# Time Complexity of Shifting

I'm developing a IMHO very interesting algorithm for integer division. This algorithm uses boolean shifting(It shifts left for multiplication by 2). I'm wondering if << c is $O(c)$ or $O(n)$. I'm hearing it depends on architecture. Algorithms ideally should be architecture independent. So for the purposes of analysing the time complexity of my algorithm, should I consider << c as 1 primitive operation or c primitive operations.

I've looked at the other questions here, and they are saying it depends on the machine.

I'm using RAM model of computation, but don't want to assign << c a cost of 1 when it may in fact cost c.

• Of course it depends on the architecture. You have to check your CPU vendors manual to see how many cycles a shift takes. – adrianN Dec 28 '16 at 14:45

lsl $rd,$rs,#$offset  where rd is the registry in which load the variable, rs is the registry in which store the result and #offset is the number of bit to shift. So you could consider the left or right shift operation as taking$O(1)$, because this operation takes only a single CPU's clock. • Sorry, don't know what 'WLOG' means. – Tobi Alafin Dec 28 '16 at 13:22 • Without loss of generality – darioSka Dec 28 '16 at 13:23 In the RAM machine model, operations on "machine words" cost$O(1)$, where a machine word has width$O(\log n)$bits,$n$being the length of the input (in bits). This reflects the instruction sets of real world CPUs, as mentioned in the other answers. While the RAM machine model makes sense for many algorithmic tasks, it is not a good fit for large integer arithmetic, since in practice the machine word is fixed while the integer length is growing. Instead, one usually measures bit operations. Shifting an$m$-bit integer by$c$bits takes$O(m+c)$bit operations. These issues are explored in a paper of Martin Fürer, the inventor of the fastest known integer multiplication algorithm (which is also the fastest known integer division algorithm). • Fürer's algorithm is no longer the fastest known. ​ ​ – user12859 Dec 29 '16 at 0:49 • I disagree with your analysis of shifting. To shift an$m$bit integer by$c$bits, I would need at most$c$bit operations. To expand, shifting right by$c$bits, is deleting the$c$most significant bits. Shifting left by$c$bits, is simply adding$c$trailing 0s, so all in all$c$operations. I guess the time complexity of shifting is$O(c)\$ then. – Tobi Alafin Dec 29 '16 at 15:41