Today s exercises. 5.17: Football Pools. 5.18: Cells of Line and Hyperplane Arrangements. Inclass: PPZ on the formula F

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1 Exercise Session slide 1 Today s exercises 5.17: Football Pools 5.18: Cells of Line and Hyperplane Arrangements Inclass: PPZ on the formula F 6.1: Harmonic vs. Geometric Mean 6.2: Many j-isolated Satisfying Assignments 6.3: Many Isolated Satisfying Assignments 6.4: Make it Hard for PPZ

2 Exercise Session slide : Football Pools An outcome (or prediction) of the matches can be mapped to an element in {0, 1} n. The number of incorrect guesses for a given prediction is exactly its hamming distance from the outcome. As any outcome is possible, the smallest number of bets needed to guarantee that at least one of the predictions has at most r incorrect guesses exactly corresponds to the problem of finding the smallest code C over {0, 1} n of covering radius r or smaller.

3 Exercise Session slide : Cells of Line and Hyperplane Arrangements 1. Proof by induction. True for n = 1. Let n > 1. Consider the arrangement formed by the union of n 1 lines which partition R 2 into 1 + (n 1) + ( ) n 1 2 cells by the induction hypothesis. Add the a new line l. Since l does not pass through any of the points of intersection between line pairs in the union of n 1 lines and l is not parallel to any of these n 1 lines, l intersects each line in a unique point. Thus, the number of cells increases by exactly n. Together, 1 + (n 1) + ( n 1) ( n + n = 1 + n ).

4 Exercise Session slide 4 2. Outer induction on d and inner induction on n. Base case d = 2 is true for all n. Let d > 2. Base case is again n = 1. Let n > 1. By the (inner) induction hypothesis there are 1 + ( n 1) ( n 1) d many cells in any n 1 hyperplane arrangement in d dimensions. Add the n-th hyperplane and project all the n 1 hyperplanes onto it to form a d 1 dimensional arrangement which by the (outer) induction hypothesis has 1 + ( n 1) ( n 1) d 1 many cells of dimension d 1. The main observation is that these cells correspond exactly to the newly formed cells in d dimensions

5 Exercise Session slide 5 upon adding the n-th hyperplane. Together the number of cells in total is ( 1 + ( n 1) ( n 1) ) (1 + ( n 1) ( n 1) ) + +, 1 d 1 d 1 which simplifies to 1 + ( n 1 1 repeatedly using the identity ( n i ) + + ( n 1 ) ( = n 1 i 1 d ) ) + ( n 1 i ) for 0 i n.

6 Exercise Session slide 6 Inclass: PPZ on the formula F Consider the formula F on variables {x 1, x 2,..., x n } containing all clauses on x 1, x 2, x 3 with at least one positive literal all clauses { x 1, x 2, x i } for 4 i n Question: What is the success probability of PPZ on F?

7 Exercise Session slide 7 Inclass: PPZ on the formula F (2) Notice that (x 1 1, x 2 1,..., x n 1) is a satisfying assignment. Claim. The success probability is 1/4 ( ) n 2. Proof: With probability 1 n(n 1), x 1 and x 2 are the first two variables in the random permutation (in some order). Supposing that x 1 and x 2 are set correctly (which happens with probability 1/4), all other variables are forced.

8 Exercise Session slide 8 6.1: Harmonic vs. Geometric Mean Use Jensen s inequality to prove this statement. Recall Jensen s inequality which says for a convex function f : R R 0, E[f(x)] f(e[x]) for x chosen according to any probability distribution with finite support over R. Appendix A.5 of the lecture notes contains this application.

9 Exercise Session slide 9 6.2: Many j-isolated Satisfying Assignments One possible way is to use the example from the lecture notes over j variables (which will be critical in every assignment) and augment it with n j dummy variables not appearing in the formula (which will be non-critical in every assignment).

10 Exercise Session slide : Many Isolated Satisfying Assignments In the example formula there are t = 2 n n/k satisfying assignments all of which are isolated. PPZ has a success probability of finding each one of them with probability at least 2 (n n/k). After one round the probability that none of them were found is at most 1 2 n n/k 2 (n n/k) = 0. Thus, the probability that PPZ does not find a satisfying assignment with this input formula is zero.

11 Exercise Session slide : Make it Hard for PPZ Since we, if possible, do not want to sum over multiple satisfying assignments, let us go for a formula with a unique isolated solution. In a unique solution, each variable has a critical clause. To minimize the success probability, we want to have exactly one critical clause per variable. Moreover, for simplicity, we would like not to have long-range influences and dependencies between the variables, so the formula should consist of components as independent as possible.

12 Exercise Session slide : Make it Hard for PPZ (2) A good example is the formula F consisting of n/k independent components, where each component is a maximal satisfiable formula over k variables. This way, there is a unique solution: each component has one satisfying assignment and composing these is the only way to get a satisfying assignment for F.

13 Exercise Session slide : Make it Hard for PPZ (3) And in each component, as long as less than k 1 variables are set, there cannot arise any unit clauses. Once k 1 variables are set in a component, there is always a unit clause forcing the last variable. Therefore exactly n n/k variables are being guessed, yielding exactly the success probability desired.

CSC 344 Algorithms and Complexity. Proof by Mathematical Induction

CSC 344 Algorithms and Complexity Lecture #1 Review of Mathematical Induction Proof by Mathematical Induction Many results in mathematics are claimed true for every positive integer. Any of these results