4 ms·
It's the same just with redundant representations (the 2-bit repr is the count of ones in the 3-bit repr) '00' -> '000' '01' -> '001', '010', '100' '10'
by oshepherd 9y ago
It's the same just with redundant representations (the 2-bit repr is the count of ones in the 3-bit repr)
'00' -> '000'
'01' -> '001', '010', '100'
'10' -> '011', '101', '110'
'11' -> '111'
this kind of transformation is truly bread & butter in hardware; we regularly numbers between binary counts, mask/number-of-set-bits and one-hot representations for optimisation purposes.
- Sniffnoy 9y agoThat's the obvious attempt at an equivalence one would come up with on being told that they're the same, but, as I stated, it doesn't work. As an example: In the 2-bit saturating counter, if you start at 00, if you see a 1 and then a 0 you're back to where you started. Whereas in the bit-shift register, if you start at 000 and see a 1 and then a 0, you're now at 010, which would correspond to 01 rather than 00.