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Dynamic Programming Code

Dynamic Programming Code. Straightforward top-down rod cutting. CUT-ROD(p, n) If n == 0 Return 0 q = -1 For I = 1 to n q = max (q, p[ i ] + CUT-ROD(p, n- i )) Return q. Memoization in rod cutting. MEMOIZED-CUT-ROD(p, n) let r[0..n] be a new array f or I = 0 to n r[ i ] = -1

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Dynamic Programming Code

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  1. Dynamic Programming Code

  2. Straightforward top-down rod cutting • CUT-ROD(p, n) • If n == 0 • Return 0 • q = -1 • For I = 1 to n • q = max (q, p[i] + CUT-ROD(p, n-i)) • Return q

  3. Memoization in rod cutting • MEMOIZED-CUT-ROD(p, n) • let r[0..n] be a new array • for I = 0 to n • r[i] = -1 • return MEMOIZED-CUT-ROD-AUX(p, n, r) • MEMOIZED-CUT-ROD-AUX (p, n, r) • if r[n] >= 0 • return r[n] • if n == 0 • q = 0 • else • q = -1 • For I = 1 to n • q = max (q, p[i] + MEMOIZED-CUT-ROD-AUX(p, n-i, r)) • r[n] = q • return q

  4. Bottom up memoization • BOTTOM-UP-CUT-ROD(p, n) • let r[0..n] be a new array • r[0] = 0 • for j = 1 to n • q = -1 • for i = 1 to j • q = max (q, p[i] + r[j – i]) • r[j] = q • return r[n] • How do we know where to make the cuts?

  5. Bottom up memoization • EXTENDED-BOTTOM-UP-CUT-ROD(p, n) • let r[0..n] and s[0..n] be a new array • r[0] = 0 • for j = 1 to n • q = -1 • for i = 1 to j • if q < p[i] + r[j – 1] • q = p[i] + r[j – 1] • s[j] = i • r[j] = q • return r[n] • How do you print the solution?

  6. Bottom up memoization • EXTENDED-BOTTOM-UP-CUT-ROD(p, n) • let r[0..n] and s[0..n] be a new array • r[0] = 0 • for j = 1 to n • q = -1 • for i = 1 to j • if q < p[i] + r[j – 1] • q = p[i] + r[j – 1] • s[j] = i • r[j] = q • return r[n] • PRINT-CUT-ROD-SOLUTION(p, n) • (r, s) = EXT-BUCR(p, n) • while n > 0 • print s[n] • n = n – s[n]

  7. LCS Theorem Let Z = z1, . . . , zk be any LCS of X and Y . 1. If xm= yn, then zk= xm= ynand Zk-1 is an LCS of Xm-1 and Yn-1. 2. If xmyn, then either zkxmand Z is an LCS of Xm-1 and Y . 3. or zkynandZ is an LCS of X and Yn-1.

  8. LCS-LENGTH (X, Y) • m← length[X] • n← length[Y] • for i← 1 to m • do c[i, 0] ← 0 • for j ← 0 to n • do c[0, j ] ← 0 • 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 • else if c[i1, j ] ≥ c[i, j1] • then c[i, j ] ← c[i 1, j ] • else c[i, j ] ← c[i, j1] • return c

  9. LCS-LENGTH (X, Y) • m← length[X] • n← length[Y] • for i← 1 to m • do c[i, 0] ← 0 • for j ← 0 to n • do c[0, j ] ← 0 • 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

  10. PRINT-LCS (b, X, i, j) • if i = 0 or j = 0 • then return • if b[i, j ] = “ ” • then PRINT-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)

  11. Matrix-chain Multiplicationcoitweb.uncc.edu/~ras/courses/Matrix-Mult.ppt • Suppose we have a sequence or chain A1, A2, …, An of n matrices to be multiplied • That is, we want to compute the product A1A2…An • There are many possible ways (parenthesizations) to compute the product

  12. Matrix-chain Multiplication …contd • Example: consider the chain A1, A2, A3, A4 of 4 matrices • Let us compute the product A1A2A3A4 • There are 5 possible ways: • (A1(A2(A3A4))) • (A1((A2A3)A4)) • ((A1A2)(A3A4)) • ((A1(A2A3))A4) • (((A1A2)A3)A4)

  13. Matrix-chain Multiplication …contd • To compute the number of scalar multiplications necessary, we must know: • Algorithm to multiply two matrices • Matrix dimensions • Can you write the algorithm to multiply two matrices?

