110 likes | 122 Views
Approximation Algorithms for NP-hard Combinatorial Problems. Probabilistic method. Magnús M. Halldórsson Reykjavik University. Max Cut : Random split. Flip a coin for each vertex What is the probability that a given edge is cut?. Turán bound. Domatic partition.
E N D
Approximation Algorithms for NP-hard Combinatorial Problems Probabilistic method Magnús M. Halldórsson Reykjavik University
Max Cut : Random split • Flip a coin for each vertex • What is the probability that a given edge is cut?
Domatic partition • Partition the vertices of a graph into the largest possible number of dominating sets • Application: Lifetime maximization A B F G C E D
Icosahedron • Domatic number is at most + 1.
Very simple randomized algorithm Use L = (+1)/3 ln n colors. Each node selects one of the L colors independently at random. • This results in a valid domatic partition, with high probability. • (If it fails, we just repeat). • It can be „derandomized“ into a greedy algorithm
Correctness (Partition is domatic) All nodes have all L colors in their nborhood Pr[Coloring is not a domatic partition] color node v Pr[v is missing color ] Pr[Coloring is a proper domatic partition] =1 – Pr[Coloring is not valid domatic partition]
d(v) Particular node v and color • Pr[the color of v is not ] = 1- 1 /#colors • Pr[N[v] misses ] = Pr[ui is not ], i=0..d(v) = (1 – 3ln n/(+1))d(v)+1 exp(-3ln n/( +1) ( + 1)) = exp(-3 ln n) = 1/n3
All nodes, all colors Pr[Invalid domatic partition] = Pr[Some node misses some color] color vV Pr[v misses certain color ] n2 1/n3 = 1/n Pr[Proper domatic partition] 1 – 1/n
More on Domatic partition • Know DN(G) +1 • Saw DN(G) ( +1)/3ln n • Also DN(G) ( +1)/3ln (Lovász Local Lemma) • Even DN(G) ( +1)/ln () • Computationally hard to determine DN(G) within 0.99 ln factor! • [Feige, H, Kortsarz, Srinivasan, STOC´00]
Derandomization • Method of conditional expectation • Order the random events in a linear order • For each event, there are several choices. • The expectation of all the choices is X (given the previous events) • Then, there is some choice that yields a benefit of X • This gives a greedy algorithm