Linear Programming

Linear Programming (2) Combinatorial rounding Pipage rounding

Weighted Vertex Cover V

LP-Relaxation Rounding A Performance Ratio: 2 (xi< 2 xi*)

Half-integerality Theorem Proof

Min-2Sat • Given a 2-CNF Boolean formula, find a satisfying assignment (if exists) with the min # of true variables. • Vertex Cover (unweighted) is a subproblem of Min-2Sat: For every edge {vi, vj} of the vertex cover problem, define a clause (xi + xj) for the Min-2Sat problem.

Integer Linear Program

Relax to Linear Program Rounding

What about xi* = ½ ? • For instance, what if the input has both (xi + xj) and (xi + xj)? • Note: If xi* = ½, then xj* > ½. So, after the initial rounding, only those clauses with xi* = xj* = ½ are left. • These clauses form a new formula for which any assignment has p.r. = 2. • There is a simple polynomial-time algorithm for 2Sat. - -

Algorithm for 2-Sat • Convert the 2-Sat formula F with variables x1, … , xn to a digraph: V = {x1, x2, … , xn}, For each (yi + yj), generate two edges (yi, yj), (yj, yi). • If xi and xi are strongly connected, then F is not satisfiable. • O/w, if xixi, then assign all y reachable from xi as TRUE. • Likewise for the case xixi _ _ _ _ _

Example _ _ x1 TRUE x2TRUE x3 TRUE x1TRUE _ x1TRUE

Scheduling on Unrelated Parallel Machines Integer program

LP-Relaxation

Combinatorial Property of x* Bipartite graph H

Rounding

Prune t < T

How to find T*?

Proof of the Combinatorial Property Case 1: H is connected. Fact: H has at most m+n edges.

Proof • x* has mn components • x* is an extreme point (a vertex), and so must satisfy exactly mn equations (active constraints). • For equations xij = 0 or xij = 1, xij must be integral. • At most m+n active constraints allow xij to be fractional. • So, H has at most m+nedges.

Each job has degree > 2, and so must have a child.

Case 2.H is not connected

Max Weighted Hitting • Input: E ={1, 2, … , n}, C = {S1, S2, … , Sm}, Si subset of E,w(Si), p > 0. • Find A subset of E, |A|=p, with the max weight {w(B) | B subset of E, A hits B}.

Associated Integer Program not linear

Relaxation:

Add new variables to make it linear

Pipage Rounding IDEA of Rounding: 1) Choose two variables xjand xk and round at least one of them to 0 or 1. 2) Balance the total = p (e.g., if xjis rounded down to 0, then xk is rounded up by xj). 3) Make sure the new value of L(x) does not decrease ?? Use a companion function F(x) instead of L(x)

Use an alternative integer program also not linear

(over x {0,1}) Relation: F(x) = L(x) F(x) > (1 – 1/e) L(x) (over 0 < x < 1) Proof Assume | Sj| = k. geom-mean arith-mean ineq satisfies Note and f (1) > 1 – 1/e

Pipage Rounding Algorithm

Lemma Proof For a specific term about Sl, if it contains zero or one of xi and xk then it is a constant or linear wrt . If it contains both xiand xk, then it is of the form with a> 0, and so it is convex.

So, In other words, F(x) is nondecreasing during the rounding processes. It means the final approx. sol x satisfies and p.r. is at most e/(e-1) ~1.58. A

Linear Programming