Dynamic Programming

[Virtual Presenter] Dynamic programming is a powerful technique used to solve complex problems by breaking them down into smaller sub-problems. It's a method that allows us to solve problems more efficiently by avoiding redundant computations. In this video, we'll explore the basics of dynamic programming and how it can be applied to real-world problems. We'll start with the fundamental concepts and build our way up to more advanced topics. So let's get started!.

[Audio] For an optimization problem to be suitable for a dynamic-programming solution, there are two essential requirements. The problem should have optimal substructures, meaning that each subproblem should have an optimal solution, which can be used to solve larger problems. Without this property, a divide-and-conquer approach would be needed instead. Additionally, the problem should have overlapping subproblems, where smaller subproblems may appear multiple times during the solution process, allowing previously computed results to be reused. With these two properties, complex optimization problems can be efficiently solved using dynamic programming..

[Audio] The development of a dynamic-programming algorithm consists of three essential components. The recurrence relation defines the value of an optimal solution, capturing the essence of the problem by breaking it down into smaller parts. Tabular computation computes the value of an optimal solution using the recurrence relation, typically done in a bottom-up manner by filling in a table with values. Traceback delivers an optimal solution by tracing back through the table to construct the desired outcome. These components work together seamlessly to yield an efficient and effective dynamic-programming algorithm..

[Audio] The final step in the dynamic programming algorithm is to construct an optimal solution from the computed information. This involves using the previously computed values to build the optimal solution. By combining these values, we can create the optimal solution that satisfies the original problem constraints..

[Audio] When comparing two DNA strings, we can use the brute-force approach to find their longest common subsequence. This involves checking each possible subsequence of one string against the other. In this example, we have two DNA strings, X and Y, where X is and Y is . The longest common subsequence between these two strings is X = A B C B D A B and Y = B D C A B A. We can see that both strings share the same sequence of letters, but in different orders. By using the brute-force approach, we can identify this shared sequence, which is essential in understanding the similarity between the two DNA strings..

[Audio] Given two sequences, we can identify a maximum length common subsequence by examining all possible subsets of one sequence against the other. This process involves comparing each element in the first sequence with every element in the second sequence, considering only those pairs where the elements match. By doing so, we can determine the longest common subsequence between the two given sequences..

[Audio] In this example, we have two sequences X and Y, where X consists of the letters A, B, C, B, D, A, and B, and Y consists of the letters B, D, C, A, B, and A. By examining these sequences, we can identify the longest common subsequences between them. Two such subsequences are found to be B, C, B, A, and B, D, A, B, both having a length of four. However, another subsequence, B, C, A, does not qualify as a longest common subsequence because it has a shorter length than the previously mentioned subsequences..

[Audio] For every subsequence of X, we need to check whether it's a subsequence of Y. This process involves scanning Y for each letter of the subsequence, starting from the beginning. With m being the length of X, there are 2m such subsequences to check. Each one requires us to scan Y, which takes O(n) time. Therefore, the running time of this brute-force approach is O(n2m)..

[Audio] The LCS algorithm begins by defining the length of the longest common subsequence between two given sequences X and Y. This is accomplished by considering the prefixes Xi and Yj of lengths i and j respectively. The length of the LCS of these prefixes is denoted by c[i,j]. The length of the LCS of X and Y is then calculated as c[m,n], where m and n represent the lengths of X and Y respectively. The algorithm employs a recursive approach to calculate the length of the LCS, taking into account whether the current elements of X and Y align or not. If they do, the length of the LCS is incremented by one; otherwise, it is determined by comparing the lengths of the LCS of the prefixes Xi-1 and Yj-1..

[Audio] The initial conditions are considered. With i and j both set to zero, empty substrings of x and y are being dealt with. The longest common substring between these two strings will always be empty because there is nothing to compare. In this case, the value of c[0,0] is simply 0, indicating that the longest common substring has no length. Additionally, when comparing an empty string to any other string, the result is also an empty string. Therefore, for all values of i and j, c[0,j] and c[i,0] are also equal to 0. This establishes the basis for our dynamic programming approach, enabling us to build upon these fundamental cases as we progress..

[Audio] When calculating the value of c[i,j], we consider two distinct scenarios. The first scenario arises when x[i] equals y[j]. In this situation, we can conclude that one more symbol in the strings X and Y aligns, resulting in the length of the longest common subsequence (LCS) between Xi and Yj being equal to the length of the LCS between the smaller strings Xi-1 and Yi-1, plus one..

[Audio] When we encounter a mismatch between the current symbols in the strings X and Y, we need to consider the lengths of the longest common subsequences of the shorter prefixes. We have two options: either extend the longest common subsequence by considering the previous elements, or ignore the mismatch and continue exploring the possibilities. The key insight is that we cannot simply take the length of LCS(Xi-1, Yj-1), because doing so would mean ignoring the possibility of extending the longest common subsequence by considering the current symbols. By considering both LCS(Xi, Yj-1) and LCS(Xi-1, Yj), we ensure that we explore all possible extensions of the longest common subsequence, even when there are mismatches between the current symbols..