  14. Algorithm to Multiply 2 Matrices Input: Matrices Ap×q and Bq×r (with dimensions p×q and q×r) Result: Matrix Cp×r resulting from the product A·B MATRIX-MULTIPLY(Ap×q , Bq×r) 1. for i ← 1 top 2. for j ← 1 tor 3. C[i, j]← 0 4. for k ← 1 toq 5. C[i, j]← C[i, j] + A[i, k]· B[k, j] 6. returnC Scalar multiplication in line 5 dominates time to compute CNumber of scalar multiplications = pqr

  15. Matrix-chain Multiplication …contd • Example: Consider three matrices A10100, B1005, and C550 • There are 2 ways to parenthesize • ((AB)C) = D105·C550 • AB  10·100·5=5,000 scalar multiplications • DC  10·5·50 =2,500 scalar multiplications • (A(BC)) = A10100·E10050 • BC  100·5·50=25,000 scalar multiplications • AE  10·100·50 =50,000 scalar multiplications Total: 7,500 Total: 75,000

  16. Matrix-chain Multiplication …contd • Matrix-chain multiplication problem • Given a chain A1, A2, …, An of n matrices, where for i=1, 2, …, n, matrix Ai has dimension pi-1pi • Parenthesize the product A1A2…An such that the total number of scalar multiplications is minimized • Brute force method of exhaustive search takes time exponential in n

  17. Dynamic Programming Approach • The structure of an optimal solution • Let us use the notation Ai..j for the matrix that results from the product Ai Ai+1 … Aj • An optimal parenthesization of the product A1A2…An splits the product between Akand Ak+1for some integer k where1 ≤ k < n • First compute matrices A1..k and Ak+1..n ; then multiply them to get the final matrix A1..n

  18. Dynamic Programming Approach …contd • Key observation: parenthesizations of the subchains A1A2…Ak and Ak+1Ak+2…An must also be optimal if the parenthesization of the chain A1A2…An is optimal (why?) • That is, the optimal solution to the problem contains within it the optimal solution to subproblems

  19. Dynamic Programming Approach …contd • Recursive definition of the value of an optimal solution • Let m[i, j] be the minimum number of scalar multiplications necessary to compute Ai..j • Minimum cost to compute A1..n is m[1, n] • Suppose the optimal parenthesization of Ai..jsplits the product between Akand Ak+1for some integer k where i ≤ k < j

  20. Dynamic Programming Approach …contd • Ai..j= (Ai Ai+1…Ak)·(Ak+1Ak+2…Aj)= Ai..k· Ak+1..j • Cost of computing Ai..j = cost of computing Ai..k + cost of computing Ak+1..j + cost of multiplying Ai..k and Ak+1..j • Cost of multiplying Ai..k and Ak+1..j is pi-1pk pj • m[i, j ] = m[i, k] + m[k+1, j ] + pi-1pk pj for i ≤ k < j • m[i, i ] = 0 for i=1,2,…,n

  21. Dynamic Programming Approach …contd • But… optimal parenthesization occurs at one value of k among all possible i ≤ k < j • Check all these and select the best one 0 if i=j m[i, j ] = min {m[i, k] + m[k+1, j ] + pi-1pk pj}if i<j i≤ k< j

  22. Dynamic Programming Approach …contd • To keep track of how to construct an optimal solution, we use a table s • s[i, j ] = value of k at which Ai Ai+1 … Ajis split for optimal parenthesization • Algorithm: next slide • First computes costs for chains of length l=1 • Then for chains of length l=2,3, … and so on • Computes the optimal cost bottom-up

  23. Algorithm to Compute Optimal Cost Input: Array p[0…n] containing matrix dimensions and n Result: Minimum-cost table m and split table s MATRIX-CHAIN-ORDER(p[ ], n) for i ← 1 ton m[i, i]← 0 for l ← 2 ton for i ← 1 ton-l+1 j← i+l-1 m[i, j]←  for k ← itoj-1 q ← m[i, k] + m[k+1, j] + p[i-1] p[k] p[j] ifq < m[i, j] m[i, j]← q s[i, j]← k returnm and s Takes O(n3) time Requires O(n2) space

  24. Constructing Optimal Solution • Our algorithm computes the minimum-cost table m and the split table s • The optimal solution can be constructed from the split table s • Each entry s[i, j ]=k shows where to split the product Ai Ai+1 … Ajfor the minimum cost

  25. Example • Show how to multiply this matrix chain optimally • Solution on the board • Minimum cost 15,125 • Optimal parenthesization ((A1(A2A3))((A4 A5)A6))

  26. LPS (problem set 4) • A palindrome is a nonempty string over some alphabet that reads the same forward and backward. Examples of palilndromes are all strings of length 1, "civic", "racecar", and "aibohphobia" (fear of palindromes). • Give an efficient algorithm to find the longest palindrome that is a subsequence of a given input string. For example, given the input "character", your algorithm should return "carac". What is the running time of your algorithm?

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