Todays exercises 7.1: Covering Radius Example 7.2: Random Covering - PowerPoint PPT Presentation

Exercise Session 10 17.05.2016 slide 1 Today’s exercises • 7.1: Covering Radius Example • 7.2: Random Covering Code • 7.3: Exact Radius • In Class: Exam Questions from SAT14 Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 2 7.1: Covering Radius Example Map any word w ∈ { 0 , 1 } 3 m to its signature ϕ ( w ) := ( a, b, c ) ∈ { 0 ..m } 3 where a, b and c are the number of ones in w within the first, second and third m bits. We now have, because of the various words in the code C ,  a + b + c     a + ( m − b ) + ( m − c )   d ( w, C ) ≤ ( m − a ) + b + ( m − c )     ( m − a ) + ( m − b ) + c   and thus, by adding the four inequalities, 4 d ( w, C ) ≤ 6 m. Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 3 7.2: Random Covering Code Choose t := ⌈ (log e 2) n · 2 n / vol ( n, r ) ⌉ many elements independently and uniformly at random from { 0 , 1 } n . Then the probability that these elements do not form a covering code of radius at most r is at most � t � 1 − vol ( n, r ) 2 n · < 1 , 2 n where the second term in the product is the probability that a given element of { 0 , 1 } n is not covered and the inequality follows by our choice of t and 1 + x < e x for x � = 0 . Therefore there is such a covering code of size t . Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 4 7.3: Exact Radius Let C be the code of radius r ′ < r . Pick a point w ∈ { 0 , 1 } n which maximizes the distance d ( w, C ) . We clearly have d ( w, C ) = r ′ . Now consider all codewords v ∈ C at distance exactly d ( v, w ) = r ′ and move all of them “away” from w (e.g. by changing the first coordinate where v and w are the same), producing a code C ′ . Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 5 7.3: Exact Radius Clearly, C ′ still has at most M codewords. (why at most?) The covering radius is now at least r ′ +1 because the codeword closest to w has distance r ′ + 1 . On the other hand, the covering radius is at most r ′ + 1 as well, as before it was r ′ and we have moved codewords by only one position, so no distance can have increased by more than one. Repeat the process until we have covering radius r . Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 6 In Class: Exam Questions from SAT14 1a. F is 2 -satisfiable and therefore by Theorem 2.6 the statement is true. The number of n clauses is 2 n whereas the number of n − 1 1b. � n 2 n − 1 so there are more n − 1 clauses than n clauses. � clauses is n − 1 1d. True because we can obtain a partition of the cube { 0 , 1 } n with faces φ i of dimension n − c i for all i = 1 , . . . , m where n = max m i =1 c i . This gives an unsatisfiable CNF-formula. Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 7 In Class: Exam Questions from SAT14 (2) Take a random assignment α over { 0 , 1 } vbl ( F ) . 3a. This satisfies exactly 3 4 m + 7 8 m = 13 16 (2 m ) clauses of F in expectation. 3b. Suppose the values for x 1 , . . . , x i ∈ vbl ( F ) are already set to t 1 , . . . , t i respectively. Let x i +1 ∈ vbl ( F ) . Let X be the expected number of clauses in F satisfied by a random assignment. Notice that E [ X | x 1 ← t 1 , . . . , x i ← t i ] = E [ X | x 1 ← t 1 , . . . , x i ← t i , x i +1 ← 0] P [ x i +1 = 0] + E [ X | x 1 ← t 1 , . . . , x i ← t i , x i +1 ← 1] P [ x i +1 = 1] . Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 8 In Class: Exam Questions from SAT14 (3) Clearly at least one of the two terms E [ . . . ] on the right must be at least as large as the term on the left. Further these two terms are polynomial time computable. So we set x i +1 to t i +1 accordingly and continue. 4a. The set { 1 n , 2 n , 3 n } is a covering code for { 1 , 2 , 3 } n of covering radius at most 2 n/ 3 . 4b. Let χ be some arbitrary vertex coloring of G = ( V, E ) . Let χ have distance at most r to χ ∗ , a proper coloring of G . Hamming Balls and k -SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 9 In Class: Exam Questions from SAT14 (4) Suppose χ is not a proper coloring because some edge { u, v } is as- signed the same color, say 1 . Then, among the four colorings obtained by changing exactly one of u or v to the values 2 or 3 , at least one of the colorings must have distance at most r − 1 to χ ∗ . This leads to a O (4 2 n/ 3 poly ( n )) time deterministic algorithm for 3 - coloring on n vertex graphs. Hamming Balls and k -SAT Algorithms Chidambaram Annamalai