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CSCA67H3 (14)
Anna (1)

# _w4.pdf

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School
University of Toronto Scarborough
Department
Computer Science
Course
CSCA67H3
Professor
Anna
Semester
Fall

Description
Cliques in Graphs Deﬁnition. A clique in a graph is a set S of vertices such that ev- ery pair of vertices in S are adjacent. If the clique has n vertices, it is denoted by Kn. ▯▯ Finding maximal sized cliques in large graphs is a challenging problem in data mining, i.e., analyzing data clustering. Maximal clique detection arises in bioinformatics, analysis of so- cial networking patterns, web page clustering etc. Cliques and 3-SAT Recall 3-SAT: Given a set of clauses C = {C ,C1,..2,C k} each of size 3, over a set of variables X = {x 1x ,2..,x n }, does there exist a truth assignment that satisﬁes C? 1 3-SAT and Cliques Suppose we have a 3-SAT formula: F :( a OR b OR d) AND (b OR !c OR d) AND (!a OR c OR !d) AND (a OR !b OR !c) Consider the following algorithm to construct a graph from F: 1. Create a vertex for each literal in the formula. Use the literal to label the vertex. 2. Group the vertices corresponding to a clause together. 3. Connect two vertices in the graph if they are: • In different clauses • and are not a negation of each other. Example ▯ ▯ ▯ ▯ ▯ ▯▯ ▯▯ ▯ ▯▯ ▯▯ ▯ ▯▯ 2 3-SAT and Cliques Q. If there is a clique in our graph equal to the number of clauses in F (so 4), what does this tell us about F? A. ▯ ▯ ▯ ▯ ▯ ▯▯ ▯▯ ▯ ▯▯ ▯▯ ▯ ▯▯ F :( a OR b OR d) AND (b OR !c OR d) AND (!a OR c OR !d) AND (a OR !b OR !c) Q. Notice that there are several possib4e K s in G. What does this mean? A. Q. If there does not exist a clique of size equal to the number of clauses then what does this tell about F? A. 3 Clique Q. Why? A. Deﬁnition. The problem known as Clique is the following: Given a graph G and a natural number k, does there exist a clique of size k in G? Q. What have we shown about Clique and 3-SAT? A. Q. Why? A. This technique of showing that one problem is as hard as another is called a reduction. 4 Proof By Induction Proof by Induction is a very powerful tool when used correctly. Visual: The Domino Argument Think of a row of dominos. If set up properly, when the i th st domino falls, so too will the i +1 domino for all values of i. Therefore, when the ﬁrst domino is knocked over so too are all the dominos. We use it to prove a statement S(n) is true for all natural numbers n larger than b where b is a natural number. Note: The natural numbers, denoted N, are the “counting num- bers” *: 0,1,2,3,...,i,i +1 ,i +2 ,... *In computer science we usually include 0 in N. 5 Inductive Proof Structure There are three steps to an inductive proof that: for aln ≥ b wheren ∈ N, S(n ) is true. 1. Base Case: Prove that S(·) holds for a set B of “smallest” consecutive values. 2. Inductive Hypothesis: Assume that S(k) holds for all values of k ∈ N such that db , S(n) holds. Q. How does each step relate to our domino analogy? A. • • 6 Mmmmm....CHOCOLATE! Suppose we have a chocolate bar consisting of a number of squares arranged in a rectangular pattern. The task is to split the bar into small squares with a minimum number of breaks. Q: How many breaks will it take (assume we only break along the lines)? Make an educated guess, and prove it by induction. Dimensions Breaks 1x1 0 1x2 1 2x2 3 3x2 5 4x3 11 Formula: l × w chocolate bar needs 7 Proof By Induction Prove that an l × w chocolate bar needs Deﬁne S(n): If n ≥ 1 then a chocolate bar with n squares requires n − 1 breaks. Prove that ∀n ∈ N where n ≥ 1,S (n) is true. The symbol ∀ means “for all”. Stay tuned....we will come back to this. 8 Another Example Prove that ∀n ∈ N,n(n + 5) is divisible by 6. Deﬁne P(n): Let P(n) be ∃m ∈ N,n(n + 5) = 6m Prove ∀n ∈ N,P (n) by simple induction. Base Case. ▯ n =0 . Let m =0 , then 0=0 · 6. Inductive Hypothesis. Let k ∈ N. Suppose P(k). Inductive Step. Prove that if P(k) is true then P(k+1) is true. (k + 1)((k + 1) + 5) =( k + 1)(k +2 k + 1 + 5) 2 2 =( k +2 k + 6) + k(k +2 k + 1 + 5) =( k +2 k +6+2 k + k)+ k(k + 5) 2 =3( k +3 k + 6) + k(k + 5) 2 =3 k(k + 1) + 6 + k(k + 5) Notice that the ﬁrst term is divisble by 6 since it is a multiple of 3 and further either k or (k + 1) is even, or divisible by 2. The second term is divisible by 6 and by the induction hypothesis, so is the third term. 9 Stamp Example – Simple Induction Given an unlimited supply of 4-cent and 7-cent stamps, prove that there exists a combination of stamps to make any amount of postage that is 18-cents or more. Deﬁne P(n), what we are proving: In English: If n ≥ 18, then postage of exactly n cents can be made using
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