# Raincomplex's State Transition Graph Dec 20, 2011

(for context, see everything2)

The image in question is a state transition graph I drew (pencil, 5cm x 5cm), with six nodes, transcribed below.

0 --024--> 0 0 --135--> 3 1 --024--> 0 1 --135--> 3 2 --024--> 1 2 --135--> 4 3 --024--> 1 3 --135--> 4 4 --024--> 2 4 --135--> 5 5 --024--> 2 5 --135--> 5

The rules describe a one-dimensional six-state cellular automaton with an asymmetric neighborhood, with one extra state not shown above (called 6). Oops. I mean, I totally omitted it so as to preserve the beautiful symmetry of the graph. The rules for 6 are: `6 --0246--> 6` and `6 --135--> 4`. In each rule, the left number is the current cell, the center numbers are allowed values for the cell just to its left, and the right number is the value of the current cell in the next generation.

A state is a finite sequence of cells valued 0-5. To the left of this sequence is an infinite number of 0s, and to its right an infinite number of 6s. (Below I indicate this with a single 0 and 6 on either side of the state.) So suppose we begin with the state `31`:

Generation 1: 0316 03 a 3 with a 0 to the left becomes a 1 1 (3 --024--> 1) 31 a 1 with a 3 to the left becomes a 3 3 (1 --135--> 3) 16 a 6 with a 1 to the left becomes a 4 4 (6 --135--> 4) so, Generation 2: 01346 Generation 3: 0456 Generation 4: 02246 Generation 5: 01126 Generation 6: 0346 Generation 7: 0156 Generation 8: 0546 Generation 9: 0256 Generation 10: 01246 Generation 11: 0426 Generation 12: 0216 Generation 13: 01046 Generation 14: 0326 Generation 15: 0146 Generation 16: 056 Generation 17: 0246 Generation 18: 0126 Generation 19: 046 Generation 20: 026 Generation 21: 016 Generation 22: 046 Generation 23: 026 Generation 24: 016 Generation 25: 046

You'll notice that the numbers "travel" to the right, eating into the infinite 6s there. At the end, the state begins to cycle (albeit offset by one cell each cycle) between 4, 2, and 1. Does this seem familiar? If we treat the states as integers in base 6, ignoring the 6s:

Generation 1: 0316 = 19 Generation 2: 01346 = 58 Generation 3: 0456 = 29 Generation 4: 02246 = 88 Generation 5: 01126 = 44 Generation 6: 0346 = 22 Generation 7: 0156 = 11 Generation 8: 0546 = 34 Generation 9: 0256 = 17 Generation 10: 01246 = 52 Generation 11: 0426 = 26 Generation 12: 0216 = 13 Generation 13: 01046 = 40 Generation 14: 0326 = 20 Generation 15: 0146 = 10 Generation 16: 056 = 5 Generation 17: 0246 = 16 Generation 18: 0126 = 8 Generation 19: 046 = 4 Generation 20: 026 = 2 Generation 21: 016 = 1 Generation 22: 046 = 4 Generation 23: 026 = 2 Generation 24: 016 = 1 Generation 25: 046 = 4

When decreasing, the numbers are being divided by two (`n _{i+1} = n_{i}/2`). When increasing, they are following the formula

`n`. This is the hailstone sequence, which if the Collatz conjecture is true, does always end with the (4, 2, 1) cycle, for any initial state ≥ 1.

_{i+1}= 3n_{i}+1Why does this work? In short, because `3n = 6(n/2)`. That is, in base 6, multiplying by three is the same thing as dividing by two and then shifting one digit to the left (i.e., adding a 0 on the right side). The rules at the top of this writeup describe division by two in base 6*. Thus to multiply by three and add one**, we simply need to move everything a digit to the left and add 4***, then divide by two as normal, according to the rules.

* The derivation of which is tedious beyond the scope of this writeup.

** as per the Collatz function when the argument is odd (which in base 6 occurs when it ends in 1, 3, or 5)

*** The rules perform truncating division for odds---adding 4 makes up for both the truncation (1/2 in base 6 is ".3") and the "add one" part of `3n+1`. This addition is the sole reason for the extra symbol (6).

Does this get us closer to a solution to the conjecture? Who knows, and probably not. The whole thing is so simple and yet something flabbergastingly impenetrable is happening. It's very easy to restate in other ways, and so far that hasn't helped anybody figure it out.