Lecture Notes 6: Approximations for MAX-SAT. 2 Notes about Weighted Vertex Cover Approximation

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1 Algorithmic Methods //00 Lecture Notes 6: Approximations for MAX-SAT Professor: Yossi Azar Scribe:Alon Ardenboim Introduction Although solving SAT is nown to be NP-Complete, in this lecture we will cover some algorithms that give an approximated solution of the weighted version that comes close to the maximal satisfaction of the clauses within a constant factor. Towards the end of the lecture we will start discussing about the integer multi-commodity max-flow problem and give a linear programming algorithm that approximates it. Notes about Weighted Vertex Cover Approximation On the last lecture, we gave a linear programming algorithm that approximates the weighted version of the VC problem.. Reminder: Weighted-VC In the weighted version of the VC problem you are given a graph G = V, E) as input, and a weight function on the vertices w : V R +. The goal is to find a subset S V s.t. u, v) E u S or v S and the weight of the cover s S ws) minimized.. Reminder: Approximation via Linear Programming To find an approximation to the optimal solution, we defined an LP problem - we defined a variable X v for every v V, and our goal function was: minimize v V wv)x v subect to: X v + X u X v 0 u, v) E v V For each X v we rounded the result; those who got a value were rounded to, the others were rounded to 0. This way we got a -approximation of the problem. For more details, chec the notes of Lecture

2 .3 Notes About the Approximation. Possible Values: After solving the linear program, we get that every X v is assigned one of the values {0,, }.. Tight Bound: The bound we get on approximating the weighted-vc using linear programming and rounding is tight. That is, it is possible to construct an example for which ALGG, w) OP T G, w). Let s tae a loo at a graph G = V, E) which is a circle on n vertices, and let s set wv) = for every v V. The optimal solution in for the LP problem would set X v = for each v V, thus giving a total weight of n ) = n, while the optimal solution for the problem would use n vertices for the cover, while giving us a total weight of n. After the rounding of the LP problem, we will set each X v =, and will get a weight of n. V V V 3 V n- V n- V 4 V V 5 3 Approximating MAX-SAT In the classic SAT problem we are given a conunctive normal form CNF) formula C that s comprised of a collection of clauses {C,..., C m }. Each clause C is a disunction of literals X X X ). Each one of the variables in the formula can be assigned a boolean value T rue/f alse), and our goal is to give an assignment to the variables that satisfies all the clauses. However, this problem is the most basic problem nown to be NP-Complete. If we loo at the -SAT problem where every clause contains at most literals, we can chec if it is satisfiable and find an assignment that satisfies it, if exists, in polynomial time. For any >, -SAT is nown to be NPC. Let s loo at the optimization problem. 3. MAX-SAT Definition We are given a CNF formula C that s comprised of a collection of clauses {C,..., C m } and a weight function w : C R + on the clauses, and our goal is to give an assignment to the variables σ : V ar {T rue, F alse} so that the weight of satisfied clauses is maximal. Let s denote the weight of the th clause as w and define an indicator function α : C σ {0, } in which αc, σ) = C is satisfied by assignment σ. The weight that 6 -

3 assignment σ gives to the formula is now defined: wc, σ) = m w αc, σ) = Although this problem seems easier than the classic SAT problem since we don t need to satisfy all the clauses, MAX--SAT is actually nown to be NP-Hard. Let s try and give an approximation of the problem. 3. First Attempt - Guessing the Assignment Let s assign independently each variable a boolean value in a uniform way. Each variable would be given T rue w.p. and F alse w.p.. Let s see what s the expectation of the the weight of the clauses that would get satisfied by this assignment. From the linearity of expectation we get that: E σ [w C, σ)] = w E σ [α c, σ)] since each clause isn t satisfied w.p. l : = w ) l since each clause has at least one literal: w OP T C, w) If we now that each clause contains at least different literals, then a random assignment would give us an approximation factor of. So with a random assignment we re expected to get as close to the optimal solution as a factor. Can we achieve this result deterministically? 3.3 De-randomization We ll use the conditioned expectation method to give a deterministic algorithm that would achieve an approximation factor. We now that: E σn [wc, σ n )] = E σ n [wc, σ n ) X = F alse] + E σ n [wc, σ n ) X = T rue] From that we now that at least one assignment of X wouldn t decrease the expectation of the total weight of the formula. If we could calculate the expectation of the formula when some of the variables are assigned a value and some are set randomly, we could choose the assignment of X which gives us the higher expectation and continue assigning values to all the variables until we have an assignment that gives us an approximation factor of. Fortunately, we can. All we need to do is to calculate E[αc, σ)] for every c C. Let s 6-3

