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.
I count 6 edges connected to these vertices (circled) |dw:1442027885574:dw|
Oh, ok. So I just erase one line ?
but then the outer vertices would only have 4 edges
Oh, that's right. Do I use the 2nd graph I have instead?
that won't work either because of the same reason as above that bridge edge makes 2 nodes have 6 connecting edges
this seems to work though |dw:1442028587295:dw|
|dw:1442028753021:dw| the number in each circle represents the degree of each node
where is the curved not allowed to be plugged in?
what do you mean?
I was trying to do this problem: Construct a simple graph on 6 vertices with 12 edges that does not contain K4 as a subgraph. and I did a graph like...|dw:1442028768219:dw|
I was told this was impossible to do.
so I'm reading that K4 is a complete graph with 4 vertices shown on page 3 of this pdf http://www.dehn.wustl.edu/~blake/circles/talks/2009-jan18-Russ_Woodroofe-Graph_Theory.pdf is that the notation your teacher is using?
Oh, I did the problem. I was just asking about the curves. Is there a rule in which they can't be used for the graphs?
I'm still not sure what you mean? In my drawing? or the one you posted?
Maybe I heard wrong...
the edges can be curved, yes http://www.math.uvic.ca/faculty/gmacgill/guide/M222Graphs.pdf quoting that PDF ` Graphs are usually represented pictorially with a point (or dot) in the plane corresponding to each vertex and a line segment (or curve of some sort) joining the corresponding points for each pair of adjacent vertices.`
Hmm, can you check another problem for me?
Since it's a cycle, I'm sure it's supposed to return to its starting point, no?
yeah you have to make a circuit and not repeat vertices
so they want you to find a circuit with 20 edges where you don't repeat vertices
your answer looks great
Thank you for checking my answers.