# Longest common subsequence problem

A strand of DNA can be represented as a string over the finite set {A,C,G,T}

We want to know how similar two strings of DNA are. Our measure is the length of the longest common subsequence

In this problem we allow gaps between letters to form a common subsequence, so

ABCBDAB and BDCABA have a common subsequence of BCA

## Formal definition

Subsequence

Given a sequence $X=\langle x_1,...,x_m \rangle$, another sequence $Z=\langle z_1,...,z_k \rangle$ is a subsequence of X if $z_1=x_{i_1},...,z_k=x_{i_k}$ for some $i_1<i_2<...<i_k$

Common subsequence
A common subsequence of X and Y is a subsequence of both X and Y

# Dynamic programming for longest common subsequence

## Step 1: Characterizing a longest common subsequence

Theorem: Optimal substructure:

Let $Z=\langle z_1,...,z_k \rangle$ be an LCS of $X=\langle x_1,...,x_m \rangle$ and $Y=\langle y_1,...,y_n \rangle$

1. If $x_m=y_n$, then $z_k=x_m=y_n$ and $Z[1..k-1]$ is an LCS of $X[1..m-1]$ and $Y[1..n-1]$

2. If $x_m\neq y_n$, then $z_k\neq x_m$ implies that Z is an LCS of $X[1..m-1]$ and $Y$

3. If $x_m\neq y_n$, then $z_k\neq y_n$ implies that Z is an LCS of $X$ and $Y[1..n-1]$

### Proof

1. If $z_k\neq x_m$ then appending $x_m=y_n$ to Z yields a common subsequence longer than Z. This is a contradiction, this $z_k=x_m=y_n$

Then $Z[1..k-1]$ is a common subsequence of $X[1..m-1]$ and $Y[1..n-1]$. Suppose there is a longer one, way W. Again appending $x_m$ to W wields a common subsequence of X and Y longer than Z. This is a contradiction, thus $Z[1..k-1]$ is an LCS of $X[1..m-1]$ and $Y[1..n-1]$

2. If $z_k\neq x_m$, then Z is a common subsequence of $X[1..m-1]$ and Y. Suppose there is a longer one, say W. Then W is also a common subsequence of X and Y but is longer than Z. This is a contradiction, thus Z is an LCS of $X[1..m-1]$ and Y

3. By symmetry

## Step 2: A recursive solution

Let $c[i,j]$ be the length of an LCS of $X[1..i]$ and $Y[1..j]$

The theorem then yields

$$c[i, j]=\left\{\begin{array}{ll}{0} & {\text { if } i=0 \text { or } j=0} \\ {c[i-1, j-1]+1} & {\text { if } i, j>0 \text { and } x_{i}=y_{j}} \\ {\max \{c[i, j-1], c[i-1, j]\}} & {\text { if } i, j>0 \text { and } x_{i} \neq y_{j}}\end{array}\right.$$

Unlike rod cutting and matrix chain multiplication problems, this time we can readily rule out some subproblems (those where $x_i=y_j$)

## Step 3: Computing the length of an LCS

Let us use a bottom up approach

Input: $X=\langle x_1,..,x_m \rangle$ and $Y=\langle y_1,...,y_n \rangle$

The algorithm stores the values $c[0..m,0..n]$ and also maintains the table $b[1..m,1..n]$ where $b[i,j]$ “points” to the next pair (i,j) to consider while reconstructing the LCS

# Algorithm

LCS(X,Y)

Let b[1..m,1..n] and c[0..m, 0..n] be new tables
for i = 1 to m:
c[i] = 0
for j = 0 to n:
c[j] = 0
for i = 1 to m:
for i = 1 to n:
if x_i==y_i:
c[i][j]=c[i-1][j-1]+1
b[i][j]="↖"
else if c[i-1][j]>=c[i][j-1]:
c[i][j] = c[i-1][j]
b[i][j]="↑"
else:
c[i][j]=c[i][j-1]
n[i][j]="⟵"
return c and b # Constructing an LCS

PRINT-LCS(b,X,i,j)

if i==0 or j==0:
return
if b[i,j] == "↖":
PRINT-LCS(b,X,i-1,j-1)
print x[i]
else if b[i][j] == "↑":
PRINT-LCS(b,X,i-1,j)
else
PRINT-LCS(b,X,i,j-1)



Initial call: PRINT-LCS(b,X,m,n)