4 loo at a clause c C and let s assume that variables X,..., X l were assigned a value, and variables X l+,..., X n are set randomly. If one of c s literals satisfied c than it s clear that E[αc, σ n l )] =. Otherwise, if c has different literals that weren t assigned a value, E[αc, σ n l )] =. Example: We are given the the following set of clauses: X X X X X X 3 X X X 3 X X X 3 where each clause has a weight of. We re expected to get a weight of: = 37 8 If we set X = F alse we get an expectation of: = 4 If we set X = T rue we get an expectation of: = 33 4 Therefore, we ll set X = F alse. Let s continue to setting the second variable. If we set X = F alse we get an expectation of: = 4 If we set X = T rue we get an expectation of: = 4 Notice that if we set X = T rue we decide which of the clauses are satisfied deterministically, and we get that the total weight is higher than the initial expectation. 3.4 Second Attempt - Guessing with a Hunch Let s tae a formula C and try to merge all the clauses with a single literal of the same variable. For instance, if we have four clauses: with weights: C = X C = X C 3 = X C 4 = X w = 7 w = w 3 = 4 w 4 = 6 we can at first merge C and C 3 to a single clause C X = X with weight w X = 7+4 = 59 and C and C 4 to a single clause C X = X with weight w X = + 6 = 8. We can now again combine the two clauses into a single clause C X = X with weight w X = 59 8 = 3. Note that every assignment of X would give us a weight smaller by 8. Notice that although we reduce the weight of the formula, every approximation factor we get with the new formula 6-4

5 would also apply to the old formula. Let s denote the new formula after merging all the variables as C, set the weight shift constant as β > 0 and the approximation factor of C as α, we can see that: ALG w C) OP T w C) = ALG w C) + β OP T w C) + β ALG w C) OP T w C) α Our approximation algorithm would first transform C into a reduced formula C where each variable can appear only once in a clause with a single literal. Afterwards, the algorithm will guess the assignment for each variable independently, but with a hunch. Every literal in C that appears by himself in a clause would be set to T rue w.p. h >, and therefore, the complement to this literal would be set to T rue w.p. h. Every variable that doesn t have a literal of his in a clause of size would be assigned a value randomly as before. Let s calculate the expectation of the indicator function α for each clause with the new algorithm. For clauses with a single literal: E[αC, σ)] = h and for clauses with different literals: Therefore: E[αC, σ)] h h E[αC, σ)] min{h, h } Because h is monotonically increasing and h is monotonically decreasing, we get that that the expectation reaches a maximum when h = h h = A de-randomization using the conditioned expectation method could be applied as before to give us a deterministic algorithm with an approximation factor of Approximation using Linear Programming Idealistically, we would lie to define the following integer LP problem and find the optimum value - let s define an indicator X i {0, } for each variable X i i m) and an indicator Z {0, } for each clause C n), and set the goal function to be: max w Z subect to: Z i C + X i + X i ) i C where C + are the variables that appear in C without negation, and C are the variables that appear in C with negation. If we could solve such a problem in polynomial time, we could do all sorts of cool stuff lie solve 3-SAT and show that P=NP. Since ILP is an NP-Complete problem, we ll solve a relaxation of the problem. The goal function is the same and the constrains are the same, but the possible values of the variables are: 0 X i 0 Z 6-5

