Saturday 9 March 2024

The solutions to 𝚙𝚘𝚙𝚌𝚗𝚝(𝚡) < 𝚝𝚣𝚌𝚗𝚝(𝚡) and why there are Fibonacci[n] of them below 2ⁿ

popcnt(x) < tzcnt(x) asks the question "does x have fewer set bits than it has trailing zeroes". It's a simple question with a simple answer, but cute enough to think about on a Sunday morning.[1]

Here are the solutions for 8 bits, in order: 0, 4, 8, 16, 24, 32, 40, 48, 64, 72, 80, 96, 112, 128, 136, 144, 160, 176, 192, 208, 224[2]

In case you find decimal hard to do read (as I do), here they are again in binary: 00000000, 00000100, 00001000, 00010000, 00011000, 00100000, 00101000, 00110000, 01000000, 01001000, 01010000, 01100000, 01110000, 10000000, 10001000, 10010000, 10100000, 10110000, 11000000, 11010000, 11100000

Simply staring at the values doesn't do much for me. To get a better handle on what's going on, let's recursively (de-)construct the set of n-bit solutions.

The most significant bit of an n-bit solution is either 0 or 1:

  1. If it is 0, then that bit affects neither the popcnt nor the tzcnt so removing it must yield an (n-1)-bit solution.
  2. If it is 1, then removing it along with the least significant bit (which must be zero, there are no odd solutions since their tzcnt would be zero) would decrease the both popcnt and the tzcnt by 1, yielding an (n-2)-bit solution.

This "deconstructive" recursion is slightly awkward. The constructive version would be: you can take the (n-1)-bit solutions and prepend a zero to them, and you can take the (n-2)-bit solutions and prepend a one and append a zero to them. However, it is less clear then (to me anyway) that those are the only n-bit solutions. The "deconstructive" version starts with all n-bit solutions and splits them into two obviously-disjoint groups, removing the possibility of solutions getting lost or being counted double.

The F(n) = F(n - 1) + F(n - 2) structure of the number of solutions is clear, but there are different sequences that follow that same recurrence that differ in their base cases. Here we have 1 solution for 1-bit integers (namely zero) and 1 solution for 2-bit integers (also zero), so the base cases are 1 and 1 as in the Fibonacci sequence.

This is probably all useless, and it's barely even bitmath.

[1] Or whenever, but it happens to be a Sunday morning for me right now.
[2] This sequence does not seem to be on the OEIS at the time of writing.