1. Longest Common SubsequenceBy
Richard Simpson

2. Subsequence of a string
A subsequence of a sequence(or string) X is and sequence obtained from X by removing 0 or more of the characters of X. The remaining characters are kept in the same order.
Example X=“abfrdfdvredffgfd”
Removing the red characters gives us
“abrdfredfffd”
which is a subsequence of X.

3. Common Subsequence
A common subsequence of sequences X and Y is a just a string that is a subsequence of both X and Y.
There may be many subsequences of X and Y. A subsequence of this set that has maximal length is called the longest common subsequence.
There may be several of these maximal subsequences.

4. How many subsequences are there for a subseq of length n?
This is related to the number of subsets of a set. How many are there?
2 to the n
Why?
Every time you add a new character you double the number of subsets. Do it by hand and check it out.
SO! Brute force is out!!!

5. ith Prefix Let X be a sequence
Then Xi is defined as the first i characters of the sequence X.d
For example let X= “asdcfrfgtr” then X4 is the sequence “asdc”
We will use these prefix's to build the LCS

6. Optimal Substructure of an LCS
Let C = x1x2x3…xm and Y = y1y2y3…yn and let Z= z1z2z3…zk be any LCS of X and Y.
If xm=yn then zk=xm=yn and Zk-1 is and LCS of Xm-1 and Yn-1
If xm≠yn then zk ≠ xm implies that Z is an LCS of Xm- 1 and Y
If xm≠yn then zk ≠ ym implies that Z is an LCS of X and Yn-1

7. Meaning? If xm=yn then zk=xm=yn and Zk-1 is and LCS of Xm-1 and Yn-1
X = acdggdcagagccda
Y = acdgadcggagdada
Z = acdgdcgagda
This is similar for the 3rd case

8. Meaning? If xm≠yn then zk ≠ xm implies that Z is an LCS of Xm-1 and Y
X = acdggdcagagccdc
Y = acdgadcggagdada
Z = acdgdcgagda

9. Recursive View
Let c[i,j] be the length of the LCS of prefixes Xi and Yj the the optimal substructure of the LCS problem gives us 𝑐 𝑖,𝑗 = 0 𝑖𝑓 𝑥=0 𝑜𝑟 𝑗=0 𝑐 𝑖−1,𝑗−1 +1 𝑖𝑓 𝑖,𝑗>0 𝑎𝑛𝑑 𝑥 𝑖 = 𝑦 𝑗 max⁡(𝑐 𝑖,𝑗−1 ,𝑐 𝑖−1,𝑗 𝑖𝑓 𝑖,𝑗>0 𝑎𝑛𝑑 𝑥 𝑖 ≠ 𝑦 𝑗

10. The Algorithm
# Initialize the arrays m=length[X]; n=length[Y] For i=1 to n do c[i,0]=0 For j=1 to n do c[0,j]=0

11. Continued
For i=1 to m do for j=1 to n do if xi=yj then c[i,j] = c[i-1,j-1]+1 b[i,j] = “↖” else if c[i-1,j]>=c[i,j-1] then c[i,j]=c[i-1,j] b[i,j]=“↑” else c[i,j]= c[I,j-1] b[i,j]=“←” Return c and b

12. The Table

13. Print it
Print-LCS(b,X,i,j) if i=0 or j=0 return if b[I,j]=“↖” then Pirnt-LCS(b,X,i-1,j-1) print xi elseif b[i,j]=“↑” then Print-LCS(b,X,i-1,j) else Print-LCS(b,X,i,j-1)