1. Chapter 15: Dynamic Programming

2. Matrix-chain multiplication
Compute the product of n matrices A1,A2,..,An
Where to put the parentheses
(A1( A2 (A3 A4)))
(A1(( A2 A3) A4))
((A1 A2) (A3 A4))
((A1 (A2 A3)) A4)
(((A1 A2) A3) A4)
How many parenthesizations?
1 if n=1
P(n) = or
P(1) P(n-1) + P(2) P(n-2) + … + P(n-1)P(1)
Known as Catalan numbers. Grow as OMEGA(4^n / n^{3/2} )
Observation Suppose that an optimal parenthesization splits between A_k and A_{k+1}. Then parenthesization of A_1 through A_k must be optimal and so is for A_{k+1} through A_n
The fact that this type of observation is correct is crucial for the ability to use dynamic programming

3. Recursive presentation of solution
0 if i=j Min[i,j] = or min {m[i,k] + m[k+1,j] + p_{i-1}p_kp_j} if i< j { i <= k < j} Understanding the complexity: How many pairs i < j for a given difference k (=j-i) Zoom on n/4 < k < 3n/4 For a given k, O(n) pairs each requiring min of O(n) values Since the number of values of k is O(n) the complexity is O(n^3)

4. Matrix-chain multiplication
# of operations
Where to split
A1 30X35
A2 35X15
A3 15X5
A4 5X10
A5 10X20
A6 20X25
6 matrices :in which order to multiply
m[2,5] is the minimum of 3 candidates:
m[2,2] + m[3,5] +p1p2p5 = X15X20=13000
m[2,3] + m[4,5] + p1p3p5 = X5X20=7125
m[2,4] + m[5,5] + p1p4p5 = X10X20=11375
=7125

5. Longest common subsequence (LCS)
Z = {B,C,D,B} is a subsequence of X = {A,B,C,B,D,A,B} Given two sequences X and Y, and a subsequence Z, Z is a common subsequence of X and Y if it is a subsequence of X and a subsequence of Y Given two sequences X and Y, the LCS problem is to find the maximum-length common subsequence of X and Y

6. Theorem 15.1: Optimal substructure of an LCS
Let X = <x_1,x_2,..,x_m> and Y=<y_1,y_2,..,y_n> and Z=<z_1,z_2,..,z_k> be sequences, where Z is any LCS of X and Y. Then:
If x_m=y_n then z_k= x_m=y_n and Z_{k-1} is an LCS of X_{m-1} and Y_{n-1}
If x_m != y_n then
2. z_k != X_m implies that Z is an LCS of X_{m-1} and Y
3. z_k != Y_n implies that Z is an LCS of X and Y_{n-1}
This structure that the following Recursive solution:
The LCS-length of <x_1,..x_i> and <y_1,..,y_j> is:
if i=0 or j=0
c[i,j] = c[i-1,j-1] if i,j >0 and x_i = y_j
max(c[i-1,j],c[i,j-1]) if i,j >0 and x_i != y_j

7. LCS-length table Complexity O(nm) X = <A,B,C,B,D,A,B>,
Y = <B,D,C,A,B,A>
Complexity O(nm)