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how to build 22 cyclces exactly with 5 vertices
best i can figure is a pentagon. but im only getting 12, maybe you can work off that
you 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
do you know the brute force algorithm? id be interested in seeing how you go about doing that.
No 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
you have to also make it follow the official 'graph' definition, no self loops and no multi edges
that can be added as conditions when generating different edge connection matrices
I think this is related to the Euler characteristic equation V-E+F=2
what do you mean
how would you incorporate faces into this
Hmmm I don't know I was thinking the faces would be our cycles or something.
thats kind of interseting lol
finding the actual solution for total number of possible cycles given vertices is still an unsolved question
Well like if you have a loop and you bridge it, you create 2 more loops, can we somehow just build off of this idea?
I 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