1. Design and Analysis of Approximation Algorithms
Ding-Zhu Du

2. Text Books
Ding-Zhu Du and Ker-I Ko, Design and Analysis of Approximation Algorithms (Lecture Notes).
Chapters 1-8.

3. Schedule Introduction Greedy Strategy Restriction Partition
Guillotine Cut
Relaxation
Linear Programming
Local Ratio
Semi-definite Programming

4. Rules
You may discuss each other on 5 homework assignments. But, do not copy each other.
Each homework is 10 points. 4 top scores will be chosen.
Midterm Exam (take-home) is 30 points.
Final Exam (in class) is 30 points.
Final grade is given based on the total points (A ≥ 80; 80 > B ≥ 60 ; 60 > C ≥ 40).

5. Chapter 1 Introduction Computational Complexity (background)
Approximation Performance Ratio
Early results

6. Computability Deterministic Turing Machine
Nondeterministic Turing Machine
Church-Turing Thesis

7. Deterministic Turing Machine (DTM)
Finite Control
tape
head

8. p h B e a l a
The tape has the left end but infinite to the right. It is divided into cells. Each cell contains a symbol in an alphabet Γ. There exists a special symbol B which represents the empty cell.

9. a
The head scans at a cell on the tape and can read, erase, and write a symbol on the cell. In each move, the head can move to the right cell or to the left cell (or stay in the same cell).

10. a
The head scans at a cell on the tape and can read, erase, and write a symbol on the cell. In each move, the head can move to the right cell or to the left cell (or stay in the same cell).

11. The finite control has finitely many states which form a set Q
The finite control has finitely many states which form a set Q. For each move, the state is changed according to the evaluation of a transition function
δ : Q x Γ → Q x Γ x {R, L}.

12. b a p q
δ(q, a) = (p, b, L) means that if the head reads symbol a and the finite control is in the state q, then the next state should be p, the symbol a should be changed to b, and the head moves one cell to the left.

13. a b p q
δ(q, a) = (p, b, R) means that if the head reads symbol a and the finite control is in the state q, then the next state should be p, the symbol a should be changed to b, and the head moves one cell to the right.

14. s
There are some special states: an initial state s and an final states h.
Initially, the DTM is in the initial state and the head scans the leftmost cell. The tape holds an input string.

15. Otherwise, the input string is rejected.x h
When the DTM is in the final state, the DTM stops. An input string x is accepted by the DTM if the DTM reaches the final state h.
Otherwise, the input string is rejected.

16. The DTM can be represented by
M = (Q, Σ, Γ, δ, s)
where Σ is the alphabet of input symbols.
The set of all strings accepted by a DTM $M$ is denoted by L(M). We also say that the language L(M) is accepted by M.

17. The transition diagram} of a DTM is an alternative way to represent the DTM.
For M = (Q, Σ, Γ, δ, s), the transition diagram of M is a symbol-labeled digraph G=(V, E) satisfying the following:
V = Q (s = , h = )
E = { p q | δ(p, a) = (q, b, D)}.
a/b,D

18. M=(Q, Σ, Γ, δ, s) where Q = {s, p, q, h}, Σ = {0, 1}, Г = {0, 1, B}.
0/0,R; 1/1,R
0/0,R
0/0,R
B/B,R s p q h
1/1,R
M=(Q, Σ, Γ, δ, s) where Q = {s, p, q, h},
Σ = {0, 1}, Г = {0, 1, B}.
δ B
s (p, 0, R) (s, 1, R)
p (q, 0, R) (s, 1, R)
q (q, 0, R) (q, 1, R) (h, B, R)
L(M) = (0+1)*00(0+1)*.

19. Nondeterministic Turing Machine (NTM)
Finite Control
tape
head

20. p h B e a l a
The tape has the left end but infinite to the right. It is divided into cells. Each cell contains a symbol in an alphabet Γ. There exists a special symbol B which represents the empty cell.

21. The finite control has finitely many states which form a set Q
The finite control has finitely many states which form a set Q. For each move, the state is changed according to the evaluation of a transition function
δ : Q x Γ → 2^{Q x Γ x {R, L}}.

22. Church-Turing Thesis
Computability is Turing-Computability.

23. Multi-tape DTM Input tape (read only) Storage tapes Output tape
(possibly, write only)

24. Time of TM TimeM (x) = # of moves that TM M takes on input x.
TimeM(x) < infinity iff x ε L(M).

25. Space
SpaceM(x) = # of cell that M visits on the work (storage) tapes during the computation on input x.
If M is a multitape DTM, then the work tapes do not include the input tape and the write-only output tape.

26. Time Bound
M is said to have a time bound t(n) if for every x with |x| < n,
TimeM(x) < max {n+1, t(n)}

27. Complexity Class
A language L has a (deterministic) time-complexity t(n) if there is a multitape TM M accepting L, with time bound t(n).
DTIME(t(n)) = {L(M) | DTM M has a time bound t(n)}
NTIME(t(n)) = {L(M) | NTM M has a time bound t(n)}

28. P = U DTIME(n )
NP = U NTIME(n ) c
C>0 c
C>0

29. NP Class

30. Earlier Results on Approximations
Vertex-Cover
Traveling Salesman Problem
Knapsack Problem

31. Performance Ratio

32. Constant-Approximation
c-approximation is a polynomial-time approximation satisfying:
1 < approx(input)/opt(input) < c for MIN or
1 < opt(input)/approx(input) < c for MAX

33. Vertex Cover
Given a graph G=(V,E), find a minimum subset C of vertices such that every edge is incident to a vertex in C.

34. Vertex-Cover
The vertex set of a maximal matching gives 2-approximation, i.e.,
approx / opt < 2

35. Traveling Salesman
Given n cities with a distance table, find a minimum total-distance tour to visit each city exactly once.

36. Traveling Salesman with triangular inequality
Traveling around a minimum spanning tree is a 2-approximation.

37. Traveling Salesman with Triangular Inequality
Minimum spanning tree + minimum-length perfect matching on odd vertices is 1.5-approximation

38. Minimum perfect matching on odd vertices
has weight at most 0.5 opt.

39. Lower Bound
1+ε
1+ε 1 1
1+ε
1+ε
1+ε

40. Traveling Salesman without Triangular Inequality
Theorem For any constant c> 0, TSP has no c-approximation unless NP=P.
Given a graph G=(V,E), define a distance table on V as follows:

41. Contradition Argument
Suppose c-approximation exists. Then we have a polynomial-time algorithm to solve Hamiltonian Cycle as follow:
C-approximation solution < cn
if and only if
G has a Hamiltonian cycle

42. Knapsack

43. 2-approximation

44. PTAS
A problem has a PTAS (polynomial-time approximation scheme) if for any ε > 0, it has a (1+ε)-approximation.

45. Knapsack has PTAS Classify: for i < m, ci < a= cG,
Sort
For

46. Proof.

48. Time

49. Fully PTAS
A problem has a fully PTAS if for any ε>0, it has (1+ε)-approximation running in time poly(n,1/ε).

50. Fully FTAS for Knapsack

51. Pseudo Polynomial-time Algorithm for Knapsak
Initially,

55. Time outside loop: O(n) Inside loop: O(nM) where M=max ci
Core: O(n log (MS))
Total O(n M log (MS))
Since input size is O(n log (MS)), this is a pseudo-polynomial-time due to M=2 3
log M

58. Thanks, End

59. Lecture 3 Complexity of Approximation
L-reduction
Two subclass of PTAS
Set-cover ((ln n)-approximation)
Independent set (n -approximation) c