Programming
Dynamic Dynamic Programming
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Outline and Reading
Matrix ChainProduct (5.3.1) The General Technique (5.3.2) 01 Knapsack Problem (5.3.3)
Dynamic Programming
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Matrix ChainProducts
Dynamic Programming is a general algorithm design paradigm.
Review: Matrix Multiplication.
Rather than give the general structure, let us first give a motivating example: Matrix ChainProducts
f B e j
C = A*B A is d e and B is e f
O(d e f ) time
e -1 k =0
e A i C i,j f
3
C[i, j ] = A[i, k ] * B[k , j ] d
d
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Matrix ChainProducts
Matrix ChainProduct:
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Enumeration Approach
Matrix ChainProduct Alg.:
Try all possible ways to parenthesize
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Greedy Approach
Idea #1: repeatedly select the product that uses the fewest operations. Counterexample:
The greedy approach is not giving us the optimal value. Dynamic Programming 6
A is 101 11 B is 11 9 C is 9 100 D is 100 99 Greedy idea #1 gives A*((B*C)*D)), which takes 109989+9900+108900=228789 ops (A*B)*(C*D) takes 9999+89991+89100=189090 ops
"Recursive" Approach
Define subproblems:
Subproblem optimality: The optimal solution can be defined in terms of optimal subproblems
Find the best parenthesization of Ai*Ai+1*...*Aj. Let Ni,j denote the number of operations done by this subproblem. The optimal solution for the whole problem is N0,n1.
There has to be a final multiplication (root of the expression tree) for the optimal solution. Say, the final multiplication is at index i: (A0*...*Ai)*(Ai+1*...*An1). Then the optimal solution N0,n1 is the sum of two optimal subproblems, N0,i and Ni+1,n1 plus the time for the last multiplication. If the global optimum did not have these optimal subproblems, we could define an even better "optimal" solution.
Dynamic Programming 7
Characterizing Equation
The global optimal has to be defined in terms of optimal subproblems, depending on where the final multiplication is at. Let us consider all possible places for that final multiplication:
N i , j = min{N i ,k + N k +1, j + d i d k +1d j +1}
i k < j
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Subproblem Overlap
Algorithm RecursiveMatrixChain(S, 0, n-1): Input: sequence S of n matrices to be multiplied Output: number of operations in an optimal parenthesization of S if i=j then return 0
k
i
j
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Subproblem Overlap
1..4 1..1
2..2 3..4
2..4
2..3 4..4
1..2
1..1 2..2
3..4
3..3 4..4
1..3
4..4
...
3..3 4..4
2..2 3..3
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Dynamic Programming Algorithm Visualization
The bottomup N i , j = min{N i ,k + N k +1, j + d i d k +1d j +1} i k < j construction fills in the N array by diagonals j ... n1 N 0 1 2 i Ni,j gets values from 0 previous entries in ith 1 row and jth column ... answer Filling in each entry in the i N table takes O(n) time. Total run time: O(n3) j Getting actual parenthesization can be done by remembering "k" n1 for each N entry
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Dynamic Programming Algorithm Visualization
A0: 30 X 35; A1: 35 X15; A2: 15X5; A3: 5X10; A4: 10X20; A5: 20 X 25
N i , j = min{N i ,k + N k +1, j + d i d k +1d j +1}
i k < j
N1, 4 = min{ N1,1 + N 2, 4 + d1d 2 d 5 = 0 + 2500 + 35 *15 * 20 = 13000, N1, 2 + N 3, 4 + d1d 3 d 5 = 2625 + 1000 + 35 * 5 * 20 = 7125, N1,3 + N 4, 4 + d1d 4 d 5 = 4375 + 0 + 35 *10 * 20 = 11375 } = 7125
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Dynamic Programming Algorithm Visualization
(A0*(A1*A2))*((A3*A4)*A5)
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Dynamic Programming Algorithm
Since Algorithm matrixChain(S): subproblems Input: sequence S of n matrices to be overlap, we don't multiplied use recursion. Instead, we Output: number of operations in an optimal construct optimal parenthesization of subproblems S "bottomup." Ni,i's are easy, so start with them Then do problems of "length" 2,3,... subproblems, and so on. Running time: O(n3)
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The General Dynamic Programming Technique
Applies to a problem that at first seems to require a lot of time (possibly exponential), provided we have:
Simple subproblems: the subproblems can be defined in terms of a few variables, such as j, k, l, m, and so on. Subproblem optimality: the global optimum value can be defined in terms of optimal subproblems Subproblem overlap: the subproblems are not independent, but instead they overlap (hence, should be constructed bottomup).
