1. Midterm: week 7 in the lecture for 2 hours
2019/4/27
chapter25

2. 2019/4/27
chapter25

3. 2019/4/27
chapter25

4. 2019/4/27
chapter25

5. Recursive Algorithm: Compute-Opt(j) if j=0 then return 0 else
return max {vj+Compute-Opt(p(j)), Compute-Opt(j-1)}
Running time: >2n/2.
(not required)
2019/4/27
chapter25

6. Index v1=2 p(1)=0 1 v2=4 p(2)=0 2 v3=4 p(3)=1 3 v4=7 p(4)=0 4 v5=26
2019/4/27
chapter25

7. (not required) It may take exponential time OPT(6) OPT(5) OPT(3)
The tree of subproblems
grows very quickly
OPT(1)
OPT(1)
It may take exponential time
2019/4/27
chapter25

8. T(n)=T(n-1)+T(n-2)>2T(n-2)>4T(n-4) > 8T(n-6)>…>2n/2T(1)
(not required)
T(n)=T(n-1)+T(n-2)>2T(n-2)>4T(n-4) > 8T(n-6)>…>2n/2T(1)
2019/4/27
chapter25

9. Weighted Interval Scheduling: Bottom-Up
Input: n, s1, s2, …, sn, f1, f2, …, fn, v1, v2, …, vn
Sort jobs by finish times so that f1f2 … fn.
Compute p(1), p(2) , …, p(n)
M[0]=0;
for j = 1 to n do
M[j] = max { vj+m[p(j)], m[j-1]}
if (M[j] == M[j-1]) then B[j]=0 else B[j]=1 /*for backtracking
m=n; /*** Backtracking
while ( m ≠0) { if (B[m]==1) then
print job m; m=p(m)
else
m=m-1 }
B[j]=0 indicating job j is not selected.
B[j]=1 indicating job j is selected.
2019/4/27
chapter25

10. Backtracking: job1, job 3, job 5
M[2]=w2+M[0]=4+0; M[3]=w3+M[1]=4+2; M[4]=W4+M[0]=7+0; M[5]=W5+M[3]=2+6;
M[6]=w6+M[3]=1+6<8;
Index
M = 2
w1=2
p(1)=0 1
w2=4
p(2)=0 2 4 2
w3=4
p(3)=1 3 2 4 6
w4=7
p(4)=0 4
w5=2
p(5)=3 2 4 6 7 5
w6=1
p(6)=3 6 2 4 6 7 8 2 4 6 7 8 8
Backtracking: job1, job 3, job 5 j: B:
2019/4/27
chapter25

11. Backtracking and time complexity
Backtracking is used to get the schedule.
P()’s can be computed in O(n) time after sorting all the jobs based on the starting times.
Time complexity
O(n) if the jobs are sorted and p() is computed.
Total time: O(n log n) including sorting.
2019/4/27
chapter25

12. Computing p()’s in O(n) time
P()’s can be computed in O(n) time using two sorted lists, one sorted by finish time (if two jobs have the same finish time, sort them based on starting time) and the other sorted by start time.
Start time: b(0, 5), a(1, 3), e(3, 8), c(5, 6), d(6, 8)
Finish time a(1, 3), b(0,5), c(5,6), d(3,8), e(6,8)
P(d)=c, p(c )=b, p(e)= a, p(a)=0, p(b)=0. (See demo7)
2019/4/27
chapter25

13. Example 2: Start time: b(0, 5), a(1, 3), e(3, 8), c(5, 6), d(6, 8)
Finish time a(1, 3), b(0,5), c(5,6), d(6,8), e(3,8)
P(d)=c, p(c )=b, p(e)= a, p(a)=0, p(b)=0.
v(a)=2, v(b)=3, v(c )=5, v(d) =6, v(e)=8.8.
Solution: M[0]=0, M[a]=2. M[b]=max{2, 3+M[p(b)]}=3.
M[c]=max{3, 5+M[p(c )]}=5+M[b]=8.
M[d]=max{8, 6+M[p(d)]}=6+M[c]=6+8=14.
M[e]=max{14, 8.8+M[p(e)]}=max{14, 8.8+M[a]}=max {14, 10.8}=14.
Backtracking: b, c, d Job: a b c d e B:
2019/4/27
chapter25

14. Longest common subsequence
Definition 1: Given a sequence X=x1x2...xm, another sequence Z=z1z2...zk is a subsequence of X if there exists a strictly increasing sequence i1i2...ik of indices of X such that for all j=1,2,...k, we have xij=zj.
Example 1: If X=abcdefg, Z=abdg is a subsequence of X. X=abcdefg, Z=ab d g
2019/4/27
chapter25

