1. The longest common subsequence (LCS) problem is the classic computer science problem.
2. The longest common subsequence problem is the problem of finding the longest subsequence common to all sequences in the set of sequences.
3. LCS is widely used by revision control systems and in biometrics.
4. DNA is represented as strings of the small alphabet, Sigma = {A, C, G, T}.
5. DNA strings can be changed by mutation by insertion of a character into string.
6. A longest common subsequence of two strings can represent the common ancestry of the two strings.

Finding a LCS:

1. Characterizing the longest common subsequences by defining optimal substructure of LCS.
2. A recursive definition: If am = bn then we should find LCS of Am-1 and Bn-1. Otherwise, compare LCS of A and Bn-1 and LCS of Am-1 and B. Then from these sequences pick up the longer sequence.
3. Compute length of LCS.
4. Construct LCS using d – table.

Example:

Using dynamic programming determine longest sequence of (A, B, C, D, B, A, C, D, F) and (C, B, A, F).

Solution:

We will build c [i, j] and d [i, j]. Initially c [i, 0] ← 0 where I represents row and c [0, j] ← 0 where j represent the column. The c table will store some values and d table will store directions.

Step 1:

The table is shown below.

Step 2:

As A [i] ≠ B [j]

If (c[0, 1] ≥ c [1, 0]) is true.

i.e. 0 ≥ 0

C [i, j] = c [i – 1, j]

C [1, 1] = c [0, 1]

C [1, 1] = 0

D [i, j] = ↑

Step 3:

Let i = 1, j = 2 then

As A [i] ≠ B [j]

If (c [i – 1, j] ≥ c [i, j - 1]) i.e. (c[0, 2] ≥ c [1, 1]) is true. i.e. 0 ≥0

C[i, j] = c [i – 1, j]

C[1, 2] = c [0, 2]

C[1, 2] = 0

D[i, j] = ↑

The table will be

Continuing in this fashion we can fill up the table as shown in figure 10: