# Algorithm of Longest Common Sequence

```LCS-LENGTH (X, Y)
1. m ← length [X]
2. n ← length [Y]
3. for i ← 1 to m
4. do c [i,0] ← 0
5. for j ← 0 to m
6. do c [0,j] ← 0
7. for i ← 1 to m
8. do for j ← 1 to n
9. do if xi= yj
10. then c [i,j] ← c [i-1,j-1] + 1
11. b [i,j] ← "↖"
12. else if c[i-1,j] ≥ c[i,j-1]
13. then c [i,j] ← c [i-1,j]
14. b [i,j] ← "↑"
15. else c [i,j] ← c [i,j-1]
16. b [i,j] ← "← "
17. return c and b.
```

## Example of Longest Common Sequence

Example: Given two sequences X [1...m] and Y [1.....n]. Find the longest common subsequences to both.

```here X = (A,B,C,B,D,A,B) and Y = (B,D,C,A,B,A)
m = length [X] and n = length [Y]
m = 7 and n = 6
Here x1= x [1] = A   y1= y [1] = B
x2= B  y2= D
x3= C  y3= C
x4= B  y4= A
x5= D  y5= B
x6= A  y6= A
x7= B
Now fill the values of c [i, j] in m x n table
Initially, for i=1 to 7 c [i, 0] = 0
For j = 0 to 6 c [0, j] = 0
```

That is:

```Now for i=1 and j = 1
x1 and y1 we get x1 ≠ y1 i.e. A ≠ B
And 	c [i-1,j] = c [0, 1] = 0
c [i, j-1] = c [1,0 ] = 0
That is, c [i-1,j]= c [i, j-1] so c [1, 1] = 0 and b [1, 1] = ' ↑  '

Now for i=1 and j = 2
x1 and y2 we get x1 ≠ y2 i.e. A ≠ D
c [i-1,j] = c [0, 2] = 0
c [i, j-1] = c [1,1 ] = 0
That is, c [i-1,j]= c [i, j-1] and c [1, 2] = 0 b [1, 2] = '  ↑  '

Now for i=1 and j = 3
x1 and y3 we get x1 ≠ y3 i.e. A ≠ C
c [i-1,j] = c [0, 3] = 0
c [i, j-1] = c [1,2 ] = 0
so c [1,3] = 0     b [1,3] = ' ↑ '

Now for i=1 and j = 4
x1 and y4 we get. x1=y4 i.e A = A
c [1,4] = c [1-1,4-1] + 1
= c [0, 3] + 1
= 0 + 1 = 1
c [1,4] = 1
b [1,4] = '  ↖  '

Now for i=1 and j = 5
x1 and y5  we get x1 ≠ y5
c [i-1,j] = c [0, 5] = 0
c [i, j-1] = c [1,4 ] = 1
Thus c [i, j-1] >  c [i-1,j] i.e. c [1, 5] = c [i, j-1] = 1. So b [1, 5] = '←'

Now for i=1 and j = 6
x1 and y6   we get x1=y6
c [1, 6] = c [1-1,6-1] + 1
= c [0, 5] + 1 = 0 + 1 = 1
c [1,6] = 1
b [1,6] = '  ↖  '
```
```Now for i=2 and j = 1
We get x2 and y1 B = B i.e.  x2= y1
c [2,1] = c [2-1,1-1] + 1
= c [1, 0] + 1
= 0 + 1 = 1
c [2, 1] = 1 and b [2, 1] = ' ↖ '
Similarly, we fill the all values of c [i, j] and we get
```

Step 4: Constructing an LCS: The initial call is PRINT-LCS (b, X, X.length, Y.length)

```PRINT-LCS (b, x, i, j)
1. if i=0 or j=0
2. then return
3. if b [i,j] = ' ↖ '
4. then PRINT-LCS (b,x,i-1,j-1)
5. print x_i
6. else if b [i,j] = '  ↑  '
7. then PRINT-LCS (b,X,i-1,j)
8. else PRINT-LCS (b,X,i,j-1)
```

Example: Determine the LCS of (1,0,0,1,0,1,0,1) and (0,1,0,1,1,0,1,1,0).

Solution: let X = (1,0,0,1,0,1,0,1) and Y = (0,1,0,1,1,0,1,1,0).

We are looking for c [8, 9]. The following table is built.

From the table we can deduct that LCS = 6. There are several such sequences, for instance (1,0,0,1,1,0) (0,1,0,1,0,1) and (0,0,1,1,0,1)