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The Probabilistic Method prepared and Instructed by Shmuel Wimer Eng. Faculty, Bar-Ilan University 1 The Probabilistic Method June 2015 The probabilistic method comprises two ideas: β’ Any random variable assumes at least one value not smaller than its expectation. β’ If an object chosen randomly from the universe satisfies a property with positive probability, there must be an object of the universe satisfying that property. Theorem. For any undirected graph πΊ π, πΈ with π vertices and π edges, there is a partition of π into π΄ and π΅ such that the edge cut-set has π 2 edges at least, namely π’, π£ β πΈ π’ β π΄ and π£ β π΅ β₯ π 2. 2 The Probabilistic Method June 2015 Proof. Consider the following experiment. Each vertex is independently and equiprobaly assigned to π΄ or π΅. The probability that the end points of an edge π’, π£ are in different sets is ½. By the linearity of expectation the expected number of edges in the cut is π 2. It follows that there must a partition satisfying the theorem.β Consider the satisfiability problem. A set of π clauses is given in conjunctive (sum) normal form over π variables. 3 The Probabilistic Method June 2015 We have to decide whether there is a truth assignment of the π variables satisfying all the clauses (POS). There is an optimization version called MAX-SAT where we seek for a truth assignment maximizing the number of satisfied clauses. This problem is NP-hard. We subsequently show that there is always a truth assignment satisfying at least π 2 clauses. This is the best possible universal guarantee (consider π₯ and π₯). Theorem: For any set of π clauses, there is a truth assignment satisfying at least π 2 clauses. 4 The Probabilistic Method June 2015 Proof: Suppose that every variable is set to TRUE or FALSE independently and equiprobaly. For 1 β€ π β€ π, let ππ = 1 if the clause is satisfied, and ππ = 0 otherwise. Due to the conjunctive form, the probability that a clause containing π literals is not satisfied is 2βπ β€ 1 2 , or 1 β 2βπ β₯ 1 2 that it is satisfied, implying π ππ β₯ 1 2. The expected number of satisfied clauses is therefore π π=1 π ππ β₯ π 2 , implying that there must be an assignment for which π π=1 ππ β₯ π 2. β 5 The Probabilistic Method June 2015 An orientation of a complete graph is called tournament. A Hamiltonian path is an π β 1 -arc uni-directed path. Theorem: (Szele 1943). There is an π-vertex tournament having at least π! 2πβ1 Hamiltonian paths. Proof: for each vertex pair we chose an arc π£π β π£π or π£π β π£π with equal probability, generating a random tournament. Let π be count the number of Hamiltonian paths in the tournament. π is a sum of π! indicator random variables for the possibility that a path is Hamiltonian. 6 The Probabilistic Method June 2015 A Hamiltonian path occurs with probability 1 2πβ1 , hence π π = π! 2πβ1 , and there must be a graph with at least π! 2πβ1 Hamiltonian paths. β 7 The Probabilistic Method June 2015 Expanding Graphs πΊ π, πΈ is called an expanding graph if there is a π > 0 such that for any π β π there is Ξ π > π π , where Ξ π is the set of πβs neighbors. . A particular type of expanding graph is a bipartite multi graph πΊ πΏ, π , πΈ called an OR-concentrator. It is defined by a quadruple π, π, πΌ, π , where πΏ = π = π, such that 1. deg π£ β€ π βπ£ β πΏ, and 2. βπ β πΏ such that π β€ πΌπ there is Ξ π > π π . In most applications it is desired to have π as small as possible and π as large as possible. 