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I got chu
i want to see this O.O
Give me y = mx + b nub
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
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
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
ohh i see now
interesting.. so this thm says a cycle must exist such that it contains all the set of k independant vertices?
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
kinda seems hard lol
but its fun
are you a professor?
or college student?
yeah it seems fun :D and i am
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
(: im a girl
@Musicdude her name is danika for any other details i know everything about this girl :D