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I got chu

i want to see this O.O

|dw:1444167912740:dw|

Give me y = mx + b nub

lol

-.-

stupid cats

Dan is mad cuz he forgot teh basics

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

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).

that is a nice one :)

what does it mean independent vertices and their neighbors

they exist on the same plane

get it?

not really a bit confused still

what does it mean 3-connected

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

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

lol its just kinnda intuitive

ohh i see now

yea

cool

YES

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

you got it

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

it all depends on the independent vertices

yeah

kinda seems hard lol

it is

but its fun

are you a professor?

or college student?

yeah it seems fun :D and i am

(: im a girl

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

lol