Also if O(V^3) is of reasonable size, Floyd-Warshall is so easy and
quick to code and be sure you haven't made a mistake that it might be
worth using anyway. "The first thing to optimise is developer time"

On Sun, Aug 15, 2010 at 10:40 PM, phars alnmr <[email protected]> wrote:
> you can use Johnson with binary min-heap implementation that is running in
> O(V*E*log(V))
>
> 2010/8/16 Mikhail Dektyarev <[email protected]>
>>
>> No. O(V^3 + V^2*log(V)) = O(V^3).
>> So Dijkstra always is not worse than Floyd (asymptotically). In practice,
>>  one operation in Floyd's algorithm is much simpler than in Dijkstra, so if
>> your graph is close to complete, Floyd's algorithm will be faster.
>>
>> On Mon, Aug 16, 2010 at 1:32 AM, Ricardo Hahn <[email protected]>
>> wrote:
>>>
>>> It depends on the graph.
>>> If the graph is complete, E = O(V²),
>>> then the dijkstra would become O(V³ + V (V²) lg V ), that is worse than
>>> Floyd Warshall
>>>
>>> On Sun, Aug 15, 2010 at 6:26 PM, Mohamed Ghoneim <[email protected]>
>>> wrote:
>>>>
>>>> Dear all,
>>>>      I have a question concerning Dijkstra and Floyd Warshall, which way
>>>> is better to get all pair shortest path, is it through using dijkstra and
>>>> looping through all the verticies and using them to find the shortest path
>>>> to any other point O(E * V + V^2 * log(V)) , or just to execute floyed
>>>> warshall on the graph O(V^3)
>>>>
>>>> Thanks
>>>> Mohamed Sayed Ghoneim
>>>>
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>>>
>>>
>>>
>>> --
>>> []'s,
>>> Ricardo
>>>
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>>
>>
>>
>> --
>> Best regards, Дектярев Михаил
>>
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