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23/4/19 chapter25 1
23/4/19 chapter25 2
23/4/19 chapter25 3
23/4/19 chapter25 4
23/4/19 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)
23/4/19 chapter25 6
1
2
3
5
4
6
Index
p(1)=0
p(2)=0
p(3)=1
p(4)=0
p(5)=3
p(6)=3
v1=2
v2=4
v3=4
v4=7
v5=2
v6=1
23/4/19 chapter25 7
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)
else
m=m-1 }
B[j]=0 indicating job j is not selected.B[j]=1 indicating job j is selected.
23/4/19 chapter25 8
1
2
3
5
4
6
Index
p(1)=0
p(2)=0
p(3)=1
p(4)=0
p(5)=3
p(6)=3
w1=2
w2=4
w3=4
w4=7
w5=2
w6=1
0 1 2 3 4 5 6
0
0
0
0
0
0
2
2 4
2 4 6
2 4 6 7
2 4 6 7 8
2 4 6 87 8
M =
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;
23/4/19 chapter25 9
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.
23/4/19 chapter25 10
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)
23/4/19 chapter25 11
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
23/4/19 chapter25 12
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
23/4/19 chapter25 13
• 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
23/4/19 chapter25 14
• 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.
23/4/19 chapter25 15
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”.
23/4/19 chapter25 16
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.
23/4/19 chapter25 17
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].
23/4/19 chapter25 18
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.
23/4/19 chapter25 19
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 }
23/4/19 chapter25 20
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
23/4/19 chapter25 21
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).
23/4/19 chapter25 22
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).
23/4/19 chapter25 23
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.
23/4/19 chapter25 24
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.
23/4/19 chapter25 25
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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} }
23/4/19 chapter25 27
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
23/4/19 chapter25 28
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.
23/4/19 chapter25 29
Summary of Week 7 • Understand the algorithms for the weighted
Interval Scheduling problem, LCS and SCS.
.
