CS2800-Probability_part_e_v.6

CS2800-Probability_part_e_v.6 - 1 Discrete Math CS 2800...

Info iconThis preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Discrete Math CS 2800 Prof. Bart Selman [email protected] Module Probability --- Part e) 1) The Probabilistic Method 2) Randomized Algorithms 2 2 The Probabilistic Method 3 The Probabilistic Method Method for providing non-constructive existence proofs: Thm. If the probability that a randomly selected element of the set S does not have a particular property is less than 1, then there exists an element in S with this property . Alternatively: If the probability that a random element of S has a particular property is larger than 0 , then there exists at least one element with that property in S. Note: We saw an earlier example of the probabilistic method when discussing the 7/8 alg. for 3-CNF. 4 Example: Lower bound for Ramsey numbers Recall the definition of Ramsey number R(k,k): Let R(k,k) be the minimal n such that if the edges of the complete graph on n nodes are colored Red and Blue, then there either is a complete subgraph of k nodes with all edges Red or a complete subgraph of k nodes with all edges Blue . R(3,3) = 6. So, any complete 6 node graphs has either a Red or a Blue triangle. (Proof: see “party problem”.) Reminder: “The party problem” Dinner party of six: Either there is a group of 3 who all know each other, or there is a group of 3 who are all strangers. Consider one person. She either knows or doesn’t know each other person. But there are 5 other people! So, she knows, or doesn’t know, at least 3 others. (GPH) Let’s say she knows 3 others. If any of those 3 know each other, we have a blue , which means 3 people know each other. Contradicts assumption. So they all must be strangers. But then we have three strangers. Contradicts assumption. The case where she doesn’t know 3 others is similar. Also, leads to constradiction. So, such a party does not exist! QED By contradiction. Assume we have a party of six where no three people all know each other and no three people are all strangers. How do we get a lower bound on R(k,k)? E.g., lower bound for R(3,3)? We need to find an n such that the complete graph on n nodes with a Red/Blue edge coloring does not contain a Red or a Blue triangle. So, R(3,3) > 5. I.e., we have a lower bound on the Ramsey Number for k =3. E.g. Can we do this for R(k,k) in general? Very difficult to construct the colorings, but… we can prove they exist for non-trivial k . Thm. For k ≥ 4, R(k,k) ≥ 2 k/2 So, e.g., k = 20 , then there exists a Red / Blue coloring of the complete graph with 1023 nodes that does not have any complete monochromatic sub graph of size 20 . (But we have no idea of how to find such a coloring!) Proof: Consider a sample space where each possible coloring of the n-node complete graph is equally likely. A sample coloring can be obtained by randomly coloring each edge. I.e., with probability ½ set edge to...
View Full Document

This note was uploaded on 10/25/2009 for the course CS 2800 at Cornell.

Page1 / 36

CS2800-Probability_part_e_v.6 - 1 Discrete Math CS 2800...

This preview shows document pages 1 - 8. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online