Dynamic Programming Code
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Transcript of 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– 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
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?
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?
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]
LCS
Theorem Let Z = z1, . . . , zk be any LCS of X and Y .1. If xm = yn, then zk = xm = yn and Zk-1 is an LCS of Xm-1 and Yn-1.2. If xm yn, then either zk xm and Z is an LCS of Xm-1 and Y .3. or zk yn and Z is an LCS of X and Yn-1.
Theorem Let Z = z1, . . . , zk be any LCS of X and Y .1. If xm = yn, then zk = xm = yn and Zk-1 is an LCS of Xm-1 and Yn-1.2. If xm yn, then either zk xm and Z is an LCS of Xm-1 and Y .3. or zk yn and Z is an LCS of X and Yn-1.
LCS-LENGTH (X, Y)1. m ← length[X]2. n ← length[Y]3. for i ← 1 to m4. do c[i, 0] ← 05. for j ← 0 to n6. do c[0, j ] ← 07. for i ← 1 to m8. do for j ← 1 to n9. do if xi = yj
10. then c[i, j ] ← c[i1, j1] + 111. 12. else if c[i1, j ] ≥ c[i, j1]13. then c[i, j ] ← c[i 1, j ]14. 15. else c[i, j ] ← c[i, j1]16. 17. return c
LCS-LENGTH (X, Y)1. m ← length[X]2. n ← length[Y]3. for i ← 1 to m4. do c[i, 0] ← 05. for j ← 0 to n6. do c[0, j ] ← 07. for i ← 1 to m8. do for j ← 1 to n9. do if xi = yj
10. then c[i, j ] ← c[i1, j1] + 111. 12. else if c[i1, j ] ≥ c[i, j1]13. then c[i, j ] ← c[i 1, j ]14. 15. else c[i, j ] ← c[i, j1]16. 17. return c
LCS-LENGTH (X, Y)1. m ← length[X]2. n ← length[Y]3. for i ← 1 to m4. do c[i, 0] ← 05. for j ← 0 to n6. do c[0, j ] ← 07. for i ← 1 to m8. do for j ← 1 to n9. do if xi = yj
10. then c[i, j ] ← c[i1, j1] + 111. b[i, j ] ← “ ”12. else if c[i1, j ] ≥ c[i, j1]13. then c[i, j ] ← c[i 1, j ]14. b[i, j ] ← “↑”15. else c[i, j ] ← c[i, j1]16. b[i, j ] ← “←”17. return c and b
LCS-LENGTH (X, Y)1. m ← length[X]2. n ← length[Y]3. for i ← 1 to m4. do c[i, 0] ← 05. for j ← 0 to n6. do c[0, j ] ← 07. for i ← 1 to m8. do for j ← 1 to n9. do if xi = yj
10. then c[i, j ] ← c[i1, j1] + 111. b[i, j ] ← “ ”12. else if c[i1, j ] ≥ c[i, j1]13. then c[i, j ] ← c[i 1, j ]14. b[i, j ] ← “↑”15. else c[i, j ] ← c[i, j1]16. b[i, j ] ← “←”17. return c and b
PRINT-LCS (b, X, i, j)1. if i = 0 or j = 02. then return3. if b[i, j ] = “ ”4. then PRINT-LCS(b, X, i1, j1)5. print xi
6. elseif b[i, j ] = “↑”7. then PRINT-LCS(b, X, i1, j)8. else PRINT-LCS(b, X, i, j1)
PRINT-LCS (b, X, i, j)1. if i = 0 or j = 02. then return3. if b[i, j ] = “ ”4. then PRINT-LCS(b, X, i1, j1)5. print xi
6. elseif b[i, j ] = “↑”7. then PRINT-LCS(b, X, i1, j)8. else PRINT-LCS(b, X, i, j1)
11-11
Matrix-chain Multiplication coitweb.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
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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:1. (A1(A2(A3A4)))
2. (A1((A2A3)A4))
3. ((A1A2)(A3A4))
4. ((A1(A2A3))A4)
5. (((A1A2)A3)A4)
11-13
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?
11-14
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 to p2. for j ← 1 to r3. C[i, j] ← 04. for k ← 1 to q5. C[i, j] ← C[i, j] + A[i, k] · B[k,
j] 6. return C
Scalar multiplication in line 5 dominates time to compute CNumber of scalar multiplications = pqr
11-15
Matrix-chain Multiplication …contd
• Example: Consider three matrices A10100, B1005, and C550
• There are 2 ways to parenthesize – ((AB)C) = D105 · C550
• AB 10·100·5=5,000 scalar multiplications• DC 10·5·50 =2,500 scalar multiplications
– (A(BC)) = A10100 · E10050
• BC 100·5·50=25,000 scalar multiplications• AE 10·100·50 =50,000 scalar multiplications
Total: 7,500
Total: 75,000
11-16
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-1pi
– 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
11-17
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 Ak and Ak+1 for 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
11-18
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
11-19
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..j splits the product between Ak and Ak+1 for some integer k where i ≤ k < j
11-20
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
11-21
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
m[i, j ] =0 if i=j
min {m[i, k] + m[k+1, j ] + pi-1pk pj } if i<ji ≤ k< j
11-22
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 … Aj is 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
11-23
Algorithm to Compute Optimal CostInput: Array p[0…n] containing matrix dimensions and nResult: Minimum-cost table m and split table s
MATRIX-CHAIN-ORDER(p[ ], n)
for i ← 1 to nm[i, i] ← 0
for l ← 2 to nfor i ← 1 to n-l+1
j ← i+l-1m[i, j] ← for k ← i to j-1
q ← m[i, k] + m[k+1, j] + p[i-1] p[k] p[j]if q < m[i, j]
m[i, j] ← qs[i, j] ← k
return m and s
Takes O(n3) time
Requires O(n2) space
11-24
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 … Aj for the minimum cost
11-25
Example
• Show how to multiply this matrix chain optimally
• Solution on the board– Minimum cost 15,125– Optimal parenthesization
((A1(A2A3))((A4 A5)A6))
Matrix Dimension
A1 30×35
A2 35×15
A3 15×5
A4 5×10
A5 10×20
A6 20×25
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?