15. Definition 2: Given two sequences X and Y, a sequence Z is a common subsequence of X and Y if Z is a subsequence of both X and Y.
Example 2: X=abcdefg and Y=aaadgfd. Z=adf is a common subsequence of X and Y.
X=abc defg
Y=aaaadgfd
Z=a d f
2019/4/27
chapter25

16. Longest common subsequence may not be unique. Example: abcd acbd
Definition 3: A longest common subsequence of X and Y is a common subsequence of X and Y with the longest length. (The length of a sequence is the number of letters in the seuqence.)
Longest common subsequence may not be unique.
Example: abcd
acbd
Both acd and abd are LCS.
2019/4/27
chapter25

17. Longest common subsequence problem
Input: Two sequences X=x1x2...xm, and
Y=y1y2...yn.
Output: a longest common subsequence of X and Y.
Applications:
Similarity of two lists
Given two lists: L1: 1, 2, 3, 4, 5 , L2:1, 3, 2, 4, 5,
Length of LCS=4 indicating the similarity of the two lists.
Unix command “diff”.
2019/4/27
chapter25

18. Longest common subsequence problem
Input: Two sequences X=x1x2...xm, and
Y=y1y2...yn.
Output: a longest common subsequence of X and Y.
A brute-force approach
Suppose that mn. Try all subsequence of X (There are 2m subsequence of X), test if such a subsequence is also a subsequence of Y, and select the one with the longest length.
2019/4/27
chapter25

19. Charactering a longest common subsequence
Theorem (Optimal substructure of an LCS)
Let X=x1x2...xm, and Y=y1y2...yn be two sequences, and
Z=z1z2...zk be any LCS of X and Y.
1. If xm=yn, then zk=xm=yn and Z[1..k-1] is an LCS of X[1..m-1] and Y[1..n-1].
2. If xm yn, then zkxm implies that Z is an LCS of X[1..m-1] and Y.
2. If xm yn, then zkyn implies that Z is an LCS of X and Y[1..n-1].
2019/4/27
chapter25

20. The recursive equation
Let c[i,j] be the length of an LCS of X[1...i] and Y[1...j].
c[i,j] can be computed as follows:
if i=0 or j=0,
c[i,j]= c[i-1,j-1] if i,j>0 and xi=yj,
max{c[i,j-1],c[i-1,j]} if i,j>0 and xiyj.
Computing the length of an LCS
There are nm c[i,j]’s. So we can compute them in a specific order.
2019/4/27
chapter25

21. The algorithm to compute an LCS
1. for i=1 to m do
c[i,0]=0;
3. for j=0 to n do
c[0,j]=0;
5. for i=1 to m do
for j=1 to n do {
if x[i] ==y[j] then
c[i,j]=c[i-1,j-1]+1;
b[i,j]=1;
else if c[i-1,j]>=c[i,j-1] then
c[i,j]=c[i-1,j]
b[i,j]=2;
else c[i,j]=c[i,j-1]
b[i,j]=3; }
2019/4/27
chapter25

22. Example 3: X=BDCABA and Y=ABCBDAB.
2019/4/27
chapter25

23. Constructing an LCS (back-tracking)
We can find an LCS using b[i,j]’s.
We start with b[n,m] and track back to some cell b[0,i] or b[i,0].
The algorithm to construct an LCS (backtracking)
1. i=m
2. j=n;
3. if i==0 or j==0 then exit;
4. if b[i,j]==1 then {
i=i-1;
j=j-1;
print “xi”; }
5. if b[i,j]== i=i-1
6. if b[i,j]== j=j-1
7. Goto Step 3.
The time complexity: O(nm).
2019/4/27
chapter25

24. Remarks on weighted interval scheduling
it takes long time to explain. (50+13 minutes)
Do not mention exponent time etc.
For the first example, use the format of example 2 to show the computation process (more clearly).
2019/4/27
chapter25

25. Shortest common supersequence
Definition: Let X and Y be two sequences. A sequence Z is a supersequence of X and Y if both X and Y are subsequence of Z.
Shortest common supersequence problem:
Input: Two sequences X and Y.
Output: a shortest common supersequence of X and Y.
Example: X=abc and Y=abb. Both abbc and abcb are the shortest common supersequences for X and Y.
2019/4/27
chapter25

