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Often, the [[probabilistic method]] is used to prove the existence of mathematical objects with some desired combinatorial properties. The proofs in that method work by showing that a random object, chosen from some probability distribution, has the desired properties with positive probability. Consequently, they are [[nonconstructive proof|nonconstructive]] — they don't explicitly describe an efficient method for computing the desired objects.
The method
The method is particularly relevant in the context of [[randomized rounding]] (which uses the probabilistic method to design [[approximation algorithm]]s).
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Unfortunately, in most applications, the conditional probability of failure is not easy to compute efficiently. There are two standard and related techniques for dealing with this:
=== Using a conditional expectation ===
In this case, to keep the conditional probability of failure below 1, it suffices to keep the conditional expectation of ''Q'' below (or above) the threshold. To do this, instead of computing the conditional probability of failure, the algorithm computes the conditional expectation of ''Q'' and proceeds accordingly: at each interior node, there is some child whose conditional expectation is at most (at least) the node's conditional expectation; the algorithm moves from the current node to such a child, thus keeping the conditional expectation below (above) the threshold.
=== Using a pessimistic estimator ===
== Example using conditional expectations ==
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In this case, the conditional probability of failure is not easy to calculate. Indeed, the original proof did not calculate the probability of failure directly; instead, the proof worked by showing that the expected number of cut edges was at least |''E''|/2.
Let random variable ''Q'' be the number of edges cut. To keep the conditional probability of failure below 1, it suffices to keep the conditional expectation of ''Q'' at or above the threshold |''E''|/2.
Given that some of the vertices are colored already, what is this conditional expectation? Following the logic of the original proof, the conditional expectation of the number of cut edges is
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4. Otherwise, color ''u'' black.
Because of its derivation, this deterministic algorithm is guaranteed to cut at least half the edges of the given graph.
== Example using pessimistic estimators ==
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Clearly the process computes an independent set. Any vertex ''u'' that is considered before all of its neighbors will be added to ''S''. Thus, letting ''d''(''u'') denote the degree of ''u'', the probability that ''u'' is added to ''S'' is at least 1/(''d''(''u'')+1). By [[Expected value#Linearity|linearity of expectation]], the expected size of ''S'' is at least
: <math>\sum_{u\in V} \frac{1}{d(u)+1} ~\ge~ \frac{|V|}{D+1}.</math>
(The inequality above follows because 1/(''x''+1) is [[Convex function|convex]] in ''x'', so the left-hand side is minimized, subject to the sum of the degrees being fixed at 2|''E''|, when each ''d''(''u'') = ''D'' = 2|''E''|/|''V''|.) ''QED''
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== Further reading ==
The method of conditional rounding is
* {{Cite book |first1=Noga |last1= Alon |authorlink1=Noga Alon
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