Link 1: http://i.imgur.com/bpRCfjq.jpg
Link 2: Euler Circuit: http://i.imgur.com/PW7j5q3.jpg
Okay the Euler circuit is pretty simple i think. I will update my answer later when i can upload it.
Each vertex has an even degree. and there are multiple ways to draw the circuit. But as along as i used every edge at least once AND ONLY ONCE (even though i touched vertices MULTIPLE times) It is a Euler circuit.
So is this correct? -
For 1.a, that is just the number n points taken 2 at a time.Originally Posted by outkast32
For 1.b, each point in one cell is joined with each point in the other cell.
In #3, there is an Euler Circuit provided the graph is connected and each vertex is of even degree.
Okay so for 1a. Am i getting this right? as a satisfactory answer:
There are as many edges in K_n as there are subsets of both endpoints (2) that can be chosen from a set of n vertices. The number of ways to select 2 vertices from n vertices is n(n-1) / 2 aka n choose 2.
Thus the # of edges in K_n is n(n-1)/2
Is that how to "show" this?
Then for 1b.
the one part of graph has m vertices each with degree n an the other part has n vertices each with degree m..
i see that m*n = e, if make my own little graph but i dont quite see how that comes from what i said above.
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