But in another article about Cholesky decomposition ( http://en.wikipedia.org/wiki/Cholesky_decomposition ) it says that the complexity of Cholesky Decomposition is about n^3/3 FLOPs. I think here FLOPs means Floating point Operations. Am I right?
"Floating point Operations" include floating point multiplications and addition, is there anything else is considered as floating point operations?
Also, I think floating point multiplication takes more time than floating point additions, right? So FLOPs is not a good measure to indicate the computational complexity because different floating point operations take different time.
On Oct 6, 10:54 am, Sean <guo.xiaoy...@gmail.com> wrote:
> I think here FLOPs means Floating point Operations. Am I right?
Yes
> "Floating point Operations" include floating point multiplications and > addition, is there anything else is > considered as floating point operations?
Sometimes divisions also are included, but in most linear algebra algorithms, the divisions are outnumbered by an order of magnitude by the multiplications and additions.
> Also, I think floating point multiplication takes more time than > floating point additions, right? So FLOPs is not > a good measure to indicate the computational complexity because > different floating point operations take different time.
That would be processor dependent. On several modern processors, the relevant floating point operation computes a * b + c, so a + c is computed as a * 1 + c, while a * b is computed as a * b + 0. Thus, a*b, a+c, and a*b+c all take the same time.
In article <66f27409-e0da-46b4-86c1-5c0e56442...@m36g2000hse.googlegroups.com>,
Sean <guo.xiaoy...@gmail.com> wrote: >Also, I think floating point multiplication takes more time than >floating point additions, right? So FLOPs is not >a good measure to indicate the computational complexity because >different floating point operations take different time.
Complexity mean asymptotic complexity. Even if multiplication takes ten times as long as additions, n^3 additions will eventually take longer than n^2 multiplications. The time taken for the total will have the same order as the total count, because both with be dominated by the same operation.
-- Richard -- Please remember to mention me / in tapes you leave behind.
