Dynamic programming is an algorithm design technique that solves problems by breaking them down into smaller overlapping subproblems and storing the solutions to subproblems to avoid recomputing them. It involves building up a solution using previously found subsolutions and working in a bottom-up fashion to solve larger subproblems from solutions to smaller subproblems. Examples where dynamic programming has been applied include computing Fibonacci numbers, finding the shortest paths in a graph, and solving optimization problems like the knapsack problem.

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Algorithm chapter 8

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  Dynamic Programming
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  Dynamic ProgrammingDynamic Programming is a general algorithm design technique.“Programming” here means “planning”. Invented by American mathematician Richard Bellman in the 1950s tosolve optimization problems Main idea: solve several smaller (overlapping) subproblems record solutions in a table so that each subproblem is only solved once final state of the table will be (or contain) solution Dynamic programming vs. divide-and-conquer partition a problem into overlapping subproblems and independent ones store and not store solutions to subproblems
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  Example: Fibonacci numbers•Recall definition of Fibonacci numbers: f(0) = 0 f(1) = 1 f(n) = f(n-1) + f(n-2)• Computing the nth Fibonacci number recursively (top-down): f(n) f(n-1) + f(n-2)f(n-2) + f(n-3) f(n-3) + f(n-4) ...
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  Example: Fibonacci numbers Computing the nth fibonacci number using bottom-up iteration:• f(0) = 0• f(1) = 1• f(2) = 0+1 = 1• f(3) = 1+1 = 2• f(4) = 1+2 = 3 ALGORITHM Fib(n)• f(5) = 2+3 = 5 F[0] 0, F[1] 1• for i 2 to n do• F[i] F[i-1] + F[i-2]• return F[n]• f(n-2) =• f(n-1) = extra space• f(n) = f(n-1) + f(n-2)
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  Examples of DynamicProgramming Algorithms Computing binomial coefficients Warshall’s algorithm for transitive closure Floyd’s algorithms for all-pairs shortest paths Constructing an optimal binary search tree Some instances of difficult discrete optimizationproblems: knapsack
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  Computing Binomial Coefficients A binomial coefficient, denoted C(n, k), is the number of combinations of k elements from an n-element set (0 ≤ k ≤ n). Recurrence relation (a problem 2 overlapping problems) C(n, k) = C(n-1, k-1) + C(n-1, k), for n > k > 0, and C(n, 0) = C(n, n) = 1 Dynamic programming solution: Record the values of the binomial coefficients in a table of n+1 rows and k+1 columns, numbered from 0 to n and 0 to k respectively. 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 …
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  Transitive ClosureThe transitiveclosure of a directed graph with n vertices can be defined as the nxn matrix T = {tij}, in which the element in the ith row (1 ≤ i ≤ n) and the jth column (1 ≤ j ≤ n) is 1 if there exists a nontrivial directed path (i.e., a directed path of a positive length) from the ith vertex to the jth vertex; otherwise, tij is 0.Graph traversal-based algorithm and Warshall’s algorithm 3 3 1 1 4 4 0 0 1 0 2 0 0 1 0 2 1 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 Adjacency matrix 1 1 1 1 0 1 0 0 Transitive closure
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  Warshall’s Algorithm• Mainidea: Use a bottom-up method to construct the transitive closure of a givendigraph with n vertices through a series of nxn boolean matrices: R(0),…, R(k-1), R(k) , …, R(n)The question is: how to obtain R(k) from R(k-1) ? R(k) : rij(k) = 1 in R(k) , iff there is an edge from i to j; or there is a path from i to j going through vertex 1; or there is a path from i to j going through vertex 1 and/or 2; or ... there is a path from i to j going through 1, 2, … and/or k 3 3 3 3 3 1 1 1 1 1 4 4 4 4 2 4 2 2 2 2 R(0) R(1) R(2) R(3) R(4) 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 1 0 1 0 0 1 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 1 0 0 1 1 1 1 1 1 1 1 1 1 1 1Does not allow an Allow 1 to be an Allow 1,2 to be an Allow 1,2,3 to be an Allow 1,2,3,4 to be anintermediate node intermediate node intermediate node intermediate node intermediate node
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  Warshall’s AlgorithmIn thekth stage: to determine R(k) is to determine if a path existsbetween two vertices i, j using just vertices among 1,…,k {rij(k) = 1: rij(k-1) = 1 or (path using just 1 ,…,k-1) (rik(k-1) = 1 and rkj(k-1)) = 1 (path from i to k and from k to i using just 1 ,…,k-1) k kth stagei Rule to determine whether rij (k) should be 1 in R(k) : • If an element rij is 1 in R(k-1) , it remains 1 in R(k) . j • If an element rij is 0 in R(k-1) ,it has to be changed to 1 in R(k) iff the element in its row i and column k and the element in its column j and row k are both 1’s in R(k-1).
