Recursive Algorithm: Compute-Opt(j) if j=0 then return 0 else

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New Algorithm:

Algorithm:

Input: e1<=e2<=e3…<=en and f(e1), f(e2), …, f(en)

for (p=0; p<n; p++) for (i=1; i<=n; i++) if (i+p<=n){ compute c(i, i+p) using the formula (in O(n) time) }

Total running time: O(n3)

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Algorithm (including backtracking):

Algorithm:

Input: e1<=e2<=e3…<=en and f(e1), f(e2), …, f(en)for (p=0; p<n; p++) for (i=1; i<=n; i++) if (i+p<=n){ compute c(i, i+p) using the formula (in O(n) time) b(i, i+p)=k, where e_k is the root to min c(i, j). }

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New Algorithm:// backtracking Let Q be a Queue.Q=empty;print b(1, n) as the root;Q.add (1, n);while (Q != empty){ (i,j)=Q.dequeue(); if (i<=b(i,j)-1) then attach b(i, b(i,j)-1) as the left child of b(i, j) Q.add(i, b(i,j)-1); if (b(i,j)+1<=j) then attach b(b(i,j)+1, j) as the right child of b(i, j). Q.add(b(i,j)+1, j);}

Best binary search tree is:e1->e2->e3->e4

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Using array:C(1, 1), C(2, 2), C(3,3), C(4,4) --size 1C(1, 2), C(2, 3), C(3, 4) --size 2C(1, 3), C(2, 4) -size 3C(1, 4) final result --size 4.

Need to compute O(n2) C(i,j),each takes O(n) time.

This is a new approach: dynamic programming approach

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p(1)=0

p(2)=0

p(3)=1

p(4)=0

p(5)=3

p(6)=3

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Running time for Compute-Opt(n)f(n) = f(n-1)+f(p(n))+O(1)

=f(n-1)+f(n-2) +O(1) (assume that p(n)=n-2) >2f(n-2) (ignoring O(!) and use f(n-2) to replace f(n-1)

=4f(n-4)

=8f(n-6)

>O(2 n/2).

This is exponential in terms of n.

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Weighted Interval Scheduling: Bottom-Up Input: n, s1, s2, …, sn, f1, f2, …, fn, v1, v2, …, vn

Sort jobs by finish times so that f1f2 … 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)

m=m-1 }

B[j]=0 indicating job j is not selected.B[j]=1 indicating job j is selected.

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p(1)=0

p(2)=0

p(3)=1

p(4)=0

p(5)=3

p(6)=3

0 1 2 3 4 5 6

2 4 6 7

2 4 6 7 8

2 4 6 87 8

j: 0 1 2 3 4 5 6

B: 0 1 1 1 1 1 0Backtracking: job1, job 3, job 5

M[j] = max { vj+m[p(j)], m[j-1]}

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;

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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.

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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(6,8), e(3,8)

P(d)=c, p(c )=b, p(e)= a, p(a)=0, p(b)=0. (See demo7)

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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: 1 1 1 1 0

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Summary of Dynamic programming

1. Define an array, e.g., c(i, j) or M(i), etc

2. Find equations to compute c(i, j) or d(i), etc

3. Give the order to computer all cells.

Remark: The solution for big size is obtained from solution(s) of smaller size.

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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

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• 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

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• 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 sequence.)

• Longest common subsequence may not be unique.

• Example: abcd acbd Both acd and abd are LCS.

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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”.

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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 mn. 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.

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Charactering a longest common subsequence

• Theorem (Optimal substructure of an LCS)

• Let X=x1x2...xi, and Y=y1y2...yj be two sequences, and • Z=z1z2...zk be any LCS of X and Y.• 1. If xi=yj, then zk=xi=yj and Z[1..k-1] is an LCS of X[1..m-

1] and Y[1..n-1]. • 2. If xi yj, then zkxi implies that Z is an LCS of X[1..i-

1] and Y.• 2. If xi yj, then zkyjimplies that Z is an LCS of X and

Y[1..j-1].

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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: 0 if i=0 or j=0,c[i,j]= c[i-1,j-1]+1 if i,j>0 and xi=yj, max{c[i,j-1],c[i-1,j]} if i,j>0 and xiyj.

Computing the length of an LCS• There are nm c[i,j]’s. So we can compute them

in a specific order.

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The algorithm to compute an LCS • 1. for i=1 to m do • 2. c[i,0]=0;• 3. for j=0 to n do• 4. c[0,j]=0;• 5. for i=1 to m do• 6. for j=1 to n do• 7. { • 8. if x[i] ==y[j] then• 9. c[i,j]=c[i-1,j-1]+1;• 10 b[i,j]=1; • 11. else if c[i-1,j]>=c[i,j-1] then • 12. c[i,j]=c[i-1,j]• 13. b[i,j]=2;• 14. else c[i,j]=c[i,j-1] • 15. b[i,j]=3;• 14 }

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Example 3: X=BDCABA and Y=ABCBDAB.

xi 0 0 0 0 0 0 0

A 0 0 0 0 1 1 1

B 0 1 1 1 1 2 2

C 0 1 1 2 2 2 2

B 0 1 1 2 2 3 3

D 0 1 2 2 2 3 3

A 0 1 2 2 3 3 4

B 0 1 2 2 3 4 4

yi B D C A B A

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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=m2. 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]==2 i=i-16. if b[i,j]==3 j=j-17. Goto Step 3.

• The time complexity: O(nm).

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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).

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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 subsequences 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.

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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]+1 if i,j>0 and xi=yj,

min{c[i,j-1]+1,c[i-1,j]+1} if i,j>0 and xiyj.

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The pseudo-codes for i=0 to n do c[i, 0]=i; for j=0 to m do c[0,j]=j; for i=1 to n do for j=1 to m do 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} }

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Example SCS for X=BDCABA and Y=ABCBDAB.

xi 0 1 2 3 4 5 6

A 1 2 3 4 4 5 6

B 2 2 3 4 5 5 6

C 3 3 4 4 5 6 7

B 4 4 5 5 6 6 7

D 5 5 5 6 7 7 8

A 6 6 6 6 7 8 8

B 7 7 7 7 8 8 9

yi B D C A B A

7+6-4 (LCS)=9 (SCS) see slide 23

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• Question 1: For the weighted interval scheduling problem, there are eight jobs with starting time and finish time as follows: j1=(0, 6), j2=(2, 3), j3=(3, 5), 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=3.8, v2=2.0, v3=3.0, v4=3.0, v5=6.5, v6=2.5, v7=13.0, and v8=6.0.

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.)

• Question 2: Let X=aabbacab and Y=baabcbb. Find the longest common subsequence for X and Y. (Backtracking process is required.)

• Question 3. Let X=aabbacab and Y=baabcbb. Find the shortest common supsequence for X and Y. (Backtracking process is required.)

Assignment 2: Due Week 11, Friday

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Summary of Week 7 • Understand the algorithms for the weighted

Interval Scheduling problem, LCS and SCS.

Recursive Algorithm: Compute-Opt(j) if j=0 then return 0 else

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