Re: Three quesition about graph coloring


milochen wrote:
I am sorry about that I forget that some symbol couldn't use here.

so,I repost about my question.

X(G):= chromatic number of simple graph G
A ^S^ B:= the intersection of A and B
A vSv B:= the union of A and B

Define a function CLRS: {graphs} -> { sets of function(s)}
s.t. G |-> {f | f:V(G)->{1,2,...,X(G)} is proper coloring}


Suppose G, H be any simple graph...
Is "CLRS(G)=CLRS(H) <-> G is isomophic to H " ?

If V(G) is not equal to V(H), then CLRS(G) cannot equal CLRS(H); in
fact, they are disjoint. (CLRS(G) contains functions with domain V(G),
and CLRS(H) contains functions with domain V(H).)

So let's assume V(G) = V(H); then CLRS(G) = CLRS(H) looks equivalent.

Is always exist simple graph F s.t. CLRS(F)=CLRS(G) ^S^ CLRS(H)?

F = G union H ? (V(F) := V(G) = V(H), E(F) = E(G) union E(H).) This
shouldn't be too difficult to prove.

Is always exist simple graph F s.t. CLRS(F)=CLRS(G) vSv CLRS(H)?

Well, F = G intersected H doesn't work, even if n=3. Let V(G) = V(H) =
{1,2,3}, and G has edge (1,2), and H edge (1,3). Then the coloring c =
(1,1,1) is in CLRS(F) but not in CLRS(F) or CLRS(G). The answer to this
might be NO, and these graphs G and H might provide a counterexample.

At first, I never know the discussion for set of coloring.
I just get the idea, but I'm not ensure whether it's good or bad.
I just get a new thinking about this.

Is any book ever talk about something like this?

Not any that I know of right away. Usually colorings are only
considered for one graph.

--- Christopher Heckman

.

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