8 The Probabilistic Method June 2015 Of particular interest are those graph where πΌ, π and π are constants independent of π and π > 1. Those are strict requirements and it is not trivial task to construct such graphs. We rather show that such graphs exist. We show that a random graph chosen from a suitable probability space has a positive probability of being an π, 18, β , 2 OR-concentrator. (The constants are arbitrary, other combinations are possible.) Theorem: There is an integer π0 such that for all π > π0 there is an π, 18, β , 2 OR-concentrator. 9 The Probabilistic Method June 2015 Proof: The proof is carried out in terms of π, π, and πΌ, while the constants are pinned at the end of the proof. Consider a random πΊ πΏ, π , πΈ , where π£ β πΏ choses its π neighbors Ξ π£ β π independently and uniformly with replacements, and avoid multi edges. Let ππ be the event that for π β πΏ, π = π there is Ξ π < ππ , namely, an OR-concentrator does not exist. Plan: We shall first bound Pr ππ , and then sum over all the values of π β€ πΌπ. We thus obtain an upper bound on the probability that the random πΊ fails to be an ORconcentrator with the parameters we seek. 10 The Probabilistic Method June 2015 Consider π β πΏ, π = π and any π β π , π = ππ . There π π are ways to choose π and ways to choose π. π ππ πΏ π π π π π π ππ π ππ Since ππ edges emanate from π, the probability that π contains all of the at most ππ β₯ π€ π is ππ π ππ . 11 The Probabilistic Method June 2015 The probability of the event that all the ππ edges emanating from some π vertices of πΏ fall within any ππ vertices of π is bounded by Pr ππ π β€ π π ππ ππ ππ π π Substituition of the inequality β€ π Pr ππ ππ β€ π π ππ ππ ππ ππ π ππ = π π ππ π π obtains πβπβ1 π π 1+π π πβπ Using πΌ = 1 3 and s β€ πΌπ, there is 12 The Probabilistic Method June 2015 1 3 Pr ππ β€ π πβπβ1 π 1+π π πβπ π 3 β€ π π 3π π+1 Using π = 2 and π = 18, there is Pr ππ β€ Let π = 2 3 18 2 3 π 18 3π 3 3π 3 , so that π < 12 , there is π 3β₯π β₯1 Pr ππ β€ π β₯1 ππ = π 1βπ < 1, showing that the desired OR-concentrator exists. β 13 The Probabilistic Method June 2015 Crossing Number The crossing number ππ πΊ of a graph πΊ is the smallest number of edge crossings in a planar embedding of πΊ. In VLSI it is the number of jumpers (via) required to layout a circuit. For a planar graph πΊ π, πΈ , π = π, πΈ = π there is ππ πΊ = 0. Euler formula for planar graph states π β π + π = 2. Since a face comprises a least 3 edges, and each edge is shared by two faces, there is 0 = π β π + π β 2 β€ π β π 3 β 2. 14 The Probabilistic Method June 2015 Since ππ πΊ = 0 for a planar πΊ, for any πΊ there is ππ πΊ β₯ π β 3π + 6 for π β₯ 3. (Homework) Stronger lower bound can be derived with the aid of expectation. Lemma: (The Crossing Lemma, proof by N. Alon). Let πΊ be a simple graph with π β₯ 4π. Then ππ πΊ β₯ 1 π3 . 2 64 π Proof: Let πΊ be a planar embedding of πΊ yielding ππ πΊ . Let π β π be obtained by choosing π£ β π randomly with probability π β 4π π. Let π» β πΊ π and π» β πΊ π . 15 The Probabilistic Method June 2015 π» is a planar embedding of π» imposed by πΊ. Let π, π and π be the random variables of the number of vertices, number of edges and the number of crossings in π», respectively. It follows from the trivial lower bound that π β ππ π» β₯ ππ π» β₯ π β 3π + 6 . By linearity of expectation there is πΈ π β₯ πΈ π β 3πΈ π . There is πΈ π = ππ and πΈ π = π2 π (an edge is defined by its two end vertices). Since a crossing is defined by four vertices, there is πΈ π = π4 ππ πΊ = π4 ππ πΊ . 16 The Probabilistic Method June 2015 All in all there is π4 ππ πΊ β₯ π2 π β 3ππ. Dividing by π4 yields ππ πΊ β₯ ππβ3π π3 = π 4π π 3 = 1 π3 . 