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The 0/1 Knapsack Problem
Given: A set S of n items, with each item i having
Goal: Choose items with maximum total benefit but with weight at most W. If we are not allowed to take fractional amounts, then this is the 0/1 knapsack problem.
wi a positive weight bi a positive benefit
In this case, we let T denote the set of items we take Objective: maximize Constraint:
b
iT
i
w
iT
i
W
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Dynamic Programming
Example
Given: A set S of n items, with each item i having
Goal: Choose with items maximum total benefit but with weight at most W. "knapsack" Items:
1 2 3 4 5
bi a positive "benefit" wi a positive "weight"
box of width 9 in
Weight: Benefit:
4 in $20
2 in $3
2 in $6
6 in $25
2 in $80
Solution:
Dynamic Programming
item 5 ($80, 2 in) item 3 ($6, 2in) item 1 ($20, 4in)
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A 0/1 Knapsack Algorithm, First Attempt
Sk: Set of items numbered 1 to k. Define B[k] = best selection from Sk. Problem: does not have subproblem optimality:
Consider set S={(3,2),(5,4),(8,5),(4,3),(10,9)} of (benefit, weight) pairs and total weight W = 20
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A 0/1 Knapsack Algorithm, Second Attempt
Sk: Set of items numbered 1 to k. Define B[k,w] to be the best selection from Sk with weight at most w Good news: this does have subproblem optimality.
B[k - 1, w] if wk > w B[k , w] = else max{B[k - 1, w], B[k - 1, w - wk ] + bk }
I.e., the best subset of Sk with weight at most w is either
the best subset of Sk1 with weight at most w or the best subset of Sk1 with weight at most w- k plus item k w
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0/1 Knapsack Algorithm
Consider set S={(1,1),(2,2),(4,3),(2,2),(5,5)} of (benefit, weight) pairs and total weight W = 10
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0/1 Knapsack Algorithm
Trace back to find the items picked
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0/1 Knapsack Algorithm
Each diagonal arrow corresponds to adding one item into the bag Pick items 2,3,5 {(2,2),(4,3),(5,5)} are what you will take away
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0/1 Knapsack Algorithm
B[k - 1, w] if wk > w B[k , w] = else max{B[k - 1, w], B[k - 1, w - wk ] + bk } Algorithm 01Knapsack(S, W): Recall the definition of Input: set S of n items with benefit
B[k,w] Since B[k,w] is defined in terms of B[k- 1,*], we can use two arrays of instead of a matrix Running time: O(nW). Not a polynomialtime algorithm since W may be large This is a pseudopolynomial time algorithm
k
1
n
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Longest Common Subsequence
Given two strings, find a longest subsequence that they share substring vs. subsequence of a string
Consider ababc and abdcb alignment 1 ababc. abd.cb
Substring: the characters in a substring of S must occur contiguously in S Substring: the characters can be interspersed with gaps.
the longest common subsequence is ab..c with length 3
alignment 2 aba.bc abdcb.
the longest common subsequence is ab..b with length 3
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Longest Common Subsequence
Let's give a score M an alignment in this way,
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2
1 2 3
j
Longest Common Subsequence
Subproblem optimality Consider two sequences
Let the optimal alignment be x1x2x3...xn1xn y1y2y3...yn1yn There are three possible cases for the last pair (xn,yn):
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2
1 2 3
j
Longest Common Subsequence
Mi,j = MAX {Mi-1, j-1 + S (ai,bj) (match/mismatch) Mi,j-1 + 0 (gap in sequence #1) Mi-1,j + 0 (gap in sequence #2) } Mi,j is the score for optimal alignment between strings a[...