问题
I have come across this in multiple sources (online and books) - Running time of square matrix multiplication is O(n^3) for matrices of size nXn. (example - matrix multiplication algorithm time complexity)
This statement would indicate that the upper bound on running time of this multiplication process is C.n^3 where C is some constant and n>n0 where n0 is some input beyond which this upper bound holds true. (http://en.wikipedia.org/wiki/Big_O_notation and What is the difference between Θ(n) and O(n)?) Problem is, i cannot seem to derive the values of constants C and n0.
My questions -
Can someone provide a mathematical proof for the statement 'big Oh of square matrix multiplication is O(n^3)' ?
what are the values of C and n0 ?
回答1:
There are 3 for loops within each other going from 0 to n-1 (or 1 to n) each (as can be seen in the link you provided, even though it's not completely correct), this results in O(n3). Formalizing it into a proper proof should be easy enough.
a) For a formal proof, running time needs to be defined in terms of some set of operations, commonly taken to be any arithmetic operation. Inside the 3 for loops there are 2 arithmetic operations (1 multiplication, 1 addition), thus we get
2.n3, thus C = 2.b) n0 = 0 because this holds true from n = 1
Note that, since big-O is just an upper bound, we can also say the complexity of this algorithm is O(nk) for any k >= 3 (the same would not be true if we use big-Theta notation). We can also take C and n0 to be any value greater than 2 and 0 respectively (since the requirement isn't to find the smallest possible values).
来源：https://stackoverflow.com/questions/13507302/why-is-the-time-complexity-of-square-matrix-multiplication-defined-as-on3