6 After solving the LP problem we get X i = P i for every variable. We ll do a probabilistic rounding of the result. For each X i, independently, we ll set X i = T rue w.p. P i and X i = F alse w.p. P i. The probability that αc ) = 0 for a certain C is: E[αC ) = 0] = P i ) P i i C + i C from the inequality of arithmetic and geometric means: i C + P i ) + i C ) P i from the constraint on Z : From that we get that: Z If we can show that for some constant β: ) = Z ) E[αC ) = ] Z ) Z ) β Z then we get a β-approximation of MAX-SAT since: E[wC, σ)] w E[αC ) = ] = Claim: The inequality Z w Z β = β w Z β OP T relaxed C, σ) β OP T C, σ) ) ) ) Z holds for 0 Z. Proof. When Z = 0 we get that 0 0. When Z = we get: ) ) ) It s easy to see that the second derivative of the left hand side of the inequality is negative when 0 Z and therefore it is concave there, and because the right hand side of the inequality is linear in Z, we get the following: 6-6

7 Left-hand side Right-hand side 0 Z From that, it is easy to see that the inequality holds when 0 Z. Since β = ) e 0.63, we get an approximation factor of β 0.63 using linear programming and rounding for the MAX-SAT problem. 3.6 Combining Linear Programming and Random Assignment Let s notice that random assignment gives us a good approximation for clauses containing many literals, while linear programming gives us a good approximation for clauses containing few literals. We ll try combining them both - with probability we ll give all the variables the value they get using LP and rounding, and with probability we ll give all the variables a random assignment. Let s try and calculate the expected weight that the assignment gives on the formula note that = ): E[wC, σ combined )] = E[wC, σ LP )] + E[wC, σ R)] = w Z β + w ) w Z β + )) If we ll show that: β + ) 3 4 we ll get a 3 4 approximation since: E[wC, σ combined )] w Z β + )) 3 w Z 3 OP T C, σ) 4 4 Claim: The following inequality holds: ) ) + )

8 Proof. Let s first rearrange the equation: ) + For = we get: For = we get: For 3 it s clear that ) e and 8, and that gives us: ) + e + 8 Great! We got a 3 4 approximation when selecting the assignment of the LP approximation w.p. and the random assignment w.p.. Can we improve the approximation factor if we choose which assignment to give to each variable w.p.? Apparently, \ gives us the best approximation factor. If we cannot improve the approximation factor, can we at least reach the approximation deterministically? This time the answer is positive. De-randomization, although a little more complicated than the previous ones, wors here as well. 4 Multi-Commodity Flow Consider the following problem - you are given a graph G = V, E) with a capacity function on the edges c : E R + and a request set s i, t i ). For each request we get a value v i if we find a path with capacity from s i to t i. A feasible solution to this problem would be to find a path P i for each request or set P i = if we choose not to serve the request, and the value for this solution would be i P i v i. Our goal is obviously to maximize this value. Let s denote l e = {i e P i }. We want to ensure that the feasible solution would maintain l e c e for every e E. 4. Linear Programming Approximation - Phase We ll mae some relaxing assumptions:. We can replace the path from s i to t i with a flow from s i to t i.. We can serve a fraction of the request, and we ll get a fraction of the appropriate v i. Let s define some new variables. X i e will denote the flow for request i on an edge e and X i would denote the total amount of flow for request i. The goal function for the LP problem is: maximize i v i X i 6-8

9 subect to: v e=u,v) e=s i,v) X i e X i e Xe i c e e E Mae sure each edge isn t exceeding its capacity) i v e=v,u) e=v,s i ) X i e = 0 i u V \ {s i, t i } Incoming flow equals to outgoing flow) X i e = X i i What leaves the source minus what enters it) As well as 0 X i and 0 X i e. After we get a flow for each request, we ll disassemble the flow from s i to t i into at most E different paths. The way of doing it is to tae the flow graph for each request, which is a directed a-cyclic graph DAG), and start removing paths from it, one by one. For every path we remove, we deduce it s weight, which is the weight of the lightest edge in the path, from each edge in the path, and continue on to the next path. Since every path removal removes at least one edge, we can re-iterate at most E times. We ll continue on to phase of the approximation algorithm at the beginning of lesson

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