If we have to prepare n:1 MUX using only m:1 MUXs where n>m. How many levels will be required?
For example preparing 64:1 MUX with only 16:1 MUXs, I will need two levels: first level will be four 16:1 MUX and second one will be single 16:1 MUX. The same is with preparing 32:1 MUX with only 8:1 MUXs.
So in both above we need 2 levels. I was trying to find if there is any formula that fits that 2 levels count.
I flelt it should be $\lceil log_mn\rceil$, but this does not seem to be the case as $\lceil log_8{32}\rceil =1$ while $\lceil log_{16}{64}\rceil =2$