64 π2 β The Crossing Lemma is useful in combinatorial geometry. Consider π points in the plane and lines passing through each pair of points. π Some of those at most distinct lines might pass 2 through more than two points. 17 The Probabilistic Method June 2015 Given π β₯ 2, how many lines can pass through at least π points? If π is a perfect square and the point are on a π × π grid, there are 2 π + 2 lines passing through π points. π π 18 The Probabilistic Method June 2015 Is there a configuration of π points in the plane yielding more lines passing through at least π points? Theorem: (Szemerédi and Trotter 1983). Let π be a set of π points in the plane, and let π be the number of lines passing through at least π + 1 points of π , 1 β€ π β€ 2 π. Then π < 32π2 π 3 . Proof: Form a graph πΊ with vertex set π. πΊβs edges are the segments between consecutive points of the π lines. πΊ has therefore at least ππ edges and its π crossing number is at most . 2 19 The Probabilistic Method June 2015 If it happens that ππ < 4π, because 1 β€ π β€ 2 π, there is π < 4π π β€ 16π2 π 3 < 32π2 π 3 . Otherwise ππ β₯ 4π, and the Crossing Lemma applies (π = ππ). π β₯ ππ πΊ 2 π3. β It follows from the lemma that π 2 2 > β₯ ππ 20 3 64π2 , yielding again π β€ 32π2 The Probabilistic Method June 2015 Properties of Almost All Graphs Theorem: (Gilbert 1959). Let πΊ be a random graph whose edges have constant probability π. Almost every such graph is connected. Proof: Let us denote the graph by πΊ π , having π vertices. πΊ π can get disconnected by vertex bipartition followed by deletion of the two-sided edges. Plan: We obtain an upper bound the probability ππ that πΊ π is disconnected, by choosing π β π and summing the probabilities π π, π = β over all π, π partitions . 21 The Probabilistic Method June 2015 Let π = π. There are π π β π possible edges in π, π , so π π, π = β = 1 β π π πβπ . By considering all π, 1 πβ1 π there is ππ β€ 2 π=1 1 β π π πβπ . π This inequality is symmetric in π and π β π, so there is π 2 π ππ β€ π=1 1 β π π πβπ . π π There is < ππ . Also, since in the above summation π there is π β π β₯ π 2 and 1 β π < 1 , there is 1 22 The Probabilistic Method June 2015 All in all there is ππ β€ π 2 π=1 π 1βπ π 2 π. For π large enough there is π 1 β π < β π=1 π 1βπ π 2 π = π 2 <1, so ππ π 1βπ π 2 . π 2 1βπ 1βπ We conclude that with π β β, there is ππ β 0, which means that for large enough graphs with constant edge probability the graphs is almost surely connected. β 23 The Probabilistic Method June 2015 Markovβs Inequality and Random Graphs Let πΊπ , ππ , π β₯ 1 be a probability space, πΊπ is a sample space and ππ βΆ πΊπ β 0,1 a probability function satisfying πβπΊπ ππ π = 1. We subsequently explore the existence of few properties in random large graphs. Large means π πΊ = π β β, whereas the probability π of an edge depends on π and satisfies π π β 0. ππ,π denotes the probability space of such graphs. 24 The Probabilistic Method June 2015 Markovβs Inequality states that if π is a nonnegative random variable and π‘ β β, π‘ > 0, then πΈ π π πβ₯π‘ β€ π‘ Markovβs Inequality is applied to show that a πΊ β ππ,π almost surly has a particular property for a certain π. It is obtained by setting π = ππ and π‘ = 1. Corollary: Let ππ β β be a nonnegative random variable in a probability space πΊπ , ππ , π β₯ 1. If πΈ ππ β 0 as π β β, then π ππ = 0 β 1 as π β β. 25 The Probabilistic Method June 2015 Asymptotic Behavior of Graphs Example: We are interested in the number π of triangles in πΊ β ππ,π . π can be expressed as the sum π= ππ βΆ π β π, π = 3 , where ππ is the indicator random variable for the event π΄π that πΊ π is a triangle. ππ = 1 if π imposes a triangle and ππ = 0 otherwise. By the expectation definition there is πΈ ππ = π ππ = 1 . There is π π΄π = π3 . 