giorgi_pkhaladze's blog

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By giorgi_pkhaladze, history, 3 years ago, In English

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3 years ago, hide # |
 
Vote: I like it +10 Vote: I do not like it

Let, $$$n$$$ = no of nodes, $$$m$$$ = no of edges Finding all pair shortest path using belmenford takes $$$O(n^2m)$$$ but floyed warshal takes $$$O(n^3)$$$ time. floyed is very easy to write and code is short.

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3 years ago, hide # |
 
Vote: I like it -36 Vote: I do not like it

Floyd can be optimized to n^3/32

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3 years ago, hide # |
 
Vote: I like it +23 Vote: I do not like it

What do you mean by $$$O(V^3)$$$ is mostly more than $$$O(V^2E)$$$?

What if $$$E\gg V$$$?

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3 years ago, hide # |
 
Vote: I like it 0 Vote: I do not like it

It's useful when E is much bigger than V

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