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- dan815

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

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- dan815

|dw:1442366092655:dw|

- dan815

how to build 22 cyclces exactly with 5 vertices

- dan815

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- anonymous

best i can figure is a pentagon. but im only getting 12, maybe you can work off that

- dan815

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

- dan815

|dw:1442368797479:dw|

- anonymous

do you know the brute force algorithm? id be interested in seeing how you go about doing that.

- dan815

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

- dan815

you have to also make it follow the official 'graph' definition, no self loops and no multi edges

- dan815

that can be added as conditions when generating different edge connection matrices

- Empty

I think this is related to the Euler characteristic equation V-E+F=2

- dan815

what do you mean

- dan815

how would you incorporate faces into this

- Empty

Hmmm I don't know I was thinking the faces would be our cycles or something.

- dan815

thats kind of interseting lol

- dan815

finding the actual solution for total number of possible cycles given vertices is still an unsolved question

- Empty

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?

- dan815

@ganeshie8 I think the solution is the K5 graph with 1 edge missing, can you confirm with a program or something

- dan815

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

- dan815

|dw:1442450749719:dw|

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