3 ms·
What I said is correct, but I suppose ambiguous. What I mean to say is that counting MSB zeros does specify that the number will be below a threshold, just not
by ars_technician 13y ago
What I said is correct, but I suppose ambiguous. What I mean to say is that counting MSB zeros does specify that the number will be below a threshold, just not an arbitrary threshold like the one you gave in your example.
For a more concrete example, if the requirement is that a 16-bit word is less than 4096, you only have to verify that it leads with 4 zeros.
- letstryagain 13y agoYes but Bitcoin uses 'below number X' which has more precision than just counting leading zeroes.