Chapter 15: Dynamic Programming

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

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)

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 = 0+2500+35X15X20=13000 m[2,3] + m[4,5] + p1p3p5 = 2625+1000+35X5X20=7125 m[2,4] + m[5,5] + p1p4p5 =4375+0+35X10X20=11375 =7125

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

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: 0 if i=0 or j=0 c[i,j] = c[i-1,j-1] + 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

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