n+9)

\forall k:{\frac {h(n)-45}{65}}<k<{\frac {h(n)-22}{38}}

Notably, when 8 divides (n+1) then this rule can potentially be applied repeatedly.

Ex: if n = 7, then we get:

\forall k\in [101,172]:C(3273+8k)\to C(3273+8(k+2))

And we see this starting with $C(4137)=C(3273+8\cdot 108)$ which repeats this rule until we get to $C(4665)=C(3273+8\cdot 174)$ .

And as n gets way bigger, these ranges of repeat will increase exponentially.

@@ Line 3: / Line 3: @@
 https://bbchallenge.org/1RB2LC1RC_2LC---2RB_2LA0LB0RA
-This is a [[BB(3, 3)]] [[holdout]] under active exploration. It simulates a complex set of Collatz-like rules with two decreasing parameters and seems as if it may be a new [[Cryptid]] (and perhaps even one that "probviously" halts! But this is really speculation at this point.)
+This is a [[BB(3, 3)]] [[holdout]] under active exploration. It simulates a complex set of Collatz-like rules with two decreasing parameters and seems as if it may be a new [[Cryptid]] (and perhaps even one that "[[probviously]]" halts! But this is really speculation at this point.)
+This is holdout #758 on Justin's 3x3 mugshots. And if you start in state C it is a [[permutation]] of #153: <code>1RB0LB0RC_2LC2LA1RA_1RA1LC---</code>.
 NOTE: These rules are under active development and may have mistakes or typos.
 == dyuan01's Rules ==
@@ Line 85: / Line 86: @@
    (a, b, c) -> (a-2, b+3, c-2)
 </pre>
 === Phases ===
 We can think of this going through two different phases. "Even Phase" (where <code>a</code> is even) and "Odd Phase" (where <code>a</code> is odd).
@@ Line 118: / Line 117: @@
 == Repeated (0, b, 2c) ==
-Let <math>f(n) = 3n+4</math>, then <math display="block">(0, b, 2c) \to (0, f(b), 2(c - b - 1))</math>.
+Let <math>f(n) = 3n+4</math>, then<math display="block">(0, b, 2c) \to (0, f(b), 2(c - b - 1))</math> Let<math display="block">h(n) = f^n(1) + 1 = 3^{n+1} - 1</math><math display="block">g(n) = \sum_{k=0}^{n-1} h(k) = \frac{3}{2} (3^n - 1) - n</math>
-Let
-<math display="block">h(n) = f^n(1) + 1 = 3^{n+1} - 1</math>
-<math display="block">g(n) = \sum_{k=0}^{n-1} h(k) = \frac{3}{2} (3^n - 1) - n</math>
-Then if <math>c > g(n)</math>:
-<math display="block">(0, 1, 2c) \to (0, f^n(1), 2 (c-g(n))) \to (2 h(n), 1, 2 (c-g(n)))</math>
+Then if <math>c > g(n)</math>:<math display="block">(0, 1, 2c) \to (0, f^n(1), 2 (c-g(n))) \to (2 h(n), 1, 2 (c-g(n)))</math>
 == Repeated (0, 1, 2c) ==
@@ Line 135: / Line 127: @@
 Let <math>C(n) = (0, 1, 2n)</math> = <code>0^inf 1 <A2 22 (20)^2n 0^inf</code>
-<math display="block">C(g(n) + 8k+1) \to C(g(n) + 8k+1 + n+9)</math>
+<math display="block">C(g(n) + 8k+1) \to C(g(n) + 8k+1 + n+9)</math><math display="block">\forall k: \frac{h(n) - 45}{65} < k < \frac{h(n) - 22}{38}</math>
-<math display="block">\forall k: \frac{h(n) - 45}{65} < k < \frac{h(n) - 22}{38}</math>
 Notably, when 8 divides (n+1) then this rule can potentially be applied repeatedly.
-Ex: if n = 7, then we get:
+Ex: if n = 7, then we get:<math display="block">\forall k \in [101, 172]: C(3273 + 8k) \to C(3273 + 8(k+2))</math>
-<math display="block">\forall k \in [101, 172]: C(3273 + 8k) \to C(3273 + 8(k+2))</math>
 And we see this starting with <math>C(4137) = C(3273 + 8 \cdot 108)</math> which repeats this rule until we get to <math>C(4665) = C(3273 + 8 \cdot 174)</math>.
 And as n gets way bigger, these ranges of repeat will increase exponentially.