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Randomized Algorithms CS648

Randomized Algorithms CS648. Lecture 23 Probabilistic methods - II. Its solution can be viewed as an application of Probabilistic method. 3-SAT Problem. 3-SAT Problem. Notations : A Boolean variable: a variable that can take value true or false .

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Randomized Algorithms CS648

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  1. Randomized AlgorithmsCS648 Lecture 23 Probabilistic methods - II

  2. Its solution can be viewed as an application of Probabilistic method. 3-SAT Problem

  3. 3-SAT Problem Notations: • A Boolean variable: a variable that can take value true or false. • A term: a Boolean variable or its negation. • A clause: Disjunction of 3disjoint terms. Let ,…, be Boolean variables. Examples of a term: Examples of a clause: Note that is not a clause.

  4. 3-SAT Problem Problem: Givenclauses ,…, defined over Boolean variables ,…,, is there any assignment of true/false to the variables that will satisfy each clause. Why is this problem difficult ? Setting and to true and to false • the first 3 clauses get satisfied but 4th clause is notsatisfied. So trying to satisfy a subset of clause, one might render many others dissatisfied.

  5. 3-SAT Problem Problem: Givenclauses ,…, defined over Boolean variables ,…,, is there any assignment of true/false to the variables that will satisfy each clause. Results known: 3-SAT Problem NP-complete. • It is unlikely to have any polynomial time solution.

  6. Max 3-SAT Problem Problem: Givenclauses ,…, defined over Boolean variables ,…,, compute the maximum number of clauses that can be satisfied simultaneously ? Results known: Max3-SAT Problem NP-complete. • It is unlikely to have any polynomial time solution. The power of Randomization An approximation algorithm: There is a very simple randomized algorithm that will satisfy at least clauses.

  7. Max 3-SAT Problem Monte Carlo Algorithm: For each to assign value true/false to randomly uniformly and independently; return all the clauses that are satisfied. Analysis: : the number of clauses that are satisfied. 

  8. Max 3-SAT Problem Las Vegas Algorithm: Repeat { For each to { assign value true/false to randomly uniformly and independently; } Let be the number of clauses that are satisfied. } Until Analysis: : the probability that a single iteration of Repeat loop is successful  Expected running time:

  9. Getting a lower bound on : the number of clauses that are satisfied.  Alternate formulation of : Let : probability that there are exactly clauses that are satisfied. Question: What is the relation between and ’s ? Answer:

  10. Getting a lower bound on Express as where. So. If then is 1. For all other cases, is a fraction with denominator . So Aim: To get a lower bound on Let be the largest integer   

  11. Max 3-SAT Problem Las Vegas Algorithm: Repeat { For each to { assign value true/false to randomly uniformly and independently; } Let be the number of clauses that are satisfied. } Until Analysis: : the probability that a single iteration of Repeat loop is successful  Expected running time:

  12. Max 3-SAT Problem Theorem: There is an time Las Vegas algorithm for approximate Max 3-SAT Problem. It computes an assignment which satisfies at least fraction of clauses. Question: Is the best approximation factor that can be achieved for this problem ? Answer: Yes, indeed. It has also been proved that if P≠NP, then there can not be any approx. algorithm that can achieve a factor better than . So the simple randomized algorithm is also the best possible algorithm. Isn’t it very inspiring ?

  13. Probabilistic Method:Alteration

  14. Alteration Suppose we wish to show the existence of a structure with desired properties. The following method is sometimes quite useful • Form a random instance of the structure. • This structure will not have the desired property but it will be very close to having the desired property. • Slightly alter the random instance and the resulting structure will have the desired property.

  15. An interesting problem in Combinatorial Geometry

  16. A conjecture: If points are placed in a unit square, the smallest of the triangles will have area. The conjecture was disproved in 1982. Theorem: It is possible to place points in a unit square so that each triangle has area .

  17. Theorem (we shall prove): It is possible to place points in a unit square so that each triangle has area at least .

  18. Select points randomly uniformly. • : the no. of triangles with area less than. • Calculate and make useful inference… Suppose are3 points selected randomly uniformly from unit square. Let probability that triangle has area less than. Question: What is relation between and ? Answer:

  19. Probability(triangle has area less than ) Question: Given that |, what is the prob. that triangle has area ? Answer: less than

  20. Probability(triangle has area less than ) Question: Given that |, what is the prob. that triangle has area ? Answer: less than Question: P() = ? Answer: Question: What is P(triangle has area )? Answer:

  21. Probability(triangle has area less than ) Lemma: If Suppose are3 points selected randomly uniformly from unit square, then the prob. that triangle has area less than is bounded by . : the no. of triangles with area less than.  Question: What is if points are selected ? Answer:

  22. Select points randomly uniformly. • : the no. of triangles with area less than. • Note that • Remove one point from each triangle with area less than . Expected number of points left: . There is no triangle formed by these points with area less than .

  23. An interesting problem in Graph Theory

  24. Dense graphs with large girth : undirected unweighted graph on vertices and edges. Girth of : Length of smallest cycle in . It can be observed that if a graph is dense, its girth will be small. For example, a complete graph has girth . In fact the following result is well known in graph theory. Theorem: If a graph has or more edges, its girth must be at most . Question: Is the result mentioned in the theorem above tight ? Answer: Yes, there are graphs which have () edges and girth at least .

  25. Dense graphs with large girth Theorem: For each , there are graphs on vertices having at least edges and girth at least . We shall prove this theorem using the method of alteration. We shall provide just a sketch. The remaining details are straightforward and you are encouraged to fill in these details. The following Lemma will be useful. Lemma: The number of cycles of length in a complete graph on vertices is

  26. Dense graphs with large girth • Construct a random graph where we add each edge randomly independently with probability . • What should be the value of so that expected number of edges is close to ? • Show that the expected number of cycles of length in the random graph will be less than . • Show that expected number of cycles of length less than is less than . • Remove one edge from each cycle of length less than . • Show that there will still be edges left (choose sufficiently large value of ). • We are done. choose

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