1. Least common subsequence:
Biological applications often need to compare the DNA of two (or more) different organisms.
S1 = {ACCGGTCGAGTGCGCGGAAGCCGGCCGAA}
S2 = {GTCGTTCGGAATGCCGTTGCTCTGTAAA}
none are substrings of each other 
we could say that both are similar if the number of changes to turn one into the other is small
Find a 3rd strand in which the letters in S3 appear in S1 and S2 in the same order but not necessarily consecutively.
CSC317

2. CSC317 More formally: Given a sequence , another sequence is a
subsequence of X if there exists a subsequence of indices of X such that for
all j = 1,2,…,k we have
Example: is a subsequence of with index sequence
CSC317

3. More formally:
Given two sequences X and Y we say that a sequence Z is a common subsequence
of X and Y if Z is a subsequence of both X and Y
Example: If and the sequence
is a common subsequence of X and Y.
Is it the longest subsequence? No
longest common subsequences
CSC317

4. In the longest-common-subsequence problem, we are given two sequences
and and want to find a maximum length
common subsequence of X and Y .
Great. How are we going to do that using dynamic programming???
Steps:
Characterizing a longest common subsequence
Recursive solution
Computing the length of LCS
Constructing a LCS
CSC317

5. CSC317 Step 1: Characterizing a longest common subsequence
Brute force solution: We simply enumerate all subsequences of X and
check each subsequence to see whether it is also a subsequence
of Y , keeping track of the longest subsequence we find 
Actually  because we would need to run through 2m subsequences … (sucks)
But, does the LCS problem have an optimal-substructure property (dynamic
programming, anyone)?
Some definitions: Given a sequence we define the ith prefix of X as
. Example: if ,
CSC317

6. CSC317 Theorem (no proof): Optimal substructure of an LCS
Let and be two sequences and be
any LCS of X and Y then:
1.) if xm = yn, then zk= xm = ym and Zk-1 is and LCS of Xm-1 and Yn-1
2.) if xm ≠ yn, then zk ≠ xm implies that Z is a LCS of Xm-1 and Y.
3.) if xm ≠ yn, then zk ≠ yn implies that Z is a LCS of X and Yn-1.
But what does that tell us?
A LCS of two sequences contains within it an LCS of prefixes of the two
sequences. Thus, the LCS problem has an optimal-substructure property, meaning we
could use a recursive solution!
CSC317

7. CSC317 Step 2: A recursive algorithm
To find an LCS of X and Y , we may need to find the LCSs of X and Yn-1 and of Xm-1 and Y . Furthermore, each of these subproblems has the subsubproblem of finding an LCS of Xm-1 and Yn-1.
c[i,j] is the length of an LCS of the sequences Xi and Yj. The optimal substructure of the LCS problem gives the recursive formula
if i = 0 or j = 0
if i, j > 0 and xi = yj
if i, j > 0 and xi ≠ yj
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8. Step 3: Computing the length of a LCS
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9. Step 3: Computing the length of a LCS
AB C BDAB
BDCAB A
BCBA
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10. AB C BDAB BCBA BDCAB A CSC317
Step 4: Constructing a LCS (Backtracking)
AB C BDAB
BDCAB A
BCBA
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