Question

If I have a number greater than 20 but that can be divided evenly by 2 without any remainder, I want to determine what number gets me closest to 20. For example:

For 2048, dividing by 2 enough times would get me to 16 which is the closest I can get to 20. If the number is 800, the closest is 25.

I can write a loop and just keep dividing and comparing the range and pick the value that is closest. Is there maybe a simpler way, possibly through shifting bits?

EDIT: When I say it evenly divides by 2, I mean it divides all the way down to 2 as well. A number of 70 would only divide down to 35 evenly. A number like 2048 or 1024 will divide evenly all the way to 2.

Sample numbers: 2048, 1920, 1600, 1536, 1080..640, 352, 320, 176. These are typical image sizes from cameras.

Answer 1

If your input number is x, I think you want x/2^[(log x/14)/log 2], assuming you want your target number to be in the interval [14,27].

In java code, Math‘s log function will come in handy (although base-2 logarithm would be even better), and you also need an integer cast (or somehow find the largest integer smaller than the expression in []).

What this does: Let x be your input and y be the number you want to find. Then, x=y*2^n for yet unknown n, while y is around 20 (see above). Obviously, n is the base-2 logarithm of x/y. Now, if you pick the smallest possible y, call it y', the integer part of the base-2 logarithm of x/y' is still the n we are looking for, unless x/y' differs from x/y by a factor of more than 2, which by assumption of repeated division by 2 it cannot. Thus, we have n and hence y=x/2^n.