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Interview Question


Country: United States



Comment hidden because of low score. Click to expand.
5
of 5 vote

construct the Graph like this：set a edge between the two people who dislike each other.We can consider : if A dislikes B,it means that B dislikes A too.Then the minimum number of the homes is the minimum color number of the graph.

- notbad July 17, 2012 | Flag Reply
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0
of 0 votes

This is correct. Unfortunately coloring is NP-complete. This problem is too, because every coloring problem can be reduced to an instance of this problem.

- eugene.yarovoi July 17, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 vote

@ eugene.yarovoi ....thats correct ..otherwiz all NP hard prblms will be solvable...

- atul gupta July 17, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
-2
of 4 vote

1. Invert the graph.
2. Count the no. of connected sets.

- Anonymous July 17, 2012 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Super!... Draw undirected edge between two nodes where one people dislike other..

modify the adjacency matrix in which you have to set 1 where the value is 0 and vice versa

draw the graph again and now, no. of connected graph gives minimum number of home

- cobra July 17, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

This approach is completely invalid. Suppose I have 3 people, and 1 dislikes 3. The inversion of that graph is still 1 connected component. This problem is the same as graph coloring, which is NP-complete.

- eugene.yarovoi July 17, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

Suppose the dislike graph forms two disjoint sets, in this case the inverse graph is always a connected(so every node is taken into a room).

- SantiagoYemmiganur July 17, 2012 | Flag
Comment hidden because of low score. Click to expand.
0
of 0 votes

@ eugene.yarovoi ....thats correct ..otherwiz all NP hard prblms will be solvable...

- atul gupta July 17, 2012 | Flag


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