[Audio] When computing the length of the longest common subsequence, we consider two cases. If the current elements xi and yj are equal, we increment the length by one and move diagonally down and left. This is because we have found a matching pair. On the other hand, if they are not equal, we choose the maximum length between moving down and left, or moving right and left. This decision is based on whether the previous elements were equal or not. The process continues until we reach the boundaries of the matrices. The final result represents the length of the longest common subsequence..

[Audio] . 14 Additional Information 0 if i,j = 0 c[i, j] = c[i-1, j-1] + 1 if xi = yj max(c[i, j-1], c[i-1, j]) if xi  yj 0 0 0 0 0 0 0 0 0 0 0 yj: D A C F A B xi j i 0 1 2 n m 1 2 0 A matrix b[i, j]: • For a subproblem [i, j] it tells us what choice was made to obtain the optimal value • If xi = yj b[i, j] = “ ” • Else, if c[i - 1, j] ≥ c[i, j-1] b[i, j] = “  ” else b[i, j] = “  ” 3 3 C D b & c: c[i,j-1] c[i-1,j].

[Audio] The longest common subsequence between two sequences X and Y is computed using a dynamic programming approach. A matrix c is initialized with zeros, where c[i, j] represents the length of the longest common subsequence between the prefixes Xi and Yj. The algorithm iterates through the sequences, updating the values in the matrix based on whether the current elements xi and yj match or not. If they match, the value is incremented by one; otherwise, it takes the maximum value from the previous row or column. The final value stored in the bottom-right corner of the matrix represents the length of the longest common subsequence. This solution has a running time complexity of O(mn)..

[Audio] The example illustrates how dynamic programming can be applied to solve a problem. In this case, we have two sequences X and Y, and we want to find the longest common subsequence..

[Audio] When we start constructing the Longest Common Subsequence, we begin at b[m, n]. We follow the arrows from there. As we move along, we keep track of the elements that belong to the LCS. Whenever we encounter a blank space in b[i, j], it means that xi equals yj, and that element becomes part of the LCS. We continue this process until we reach the bottom-right corner of the table..

[Audio] The function PRINT-LCS prints the longest common subsequence between two strings X and Y by recursively calling itself with smaller values of i and j until it reaches the base case where either i or j is zero. If the current character in both strings matches, it prints the character and calls itself with smaller values. If not, it chooses the larger value and calls itself with that value. The running time of this function is O(m + n), where m and n are the lengths of the input strings..

[Audio] If we only need the length of the LCS, we can improve the code by working only on two rows of C at a time. Specifically, we consider the current row being computed and the previous row. By storing only these two rows, we can reduce the asymptotic space requirements. This optimization allows us to compute the length of the longest common subsequence more efficiently..

[Audio] The LCS algorithm calculates the values of each entry of the array c[m,n]. Since each c[i,j] is calculated in constant time, and there are m*n elements in the array, we can conclude that the running time of the LCS algorithm is O(m*n). This means that the time taken by the algorithm increases linearly with the size of the input. In other words, as the size of the input increases, the time taken by the algorithm also increases proportionally..

Rod Cutting Problem 21.

[Audio] Consider a rod of four inches in length. We can either not cut it at all, or we can cut it into two parts of lengths one inch each, or we can cut it into three parts of lengths one, one, and two inches respectively. The prices for these different ways of cutting the rod are given in the table. Our goal is.

[Audio] The general solution for the rod cutting problem involves breaking down the problem into smaller sub-problems, solving each one recursively, and storing the results to avoid redundant computation. This approach allows us to find the optimal solution by considering all possible combinations of cuts and their corresponding values..

[Audio] The recursive top-down implementation we have seen so far may not be efficient because it involves repeated computations. Imagine we need to cut a rod of length eight. We would first compute the maximum value for a rod of length seven, then for a rod of length six, and so on. But when we need to cut a rod of length nine, we would again compute the maximum values for rods of lengths eight, seven, six, and so on. This means we are repeating some computations..

[Audio] To solve this problem, we use dynamic programming to find the optimal way to cut a rod into smaller pieces to maximize profit. We define the problem's structure, where each subproblem represents a possible way to cut the rod. We recursively define the value of an optimal solution, considering all possible ways to cut the rod. We compute the value of an optimal solution in a bottom-up fashion, using memorization to store previously computed values. This avoids redundant computations and efficiently finds the optimal solution. By constructing an optimal solution from these computed values, we determine the best way to cut the rod to maximize profit..

