Using Resolved Probability Bias to Gain a Superior Understanding of the Collatz Conjecture

The Collatz Conjecture is named after Lothar Collatz, a mathematician who introduced the idea in 1937.?It is also known as the 3N + 1 problem.?A more elaborate description of the problem can be found at https://en.wikipedia.org/wiki/Collatz_conjecture ).

The problem involves starting with any arbitrary integer and then applying a succession of 2 possible arithmetic operations.?If the number is odd, then multiply by 3 and add 1.?If the number is even, then divide by 2.?The conjecture states that for any arbitrarily large integer N, the series will eventually converge to 1.?

While an accepted mathematical proof does not yet exist for the conjecture, we can use the concept of Resolved Probability Bias to show that every series beginning with any arbitrary large integer N will always converge to 1. An explanation of the problem and an understanding of the solution can be found below.

Consider the set of positive integers and start with any number N0.?Successively apply the following function to the ith number Ni??where i ranges from 0 to infinity:

If Ni is odd, then Ni+1 = 3 * Ni + 1???

if Ni is even, then Ni+1 = Ni / 2

Prove that for any arbitrary starting point N0, the sequence will always reach its stable state of a continuous 3 node loop of 4 → 2 → 1

Or at least gain a good understanding of why this will always hold true for any arbitrarily large number N0 as a starting point. ?

?To begin understanding this conjecture, consider the following simple statements:

1. An odd number plus an odd number will always result in an even number.

2. An odd number plus an even number will always result in an odd number.

3. An odd number plus 1 will always result in an even number.

4. An odd number multiplied by 3 will always result in an odd number.

A number is deemed to be Odd Parity when its last digit is in the set of {1, 3, 5, 7, 9}.

A number is deemed to be Even Parity when its last digit is in the set of {0, 2, 4, 6, 8}.

Whenever Ni is odd, then Ni+1 will be even and larger than Ni.

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Let us consider the expected probability of the Ni+1 number becoming greater.?At any random number N, the expected probability of it being odd will be 50% (5/10).

For the Odd Parity function, at any random Ni node, the expected value will be approximately 1.5 Ni.

For the sake of simplicity, let us consider the +1 part of the Odd Parity function to be negligible in terms of value.?The primary effect of the +1 in the term is to negate the parity of the resulting N.?The primary purpose of the +1 term is to change the parity of the result to even and in this regard is incredibly important.?In terms of adding value to the size of the Ni+1 result, it is negligible, especially for an arbitrarily large number.

For the even portion of the function, a simplistic case can be made that the expected value of the Even Function is 1.?(Divide by 2, 50% of the time.)

Thus one can erroneously conclude that the expected value of the combined function (Odd and Even Parity) should deliver an expected value of 1.5.?In other words, the 3N + 1 problem should deliver a series that is constantly increasing by an average factor of approximately 1.5.

Why then does the function always eventually decrease and resolve to a steady state of 4 → 2 → 1 ??

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To answer this, and to gain an understanding of the dynamics of the conjecture, we need to introduce the concept of expected probability skew or resolved probability bias.

(Expected Probability Skew or Resolved Probability Bias should be thought of as new terms and should not be confused with existing terminology definitions.)

The concept of resolved probability bias is that for some random number results Ni+1, the successive results Ni+2, Ni+3 are not completely random but instead have a bias or skew attached to them.

For example, in our even parity series of {0, 2, 4, 6, 8}, the function will divide by 2 and thus we erroneously assume that 50% of the time we are dividing by 2.

This is not the case however.

Once we reach an even number, resolved probability bias enters the system.?Half the time, the even number will result in a 2nd divide by 2. ?We always reach an even number after an odd number.

The even parity integers introduce skewness and bias into the expected results and thus the expected value.?For the combined set of numbers ending in {0, 2, 4, 6, 8}, the effect of the bias increases the probabilities of a division by 2.

Thus, the expected value for any arbitrary number N is 0.875N.

Expected Value N???????????????= N * (ExpValOddFunc * 50% + ExpValEvenFunc * 50%)

???????????????????????????????????????????????= N* (1.5 * 50% + 0.25 * 50%)

???????????????????????????????????????????????= N * 0.875

Add in the constraints that we remain in the set of integers, then for any arbitrarily large number N0, the series will converge close to 0 and we will reach a steady node state of 4 → 2 → 1.

This framework of thought can be applied to the entire series of such problems such as 3N-1 or 5N+3?and divide by 3.??It is a pedantic matter to extend it to the family of such problems.

Further insight and greater understanding of the conjecture can be gained by considering the following statements.

The primary purpose of the +1 (or -1) is to negate the parity of the successive integer.

There is an entire series of numbers that will immediately lead to the steady state of the conjecture.?By now, it should be obvious what these series of numbers are.

The entire set of integers of (1, 2, 4, 8, 16, 32, 64, … 2x} will immediately lead the sequence to the steady state solution.?Any intersection of 3N+1 and 2x will immediately lead the sequence to the steady state solution.?One can imagine that for any arbitrary large number N, there will always be an intersection of 3Ni +1 with 2x that will then lead to the steady state of the series.?

More specifically, it is as a direct result of a collision with the set of 2x that results in the success of the conjecture.?The expected value of the series is to constantly decrease to close to zero where a collision with the set of 2x becomes inevitable since all numbers are constrained to integers.?

If one accepts that the expected value for any arbitrarily large value of N0 is N0 * 87.5%, then one also needs to accept that it holds true for any number following that sequence since every successive number is also an N0?for its own starting sequence.