1. Longest common subsequence (LCS)

2. The longest common subsequence (LCS) problem
A string : A = b a c a d
A subsequence of A: deleting 0 or more symbols from A (not necessarily consecutive).
e.g. ad, ac, bac, acad, bacad, bcd.
Common subsequences of A = b a c a d and
B = a c c b a d c b : ad, ac, bac, acad.
The longest common subsequence (LCS) of A and B:
a c a d.
Subsequence need not be consecutive, but must be in order.

3. Longest common subsequence
INPUT: two strings
OUTPUT: longest common subsequence
ACTGAACTCTGTGCACT
TGACTCAGCACAAAAAC

4. Longest common subsequence
INPUT: two strings
OUTPUT: longest common subsequence
ACTGAACTCTGTGCACT
TGACTCAGCACAAAAAC

5. Longest common subsequence
INPUT: two strings
OUTPUT: longest common subsequence
ACTGAACTCTGTGCACT
TGACTCAGCACAAAAAC

6. Longest common subsequence
INPUT: two strings
OUTPUT: longest common subsequence
ACTGAACTCTGTGCACT
TGACTCAGCACAAAAAC

7. Longest common subsequence
INPUT: two strings
OUTPUT: longest common subsequence
ACTGAACTCTGTGCACT
TGACTCAGCACAAAAAC

8. Naïve /Brute Force Algorithm
For every subsequence of X, check whether it’s a subsequence of Y .
X has 2m subsequences.
Each subsequence takes Θ(n) time to check: scan Y for first letter, for second, and so on.
Time: Θ(n2m).

9. Optimal Substructure Theorem
Let Z = z1, , zk be any LCS of X x1, . . , xk and Y y1, . . , yk .
1. If xm = yn, then zk = xm = yn and Zk-1 is an LCS of Xm-1 and Yn-1.
2. If xm  yn, then either zk  xm and Z is an LCS of Xm-1 and Y .
or zk  yn and Z is an LCS of X and Yn-1.

10. Optimal Substructure Proof: (case 1: xm = yn)
Theorem
Let Z = z1, , zk be any LCS of X and Y .
1. If xm = yn, then zk = xm = yn and Zk-1 is an LCS of Xm-1 and Yn-1.
Proof: (case 1: xm = yn)
Any sequence Z’ that does not end in xm = yn can be made longer by adding xm = yn to the end. Therefore,
longest common subsequence (LCS) Z must end in xm = yn.
Zk-1 is a common subsequence of Xm-1 and Yn-1, and
there is no longer CS of Xm-1 and Yn-1, or Z would not be an LCS.

11. Optimal Substructure Proof: (case 2: xm  yn, and zk  xm)
Theorem
Let Z = z1, , zk be any LCS of X and Y .
2. If xm  yn, then either zk  xm and Z is an LCS of Xm-1 and Y .
Proof: (case 2: xm  yn, and zk  xm)
Since Z does not end in xm,
Z is a common subsequence of Xm-1 and Y, and
there is no longer CS of yn-1 and x, or Z would not be an LCS.

12. Optimal Substructure Proof: (case 2: xm  yn, and zk  yn)
Theorem
Let Z = z1, , zk be any LCS of X and Y .
2. If xm  yn, then zk  yn and Z is an LCS of X and Yn-1.
Proof: (case 2: xm  yn, and zk  yn)
Since Z does not end in yn,
Z is a common subsequence of yn-1 and x, and
there is no longer CS of Xm-1 and Y, or Z would not be an LCS.

13. Recursive Solution Sequences x1,…,xn, and y1,…,ym
LCS(i,j) = length of a longest common
subsequence of x1,…,xi and y1,…,yj
if xi = yj then
LCS(i,j) =

14. Recursive Solution Sequences x1,…,xn, and y1,…,ym
LCS(i,j) = length of a longest common
subsequence of x1,…,xi and y1,…,yj
if xi = yj then
LCS(i,j) = 1 + LCS(i-1,j-1)

15. Recursive Solution Sequences x1,…,xn, and y1,…,ym
LCS(i,j) = length of a longest common
subsequence of x1,…,xi and y1,…,yj
if xi  yj then
LCS(i,j) = max (LCS(i-1,j),LCS(i,j-1))
xi and yj cannot both be in LCS

