Collatz Conjecture Analysis (Continued)
©2014 Darel Rex Finley. This complete article, unmodified, may be freely distributed for educational purposes.
This is a continuation of the analysis of the Collatz Conjecture that begins here.
The above chart consists entirely of counting numbers that are not divisible by 2 or 3. (For reasons already discussed, these are the only numbers with which we need concern ourselves.)
The chart extends infinitely to the right and down.
Every counting number not divisible by 2 or 3 is represented in this chart exactly twice: once above the line, and once below the line.
Above the line, all counting numbers not divisible by 2 or 3 are simply listed in numerical order.
Below the line, each column consists of all counting numbers not divisible by 2 or 3 that go to their above-the-line number in one step, where “one step” means:
multiply by 3; then add 1; then divide by 2 as many times as possible
Within any one below-the-line column, the numbers are arranged in numerical order.
It so happens that the numbers in any one below-the-line column progress by the procedure: multiply by 4; then add 1. (But numbers divisible by 3 are omitted.)
To perform forward hailstone in this chart, start on any above-the-line number, then find that same number below the line, and go up to the top of the column. This diagram depicts forward hailstone going from 25 to 19 to 29 to 11:
(Note: 7 is not part of this sequence; the last up-arrow skips directly from 29 to 11; it does not visit 7.)
Forward hailstone is strictly deterministic; you have no choice about what to do next.
The conjecture claims that forward hailstone will always reach 1.
To perform reverse hailstone in this chart, start on any above-the-line number (but typically 1), then choose any number below it, and find that same number above the line. This diagram depicts reverse hailstone going from 1 to 5 to 53 to 35 to 23:
(Note: 13 is not part of this sequence; the second down-arrow skips directly from 5 to 53; it does not visit 13.)
Reverse hailstone provides you with an infinity of choices with each step, because you can choose from any of the below-the-line numbers in your current (infinitely long) column.
The conjecture claims that using reverse hailstone from 1, you can reach any number in this chart.
(Note: A formal proof of the conjecture that utilizes the above chart would have to include proofs that all the above statements about the chart are correct and/or consistent. I believe that such would be trivially easy to do.)
Ideas For A Proof
Suppose we start from any number N that is greater than 1 (and, of course, not divisible by 2 or 3).
Obviously, if you could prove that forward hailstone will, in one or more steps, reach a number less than N, then that would prove the conjecture, because the same rule would apply to that number, and so you could get to smaller and smaller numbers until you would have to reach 1.
Less obviously, if you could prove that reverse hailstone (via any of its branches) can, in one or more steps, reach a number less than N, then that too would prove the conjecture. To see that this is so: Let C be the set of all Collatz-compliant numbers, and U be the set of all Collatz-uncompliant numbers. (The conjecture claims that U is empty.) There is no way that any member of C can be connected (via forward or reverse hailstone) to any member of U without violating the compliant/uncompliant status of one or the other. Therefore, U’s smallest member cannot (in any number of steps) reach a number smaller than itself (which would have to be a member of C) via forward or reverse hailstone. So if you are able to prove that N can go to a smaller number via reverse hailstone (in any number of steps via any branches), then you have shown that U cannot have a smallest member, so U must be empty.
Similarly, if you could prove that from N, it must be possible to get to a smaller number via either forward or reverse hailstone (without specifying which for any particular N), then the conjecture is proven.
Further still: If you could prove that from N, it must be possible to get to a smaller number via either forward hailstone, reverse hailstone, or some-forward-then-some-reverse hailstone — that too would prove the conjecture.
That’s it for now!