Outline

• Approximation Algorithm for TSP sub-case
• In-Class Exercise
• A TSP Inapproximibility Result
• Approximating Set Cover

Introduction

• Recall we were talking about approximation algorithms for optimization problems associated with a given `NP`-complete problem.
• On Monday, we gave an algorithm that computes a vertex cover which is within a factor of 2 of the optimally sized vertex cover for a graph.
• We had earlier shown that the problem: Does a graph have a vertex cover of size `k`? is NP-complete.
• Before we continue with other approximation algorithms let's briefly review some defintions...
• We say an algorithm for a problem has an approximation ratio of `r(n)`, if for any input of size `n`, the cost `C` of the solution produced by the algorithm is within a factor of `r(n)` of the cost `C^star` of the optimal solution. That is, `max(C/C^star, C^star/C) le r(n)`.
• We call an algorithm that achieves an `r(n)`-approximation ratio an `r(n)`-approximation algorithm.
• Some `NP`-complete problems have a trade-off between the approximation ratio and the run time.
• An approximation scheme for an optimization problem is an algorithm that takes both an instance of the problem as well as a constant `epsilon` and then runs a `(1 + epsilon)`-approximation algorithm on the instance.
• If for any `epsilon`, the approximation scheme runs in `p`-time, then it is called a polynomial time approximation scheme.
• We say that an approximation scheme is a fully `p`-time approximation scheme if it is an approximation scheme and its run time is `p`-time in both `1/epsilon` and the instance size `n`. For example, the scheme might have a running time of `O((1/epsilon)^2n^3)`.

Approximating the Traveling Salesman Problem

• The optimization problem associated with TSP is to find a tour of least cost.
• Here is a 2-approximation algorithm for this problem when the triangle inequality holds on the distances between cities.

APPROX-TSP-TOUR(G, c)
1. Select a vertex r to be a root vertex
2. Compute the minimal spanning tree for G from root r using Prim's algorithm
3. Let L be the list of vertices visited in a pre-order tree walk of T
4. return the Hamiltonian cycle H that visits the vertices in order L.

Subroutines used by our algorithm

• Recall in a pre-order traversal of a graph starting from some node, we visit the node and then each child we have not yet visited.
• Recall Prims algorithm contructs a minimal spanning tree from a tree so far, denoted `A`, which at the start of the algorithm is the empty tree.
• We maintain a priority queue of all the vertices not in A.
• The priority, `v.key`, for a vertex `v` in the queue is the least weight of any edge connecting `v` with `A`. If no such edge exists than it is `infty`.
• Let `v.pi` be the parent of `v` in the tree. Rather than explicitly have an `A` we use this parent structure to get the tree when the algorithm terminates.
• Here is the pseudo-code:

  MST-PRIM(G, w, r) // r is a starting node to grow the tree from
  01 for each u in G.V
  02    u.key = infty
  03    u.pi = NIL
  04 r.key = 0
  05 r.pi = 0;
  06 Q = MAKE-QUEUE(G.V) //will have all vertices
  07 while Q != 0
  08     u = EXTRACT-MIN(Q)
  09     for each v in G.adj[u]
  10        if v in Q  and u.key + w(u, v) < v.key
  11            v.pi = u
  12            v.key = u.key + w(u,v) //call appropriate DECREASE-KEY
  

Analysis of APPROX-TSP-TOUR

Theorem. APPROX-TSP-TOUR is a p-time 2-approximation algorithm for TSP with triangle-inequality holding on the cost function.

Proof. The minimal spanning tree algorithm runs in time `O(|V|^2)`. The remaining step take at most `O(|G|)` time.

Let `H^star` denote the optimal tour of the vertices. Since we can obtain a spanning tree from any tour by deleting an edge, we have `c(T) le c(H^star)` where `T` is our minimal spanning tree. A full walk `F` of `T` lists the vertices when they are first visited and also whenever they are returned to after a visit to a subtree. So `c(F) = 2c(T) le 2c(H^star)`. A full walk is typically not a tour since it lists some vertices twice.

On the other, the `H` returned by the algorithm is a tour and satisfies `c(H) le c(F)`, since it is obtained by deleting vertices from the full walk and since the triangle inequality holds. We are using the triangle inequality as if we have a sequence `a b c` in the full walk and delete `b`, in our tour we want that the cost does not rise.

In-Class Exercise

• Consider the weighted graph on four vertices `\{ 1,2,3,4 \}` where the edge weight is computed as the sum of the vertices. So edge `{1,3}` have weight 4.
• Verify that the edges in this graph obey the triangle inequality.
• Show how the algorithm just given would compute a tour.
• Post your answer to In-Class Exercise Thread.

General TSP

Theorem. If `P ne NP`, then for any constant `d ge 1`, there is no `p`-time approximation algorithm with approximation ratio `d` for general TSP.

Proof. Suppose that for some number `d ge 1`, there was an approximation algorithm `A` for general TSP with the given approximation ratio. Without loss of generality, we can assume `d` is an integer. We will then show how to use `A` to solve instances of HAM-CYCLE. Since HAM-CYCLE is NP- complete, this will imply the result...

Proof of Inapproximability of General TSP cont'd

Let `G = (V, E)` be an instance of the HAM-CYCLE problem. Let `G'= (V, E')` be the complete graph on `V`. Assign a cost to each edge in `E'` as follows:
`c(u,v) = {(1,mbox(if ){u,v} in E),(d cdot |V| + 1,mbox(otherwise.)):}`

This instance `(G', c)` of the TSP optimization problem can be created in p-time in the HAM-CYCLE instance length. If the original graph has a Hamiltonian cycle, then there is a tour following its edges of cost `|V|`. On the other hand, if no such tour exists, then a tour uses at least one edge not in `E`, so has cost `(d cdot |V| +1) + (|V| - 1) > d cdot |V|`. Since our approximation algorithm needs to find a tour within a factor of `d` of the smallest one, if there is a Hamiltonian cycle in `G` when we run `A` the tour output will have cost `le d cdot|V|`. On the other hand, if the graph G does not have a hamiltonian cycle our algorithm on this instance will return a value `> d cdot|V|`.

