Lecture 6 Alterations

CSE 713: Random Graphs and Applications SUNY at Buffalo, Fall 2003

Lecturer: Hung Q. Ngo Scribe: Anusha R. Iyer

Lecture 6: Alterations Often the random object generated must be altered before the result may be obtained. This is termed as an alteration.

1 1.1

Applications Dominating Set

Recall the example from the first lecture. Let G = (V, E) be a graph on n vertices with δ(G) = δ > 1. We constructed a dominating set of G by randomly picking a set of vertices X, where each vertex belongs to X with probability p. The set X was then augmented with vertices Y ⊆ V − X such that for each y ∈ Y , both y and its neighbors are not in X. The set X ∪ Y is then a dominating set.

1.2

Ramsey Numbers

Definition 1.1. R(k, k) is the minimum natural number n such that any 2-edge coloring (with BLUE or RED) of Kn contains either a RED Kk or a BLUE Kk . Proposition 1.2. R(k, k) > n0 where n0 = 1e k2k/2 (1 + o(1)) Proof. Randomly color edges of Kn with RED/BLUE with probability ( 0 if S is not monochromatic, IS = 1 if S is a monochromatic.

1 2

each. Let

Then, summing over all k-subsets S of [n], we get #   X X n 1−(k) E[# monochromatic Kk ] = E [ IS ] = E[IS = 2 2 k S

(1)

S

We have used the following argument: if E[# monochromatic Kk ] < 1 then R(k, k) > n, which implied k R(k, k) > 2b 2 c . However, the bound was not too good. The method of alteration gives a better bound. Remove one vertex from each monochromatic Kk .

k k We have at least n − nk 21−(2) vertices left. Let n0 = n − nk 21−(2) , then R(k, k) > n0 . We then only k

need to find n that maximizes n0 (n). Elementary analysis gives n0 = 1e k2 2 (1 + O(1)).

1.3

Independence Number α(G)

Proposition 1.3. Let G = (V, E) where |V | = n and |E| ≤

1

nd 2

and d ≥ 1. Then, α(G) ≥

nd 2 .

Proof. We show there is an independent set whose size is at least nd 2 . Pick each vertex of G at random with probability p. Let X be the set of chosen vertices. The chosen set X may not necessarily be a independent set. Let Y be the set of edges both of whose endpoints are in X. For each y ∈ Y , discard one of the endpoints. The result is an independent set of size at least (|X| − |Y |). Let ( 0 if both endpoints of e are not in X Ie = 1 if both endpoints of e are in X Then E[|X| − |Y |] = E[|X|] − E[|Y |] X = np − E[ Ie ] e

= np − |E|p2 nd 2 ≥ np − p 2 Find p that maximizes E[|X| − |Y |], then E[|X| − |Y |] ≥

nd 2

as desired.

Side note: we will later discuss the Markov and Chebyshev inequalities that measure the probability of deviating from the expected value.

1.4

Extremal Set Theory

Recall from the first lecture that a hypergraph H = (V, E) has property B if it is two-colorable, i.e. there exists a two-coloring so that no edge is monochromatic. Let m(n) be the minimum number of edges m such that there is an n-uniform hypergraph with m edges and it is not 2-colorable. Our problem is to find m(n) or good bounds for m(n). To find an upper bound m for m(n), we just have to find some n-uniform
hypergraph with at least m edges which is not 2-colorable. A trivial upper bound is then m(n) ≤ 2n−1 (why?). n To find a lower bound m for m(n), we show that for any n-uniform hypergraph H, there is a positive probability that a 2-coloring of H is good. In what follows, we try to find some good lower bound. 1.4.1

Naive Approach #1

Consider an n-uniform hypergraph H = (V, E) with m edges. Randomly 2-color v ∈ V . If P[no monochromatic edge] > 0, then there exists a good coloring for the graph H. This implies m(n) > m. 1.4.2

Naive Approach #2

Take any hypergraph H = (V, E) with |E| = m. Let ( 0 if e is not monochromatic Ie = 1 if e is monochromatic.

2

Then X Ie ] E[#monochromatic edges] = E[ e

= mE[Ie ] = m21−n . Thus, if m < 2n−1 then there is a good 2-coloring, which implies m(n) > 2n−1 . Some History Erd¨os (1965) m(n) ≥ 2n−1 1 Beck (1978) m(n) ≥ Ω(2n n 3 ) n 12 ..., ... (2000) m(n) ≥ Ω(2n ( lnn ) ) Open Problem 1.4. The gap between the upper and lower bound is still very large. 1.4.3

Using Alterations

Algorithm: 1. Order all vertices randomly 2. For each v ∈ V , flip 2 coins. coin 1: head/tail with probability 21 /

1 2

coin 2: head/tail with probability p/ 1 − p Let ci (v) be the value of coin i of v. 3. Color v RED if c1 (v) = head BLUE if c1 (v) = tail 4. Let D = {v|v is in some monochromatic edge} For each v ∈ D in the random ordering, if v is still in some monochromatic edge e of the first coloring and e is still monochromatic at the point when v is considered, then switch v’s color if c2 (v) = head. We bound the probability that the coloring fails to be good: X P[the coloring is not good] ≤ Pr[e is monochromatic] e∈E

≤ 2

X

Pr[e is RED]

e∈E

Let Ae be the event that e is RED in the first place and none of the vertices in e changed color. Let Ce be the event e started with some BLUE vertices and then e was RED at the end. Pr[e is RED] ≤ Pr[Ae ] + Pr[Ce ] 1 ≤ ( )n (1 − p)n + ... 2 Let v be the last vertex of e which changed color from BLUE to RED. The reason v changed its color was because there is some f ∈ E such that f was BLUE in the first coloring, and remains blue until the 3

point v is considered. Moreover, f ∩ e = {v}, since another v 0 ∈ f ∩ e would necessarily be BLUE (v 0 ∈ f ) and necessarily be RED after the second coloring (v 0 ∈ e); however, v 0 must have changed its color before v, which means f was no longer BLUE at the point v was considered. Consequently, if Ce happens then there exists and edge f such that • e ∩ f = {v} • v was the last of e that changed color. • When v changed color, f was still BLUE. • The first coloring of f was BLUE, but the second coloring of e is RED. Let Bef be the event that e and f are related in this way. Each random coloring of V induces a random coloring σ of f ∪ e. Let i be the # of elements of e that come before v in σ. Let j be the # of elements of f that come before v in σ. p 2× 1

P[Bef |σ] ≤

(v started BLUE, turned RED)

×

(the rest of f is BLUE)

2n−1 1 × 2n−1−i j (1 − p) ×

( 21 =

p 22n−1

(1

(all of elements of e after v are RED) (the elements of f before v are unaltered)

+ p2 )j × − p)j (1 +

(every elem of e before v must be RED already or BLUE turned RED) p)j

Therefore P[Bef ] =

X

=

X

P[Bef |σ]P[σ]

σ

≤

p 1 (1 − p)j(σ) (1 + p)j(σ) 2n−1 2 (2n − 1)! σ    n−1 X n−1 X 1 n−1 p j i n−1 (1 − p) (1 + p) (i + j)!(2n − 2 − i − j)! 22n−1 (2n − 1)! i j j=0 i=0

≤ ... ≤

p 22n−1

Now X

P[Ce ] <

e

m2 p 2 22n−1

Returning to our original equation P[failing] ≤ 2[m

(1 − p)n m2 p + ] 2n 2 22n−1 1

n 2 If P[failing] < 1, then m(n) > m. After some analysis, we may arrive at m(n) > Ω(2n ( lnn ) ).

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