giorgi_pkhaladze's blog

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By giorgi_pkhaladze, history, 17 months ago, In English

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17 months ago, # |
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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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17 months ago, # |
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Floyd can be optimized to n^3/32

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17 months ago, # |
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What do you mean by $$$O(V^3)$$$ is mostly more than $$$O(V^2E)$$$?

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

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    17 months ago, # ^ |
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    Yes that why I said mostly and I think that bellman ford takes o((v-1)*v*e)

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      17 months ago, # ^ |
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      So then in these cases where $$$E\gg V$$$, the Floyd-Warshall will be much faster, so that's why it's useful

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17 months ago, # |
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It's useful when E is much bigger than V

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