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In the adjoining table, a dot indicates the combination of chemicals that cannot be placed in the same location as the other because of safety issues. If a company is dealing with the chemicals a,b,c,d,e,f,g,h,i,j listed in the table, what is the minimum number of storage locations needed by the company?

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My idea of the solution is something like this:

I start with 'a'. I can collocate it with c,d,f,g,h,i, but not with b,e,j. But then, I note that d and f cannot be collocated. How/where do I go from here?

F. A. Mala
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    Draw the incidence graph and determine whether it's bipartite. – Robert Shore Aug 04 '23 at 00:09
  • Although it does not appear to be so, what if it is bipartite? – F. A. Mala Aug 04 '23 at 00:10
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    If it's bipartite, you only need two shelves. Three is clearly sufficient. You can store F by itself and the remaining graph is bipartite. Shelf 1: ACDGI. Shelf 2: F. Shelf 3: BEHJ. – Robert Shore Aug 04 '23 at 00:14
  • How do you decide bipartite-ness from the incidence graph? – F. A. Mala Aug 04 '23 at 00:17
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    I just drew it. Start with A on one side and connect it via an edge to all "forbidden" pairs. Then repeat with B. Since B is on the other side, put all the "vertices" adjacent to B on the same side as A. Lather, rinse, repeat. It turned out to be easy. – Robert Shore Aug 04 '23 at 00:19
  • May I see what you have drawn, please? – F. A. Mala Aug 04 '23 at 00:21
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    I'm on a desktop and I drew it on a physical piece of paper on my desk, so no. You end up with a graph with ACDGI on one side, BEFHJ on the other side, and the only edge internal to a group is FJ. Just write down A. It's incident to B, E, and J, so put those vertices in the other group and draw edges. Now repeat with B. It's incident to A, D, and G so put D and G in the first group and draw edges. Keep going until you're done. – Robert Shore Aug 04 '23 at 00:23
  • Thank you for such efforts. I appreciate it. – F. A. Mala Aug 04 '23 at 00:24
  • @F.A.Mala A quick way to see that the incidence graph is not bipartite is to notice that it contains a cycle of odd length, ABDFJ. See https://math.stackexchange.com/questions/311665/proof-a-graph-is-bipartite-if-and-only-if-it-contains-no-odd-cycles. – alexg Aug 04 '23 at 10:15

1 Answers1

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Three shelves are sufficient. Shelf $1$: ACDGI. Shelf $2$: F. Shelf $3$: BEHJ.

Three shelves are necessary. Let's assume toward a contradiction that two shelves were possible. F cannot share a shelf with either D or J, so D and J must share the other shelf. D cannot share a shelf with B, so B must be on the same shelf as F. J cannot share a shelf with A so A also must be on the same shelf with F, which also must have B. But A and B cannot share a shelf. Thus, no two-shelf solution is possible.

Robert Shore
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