y j LCS-Length(X,Y) Running time: O(st) set c[i,0] s and c[0,j] s to 0 for i=1 to s for j=1 to t if x i =y j then else if

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1 Recursive solution for finding LCS of X and Y if x s =y t, then find an LCS of X s-1 and Y t-1, and then append x s =y t to this LCS if x s y t, then solve two subproblems: (1) find an LCS of X s-1 and Y; (2) find an LCS of X and Y t-1 ; whichever of these two LCS s is longer --- that s an LCS of X and Y c[i,j]= length of an LCS of the seuences X i and Y j when i = 0 or j = 0, then c[ i, j] = 0 when i > 0, j > 0, and x i = y j, then c[ i, j] = ci [ 1, j 1] + 1 when i > 0, j > 0, and x i y j, then c[ i, j] = max { c[, i j 1], ci [ 1, j] } LCS-Length(X,Y) Running time: O(st) set c[i,0] s and c[0,j] s to 0 for i=1 to s for j=1 to t if x i =y j then else if c[ i, j] = ci [ 1, j 1] + 1 c[ i, j] = max { c[, i j 1], ci [ 1, j] }

2 Proof: (1) if z k x s, then we could append x s =y t,to Z and get a CS of length k+1 ><. If there were a CS W of X s-1 and Y t-1 with length k, then we could append x s =y t to W and get a CS of X and Y of length k+1 ><. Thus, a CS of X s-1 and Y t-1 with length k is an LCS, i.e. Z k-1 is an LCS of X s-1 and Y t-1 X X s-1 Y Y t-1 Z W gsdg k (2) if z k x s, then that Z is a CS of X s-1 and Y. If there were a CS W of X s-1 and Y with length greater than k, then W would also be a CS of X s and Y ><. X Y Z r X s-1 gsdg k+1 (3) same argument as (2) Q.E.D. Y W

3 Longest common subseuence problem seuences X = <x 1, x 2,...,x s > and Y = <y 1, y 2,...,y t > Z = <z 1, z 2,...,z k > is a subseuence of X if there exists a strictly increasing seuence <i(1), i(2),...,i(k)> of indices of X such that for all j, x i(j) =z j <A,C,Z,K> is a subseuence of < C, A, F, C, C, Z, K, Y> with index seuence <2,4,6,7> or <2,5,6,7> Z is a common subseuence of X and Y if Z is a subseuence of both X and Y X=<A,B,C,B,D,A,B> Y=<B,D,C,A,B,A> a common subseuence (CS) B,C, A B, C,A a longest CS (LCS) B,C,B, A B, C, B,A Brute force -- try all common subseuence -- takes exponential time Let X i = <x 1, x 2,...,x i >, i.e. X i is the i-th prefix of X Theorem: Let Z = <z 1, z 2,...,z k > be any LCS of X and Y. 1. if x s =y t, then z k =x s =y t and Z k-1 is an LCS of X s-1 and Y t-1 2.if x s y t, then z k x s implies that Z is an LCS of X s-1 and Y 3.if x s y t, then z k y t implies that Z is an LCS of X and Y t-1

4 Dynamic programming elements: - optimal substructure of the optimal solution - overlapping subproblems, i.e. the total number of subproblems that need to be recursively computed is reasonable, although a recursive algorithm would solve the same subproblems over and over Do we have to compute bottom-up? no, can do smart recursion (book calls it memoization) once we computed a solution to a subproblem, keep that somewhere Memoized-Matrix-Chain(p) for i=1 to n for j=i to n m[i,j]= return Lookup(1,n) Lookup(i,j) if m[i,j]< then return m[i,j] if i=j then m[i,j]=0 else for k=i to j-1 =Lookup(i,k)+Lookup(k+1,j)+d i-1 d k d j if <m[i,j] then m[i,j]= return m[i,j]

5 So, compute solutions bottom up and keep them in the table i.e. compute all the m[i,i] s, then all the m[i,i+1] s, then all the m[i,i+2] s, then all the m[i,i+3] s, etc 1 n [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] [ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ][ ] 1 n [ ][ ][ w ][ ][ ][ ][ ][ ][ ] [ ][ ][ ][ ][ ][ ][ ][ ][ ] [ ][ ][ ][ ][ ][ ][ ][ ] [ ][ ][ ][ ][ ][ ][ ][ ] Matrix-Chain-Order matrix m for i=1 to n m[i,i]=0 for w=2 to n for i=1 to n-w+1 j=i+w-1 m[i,j]= for k=i to j-1 =m[i,k]+m[k+1,j]+d i-1 d k d j if <m[i,j] then m[i,j]= and s[i,j]=k Running time: O(n 3 )

6 Let the dimensions of matrix A i be d i 1 d i then dimension of A 1 x A 2 x... x A n is d 0 d n, and of C[i..k] is d i 1 the cost of multiplying C[i..k] by C[k+1..j] is d i 1 d k d j Let m[i,j] be the cost of optimal parenthesization of C[i..j]; we want m[1,n] d k Recursive solution when i = j, m[ i, j] = 0 when i j, m[ i, j] = min i k< j { m[ i, k] + m[ k+ 1, j] + d i 1 d k d j } Computing the optimal costs write recursive algorithm that computes m[i,j] for all i and j there are O(n 2 ) values m[i,j] to be computed but, recursive algorithm s running time is Ω(2 n ) why? we keep recomputing many m[i,j] s over and over and over and over Ex: to compute m[1,5], we compute m[1,4] & m[2,5] (and other subproblems) to compute m[1,4], we compute m[2,4], etc to compute m[2,5], we again compute m[2,4], etc

7 How about trying all the possible parenthesizations? there are more than 2 n of them -- too expensive! Structure of optimal parenthesization: let C[i..j] = A i x A i+1 x... x A j i j 1 [ ] n A 1 x A 2 x... x A n = (A 1 x... x A k ) x (A k+1 x... x A n ) for some k, 1 k < n 1 k k+1 n ( )( ) The cost of computing C[1..n] is the cost of computing the matrix C[1..k], plus the cost of computing matrix C[k+1..n], plus the cost of multiplying matrix C[1..k] by matrix C[k+1..n] Key observation: if the optimal parenthesization of C[1..n] contains parenthesization T[1..k] of C[1..k] and parenthesization T[k+1..n] of C[k+1..n], then T[1..k] must be an optimal parenthesization of C[1..k] and T[k+1..n] must be an optimal parenthesization of C[k+1..n]

8 TOPIC 14: Dynamic Programming Matrix-chain multiplication: given seuence (chain) of n matrices to multiply, compute product A 1 x A 2 x... x A n with minimum number of operations To multiply 2 matrices A x B = D, where A is r x s matrix and B is s x t matrix 5 3 s = 5 4 D[ i, j] = A[ i, k] B[ k, j] k = dimensions of D are r x t to compute D[i,j] costs O(s) mathematical operations ( + and x ) to compute D costs O(rst) operations Matrix multiplication is associative: A 1 x ( A 2 x A 3 ) =(A 1 x A 2 ) x A 3 But it may take a different number of operations to compute the two products ex: A 1 is 100x1 A 2 is 1x10 A 3 is 10x10 A 1 x ( A 2 x A 3 ) costs 1(10)(10)+100(1)(10) = 1,100 operations (A 1 x A 2 ) x A 3 costs 100(1)(10) + 100(10)(10) = 11,000 operations Matrix-chain multiplication: figure out the best way to parenthesize the chain