Chapter 6. Relaxation

1. Chapter 6. Relaxation I. Superstring

2. restriction f(y*) = min f(y) f(x*) = min f(x) y in Γ x in Ω Ω x* y* Γ y

3. Relaxation xA x*  x* Min f(x) x in Ω xA f(x*) <opt = min f(x) <f(xA) x in Ω

4. Relaxation (for max problems) x* max f(x) x in Ω xA f(x*) >opt = max f(x) >f(xA) x in Ω

5. Relaxation vs. Restriction • In relaxation, we compute the optimal sol. A in the relaxed form, and then modify it to get an approx. sol. A’. (Modification is part of the algorithm.) • In restriction, we compute the optimal sol. A in the restricted form, and use it as an approx. sol. (The modification from the real opt. sol. A* to the restricted form is for analysis only.)

6. Traveling Salesman Problem (Given a distance table for a complete graph) relaxed TSP tour  spanning graph  minimum spanning graph = minimum spanning tree  tour add 0.5 opt So, the p.r. = 1.5 (If satisfying  inequality)

7. Matching Shortcut MST

8. Max (Directed) HC (Given a distance table for a complete (di)graph) relaxed HC  Two or three independent matchings  maximum matching  HC a HC connecting all edges of the matching P.r. =3: The maximum matching is at least 1/3 of HC. (For Max-DHC, this is as good as the greedy algorithm.)

9. Directed TSP with  inequality (Given a distance table for a complete digraph without loop) relaxed Directed tour  1-factor (assignment)  minimum assignment  directed tour a directed tour connecting all cycles in 1-factor

10. Assignment: Collection of disjoint cycles that cover all vertices. • Minimum Assignment A (as min matching) can be found in polynomial time. • Modification: Find a cycle C over one vertex of each cycle of A. Then, form a TSP. • Cost <w(A) + w(C) = O(opt), if w(C) = O(opt), or if C is a linear approx to DTSP ?

11. TSP from Min Assignment

12. Superstring Given n strings s1, …, sn, find a shortest string s containing every si as a substring. ov(s1,s2) = maximum overlapping of s1 and s2 s1 = pref(s1,s2) ov(s1,s2) s1 pref(s1,s2) ov(s1,s2) s2

13. Superstring Directed TSP Vertices: s1, …, sn, sn+1 = ε Distance: d(si,sj) = |pref(si,sj)| Minimum assignment: r1 r2 rt=sn+1=ε ct c1 c2 d(ci) = total distance of cycle ci.

14. cycle (r1,r2, …, rt) d(cycle (r1,r2, …, rt)) = d(r1,r2) + ··· + d(rt-2,rt-1) + d(rt-1,rt) + d(rt,r1) = |pref(r1,r2)| + ··· + |pref(rt-2,rt-1)| + |rt-1| + 0 = |superstring <r1,r2,…,rt-1>| = Σ |ri| - |ov(r1,r2)| - ··· - |ov(rt-2,rt-1)|

15. <r1,r2, …, rt-1> r1 ov(r1,r2) r2 rt-1

16. Assume <r1*,r2*, …, r(t-1)*> is opt superstring and<r1,r2, …, rt-1> is obtained by greedy Algorithm on overlapping. |ov(r1*,r2*)| + ··· + |ov(r(t-2)*,r(t-1)*)| < 2(|ov(r1,r2)| + ··· + |ov(rt-2,rt-1)|) (from Chapter 2, by quadrilateral condition)

17. Greedy Algorithm S ← {s1, …, sn}, while |S| > 1 do choose s and s’ from S to maximize |ov(s,s’)| and set S ← (S – {s,s’}) U {pref(s,s’)s’}, output unique string in S.

18. |<r1,r2, …, rt-1>| = Σ |ri| - |ov(r1,r2)| - ··· - |ov(rt-2,rt-1)| <Σ |ri| - 0.5(|ov(r1*,r2*)| + ··· + |ov(r(t-2)*,r(t-1)*)|) < |<r1*,r2*, …, r(t-1)*>| + 0.5(|ov(r1*,r2*)| + ··· + |ov(r(t-2)*,r(t-1)*)|) < opt (opt superstring for s1, …, sn) + 0.5(|ov(r1*,r2*)| + ··· + |ov(r(t-2)*,r(t-1)*)|)

19. Lemma. |ov(ri,rj)| < d(ci) + d(cj) If it is true, then |ov(r1*,r2*)| + ··· + |ov(r(t-2)*,r(t-1)*)| < 2 Σ d(ci) < 2 opt Therefore, we obtain a 3-approximation:

20. 3-Approximation Algorithm Step 1. Compute a minimum assignment {c1, …, ct}. Step 2. Choose a vertex ri from ci, with rn+1=ε. Step 3. Find an ordering (r1, …, rt-1) by greedy algorithm on overlapping. Step 4. Merging c1, …, ct into a Hamiltonian cycle based on the ordering (r1, …, rt-1, rt).

21. Lemma. |ov(ri,rj)| < d(ci) + d(cj) Proof |ov(ri,rj)| > d(ci) + d(cj). Assume We want to show ρ(ci) = ρ(cj) where ρ(c) is a root of c.

22. Root For a nonempty string x, ρ(x) is the shortest string y such that m x = y for some integer m > 0. For a setc of strings, ρ(c) is a string y such that every string in c is a substring of y for some m. m

23. Property of roots (u, v nonempty) ρ(u) =ρ(v) uv = vu Proof. (by induction on |uv|) Case 1. |u| = |v|. We have u = v. Case 2. |u| > |v|. Then u = vw. So, vwv = vvw. Hence, wv = vw. Since v ≠ ε, |wv| < |uv|. By induction hypothesis, ρ(w)=ρ(v). Hence, ρ(u) = ρ(vw) = ρ(v).

24. Property I of ρ(c) for string set c There exists a cycle c’ connecting all strings in c such that d(c’) = |ρ(c)|. ρ(c) ρ(c) s1 d(s1,s2) s2 d(s2,s3) s3 d(s3,s1) s1 We may assume all si start in the first copy of (c), then use this ordering to define c'.

25. Property II of ρ(c) for string set c For any permutation (s1, …, sk) of strings in c, pref(s1, s2) pref(s2, s3) ··· pref(sk-1, sk)pref(sk, s1) is a ρ(c). s1 s2 s3 s1

26. Property III of ρ(c) for string set c Any substring of ρ(c)ρ(c), with length |ρ(c)|, is a root of c. ρ(c) ρ(c) |ρ(c)|

27. Lemma. |ov(ri,rj)| < d(ci) + d(cj) Proof Assume |ov(ri,rj)| > d(ci) + d(cj). We want to show ρ(ci) = ρ(cj) If this is true, then cycles ci and cj can be merged into one c, with ρ(c) = ρ(ci) = ρ(cj), Contradicting the minimum of assignment.

28. |ov(ri,rj)| > d(ci) + d(cj). Proof Assume Define u =prefix of ov(ri, rj) of length d(ci) v =prefix of ov(ri, rj) of length d(cj) ri u v d(ci) d(cj) v u rj d(cj) d(ci) Note: u has length d(ci) and so ov(ri, rj)= u u … = v v … uv = vu ρ(ci) = ρ(cj)