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Approximation Algorithms for NP-hard Combinatorial Problems Magnús M. Halldórsson Reykjavik University Probabilistic method.

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Presentation on theme: "Approximation Algorithms for NP-hard Combinatorial Problems Magnús M. Halldórsson Reykjavik University Probabilistic method."— Presentation transcript:

1 Approximation Algorithms for NP-hard Combinatorial Problems Magnús M. Halldórsson Reykjavik University Probabilistic method

2 Max Cut : Random split Flip a coin for each vertex What is the probability that a given edge is cut?

3 Turán bound

4 Domatic partition Partition the vertices of a graph into the largest possible number of dominating sets Application: Lifetime maximization C A FG E D B

5 Domatic number is at most  + 1. Icosahedron

6 Very simple randomized algorithm This results in a valid domatic partition, with high probability. (If it fails, we just repeat). It can be „derandomized“ into a greedy algorithm Use L = (  +1)/3 ln n colors. Each node selects one of the L colors independently at random.

7  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]

8 Particular node v and color Pr[the color of v is not ] = 1- 1 /#colors Pr[N[v] misses ] =  Pr[u i is not ], i=0..d(v) = (1 – 3ln n/(  +1)) d(v)+1  exp(-3ln n/(  +1)  (  + 1)) = exp(-3 ln n) = 1/n 3 d(v)

9 All nodes, all colors Pr[Invalid domatic partition] = Pr[Some node misses some color]   color  v  V Pr[v misses certain color ]  n 2  1/n 3 = 1/n Pr[Proper domatic partition]  1 – 1/n 

10 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]

11 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


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