Anyone know any interesting open graph theory problems, unsolved problems that we could possibly write a paper on

- dan815

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- ShadowLegendX

I got chu

- anonymous

i want to see this O.O

- ShadowLegendX

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## More answers

- ShadowLegendX

Give me y = mx + b nub

- anonymous

lol

- dan815

-.-

- dan815

stupid cats

- ShadowLegendX

Dan is mad cuz he forgot teh basics

- anonymous

are you referring to Paul Erd˝os theories or no?

- anonymous

how about this
Is it true that every k-connected (k>1) graph which does not have a Hamiltonian cycle has a cycle that contains k independent vertices and their neighbors? This is known to be true for k = 2 and 3. For example, the graph to the right is 3-connected but not Hamiltonian. And the dotted cycle shown contains 3 independent vertices (the three vertices which are lighter in color) and thier neighbors. To see that it is not Hamiltonian, notice that this graph is just the complete bipartite graph K(3,4).

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- anonymous

that is a nice one :)

- dan815

what does it mean independent vertices and their neighbors

- anonymous

they exist on the same plane

- anonymous

get it?

- dan815

not really a bit confused still

- dan815

what does it mean 3-connected

- ikram002p

my favorite is Euler trail but it already had a solution so :-\

- anonymous

vertices act as vertices. 3- means it has 3 independent vertices

- anonymous

lol its just kinnda intuitive

- anonymous

yea.. have fun i did this a while back and i really don't feel like doing it again but i had it saved so yea

- dan815

ohh i see now

- anonymous

yea

- anonymous

cool

- dan815

interesting.. so this thm says a cycle must exist such that it contains all the set of k independant vertices?

- anonymous

YES

- dan815

depending on the vertices you pick you will get a different number for k

- anonymous

you got it

- dan815

i cant say the last statement for sure, it seems like its true though

- anonymous

it all depends on the independent vertices

- dan815

yeah

- dan815

kinda seems hard lol

- anonymous

it is

- anonymous

but its fun

- anonymous

are you a professor?

- anonymous

or college student?

- dan815

yeah it seems fun :D and i am

- anonymous

Nice well im only a Senior in HS so i guess you would be able to explain to yourself more then i could at this point lol

- dan815

(: im a girl

- ikram002p

@Musicdude her name is danika for any other details i know everything about this girl :D

- anonymous

lol

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