Wikipedia:Reference desk/Archives/Mathematics/2008 October 26
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October 26
editDual Graph Problem...
editNeed comments about the Dual Graph. One good application of dual graph is to create a complementary circuit which works in the opposite way to the original circuit. I have an example that does not satisfy this rule. Please let me know why this example does not satisfy, or whether I miss something in this case.
http://commons.wikimedia.org/wiki/Image:DualGraph.JPG#file
In the above graph, the conducting path 1-3-5 or 2-3-4 between A & B in graph G is also a conducting path in dual graph G'(1*-3*-5* or 2*-3*-4*). According to the theory of dual graph, G and G' should work in a complementary way. Please someone explains this problem.
Hwang000 (talk) 09:27, 26 October 2008 (UTC)
- You've drawn the dual graph right so no problem there. I guess then you are talking about the application of Kirchhoff's circuit laws to the dual. The complementary way means it's as if you had the inverse of the resistances in the circuit, see Duality (electrical circuits) for more on this. Dmcq (talk) 11:09, 26 October 2008 (UTC)
- How is that dual graph correct? Shouldn't the dual graph have as many vertices as the graph has faces, which in this case is 3? --Tango (talk) 15:45, 27 October 2008 (UTC)
- I think those funny dotted lines are edges. Algebraist 23:24, 27 October 2008 (UTC)
- Ah, yes, that makes it work. I'd assumed they were just there to show which two points the OP wanted to get between. --Tango (talk) 23:57, 27 October 2008 (UTC)
- I think those funny dotted lines are edges. Algebraist 23:24, 27 October 2008 (UTC)
- How is that dual graph correct? Shouldn't the dual graph have as many vertices as the graph has faces, which in this case is 3? --Tango (talk) 15:45, 27 October 2008 (UTC)
No, the complementary way means that the conducting path 1-3-5 or 2-3-4 between A and B in graph G should not form a conducting path in dual of G. Therefore, the path 1*-3*-5* or 2*-3*-4* between A and B in G' shouldn't be a conducting path, but in the given example, it is a conducting path again. What is wrong in this case?? unsigned comment added by Hwang000 (talk • contribs) 10:25, 27 October 2008 (UTC)
- I'm not certain where you got this rule about conducting paths from or what it means exactly. All the dual does is swap over voltages and currents, resistance and conductance and a load of suchlike if you have them. So for instance parallel and series expression for resistors stuck together can be derived from duality if one likes. Dmcq (talk) 16:30, 27 October 2008 (UTC)
Anyway, thanks for your comments. Let me give you the reference about this rule. The author is C. L. Liu and the title of this book is "Introduction to combinatorial mathematics" published by McGraw-Hill Company in 1968. Please refer to it on pages 227-230. It is an old book, but still valuable one. I want to know whether what I have found is a counter example, or I misunderstand something here.... Hwang000 (talk) 04:37, 28 October 2008 (UTC)
- Oh sorry I assumed from current you were talking about electrical circuits and solving meshes. In the combinatorial books they normally talk about transportation networks instead so I'd guess that might be more relevant. There they have the concept of a line across the network through which all the goods must flow. Have no access to that book sorry. I'll have a look to see if there is a mention of that term somewhere else. Dmcq (talk) 12:20, 28 October 2008 (UTC)