The Set Covering Problem

• Set covering was one of the 21 `NP`-complete problems given by Karp in 1972.
• It models a variety of resource selections problems.
• An instance `(X, F)` of the set covering problem consists of a finite set `X` and a family of subsets of `X`, `F`, such that every element of `X` belongs to at least one subset of `F`. I.e., `X = cup_(S in F) S`.
• We say that a subset of `S in F` covers its elements.
• The set cover optimization problem is to find a minimum-sized subset `C subseteq F` whose members cover all of `X`. I.e., `X = cup_(S in C) S`.
• In the above picture, we have a set `X` of 12 elements, and we have a set `F = {S_1, ..., S_6}` of subsets of `X`. The set `C={S_3, S_4, S_5}` is a minimal cover.
• The NP-complete decision problem is to determine if `(X, F)` has a set cover of size `k`.

Example Use of Set Cover

• Suppose `X` represents a set of skills that are needed to solve a problem.
• `F` might be a set of people each of which have some of these skills.
• We might want to find a team `C` of as few people as possible that together have all the skills needed to solve the problem.

Greedy Algorithm For Set Covering

One can give a greedy algorithm for finding a cover by picking the set `S` at each stage that covers the greatest number of remaining elements that are uncovered.

GREEDY-SET-COVER(X, F)
1 U := X
2 C := ∅
3 while U ≠ 0
4     select an S ∈ F that maximizes {S ∩ U}
5     U := U - S
6     C := C ∪ {S}
7 return C

Approximation Result for Set Cover

Let `H(d) = sum_(i=1)^d 1/i` denote the `d`th harmonic number, defining `H(0) = 0`.

Theorem. GREEDY-SET-COVER is a polynomial-time `r(n)`-approximation algorithm, where
`r(n) = H(max{|S| : S in F})` on instances `(X,F)` or size `n`.

Proof. GREEDY-SET-COVER deletes at least one `S` from `F` in each iteration and the select step is at most quadratic time, so the algorithm will be polynomial time in the instance size.

To see that GREEDY-SET-COVER is an `r(n)`-approximation algorithm, we assign a cost of `1` to each set selected by the algorithm, distribute this cost over the elements covered for the first time, and then use these costs to derive the desired relationship between the size of an optimal set cover `C^star` and the size the cover `C` returned by the algorithm...

Proof of Approximation Result for Set Cover cont'd

Let `S_i` denote the `i` set selected by GREEDY-SET-COVER. We spread the cost of selecting `S_i`, 1, evenly among the elements covered for the first time by `S_i`. Let `c_x` denote the cost allocated to element `x in X`. If `x` is covered by `S_i`, then
`c_x = 1/(|S_i - (S_1 cup S_2 cup ... cup S_(i-1))|)`.
At each step of the algorithm, 1 unit of cost is assigned, and so
`|C| = sum_(x in X)c_x`.
The cost of the optimal cover is `sum_(S in C^star)sum_(x in S)c_x`,
and as each `x in X` is in at least one `S in C^star`, we have
`sum_(S in C^star)sum_(x in S)c_x ge sum_(x in X)c_x = |C|` (**).

We will show the theorem follows from the following claim:

Claim.`sum_(x in X)c_x le H(|S|)` for all `S in F`.

(Proof of Theorem from Claim). From (**) and the claim, we have
`|C| le sum_(S in C^star) H(|S|)`
`le |C^star| cdot H(max{|S| : S in F})`

Proof of Element Cover Cost Claim

Consider any `S in F` and `i = 1, ..., |C|`. Let
`u_i = |S - (S_1 cup S_2 cup ... cup S_(i))|`.
We define `u_0 = |S|`. Let `k` be the least index such that `u_k = 0`. At `k`, each element in S will be covered by at least one of `S_1, ... S_k`. We have `u_(i-1) ge u_i`, and that `u_(i-1) - u_i` elements of `S` are covered for the first time by `S_i`. Thus,
`sum_(x in S)c_x = sum_(i = 1)^k(u_(i-1) - u_i) cdot 1/(|S - (S_1 cup S_2 cup ... cup S_(i -1))|)`
Observe that
`|S_i - (S_1 cup S_2 cup ... cup S_(i - 1))| ge |S - (S_1 cup S_2 cup ... cup S_(i - 1))| = u_(i - 1)`
because we chose `S_i` greedily. This gives
`sum_(x in S)c_x le sum_(i = 1)^k(u_(i-1) - u_i) cdot 1/(u_(i-1))`
`= sum_(i=1)^k sum_(j = u_i + 1)^(u_(i-1))1/(u_(i-1))`
`le sum_(i=1)^k sum_(j = u_i + 1)^(u_(i-1)) 1/j` (because of the start condition of sum, `j le u_(i-1)`)
`= sum_(i=1)^k ( sum_(j = 1)^(u_(i-1)) 1/j - sum_(j = 1)^(u_(i)) 1/j)`
`= sum_(i=1)^k (H(u_(i-1)) - H(u_i))`
`= H(u_0) - H(u_k)` (telescoping series)
`= H(u_0) - H(0)`
`= H(u_0)`
`=H(|S|)`, proving the claim.