Abstract
Suppose that a random n -bit number V is multiplied by an odd constant M ≥ 3, by adding shifted versions of the number V corresponding to the 1s in the binary representation of the constant M . Suppose further that the additions are performed by carry-save adders until the number of summands is reduced to two, at which time the final addition is performed by a carry-propagate adder. We show that in this situation the distribution of the length of the longest carry-propagation chain in the final addition is the same (up to terms tending to 0 as n → ∞) as when two independent n -bit numbers are added, and in particular the mean and variance are the same (again up to terms tending to 0). This result applies to all possible orders of performing the carry-save additions.
Talk to us
Join us for a 30 min session where you can share your feedback and ask us any queries you have
Schedule a callDisclaimer: All third-party content on this website/platform is and will remain the property of their respective owners and is provided on "as is" basis without any warranties, express or implied. Use of third-party content does not indicate any affiliation, sponsorship with or endorsement by them. Any references to third-party content is to identify the corresponding services and shall be considered fair use under The CopyrightLaw.
