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}3m 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, d(w, C) ≤

            

a + b + c a + (m − b) + (m − c) (m − a) + b + (m − c) (m − a) + (m − b) + c and thus, by adding the four inequalities, 4d(w, C) ≤ 6m.

Hamming Balls and k-SAT Algorithms Chidambaram Annamalai

Exercise Session 10 17.05.2016 slide 3

7.2: Random Covering Code

Choose t := ⌈(loge 2)n·2n/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

• f radius at most r is at most

2n ·

• 1 − vol(n, r)

2n

t

< 1, 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 < ex 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. 1b. The number of n clauses is 2n whereas the number of n − 1 clauses is

n

n−1

• 2n−1 so there are more n − 1 clauses than n clauses.
• 1d. True because we can obtain a partition of the cube {0, 1}n with

faces φi of dimension n − ci for all i = 1, . . . , m where n = maxm

i=1 ci.

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)

3a. Take a random assignment α over {0, 1}vbl(F). This satisfies exactly 3

4m + 7 8m = 13 16(2m) clauses of F in expectation.

3b. Suppose the values for x1, . . . , xi ∈ vbl(F) are already set to t1, . . . , ti respectively. Let xi+1 ∈ vbl(F). Let X be the expected number of clauses in F satisfied by a random assignment. Notice that E[X | x1 ← t1, . . . , xi ← ti] = E[X | x1 ← t1, . . . , xi ← ti, xi+1 ← 0]P[xi+1 = 0] +E[X | x1 ← t1, . . . , xi ← ti, xi+1 ← 1]P[xi+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 xi+1 to ti+1 accordingly and continue.

• 4a. The set {1n, 2n, 3n} is a covering code for {1, 2, 3}n of covering

radius at most 2n/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(42n/3poly(n)) time deterministic algorithm for 3- coloring on n vertex graphs.

Hamming Balls and k-SAT Algorithms Chidambaram Annamalai