Dynamic Programming. Design and Analysis of Algorithms. Entwurf und Analyse von Algorithmen. Irene Parada. Design and Analysis of Algorithms

Transcription

1 Entwurf und Analyse von Algorithmen Dynamic Programming

2 Overview Introduction Example 1 When and how to apply this method Example 2 Final remarks

3 Introduction: when recursion is inefficient Example: Calculation of Fibonacci numbers f n recursively: f n = f n 1 + f n 2, f 1 = f 2 = 1 The solution (f n ) is the combination of the solutions of subproblems (f n 1 and f n 2 ). We are solving the same subproblems again and again!

4 Introduction: bottom-up! f n f n 1 Advantage: Each sub-problem has to be calculated only once. f 4 f 3 f 2 f 1 = 3 = 2 = 1 = 1 Disadvantage: Possibly additional storage requirement. But often... asymptomptotically same needed time. old values can be overridden.

5 Introduction: typical behaviour of subproblems Divide & Conquer Dynamic Programming

6 Introduction: dynamic programming Steps: Characterization of the structure of an optimal solution. Recursive definition of the value of an optimal solution. Computation of the value of an optimal solution in a bottom-up fashion. Construction of an optimal solution from computed information.

7 Example 1: matrix-chain multiplication Setting: We are given a sequence (chain) A 1, A 2,..., A n of n matrices to be multiplied, and we want to compute the product A 1 A 2 A n. Standard algorithm: pairs of matrices. fully parenthesized: either a single matrix or the product of two fully parenthesized matrix products. Cost of multiplying a p q matrix with a q r matrix (school method) pqr scalar multiplications. Matrix multiplication is associative: all parenthesizations yield the same. BUT... At different costs in general! A 1 (A 2 A 3 ) = (A 1 A 2 )A 3

8 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dimension p i 1 p i for i = 1,..., n, fully parenthesize the product A 1 A 2 A n such that minimizes the number of scalar multiplications needed. Note that we are not actually multiplying matrices! The goal is only to determine an order for multiplying matrices that has the lowest cost.

9 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. How NOT to solve it: try it all P (n):= number of alternative parenthesizations of a sequence of n matrices. 1 for n = 1 P (n) = n 1 P (k)p (n k) for n 2 k=1 It s the Catalan numbers! P (n) = C(n 1) with C(n) = 1 ( ) 2n n + 1 n = Θ ( 4 n n 1.5 )

10 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. How to solve it: using dynamic programming

11 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 1: structure of an optimal solution Notation: A i..j with i j denotes the matrix that results from evaluating the product A i A i+1 A j. Observation: if the problem is nontrivial, i.e., i < j, then we must split the product between A k and A k+1 for some integer k in the range i k < j. Optimal (A i A k ) (A k+1 A j ) substructure Optimal solution subchains are optimally parenthesized!

12 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 2: a recursive solution A i..j := product A i A i+1 A j. m[i, j]:= the minimum cost of computing the matrix A i..j. we are interested in m[1, n]. 0 for i = j m[i, j] = min {m[i, k] + m[k + 1, j] + p i 1p k p j } for i < j i k<j (A i A k ) (A k+1 A j )

13 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 2: a recursive solution Direct recursive algorithm takes exponential time!

14 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 3: bottom-up algorithm A i..j := product A i A i+1 A j. m[i, j]:= the minimum cost of computing the matrix A i..j. we are interested in m[1, n]. 0 for i = j m[i, j] = min {m[i, k] + m[k + 1, j] + p i 1p k p j } for i < j i k<j computing the product of j i + 1 matrices depends only on the costs of computing the products of fewer than j i + 1 matrices.

15 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 3: bottom-up algorithm Order: fill in the table m solving the problem on matrix chains of increasing length. for i := 1 to n do m[i, i] := 0 od for l := 2 to n do for i := 1 to n l + 1 do All possible starting points of the subsequence j := i + l 1 All possible sending points of the subsequence m[i, j] := Special value meaning IDK! for k := i to j 1 do All possible spliting points of the subsequence hlp := m[i, k] + m[k + 1, j] + p i 1 p k p j if hlp < m[i, j] then m[i, j] := hlp

16 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 4: constructing a solution s[i, j]:= value of k s.t. an optimal parenthesization of A i A j splits the product between A k and A k+1. for i := 1 to n do m[i, i] := 0 od for l := 2 to n do for i := 1 to n l + 1 do j := i + l 1 m[i, j] := for k := i to j 1 do hlp := m[i, k] + m[k + 1, j] + p i 1 p k p j if hlp < m[i, j] then the final matrix multiplication for computing A 1..n optimally is A 1..s[1,n] A s[1,n]+1..n. m[i, j] := hlp ; s[i, j] = k

17 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 4: constructing a solution s[i, j]:= value of k s.t. an optimal parenthesization of A i A j splits the product between A k and A k+1. for i := 1 to n do m[i, i] := 0 od for l := 2 to n do for i := 1 to n l + 1 do j := i + l 1 m[i, j] := for k := i to j 1 do hlp := m[i, k] + m[k + 1, j] + p i 1 p k p j if hlp < m[i, j] then the final matrix multiplication for computing A 1..n optimally is A 1..s[1,n] A s[1,n]+1..n. m[i, j] := hlp ; s[i, j] = k Θ(n 3 ) time Θ(n 2 ) space

