Try out some corner cases like complete graphs, etc. Does your algorithm
return 1 for a complete graph?

On Mon, May 2, 2011 at 9:09 PM, prajay shetty <[email protected]>wrote:

> thanks for the admin coorporation please post the problem
>
>
> On Mon, May 2, 2011 at 8:19 PM, prajay shetty <[email protected]>wrote:
>
>>
>> Hey all,
>>
>> Last week  i was going trough corfment this is not a gcj problem but i had
>> wrote an algorithm on np complete that is an independent set problem  which
>> has no cycle and no weight i had tried this algorithm on this type of graphs
>> , can anyone provied me with counter example if any ,I taught this was the
>> good one comunicate with you members since most of you will be good at DSA
>> atleast more than me ): thanks for any help in the problem
>>
>> so here is our graph which has following properties as an example
>> 1) node 1 is connected to node 2
>> 2) node 2 is connected to node 3
>> 3) node 3 is connected to node 1
>> 4)node 2 is  connected to node 5
>> 5) node 2 is connected to node 4
>> 6) node 5 is connectd to node 6
>> 7)node 4 is connected to node 5
>> 8) node 3 is connected to node 7
>> 9) node 7 is connected to node 8
>> 10) node 8 is connected to node 3
>>
>>
>> First by seeing the graph we have independent set  is 1,4,6,8
>> Algorithm
>>
>> 1)We genearte dfs on graph g
>> 2) We genearte bfs on g
>> 3)DFS: here we modify our dfs a bit insteda as we go through dfs  we color
>> first node as blue and we dont color the next node  and we color the next
>> node  blue this continues till we  reach a dead end.,we list all the node
>> that we marked as blue.
>> for example we get the output as 1,3,5  now we explore 3  and we dont
>> color 7 and then we color 8  like this we procceed, output is 1,3,4,6,8
>> 4) BFS: we generate bfs, then  once we got the list of nodes we maintain a
>> pair node in an array as parent child in our array for example we store
>> 1->2,3 and 2->4,5 and 3->7,8 for bfs we can just reduce the memory space by
>> working on efficient ds on this.
>> 5) then we color all the node as yellow in our bfs tree for the nodes that
>> we obtain in step 3
>> 6)now we check if the node that is in the list has a node  connected to
>> it  colored as yellow ,if yes we delete it from our final list and in this
>> way we get we remove 3 as 3  is connected to 1
>> so we get our final list as 1,4,6,8
>>
>> just want to check if there is any counter example for this ...thanks for
>> your help
>>
>>
>> Note: step 3 can be done with step 1 and step 2 can be done with step 4
>>
>>
>> --
>> Regards ,Prajay
>>
>
>
>
> --
> Regards ,Prajay
>
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-- 
Rahul Gulati
[email protected]
Department of Computer Science
Jaypee Institute of Information Technology (Deemed University)
India

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