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By Richman F. (ed.)

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If sd consists of n sets of arbitrary size Straight probabilistic methods could not achieve this result. If a set has size x then a random coloring gives discrepancy about x 1/2 which can be arbitrarily large. Combining linear algebra with the probabilistic method is very powerful. Note that all the steps are algorithmic. There is a polynomial (in m, n) time algorithm that gives the desired coloring. Open Problem. What is the maximal cn so that if ) herdisc (s4)l To show cn > 0 consider the proof of lindisc (jtf) < herdisc (s£).

The last surviving small component is an isolated point; when it is joined the graph becomes connected. Erdos and Renyi found a surprisingly precise description of when this occurs. THEOREM. Let p = p ( n ) = (In n)/n + c/n. Then Proof. Let A, be the event "/ is isolated," Xt the associated indicator random variable and X = Xr + - • - + Xa the number of isolated points. Set p = Pr [A,] = (\-pY~1 so that fji ~ e~pn = e~c/n. Then Let us show X is asymptotically Poisson. According to our earlier notation by symmetry.

Note that the following does not work: Take an independent set of size 2 Ig w, color it "1," delete it, and iterate. The problem is that the remaining G' cannot be treated as a random graph. Sparse graphs. Let G have n vertices with edge probability p = ne~l. Let X ( k ) be the number of k- cliques in G. Then For k>(2e\nn)/p, E [ X ( k ) ] « 1 and so w(G)^k. Set d = np = n\ the average degree. Then The selection procedure described for p = \ works also in this case and gives An early gem. I find the probabilistic method most striking when the statement of the theorem does not appear to call for its use.

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