Puzzle #5 - Binary-Gray counters - solution

The binary-Gray puzzle from last week generated some flow of comments and emails.
Basically, the important point to notice is the amount each counter toggles while going through a complete counting cycle.

For Gray coded counter, by definition only one bit changes at a time. Therefore, for an n stage counter we get 2^n toggling events for a complete counting cycle.

For binary coded n-bit counter, we have 2^(n+1)-2 toggling events for a complete counting cycle. you could verify this by

1. Taking my word for it (don’t - check it yourself)
2. Writing down manually the results for a few simple cases and convince yourself it is so
3. Calculate the general case, but you have to remember something about how to calculate the sum of a simple series (best way)

Anyways, given the above assumptions and the fact that per bit the Gray counter consumes 3 times more power (2 times more would also just work, but the difference would be a constant), the Gray counter will always consume more power.
3*2^n > 2^(n+1) - 2