26. Let c[i,j] be the length of an SCS of X[1...i] and Y[1...j].
Recursive Equation:
Let c[i,j] be the length of an SCS of X[1...i] and Y[1...j].
c[i,j] can be computed as follows:
j if i=0
i if j=0,
c[i,j]= c[i-1,j-1] if i,j>0 and xi=yj,
min{c[i,j-1]+1,c[i-1,j]+1} if i,j>0 and xiyj.
2019/4/27
chapter25

27. 2019/4/27
chapter25

28. The pseudo-codes for i=0 to n do c[i, 0]=i; for j=0 to m do c[0,j]=j;
if (xi == yj) c[i ,j]= c[i-1, j-1]+1; b[i.j]=1;
else {
c[i,j]=min{c[i-1,j]+1, c[i,j-1]+1}.
if (c[I,j]=c[i-1,j]+1 then b[I,j]=2;
else b[I,j]=3; }
p=n, q=m; / backtracking
while (p≠0 or q≠0)
{ if (b[p,q]==1) then {print x[p]; p=p-1; q=q-1}
if (b[p,q]==2) then {print x[p]; p=p-1}
if (b[p,q]==3) then {print y[q]; q=q-1}
2019/4/27
chapter25

29. Exercises
Exercise 1: For the weighted interval scheduling problem, there are eight jobs with starting time and finish time as follows: j1=(0, 8), j2=(2, 3), j3=(3, 6), j4=(5, 9), j5=(8, 12), j6=(9, 11), j7=(10, 13) and j8=(11, 16). The weight for each job is as follows: v1=3.5, v2=2.0, v3=3.0, v4=3.0, v5=6.5, v6=2.5, v7=12.0, and v8=8.0.
Find a maximum weight subset of mutually compatible jobs. (Backtracking process is required.) (You have to compute p()’s. The process of computing p()’s is NOT required.)
Exercise 2: Let X=abbacab and Y=baabcbb. Find the longest common subsequence for X and Y.
               Backtracking process is required.  
2019/4/27
chapter25

30. Summary of Week 6
Understand the algorithms for the weighted Interval Scheduling problem, LCS and SCS.
The “alignment of sequences” part is not taught.  
2019/4/27
chapter25

31. Alignment of sequences
An alignment:
inserting spaces into X and Y such that the two resulting sequences X’ and Y’ are of the same length.
every letter in X’ is opposite to a unique letter in Y’.
Examples: o-currence o-curr-ance abbbaa--bbbbaab
occurrence o-curre-nce ababaaabbbbba-b
The alignment value:
where X’[i] and Y’[i] are the two letters in column i of the alignment and s(X’[i], Y’[i]) is the score (weight) of these opposing letters.
There are several popular socre schemes for DNA and protein sequences.
2019/4/27
chapter25

32. Recursive equations:
c[i,j]=max{ c[i-1, j-1]+s(X[i], Y[j]), c[i, j-1]+s(_,Y[j]), c[i-1, j)+s(X[i],_)}.
Similarity Score Scheme (max):
match: 1;
mismatch or insertion or deletion: 0.
Example:
A B B C A A A A B B C A A A
A B C C A A A B C A A
The same as LCS if we use the special similarity score and maximization
2019/4/27
chapter25

33. The same as SCS if we use the special distance score and minimization
Recursive equations:
c[i,j]=min{ c[i-1, j-1]+s(X[i], Y[j]), c[i, j-1]+s(_,Y[j]), c[i-1, j)+s(X[i],_)}.
Distance Score Scheme (mix):
match: 0 insertion and deletion 1;
Mismatch 2
Example:
A B B C A A A A B B C A A A
A B C C A A A B C C A A
The same as SCS if we use the special distance score and minimization
2019/4/27
chapter25

34. A score emphasizing A-A match: (max) A-A match: 1,
Any other match or mismatch: 0.
Example:
A B B C A A A A B B C A A A
A B C C A A A B C C A A
There are 3 A-A matchs
2019/4/27
chapter25

35. Time and space complexity
Recursive equations:
c[i,j]=min{ c[i-1, j-1]+s(X[i], Y[j]), c[i, j-1]+s(_,Y[j]), c[i-1, j)+s(X[i],_)}.
c[i,j]=max{ c[i-1, j-1]+s(X[i], Y[j]), c[i, j-1]+s(_,Y[j]), c[i-1, j)+s(X[i],_)}.
Time and space complexity
Both are O(nm) or O(n2) if both sequences have equal length n.
Why?
We have to compute c[i,j] (the cost) and b[i,j] (for back-tracking). Each will take O(n2).
2019/4/27
chapter25