Graph Theory Review http://prntscr.com/8oyhbn

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Graph Theory Review http://prntscr.com/8oyhbn

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

gimme a sec i gotta see which ones i have to go over
One Mississippi...
What's a 1 factor and 2 factor of this graph

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

|dw:1444270569479:dw|
i dont know if u can repeat edges in a 2 factor or not
|dw:1444270981730:dw| |dw:1444271018570:dw|
ohh okay cycle hehe
why dont they just say that hehe
all vertices must have k-degree in a k-factor. So you can see that in a 1-factor they're all matched up. Why the hell they don't say that idk but I just learned this from wikipedia just now.
ok great
lets try to prove this lemma
There are multiple ways to 1-factor this graph but only one way to 2-factor it and of course no way to 3-factor it.
A graph G is a tree if and only if there is only 1 path that exists exactly one path between any 2 vertices
i hate proofs like this dan
if u see an if and only if, we gotta prove both ways Show if you have a tree, then only 1 path between and 2 vertices if you have only 1 path between any 2 vertcies, its a tree
i know its obvious lol but
i should practice writing this, since it might come up tmr
no this is why I hate proofs and rant about squares, let's do some other graph theory problem that's more fun.
well i gotta go over 3 other theorems in this series, they might be more involved
Alright, I can help try to figure out stuff that's not like intuitively obvious I think. But I won't try to get into proving too much, you gotta figure that part out on your own.
|dw:1444271858604:dw|
If a pseudograph G has an euler circuit, then G is connected and the degree of every vertex is even
a psedugraph is a Graph that allows multiple edges from 1 vertex to another and loops for example
|dw:1444272096126:dw|
a loop is seen as a 2 degree connection
An Euler circuit is a circuit that uses every edge of a graph exactly once. ▶ An Euler path starts and ends at different vertices. ▶ An Euler circuit starts and ends at the same vertex. Page 2.
If a pseudograph G has an euler circuit, then G is connected and the degree of every vertex is even
okay lets show this is true
Ok sure, makes sense how would we prove this
ull get stuck at a vertex without being able to return to your starting vertex if u dont have even
if you enter a point you have to leave that point so there has to be an even degree at every vertex. Proved?
ya lol
\[\Huge\square\]
hahahaha
"HUDE BOX"
no OKAY LEMME SEE what else they say
ok haha
okay next
http://prntscr.com/8oypzs
okay same thing
hmm but why did they ask this
oh wait its not obvious
u gotta show that any circuit u get is part of the euler circuit
What does it mean to be a Connected Graph?
a circuit is where u cant repeat edges, but u can repeat vertcies, for a psedugraph
connected means all the vertices are connected together no isolated vertices
|dw:1444272700171:dw|
Like there has to be a path between any 2 given vertices
then its a connected graph
Ahh ok gotcha.
i think im done, ill just read the book for this thm!! lets do quantum stuff, i got a midterm on that too day after
this one is basically just showing since its connected if u get one smaller circult u can always attach it to a bigger circuit until its an euler circuit
for example
I don't know what constitutes as proof. It has all even number of edges you can split every vertex apart so that only two edges touch it. Then if you find a vertex with 2 edges you can turn that into a single edge until you have just one loop left.
|dw:1444273055412:dw|
I don't know what constitutes as proof. It has all even number of edges you can split every vertex apart so that only two edges touch it. Then if you find a vertex with 2 edges you can turn that into a single edge until you have just one loop left. \(\Huge \square\)
what do u mean split every vertex apart
|dw:1444273207762:dw|
ah thats nice :)
Suppose though you want to prove a little deeper like, maybe a cut will sever it like this! |dw:1444273374845:dw| Proof that this will never hapen coming up:
We know the graph is connected: |dw:1444273416153:dw| Now let's say we pick one arbitrary way of cutting it so that it makes the graph no longer connected, then we know since it's connected the graph looks something like this: |dw:1444273477507:dw| So we always have a way to cut two vertices apart so that the graph remains connected. \[\huge \square\]
lets do quantum stuff yeah
okay i have these bunch of problems to work thru for preparation
hold on ill find em leme close this
@dan815 can you help me on a calculus problem please ?
sec im tryna do something rn

Not the answer you are looking for?

Search for more explanations.

Ask your own question