The loop will break once when w or h becomes 1. Here is how I got that answer:

The initial values of w and h are 512.
The loop will iterate 6 times, with i ranging from 0 to 5.
In each iteration, r is a random number between 2 and 4.
w and h are divided by r^i and rounded down to the nearest integer.
If w or h becomes 1, the loop will break.
Let’s assume the worst case scenario, where r is always 4. Then we can calculate the values of w and h in each iteration as follows:

i = 0: w = 512 / 4^0 = 512, h = 512 / 4^0 = 512
i = 1: w = 512 / 4^1 = 128, h = 512 / 4^1 = 128
i = 2: w = 128 / 4^2 = 8, h = 128 / 4^2 = 8
i = 3: w = 8 / 4^3 = 1, h = 32 / 4^3 = 1

As you can see, when i is 3, both w and h become zero, which is less than one. So the loop will break at this point. Therefore, the loop will break once in the worst case scenario.

Thank you for this! I think this update makes sense, and will merge as soon as possible.