26 The Probabilistic Method June 2015 By linearity of expectation, there is π 3 πΈ π = πΈ ππ βΆ π β π, π = 3 = π < ππ 3 . 3 Thus if ππ β 0 as π β β, thenπΈ π β 0, so π π = 1 β 0 and π π = 0 β 1. It means that if ππ β 0 as π β β, πΊ will almost surly be triangle-free. We are subsequently interested in the probability of having the independent sets in a graph of π vertices and edge probability π, not exceeding a certain size, which of course depends on π. 27 The Probabilistic Method June 2015 Theorem: (ErdΕs 1961). A random graph πΊ β ππ,π almost surely has stability number (size of maximal independent set) no larger than 2πβ1 log π . The theorem states that if the probability of an edge is fixed, it is very difficult to find an independent set of size that grows with π, even very slowly as log π. Proof: Let π β π πΊ , π = π + 1, π β β. π is pinned down later. The probability that π is an independent set is the probability that none of the vertex pairs has a connecting edge, namely, 1 β π 28 π+1 2 The Probabilistic Method . June 2015 Let π΄π be the event that π is an independent set and let ππ be the corresponding indicator random variable. There is πΈ ππ = π ππ = 1 = π π΄π = 1 β π π+1 2 . Let π be the number of independent sets of size π + 1. Then π = ππ βΆ π β π, π = π + 1 . By linearity of expectation there is πΈ π = πΈ ππ βΆ π β π, π = π + 1 = π π+1 29 1βπ π+1 2 The Probabilistic Method . June 2015 π There is β€ π+1 ππ+1 π+1 ! and 1 β π < π βπ . Substitution in πΈ π yields πΈ π β€ π π+1 βπ π+1 2 π π+1 ! = π+1 βππ 2 ππ π+1 ! Let us now pin down π, supposing π = 2πβ1 log π . Then π β₯ 2πβ1 log π, and by exponentiation there is ππ βππ 2 β€ 1, hence 1 πΈ π β€ π+1 ! 30 The Probabilistic Method June 2015 Since π β₯ 2πβ1 log π, π grows at least as fast as log π, hence πΈ π β 0 as π β β. Recall the corollary stating that if πΈ π β 0 as π β β, then π π = 0 β 1 as π β β. That means that with probability approaching 1, ππ,π has no independent set larger than π, or in other words, its stability number is at most π. β 31 The Probabilistic Method June 2015 Theorem. If π is a constant then almost every πΊ π has diameter 2 (and hence connected). Proof. Recall that the distance between two vertices is the edge length of the shortest path connecting them. The diameter of a graph is the maximum of the distance taken over all vertex pairs. Let π πΊ π count the number of unordered vertex pairs which distance is larger than 2 , hence having no common neighboring vertex. If there are none such pairs, then πΊ π is connected and has diameter 2. 32 The Probabilistic Method June 2015 π πΊ π is a random variable. If it would happen that πΈ π β 0 as π = π β β then it follows by Markovβs Inequality that the theorem holds. For two vertices π£π , π£π β π let πππ be an indicator random variable specifying that they do not share a common neighboring vertex. πππ = 1 would happen if there is no common neighboring vertex. For each of the other π β 2 vertices the probability it does not connect to either of π£π , π£π is 1 β π2 . Hence π πππ = 1 = 1 β π2 πβ2 . 33 The Probabilistic Method June 2015 π There are distinct vertex pairs. The random variable 2 π π is a sum of the random variables πππ . 2 If follows from the linearity of expectation that πΈ π π = 1 β π2 πβ2 . 2 Since π is constant while π β β, there is πΈ π β 0. Consequently, almost every πΊ π has diameter 2, and is also connected. β This theorem is stronger than Gilbertβs theorem. While the latter states that almost every πΊ π is connected, this one provides also the diameter. 34 The Probabilistic Method June 2015