Approximation Algorithm

1. Approximation ratio

Cost: the size of the solution, for example, in vertex cover, it’s the size of the cover; in TSP, it’s the total distance.

Since the approximation algorithm, the cost we have is always greater than the optimal solution.

2. Vertex Cover

• Running time: O ( ∣ V ∣ + ∣ E ∣ ) O(|V|+|E|)

  O(∣V∣+∣E∣), loop over all edges and all vertices.

• 

  ∣ A ∣ |A| ∣A∣: the number of edges chosen by the algorithm, and since one edge covers two vertices, so ∣ A ∣ = ∣ C ∣ / 2 |A|=|C|/2 ∣A∣=∣C∣/2. And since this is an approximation algorithm, and at least one of vertices chosen each time must be in C C

  C, thus ∣ C ∣ |C| ∣C∣ must be greater than the edges chosen by the algorithm.

3. Traveling Salesperson

3.1 Algorithm

• Construct a minimum spanning tree, whose sum of edge weights is as small as possible.

• Remember to skip duplicates;

• Proof to the theorem:

  • Since the distance in MIS is less than the distance in TSP; c ( T ) ≤ c ( H ∗ ) c(T)\le c(H^*)

    c(T)≤c(H∗)

  • the sum of weights in Euler tour is twice the sum in MIS, since we go through each vertex twice; c ( W ) = 2 c ( T ) c(W)=2c(T) c(W)=2c(T)

  • the sum of weight returned by the algorithm is less than the sum of Euler tour, since we will skip duplicates: c ( H ) ≤ c ( W ) c(H)\le c(W) c(H)≤c(W)

  • Above all, we have: c ( H ) ≤ 2 c ( H ∗ ) c(H)\le 2c(H^*) c(H)≤2c(H∗).

4. Set Cover

Find a set of subsets, which can cover the big set.

4.1 Greedy Algorithm

Intuition: each time pick the remaining largest subset, until the set is covered.

∣ X ∣ |X| ∣X∣: the size of the big set.

• c x : c_x: cx​: 1 over the size of the subset
• ∑ x ∈ S i − S < i c x \sum_{x\in S_i-S_{<i}}c_x ∑x∈Si​−S<i​​cx​: sum up the vertices in each subset;
• intuition: sum up the vertices in each chosen subset and sum up all the subset, it’s exactly the size of subsets returned by the algorithm

• Double count the c x c_x cx​ in C ∗ C^* C∗, and with the lemma above, we then prove the theorem. Notice: H ∣ S ∣ ≤ H ∣ X ∣ H_{|S|}\le H_{|X|} H∣S∣​≤H∣X∣​.

• Proof for the lemma:

  

  

  Vertex cover

  Choose the vertex with maximum degree