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Yep, I found. Build the adjacency matrix for W3.<br> 1) Consider (# of ways to choose i vertices), where i=1,2,3,4.<br> 2) Also consider # of places for edges in each configuration (with 1, 2, 3, 4 vertices) - build the edge or not (2 choices).<br> 3) Combine these using product & sum rules of counting.<br> Hope it helps<br>--[[User:Asuleime|Asuleime]] 22:45, 5 November 2008 (UTC) | Yep, I found. Build the adjacency matrix for W3.<br> 1) Consider (# of ways to choose i vertices), where i=1,2,3,4.<br> 2) Also consider # of places for edges in each configuration (with 1, 2, 3, 4 vertices) - build the edge or not (2 choices).<br> 3) Combine these using product & sum rules of counting.<br> Hope it helps<br>--[[User:Asuleime|Asuleime]] 22:45, 5 November 2008 (UTC) | ||
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| + | I also thought of another way of doing this. Draw the graph with all the edges. (It should be a triangle with another vertex in the center, all connected.) Counting the edges, there are 6. Consider every edge to be a binary function of either on or off, therefore there are 2^6 - 1 ways when you subtract the null value graph. | ||
Revision as of 21:31, 5 November 2008
Has anyone found a method for counting the number of subgraphs of W3 without actually drawing out every graph and counting them?
--Aoser 16:51, 3 November 2008 (UTC)
Yep, I found. Build the adjacency matrix for W3.
1) Consider (# of ways to choose i vertices), where i=1,2,3,4.
2) Also consider # of places for edges in each configuration (with 1, 2, 3, 4 vertices) - build the edge or not (2 choices).
3) Combine these using product & sum rules of counting.
Hope it helps
--Asuleime 22:45, 5 November 2008 (UTC)
I also thought of another way of doing this. Draw the graph with all the edges. (It should be a triangle with another vertex in the center, all connected.) Counting the edges, there are 6. Consider every edge to be a binary function of either on or off, therefore there are 2^6 - 1 ways when you subtract the null value graph.