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  Floyd’s Algorithm: Allpairs shortest paths All pairs shortest paths problem: In a weighted graph, dij(k) = length of thefind shortest paths between every pair of vertices. shortest path from i to j with each vertexApplicable to: undirected and directed weighted graphs; numbered no higherno negative weight. than k.Same idea as the Warshall’s algorithm : constructsolution through series of matrices D(0) , D(1), …, D(n)Example: 4 3 0 ∞ 4 ∞ 0 ∞ 4 ∞ 1 1 0 6 3 1 0 5 3 1 ∞ ∞ 0 ∞ 6 ∞ ∞ 0 ∞ 1 5 ∞ 5 1 0 6 5 1 0 4 2 3 weight matrix distance matrix
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  Floyd’s AlgorithmD(k) :allow 1, 2, …, k to be intermediate vertices.In the kth stage, determine whether the introduction of k as a new eligibleintermediate vertex will bring about a shorter path from i to j.dij(k) = min{dij(k-1) , dik(k-1) + dkj(k-1} for k ≥ 1, dij(0) = wij dik(k-1) k i dkj(k-1) kth stage dij(k-1) j
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  General CommentsThe crucialstep in designing a dynamicprogramming algorithm: Deriving a recurrence relating a solution to the problem’s current instance with solutions of its smaller ( and overlapping) subinstances.
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  The Knapsack Problemand Memory Functions (1)The problem Find the most valuable subset of the given n items that fit into a knapsack of capacity W.Consider the following subproblem P(i, j) Find the most valuable subset of the first i items that fit into a knapsack of capacity j, where 1 ≤ i ≤ n, and 1≤ j ≤ W Let V[i, j] be the value of an optimal solution to the above subproblem P(i, j). Goal: V[n, W] The question: What is the recurrence relation that expresses a solution to this instance in terms of solutions to smaller subinstances?
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  The Knapsack Problemand Memory Functions (2)The Recurrence Two possibilities for the most valuable subset for the subproblem P(i, j) It does not include the ith item: V[i, j] = V[i-1, j] It includes the ith item: V[i, j] = vi+ V[i-1, j – wi]V[i, j] = max{V[i-1, j], vi+ V[i-1, j – wi] }, if j – wi ≥ 0 { V[i-1, j] if j – wi < 0V[0, j] = 0 for j ≥ 0 and V[i, 0] = 0 for i ≥ 0
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  Memory Functions Memory functions: a combination of the top-down and bottom- up method. The idea is to solve the subproblems that are necessary and do it only once. Top-down: solve common subproblems more than once. Bottom-up: Solve subproblems whose solution are not necessary for the solving the original problem.ALGORITHM MFKnapsack(i, j)if V[i, j] < 0 //if subproblem P(i, j) hasn’t been solved yet. if j < Weights[i] value MFKnapsack(i – 1, j) else value max(MFKnapsack(i – 1, j), values[I] + MFKnapsck( i – 1, j – Weights[i])) V[i, j] valuereturn V[i, j]