3 ms·
It's still a modified base 4, because the significance of the i'th digit is 4^i, not 5^i. Edited to add: I'm also not sure whether real-life implementations ha
by less_less 3mo ago
It's still a modified base 4, because the significance of the i'th digit is 4^i, not 5^i.
Edited to add: I'm also not sure whether real-life implementations have -0 as an option. Of course -0 could be normalized to +0, but it might be cheaper not to bother if the sign is applied after the digit selection.
- numeri 3mo agoNo, balanced ternary, for example, uses {-1, 0, 1}. The system you're discussing is balanced quinary (base 5). https://en.wikipedia.org/wiki/Signed-digit_representation https://en.wikipedia.org/wiki/Signed-digit_representation
- less_less 3mo agoIt isn't balanced quinary, but rather redundant balanced quaternary (base 4). In balanced quinary (base 5), each digit has 5x the significance of the previous one, but in Booth's encoding algorithm it's 4x. If digit i has significance b^i, then b (the base of the exponentiation) is the base (or radix) of the number system. The page you linked explicitly mentions the binary version of Booth encoding as having base b=2 and three signed digits {-1, 0, 1}. The quaternary version similarly has b=4 and five signed digits {-2, -1, 0, 1, 2} ... and possibly sometimes -0 in practice, not sure.