16. Recursive Solution Sequences x1,…,xn, and y1,…,ym
LCS(i,j) = length of a longest common
subsequence of x1,…,xi and y1,…,yj
if xi = yj then
LCS(i,j) = 1 + LCS(i-1,j-1)
if xi  yj then
LCS(i,j) = max (LCS(i-1,j),LCS(i,j-1))

17. Recursive Solution Define c[i, j] = length of LCS of Xi and Yj .
We want c[m,n].

18. Constructing an LCS

19. LCS Example What is the Longest Common Subsequence of X and Y?
We’ll see how LCS algorithm works on the following example:
X = ABCB
Y = BDCAB
What is the Longest Common Subsequence
of X and Y?
LCS(X, Y) = BCB
X = A B C B
Y = B D C A B

20. ABCB BDCAB X = ABCB; m = |X| = 4 Y = BDCAB; n = |Y| = 5j i Yj B D C A B Xi A 1 B 2 3 C 4 B
X = ABCB; m = |X| = 4
Y = BDCAB; n = |Y| = 5
Allocate array c[5,4]

21. ABCB BDCAB for i = 1 to m c[i,0] = 0 for j = 1 to n c[0,j] = 0Yj B D C A B Xi A 1 B 2 3 C 4 B
for i = 1 to m c[i,0] = 0
for j = 1 to n c[0,j] = 0

22. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 C if ( Xi == Yj )A 1 B 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

23. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 B 2 3 C if ( Xi == Yj )A 1 B 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

24. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 B 2 3 CA 1 1 B 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

25. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 3 CA 1 1 1 B 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

26. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 3 CA 1 1 1 B 2 1 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

27. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 3 CA 1 1 1 B 2 1 1 1 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

28. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 3 CA 1 1 1 B 2 1 1 1 1 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

29. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 4 B

30. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 4 B

31. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 2 4 B

32. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 2 2 4 B

33. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 2 2 2 4 B

34. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 2 2 2 4 B 1

35. ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 2 2 2 4 B 1 1 2 2

36. 3 ABCB BDCAB j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 CA 1 1 1 B 2 1 1 1 1 2 3 C
if ( Xi == Yj )
c[i,j] = c[i-1,j-1] + 1
b[i,j] = “ ”
if ( c[i-1,j] >= c[i,j] )
c[i,j] = c[i-1,j]
b[i,j] = “ ”
else c[i,j] = c[i,j-1] 1 1 2 2 2 3 4 B 1 1 2 2

37. Computing the Length of an LCS

38. Analysis of an LCS O(m*n) Running time and memory: O(mn) and O(mn).
since each c[i,j] is calculated in constant time, and there are m*n elements in the array
Running time and memory: O(mn) and O(mn).

39. How to find actual LCS 3 j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 2 B 1Xi A 1 1 1 2 B 1 1 1 1 2 3 C 1 1 2 2 2 4 3 B 1 1 2 2

40. Finding LCS 3 j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 3 4 B 1 1 2 2

41. How to find actual LCS 3 j 0 1 2 3 4 5 i Yj B D C A B Xi A 1 1 1 2 B 1Xi A 1 1 1 2 B 1 1 1 1 2 3 C 1 1 2 2 2 4 3 B 1 1 2 2
Initial call is PRINT-LCS (b, X,m, n).
When b[i, j ] = , we have extended LCS by one character. So LCS = entries with in them.
Time: O(m+n) (no of characters to be checked)

42. 3 B C B LCS (reversed order): B C Bj i Yj B D C A B Xi A 1 1 1 B 2 1 1 1 1 2 3 C 1 1 2 2 2 3 4 B 1 1 2 2 B C B
LCS (reversed order):
B C B
(this string turned out to be a palindrome)
LCS (straight order):

43. LCS Example What is the Longest Common Subsequence of X and Y?
We’ll see how LCS algorithm works on the following example:
X = PRESIDENT
Y = PROVIDENCE
What is the Longest Common Subsequence
of X and Y?

45. Output: PRIDEN