well let me say I think its the more complicated way, but hey that makes it more fun. I would guess you will have to read the count on the selector, compare it to the last count, If it is higher, add 1 to the freq, if its less subtract. To cross the zero, you will have 2 choices when at 0 of at $F. from o you will go to 1 or $F, and from $F you will go to 0 or $E. So just check for that when you get to those counts.
I would think you will have to really spin the selector VERY fast to miss any counts, so I would start with that scheme and see how it works. Don't forget you may need to de-bounce the inputs.
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