## Friday, February 29, 2008

### Homework 2: Notes and Comments

Here are solution sketches and comments on the rest of problems in HW2.
1. For the naive algorithm on unweighted graphs, several answers said: "if an edge is uncovered, the algorithm will pick two vertices when the optimal algorithm picks one vertex". While this is true (and is indeed the basis for the proof), you need to show that we are not over-counting. One way is to clearly state a lower bound on OPT (the size of any maximal matching), and state why the algorithm is at most twice the lower bound.

For the randomized VC algorithm, see the solutions to problem 2b from this previous semester's course.

For the primal-dual algorithm, we write down the dual which maximizes $\sum_e y_e$ such that for each vertex $\sum_{e: v \in e} y_e \leq c_v$, and the $y_e$ values are non-negative. We start with the zero primal and dual solutions, and raise all the $y_e$'s uniformly until some constraint becomes tight (i.e., for some vertex $v$, $\sum_{e: v \in e} y_e = c_v$. Pick the vertex in the primal solution, and "freeze" all the variables involved in this constraint. Now continue raising all the unfrozen edge variables, picking vertices and freezing edges when other contraints become right etc, until all the edge variables become tight: we now have have a feasible primal vertex cover $C$. (why?) Now the cost of this solution is $\sum_{v \in C} c_v = \sum_{v \in C} \sum_{e: v \in e} y_e$ $\leq \sum_{e} \sum_{v \in C \cap e} y_e \leq 2 \sum_e y_e$, which is the dual value. (The last inequality is because each edge contains at most 2 vertices.) Hence the vertex cover costs twice the feasible dual solution.

For the greedy algorithm, if you have a star with center cost $1+\epsilon$, and $n$ leaves with costs $1, 1/2, 1/3, \ldots, 1/n$, then you will pick all the leaves instead of the center, and hence only get an $H_n$ approx. An $O(\log m) = O(\log n)$ upper bound follows from the standard set cover analysis.

2. One point often missed: this problem defined $C^*$ and $F^*$ to be the facility and connection costs for the optimal integer solution, and not for the optimal LP solution.

Also, if you claim that after filtering you can use the standard Set Cover randomized rounding to obtain a solution, you need to show how the LP solution for facility location can be mapped to a LP solution for Set Cover.

3. No surprises here.

4. Build a bipartite graph on clauses and variables, and use Hall's theorem to infer the existence of a perfect matching.

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