Longest Common Subsequence & Matrix Chain Multiplication

Longest common subsequence
&
Matrices chain multiplication
Course Title : Algorithm Design And Analysis

2 LONGESTCOMMONSUBSEQUENCE
The longest common subsequence (LCS) problem is the
problem of finding the longest subsequence common
to all sequences in a set of sequences (often just two
sequences).

abc a b c ab bc ac
cab c a b ca ab cb
LCS=ab
3

Condition Table Value Sign
If i=j=0 0 No sign
If (i,j>0) && Xi=Yj C[i-1,j-1]
Else if (i,j>0 && Xi=Yj)
If [C(i-1,j)>=C(i,j-1)]
Else [C(i-1,j)<=C(i,j-1)]
C[i-1,j]
C[I,j-1]
LONGESTCOMMONSUBSEQUENCE

Example: Find the length of the longest subsequence present in both of them…….
Seq1: ABCDEFG , Seq2: CDBAEF
FEDC
Seq1 A B C D E F G
Seq2 0 0 0 0 0 0 0 0
C 0 0 0 1 1 1 1 1
D 0 0 0 1 2 2 2 2
B 0 0 1 1 2 2 2 2
A 0 1 1 1 2 2 2 2
E 0 1 1 1 2 3 3 3
F 0 1 1 1 2 3 4 4

1. m:=length(X)
2. n:=length(Y)
3. Let B=[1……m,1…..n]& C=[0……m,0…….n]
4. For i=1 to n
5. C[i,0]:=0
6. For j=1 to n
7. C[0,j]:=0
8. For i=1 to n
9. for j=1 to n
10. If Xi=Yj
11. C[I,j]:=C[i-1,j-1]
12. B[i,j]=“ “
13. Else
14. if C[i-1,j]>=C[i,j-1]
Algorithm

15. C[i,j]:=C[i-1,j]
16. B[i,j]:=“ “
17. Else
C[i,j]:=C[i,j-1]
18.B[i,j]:=“ “
19.Return
Algorithm

Definition: if we are given a sequence/chain of n Matrices (A1,A2………Ai) where i=1,2…..n &
Ai has dimension Pi-1*Pi to be multiplied and we wish to compute the product or
parenthesis In a way that the total number of scalar multiplication is minimized.
This process is known as Matrix Chain Multiplication.
Goal of MCM: parenthesizing a chain of matrices so that it can have a dramatic impact on
the cast evaluating multiplication of matrices or the product.
Where it is being used:
 Compiler design for cost optimization.
 Database or query optimization.
Rules to remember :
-Multiplication is associative not commulative.
Matrix chain multiplication

A1 A2
(5*4) (4*6)
Dimension: A1 A2 = 5*6
=30
Cost : A1 A2 = 5*4*6
= 120
How does cost can be calculated?

If the matrices are
A1 A2 A3
(5*4) (4*6) (6*2)
 (A1 A2 )A3
 A1( A2 A3)
 (A1 A2 )A3
A1 A2 = 5*4*6
C1 = 120
(A1 A2 )A3 = 5*6*2
C2 = 60
Total Cost = C1 + C2
=120 + 60
= 180

Law :
How many parenthesizing could be possible?
2ncn
P(n-1) =
n+1
Note : if the matrices are A1 ,A2,A3 ,A4 then we take n=3
2*3c3
P(3-1) =
3+1
= 5

A1 A2 A3 A4
(5*4) (4*6) (6*2) (2*7)
P0 P1 P1 P2 P2 P3 P3 P4
2*3c3
P(3-1) =
3+1
= 5
How many parenthesizing could be possible?

m(i,i)= 0
m(j,j)= 0
m(i,j)= min(i<=k<j) {m(i,k) + m(k+1,j) + pi-1 *pk * pj}
m(1,2)= min(1<=k<2) {m(1,1) + m(1+1,2) + p0 *p1 * p2}
= min(1<=k<2) {0 + 0+ 5 *4 *6}
= 120 (for k=1)

m(2,3)= min(2<=k<3) {m(2,2) + m(3,3) + p1 *p2 * p3}
= min(1<=k<3) {0 + 0+ 4 *6 *2}
= 48 (for k= 2)
m(3,4)= min(3<=k<4) {m(3,3) + m(4,4) + p2 *p3 * p4}
= min(1<=k<4) {0 + 0+ 6 *2 *7}
= 84 (for k= 3)

m(1,3)= min(1<=k<3) {m(1,1) + m(2,3) + p0 *p1 * p3} + {m(1,2) + m(3,3) + p0 *p2 * p3}
= min(1<=k<3) {0 + 48+ 5 *4 *2} + {120 + 0 + 5 *6 *2}
= min(1<=k<3) {88,180}
= 88 (for k= 1)
m(2,4)= min(2<=k<4) {m(2,2) + m(3,4) + p1 *p2 * p4} + {m(2,3) + m(4,4) + p1 *p3 * p4}
= min(1<=k<4) {0 + 84+ 4 *6 *7} + {48 + 0 + 4 *2 *7}
= min(1<=k<4) {272,104}
= 104 (for k= 3)

m(1,4)= min(1<=k<4) {m(1,1) + m(2,4) + p0 *p1 * p4} + {m(1,2) + m(3,4) + p0 *p2 * p4} + {m(1,3) + m(4,4) + p0 *p3 * p4}
= min(1<=k<4) {0 + 104+ 5 *4 *7} + {120 + 84 + 5 *6 *7} + {88 + 0 + 5 *2 *7}
= min(1<=k<4) {244,414,158}
= 158 (for k= 3)
Short :
m(1,2) = 120
m(1,3) = 88
m(1,4) = 158
m(2,3) = 48
m(2,4) = 104
m(3,4) = 84

M[1………….m,1………..n] s[1……..n-1,2…………n]
0 120 88 158
0 48 104
0 84
0
1 1 3
2 3
3
For cost value
For the value of k
J=1 2 3 4 J = 2 3 4
i=1
2
3
4
i=1
2
3

For cost value For the value of k

158, k=3
m(i,j)
m[1,4]
m(i,k) , m(k+1,j)
m(1,3) , m(4,4)
88, k=1
m(i,j)
m[1,3]
m(i,k) , m(k+1,j)
m(1,1) , m(2,3)
A4
A1
48, k=2
m(i,j)
m[2,3]
m(i,k) , m(k+1,j)
m(2,2) , m(3,3)
A2 A3
A1 A2 A3 A4

A1 A2 A3 A4
((A1(A2A3))A4)
A1 A2 A3 A4
A1 A2 A3
A2 A3
Tree Diagram

Longest Common Subsequence & Matrix Chain Multiplication

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