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dan815
 one year ago
Best ResponseYou've already chosen the best response.0gimme a sec i gotta see which ones i have to go over

dan815
 one year ago
Best ResponseYou've already chosen the best response.0What's a 1 factor and 2 factor of this graph

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i dont know if u can repeat edges in a 2 factor or not

Empty
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444270981730:dw dw:1444271018570:dw

dan815
 one year ago
Best ResponseYou've already chosen the best response.0why dont they just say that hehe

Empty
 one year ago
Best ResponseYou've already chosen the best response.0all vertices must have kdegree in a kfactor. So you can see that in a 1factor they're all matched up. Why the hell they don't say that idk but I just learned this from wikipedia just now.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0lets try to prove this lemma

Empty
 one year ago
Best ResponseYou've already chosen the best response.0There are multiple ways to 1factor this graph but only one way to 2factor it and of course no way to 3factor it.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0A graph G is a tree if and only if there is only 1 path that exists exactly one path between any 2 vertices

Empty
 one year ago
Best ResponseYou've already chosen the best response.0i hate proofs like this dan

dan815
 one year ago
Best ResponseYou've already chosen the best response.0if 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

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i know its obvious lol but

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i should practice writing this, since it might come up tmr

Empty
 one year ago
Best ResponseYou've already chosen the best response.0no this is why I hate proofs and rant about squares, let's do some other graph theory problem that's more fun.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0well i gotta go over 3 other theorems in this series, they might be more involved

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Alright, 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0If a pseudograph G has an euler circuit, then G is connected and the degree of every vertex is even

dan815
 one year ago
Best ResponseYou've already chosen the best response.0a psedugraph is a Graph that allows multiple edges from 1 vertex to another and loops for example

dan815
 one year ago
Best ResponseYou've already chosen the best response.0a loop is seen as a 2 degree connection

dan815
 one year ago
Best ResponseYou've already chosen the best response.0An 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.

dan815
 one year ago
Best ResponseYou've already chosen the best response.0If a pseudograph G has an euler circuit, then G is connected and the degree of every vertex is even

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay lets show this is true

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Ok sure, makes sense how would we prove this

dan815
 one year ago
Best ResponseYou've already chosen the best response.0ull get stuck at a vertex without being able to return to your starting vertex if u dont have even

Empty
 one year ago
Best ResponseYou've already chosen the best response.0if you enter a point you have to leave that point so there has to be an even degree at every vertex. Proved?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0no OKAY LEMME SEE what else they say

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hmm but why did they ask this

dan815
 one year ago
Best ResponseYou've already chosen the best response.0oh wait its not obvious

dan815
 one year ago
Best ResponseYou've already chosen the best response.0u gotta show that any circuit u get is part of the euler circuit

Empty
 one year ago
Best ResponseYou've already chosen the best response.0What does it mean to be a Connected Graph?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0a circuit is where u cant repeat edges, but u can repeat vertcies, for a psedugraph

dan815
 one year ago
Best ResponseYou've already chosen the best response.0connected means all the vertices are connected together no isolated vertices

dan815
 one year ago
Best ResponseYou've already chosen the best response.0Like there has to be a path between any 2 given vertices

dan815
 one year ago
Best ResponseYou've already chosen the best response.0then its a connected graph

dan815
 one year ago
Best ResponseYou've already chosen the best response.0i think im done, ill just read the book for this thm!! lets do quantum stuff, i got a midterm on that too day after

dan815
 one year ago
Best ResponseYou've already chosen the best response.0this 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

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

Empty
 one year ago
Best ResponseYou've already chosen the best response.0I 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. \(\Huge \square\)

dan815
 one year ago
Best ResponseYou've already chosen the best response.0what do u mean split every vertex apart

Empty
 one year ago
Best ResponseYou've already chosen the best response.0Suppose 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:

Empty
 one year ago
Best ResponseYou've already chosen the best response.0We 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. \[\huge \square\]

Empty
 one year ago
Best ResponseYou've already chosen the best response.0lets do quantum stuff yeah

dan815
 one year ago
Best ResponseYou've already chosen the best response.0okay i have these bunch of problems to work thru for preparation

dan815
 one year ago
Best ResponseYou've already chosen the best response.0hold on ill find em leme close this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dan815 can you help me on a calculus problem please ?

dan815
 one year ago
Best ResponseYou've already chosen the best response.0sec im tryna do something rn
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