1. Pseudo-polynomial time algorithm (The concept and the terminology are important)
Partition Problem:
Input: Finite set A=(a1, a2, …, an} and a size s(a) (integer) for each aA.
Question: Is there a subset A’A such that
 a A’ s(a) =  a A –A’ s(a)?
Theorem: Partition problem is NP-complete (Karp, 1972). Dynamic programming algorithm:
For in and j 0.5  a A s(a) , define t(i, j) to be true
if and only if there is a subset Ai of {a1, a2, …, ai} such that
 a Ai s(a)=j.
Formula:
T(i,j)=true if and only if t(i-1, j)=true or t(i-1, j-s(ai))=true.

2. Example j i 1 2 3 4 5 6 7 8 9 10 11 12 13 T F
Figure 4.8 Table of t(i,j) for the instance of PARTITION for which A={a1,a2,a3,a4,a5}, s(a1)=1, s(a2)=9, s(a3)=5, s(a4)=3, and s(a5)=8. The answer for this instance is "yes", since t(5,13)=T, reflecting the fact that s(a1)+s(a2)+s(a4)=13=26/2.

3. Backtracking i=n; W=B; if ( t(n, W)== false) then stop;
While (i> 0 ) do
{ if (t(i, W) == true) {
if (t(i-1, W)== true) then i=i-1;
else
{ W=W-s(ai); print “ai” i=i-1; } }

4. Time complexity
The algorithm takes at most O(nB) time to fill in the table, where (Each cell needs constant time to compute).
Do we have a polynomial time algorithm to solve the Partition Problem and thus all NP-complete problems?
No.
O(nB) is not polynomial in terms of the input size.
S(ai)=2n=10000…0 . (binary number of n+1 bits ).
So B is at least O(2n). The input size is O(n2) if there some ai with S(ai)=2n.
B is not polynomial in terms of n (input size) in general.
However, if any upper bound is imposed on B, (e.g., B is Polynomial), the problem can be solved in polynomial time for this special case.
(This is called pseudo-polynomial.)

5. 0-1 version

10. 35

11. Change-making problem:
Given an amount n and unlimited quantities of coins of each of the denominations
d1, d2, …, dm,
find the smallest number of coins that add up to n or indicate that the problem does not have a solution.
Solution:
Let d(i) be the minimum number of coins used for amount i.
Initial values: d(0)=0, d(i)=.
equation: d(i) = min d(i-dk)+1.
k=1, 2, …, m.

12. Change-making problem:
Initial values: d(0)=0, d(i)=.
equation: d(i) = min d(i-dk)+1
k=1, 2, …, m & i-dk≥0
d1=2 and d2=5. i=7.
i= 0, 1, 2, 3, 4, 5, 6, 7
d(i): 0, , 1, , 2, 1, 3, 2.
Backtracking: $5, $2.
(how do you get d(7)=2? d(7)=d(2)+5. Print $5 and goto d(2). How do you get d(2)? D(2)=2+d(0). Print out $2 and goto d(0). Whenever reach d(0), we stop. ) for i=3, since d(3)= , there is no solution.

13. More On Dynamic programming Algorithms
Shortest path with edge constraint:
Let G=(V, E) be a directed graph with weighted edges. Let s and v be two vertices in V. Find a shortest path from s to u with exactly k edges. Here kn-1 is part of the input.
Solution:
Define d(i, v) be the length of the shortest path from s to v with exactly i edges.
d(i, v)=min {c(w, v)+d(i-1, w)}
wV.
Initial values: d(i, s)=0, for i=0, d(i,s)=  for i=1, 2, …, d(0, v)=;
d(k, v) will give the length of the shortest path. A backtracking process can give the path.

14. 6 7 9 2 5 -2 8 -3 -4 z u v x y
i z, u, x, v, y
   
1  z z  
  u 4x 2u
3 4y v  y x.

15. Exercise: Let T be a rooted binary tree, where each internal node in the tree has two children and every node (except the root) in T has a parent. Each leaf in the tree is assigned a letter in ={A, C, G, T}. Figure 1 gives an example. Consider an edge e in T. Assume that every end of e is assigned a letter. The cost of e is 0 if the two letters are identical and the cost is 1 if the two letters are not identical. The problem here is to assign a letter in  to each internal node of T such that the cost of the tree is minimized, where the cost of the tree is the total cost of all edges in the tree. Design a polynomial-time dynamic programming algorithm to solve the problem.

16. A C
Figure 1 A

17. Assignment 4. (Due on Friday of Week 13. Drop it in Mail Box 71 or 72)
This time, the tutors and lectures can explain the questions, but we will NOT tell you how to solve the problems.
Question 1: (45 points)You are going to drive from Princeton to Palo Alto along a fixed route. There are n gas stations along the route. The locations of the n stations are given as x1=0, x2, …, xn, where xi >0 is a real number indicating the distance between the starting place (Princeton) and the i-th station. When the fuel tank is full, the car can go to C distance. The price of the fuel at the i-th station is Ci for the whole tank. Each time, when you stop at a station, you have to fulfil the tank. If the remaining oil at that time is y, say y=51%, then you have to pay (1-y)Ci at the i-th station. The fuel required from the i-th station to the (i+1)-th station is (xi+1-xi)/C1. When you start, the fuel tank is full. When you arrive the destination (the n-th station), you also have to add fuel so that the tank is full. Design a polynomial time algorithm to compute a set of stations along the route to get the fuel so that the total cost of the fuel you pay is minimized. The running time of the algorithm should be polynomial. x2 x1 xn

18. Assignment 4.
Question 2. (45 points). Suppose you have one machine and a set of n jobs a1, a2, …, an to process on that machine. Each job aj has an integer processing time tj, a profit pj and an integer deadline dj. The machine can process only one job at a time, and job aj must run uninterruptedly for tj consecutive time units. If job aj is completed by its deadline dj, you receive a profit pj, but if it is completed after its deadline, you receive a profit of 0. Give a dynamic programming algorithm to find the schedule that obtains the maximum amount of profit. What is the running time of your algorithm?
(Let d be the biggest deadline, the running time can be related to d. )
Question 3. (10 points) Give the names of two NP-complete problems.