18 Example 1: matrix-chain multiplication Matrix-chain multiplication problem: given a chain A 1,..., A n of n matrices, A i of dim. p i 1 p i i, fully parenthesize A 1 A 2 A n minimizing the cost. Step 4: constructing a solution s[i, j]:= value of k s.t. an optimal parenthesization of A i A j splits the product between A k and A k+1. the final matrix multiplication for computing A 1..n optimally is A 1..s[1,n] A s[1,n]+1..n. F UNCT ION PRINT-OPTIMAL-PARENS(s, i, j) : ST RING x, y : ST RING if i < j then else x := PRINT-OPTIMAL-PARENS(s, i, s[i, j]) y := PRINT-OPTIMAL-PARENS(s, s[i, j] + 1, j) print (, x, y, ) print A i O(n) time Call PRINT-OPTIMAL- PARENS(1, n)

19 When and how to apply it? Typically optimization problems Two key characteristics: Optimal substructure: an optimal solution to the problem contains within it optimal solutions to subproblems. Often cut & paste argument - Try to keep the subproblems simple Variants: - How many subproblems we use to solve the original problem. - How many choices in which subproblem(s) to use. We might want to store the choice we made for reconstructing the solution (step 4). Running time (informally): (# subproblems overall) (# choices we look at for each subproblem).

20 When and how to apply it? Typically optimization problems Two key characteristics: Optimal substructure: an optimal solution to the problem contains within it optimal solutions to subproblems. Overlapping subproblems: recursive algorithm revisits the same problem repeatedly. - We want the total number of distinct subproblems to be a polynomial in the input size.

21 Example 2: Longest common subsequence A subsequence of a given sequence is just the given sequence with zero or more elements left out. Formally: Given the sequences X = (x 1, x 2,..., x m ) and Z = (z 1, z 2,..., z k ), k m, Z is a subsequence of X if there exists a strictly increasing sequence i 1, i 2,..., i k of indices of X such that z j = x ij for all j = 1..k. Given two sequences X and Y, we say that a sequence Z is a common subsequence of X and Y if Z is a subsequence of both X and Y. Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y.

22 Example 2: Longest common subsequence Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y. Step 1: structure of an optimal solution i-th prefix of X, for i = 0..m := X i = (x 1, x 2,..., x i ). Let Z = (z 1,..., z k ) be any LCS of X and Y. 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 : if z k x m then Z is an LCS of X m 1 and Y. if z k y n then Z is an LCS of X and Y n 1. An LCS of two sequences contains an LCS of prefixes of the two sequences.

23 Example 2: Longest common subsequence Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y. Step 2: a recursive solution l[i, j] with 0 i m, 0 j n := length of an LCS of the sequences X i and Y j. 0 if i = 0 or j = 0 l[i, j] = l[i 1, j 1] + 1 if i, j > 0 and x i = y j max{l[i 1, j], l[i, j 1]} if i, j > 0 and x i y j

24 Example 2: Longest common subsequence Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y. Step 3: bottom-up algorithm i = 0..m The computation of l[i, j] requires only the values of l[i 1, j], l[i, j 1] and l[i 1, j 1]. Compute l ( M m n ) line by line from bottom to top j = 0..n

25 Example 2: Longest common subsequence Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y. Step 3: bottom-up algorithm l[i, j] := length of an LCS of the sequences X i and Y j. for i := 0 to m do l[i, 0] := 0 od Initialization for j := 1 to n do l[0, j] := 0 od for i := 1 to m do for j := 1 to n do if x i = y j then l[i, j] := l[i 1, j 1] + 1 First case else l[i, j] := max(l[i 1, j], l[i, j 1]) fi od od Second case Θ(mn)time O(mn) space Compute l ( M m n ) line by line from bottom to top

26 Example 2: Longest common subsequence Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y. Step 4: constructing a solution l[i, j] := length of an LCS of the sequences X i and Y j. for i := 0 to m do l[i, 0] := 0 od for j := 1 to n do l[0, j] := 0 od for i := 1 to m do for j := 1 to n do if x i = y j then l[i, j] := l[i 1, j 1] + 1 else l[i, j] := max(l[i 1, j], l[i, j 1]) fi od od We could, but there is no need to add a new matrix storing where we came from. The first case gives you one character of the sequence. In every step, either i or j or both decrease one unit O(m + n) time for reconstructing the solution.

27 Example 2: Longest common subsequence Longest-common-subsequence (LCS) problem: given two sequences X = (x 1, x 2,..., x m ) and Y = (y 1, y 2,..., y n ) find a maximum length common subsequence of X and Y. Remark If we are only interested in the length of the optimal subsequence, but not in an optimal sequence itself, the matrix l[i, j] does not need to be stored. It is enough to save only the last two lines! memory required O(min{m, n}).

28 Final remarks Two key characteristics that a problem must have for dynamic programming to be a possible (but might the one!) solution technique are optimal substructure and overlapping subproblems. Explained bottom-up approach for dynamic programming, but... a top-down recursive approach that takes advantage of the overlapping-subproblems (memoization) is also possible. Bibliography: Introduction to Algorithms by T. H. Cormen, C. E. Leiserson, R. L. Rivest (and C. Stein).