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dan815
 one year ago
draw a graph with 5 vertices that has exactly 22 cycles.
a cycle has to start and end with the same vertex, and can only intersect all other vertices only once
dan815
 one year ago
draw a graph with 5 vertices that has exactly 22 cycles. a cycle has to start and end with the same vertex, and can only intersect all other vertices only once

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dan815
 one year ago
Best ResponseYou've already chosen the best response.0how to build 22 cyclces exactly with 5 vertices

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0best i can figure is a pentagon. but im only getting 12, maybe you can work off that

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you can solve it with brute for algorithm, if u make sure to get rid of the repeating cycles, but im wondering if there is some simple way of figuring out how to construct cycles

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0do you know the brute force algorithm? id be interested in seeing how you go about doing that.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0No I dont have one yet, but i was thinking of making one if nothing else. I think it can be made with a few simple rules. Like first drawing out the edge connection matrix lets say v1 v2 v3 v4 v5 v1 0 1 1 1 0 v2 1 v3 . v4 . v5 . .......... we would check for every single possible matrix you can see the possible paths for every vertex and only traverse an edge if it doesnt lead to a vertex you have been to already, or if the vertex you have going to is the starting vertex with some condition statements you can solve for all the paths like this, and to eliminate the identical cycles we have to do a check like this if the path is v1 v2 v3 v4 v5 that is same as v2 v3 v4 v5 v1 v3 v4 v5 v1 v2 ... and so on basically like shifted

dan815
 one year ago
Best ResponseYou've already chosen the best response.0you have to also make it follow the official 'graph' definition, no self loops and no multi edges

dan815
 one year ago
Best ResponseYou've already chosen the best response.0that can be added as conditions when generating different edge connection matrices

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I think this is related to the Euler characteristic equation VE+F=2

dan815
 one year ago
Best ResponseYou've already chosen the best response.0how would you incorporate faces into this

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Hmmm I don't know I was thinking the faces would be our cycles or something.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0thats kind of interseting lol

dan815
 one year ago
Best ResponseYou've already chosen the best response.0finding the actual solution for total number of possible cycles given vertices is still an unsolved question

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Well like if you have a loop and you bridge it, you create 2 more loops, can we somehow just build off of this idea?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 I think the solution is the K5 graph with 1 edge missing, can you confirm with a program or something

dan815
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking maybe write a program t hat has all permutations of 1,2,3,4,5 then since we are missing an edge elimnate all the cases where 2 and 5 are together
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