Question

****ANY USEFUL INPUT IS WELCOME****(Don't be shy)
I would like to develop a graph theory approach to this question. http://assets.openstudy.com/updates/attachments/5538891fe4b0f0fc37d2cc09-javk-1429768508431-capture1.png

Answer 1

so we'd need to generalize the graph

Answer 2

I'm thinking something like we have 9 possible scenarios right?

Answer 3

ABC(We have edges between these sets)
AB
AC
BC
A
B
C

Actually only 7

Answer 4

So we only have three non-isomorphic graphs

Answer 5

do you agree?

Answer 6

The edges represent what exactly? An element being in those two sets?

Answer 7

Yea, I was using to represent an element in both sets

Answer 8

So How can we combine this into one graph to have the properties necessary? Any thoughts?

Answer 9

As far as the "properties" go, are we attempting to replicate the membership table?

Answer 10

the membership table. I want to create an adjacency matrix that is similar to the membership table

Answer 11

that yields the same information

Answer 12

We would need to have more points that represent different things it seems. So instead of representing the sets, what if we represented the intersection of sets.

Answer 13

Yeah, maybe the points are really the 8 disjoint regions...

Answer 14

so something like \[{D:=A\cap B\cap C}\\E:=A\cap B\\F:=A\cap C\\G:=B \cap C\]....maybe not.....

Answer 15

So adjacency matrix is \[\begin{matrix}0&0&0&0&1&1&0\\0&0&0&0&1&0&1\\0&0&0&0&0&1&1\\0&0&0&0&1&1&1\\1&1&0&1&0&0&0\\1&0&1&1&0&0&0\\0&1&1&1&0&0&0 \end{matrix}\]

Answer 16

that's a 7x7.... issue you guys had a 8x7

Answer 17

only two colors needed for a propercoloring

Answer 18

Then, again our "points" of interest were very different from these "simple" sets.

Answer 19

true, but it is close.

Answer 20

I'm wondering if this is an accurate model

Answer 21

\[E\cap F\cap G\] is extraneous

Answer 22

This is what I'm trying to model http://assets.openstudy.com/updates/attachments/5538891fe4b0f0fc37d2cc09-jtvatsim-1429770521634-partii.jpg

Answer 23

Gotta check out here, but good luck, I'll be curious to see if there's a way to make this connection. :)

Answer 24

Thanks, I'm gonna think on it

Answer 25

So I guess we could say X:=G here... so All points that are not directly adjacent to A or 1-connected to a point satisfying that are in X

Answer 26

Then because Y(our D point) is defined as being connected to G,E,F It does not satisfy the conditions of X and therefore is not a subset

Answer 27

I think that answers the question if we talk about it this way... hmmmm any thoughts?

Answer 28

maybe define X as any vertex that does not have a two or less path leading to A.

Answer 29

nope that last doesn't work

Answer 30

actually it does work because then we only get G and since there are no subsets of G that's all we should get