Solutions to selected exercise of Randomized Algorithms

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1 Solutios to selected exercise of Radomized Algorithms Zhag Qi December 5, 006 Chapter. Problem.7. We first prove that e λx λ(e x ) Whe x = 0, LHS = RHS = 0. Ad whe x > 0, (e λx ) = λe λx λe x = [λ(e x )] The left work is easy e λx λ(e x ) e λ(x x) e x λ(e x e x ) + e x (x = x x ) e λx+( λ)x λe x + ( λ)e x. This part is a bit tricky. E[f(z)] = z = z f(z)p r[z = z] f[0 ( z) + z]p r[z = z] z [z f() + zf(0)]p r(z = z) = z zp r(z = z)f() + z ( z)p r(z = z)f(0) = E(Z)f() + ( E(Z))f(0) = pf() + ( p)f(0) = E[f(x)]

2 . Here comes that fial step. P r(y > λ + µ) < E(etY ) e = i= ) E(etYi t(λ+µ) e t(λ+µ) i= ) < E(etZi (by () ad (), Z i is the Beroulli radom variable) e t(λ+µ) )µ =... = e(et e t(λ+µ) = e λ µ ( + λ λ µ )(+ µ ) e λ [e λp ( ( λp ) µ ] (p+ λ (µ = p) ) ) e λ p = e λ (here p = /) Oe ca also use Azuma s iequality, but the result will be e λ, which is a little worse tha Hoeffdig s boud. Chapter 5. Problem 5.5 For every vertex v L, we choose vertices uiformly ad radomly from R as v s eighbors. Let A be the evet of every vertex i R has degree >, ad let B be the evet of every subset of vertices i L that has fewer tha eighbors i R. P r[a] < could be verified by Cheroff boud easily. Now we are goig to prove that P r[b] <. P r[b] ( )( ( ) ( e e ) ( e + l < ) ( )/ ) ( ) 7/ Therefore, P r(a B) <.. Problem 5.6 I the Lazy Select algorithm, it is guarateed that give selectios, at least ( O( )) > selectios will be successful. Thus log radom bits are

3 eough sice the probability of failure is at most / =.. Problem 5.7. First I would like to show a wrog attempt. With the idea of add vertex oe by oe.. = # (coected subgraph of size r) ( ) P r[the r vertices are coected] r ( ) r ( ) i d r i= ( e ) ( ) r r d ( r ) r πr r e rd = d r πr/rd d r Ufortuately, this approach is wrog. The crux lies i that whe cosiderig add vertex oe by oe we have assumed that there is a order list (v, v,..., v r ), i which v j is adjacet to some v i with i < j. It is also correspodig to a spaig tree. But we did t cout the umber of such trees. The correct solutio. First, we kow that every coected subgraph with r vertices has a spaig tree. Ad we kow that the umber of spaig trees is less tha r up to isomorphism. Now we fix oe of such tree structures, label the vertices with (A, A,..., A r ) ad look for how may subgraphs of the vertices graph have such a structure. Notice that A j is adjacet with A i for some i < j. Oe ca also thik that A i is the paret of A j ad A is the root. We choose A first, there are possibilities, ad ext we choose A, there are at most d possibilities, ad A,..., A r. Therefore the total umber of such subgraphs is less tha r d r, which is less tha d r if oe likes. P r[survivig coected subgraph of size α log for some α > ] [ ( ) ] l = #(coected subgraph of size l) d l α log [ = ( ) ] l (d) l d d l α log No close formula for the umber t() of trees with vertices up to isomorphism is kow, however, we have approximatio t() βα 5/ where β 0.5 ad α

4 ) α log d = ( d = β α( log d)+ (β = /(d )) If usig d r istead of r d r, oe will get... = [ ( ) ] l (d ) l d = l α log. Problem 5.8. For ay edge e E, l α log P r[e survives] = ( ) = 9 ( ) l = α+ = o() Thus e 9 ( + ), the Lovász Local Lemma tells us that such a idepedet set exists.. Sice E( V ) =, by the Cheroff boud P r[ V > ( + ɛ) ] < [ Agai, we have E( E ) = 9 = 8 ad e ɛ ( + ɛ) (+ɛ) P r[ E < ( ɛ) 8 ] < e 8 ɛ / ] From above we kow that the P r[ V < ( + ɛ) E > ( ɛ) 8 ] is very high. Give ( + ɛ) vertices, if IS α for some α >, the maximum umber of edges we could expect i the graph is (all form cliques with equal size) α ( ( + ɛ) ) α = ( + ɛ) α8 < ( ɛ) 8 Which is very ulikely. After the radom deletio, there are still about vertices left, which is cosiderably larger tha, so the positive probability proved by Lovász Local Lemma is ideed very small..5 Problem 5.0 We choose a complete graph G = (V, E), ad delete every edge with probability. Give ay subset T V with size /, we ca obtai the expect value of c(t ) = /8 [that is, #(vertices v i T ) #(its eighbors u i / T ) (P r[e = (v i, u i )] ot deleted)].

5 Next we defie evet A T to be c(t ) 8 > f() where f() will be specified later. It is obviously that A T is a Biomial radom variable with expectatio 8. By the stadard estimates for Biomial distributio give i A.. i Book [], we obtai [ ] ( ) P r c(t ) 8 > f() < exp f ( () ( ) = e f() ) 8 8 There are ( ) subsets with size of, so A T is depedet o at most other evets (Yeah, quite large!). Now we plug everythig to the symmetric case of the local lemma, we wat ( e ( + ) e f() ) < It is true for f() = α for some α > l. Therefore we ca make f() = O( ). Aother perspective O the other had, we ca also use Martigales to show that o a radom graph G(, /), with high probability, a fixed subset T with size of reders a c(t ) with c(t ) 8 O( ). We use a geeral settig itroduced by Alo, see [] p.0. The theorem asserts that for all ɛ > 0 there exists δ > 0 so that the followig holds. Suppose Paul has a strategy for fidig Y such that every lie of questioig has total variace at most σ. The P r[ Y E[Y ] > ασ] e α (+ɛ) for all positive α with αc < σ( + ɛ)δ. Here we choose a Edge Exposure Martigale, such that a edge is deleted or ot will chage the value of c(t ) at most, so the variace of the total lie of the questio would be σ = Pi ( P i )c i < = 8. We choose ɛ =, δ =, α =, ad the [ ] P r c(t ) 8 > e Sice there are at most ( /) may subsets with / vertices i G, ad e <, we kow that a radom graph has a positive probability to satisfy all the subsets. Summary The first solutio shows that we could costruct a graph by deletig every edge from a complete graph with probability /. The secod approach shows that a radom graph G = (, /) has a positive probability to meet the requiremet. The secod result is stroger tha the first i some sese. 5

6 .6 Problem 5. See book [] Chapter 5, p.5-5. Refereces [] Noga Alo ad Joel H. Specer. The probabilistic method (ed). New York: Wiley-Itersciece,