[Audio] In this bottom-up approach, we begin by examining the base cases, where the length of the rod is either zero or one. We then construct the solutions for longer lengths by utilizing previously calculated values. This method enables us to circumvent redundant computations and guarantee that our solution is optimal. Consequently, we can efficiently resolve the issue of determining the maximum revenue that can be achieved by dividing a rod of a specified length into smaller segments..

[Audio] We start by computing the minimum cost for small lengths of rods, and then gradually build up to larger lengths. We use a table to store the results of subproblems, so that we can avoid recomputing them. This allows us to solve the problem more efficiently. The function call EXTENDED-BOTTOM-UP-CUTROD(p, 10) would compute the minimum cost for cutting a rod of length p into pieces of length 10 or less. Similarly, the function call print-cut-rod(P,10) would print the optimal way to cut the rod. By using dynamic programming, we can find the optimal solution to this problem in a more efficient manner..

[Audio] When computing the product of multiple matrices, there are many possible ways to do so. This is because the order in which the matrices are multiplied can greatly impact the efficiency of the computation. In this case, we have a sequence or chain of n matrices, denoted by A1, A2,..., An, that need to be multiplied together to produce their product. The key challenge here is finding the most efficient way to perform this multiplication, as there are many different possibilities. We will explore these options further in our discussion..

[Audio] The five different ways to compute the product A1A2A3A4 are represented by the parenthesized expressions listed above. Each expression represents a different order in which the matrices should be multiplied. We will use these expressions later when discussing the matrix-chain multiplication problem..

[Audio] The algorithm to multiply two matrices is straightforward. To multiply two matrices, we need to perform matrix multiplication. The process involves iterating over each element of one matrix and multiplying it with the corresponding elements of another matrix. We then add these products together to get the result. This process is repeated for all pairs of corresponding elements in both matrices. The resulting matrix will have the same number of rows as the first input matrix and the same number of columns as the second input matrix..

[Audio] The algorithm to multiply two matrices A and B, with dimensions p by q and q by r respectively, results in a matrix C with dimensions p by r. The process involves initializing each element of C to zero, then iterating over the rows and columns of C, performing scalar multiplications and additions to compute the final values. The dominant operation in this process is the scalar multiplication in line five, which occurs pqr times. This gives us the total number of scalar multiplications required to compute the result..

[Audio] There are two ways to parenthesize the matrix multiplication. One way is ((AB)C), which requires 7,500 scalar multiplications. Another way is (A(BC)), which requires 75000 scalar multiplications. Different parenthesizations can lead to vastly different numbers of scalar multiplications required..

[Audio] Given a chain of matrices, our goal is to find the most efficient way to multiply them together. This involves finding the optimal parenthesization of the product, one that minimizes the number of scalar multiplications required. The diagram shows the chain of matrices, with each matrix having its own dimensions. Our task is to determine the best way to group these matrices together to minimize the number of scalar multiplications needed..

[Audio] Consider the subproblem of parenthesizing the product Ai...j. This subproblem involves splitting the product into two parts: Ai...k and Ak+1...j. The optimal parenthesization splits the product at k, where i ≤ k < j. We assume that the minimum number of multiplications required to compute Ai...k is stored in m[i, k], and similarly for Ak+1...j. The minimum number of multiplications required to compute Ai...kAk...j is then the sum of these two values plus the cost of multiplying Ai...k and Ak+1...j..

[Audio] Computing the optimal costs involves computing the minimum cost for each cell in the matrix. This is done by considering all possible choices and selecting the one that yields the minimum cost. The formula used is m[i, j] = min . The length of the computation depends on whether we are dealing with a single cell or two cells. In the case of a single cell, i equals j, while in the case of two cells, j equals i plus one. Once the optimal costs have been computed, they can be used to construct the optimal solution. This is done by tracing back the path that corresponds to the minimum cost. The optimal values of k are kept in a similar matrix s..

[Audio] The dynamic programming algorithm solves an optimization problem by finding the minimum value of the expression m[i, k] + m[k+1, j] + pi-1pkpj. This expression relies on the recursive calculation of m[i, j] values using previously computed values. Since m[i, j] values depend only on previously computed values, we can construct the solution incrementally, avoiding redundant computations and reducing the problem's computational complexity..

[Audio] When computing the costs for multiplying four matrices, we start by considering the intermediate results of AB, BC, and CD. We do not need to consider the intermediate result of AC or BD because it's not necessary for our calculation. By doing so, we can reduce the complexity of the problem and make it more manageable. This approach allows us to break down the problem into smaller sub-problems, which can be solved independently and combined to obtain the final solution..

[Audio] When computing the costs in the bottom-up manner, we initially consider various combinations of matrices. Specifically, we examine A(BC), (AB)C, B(CD), and (BC)D. Notably, we do not need to consider other combinations like (AB)D or A(CD). This approach enables us to select the minimum cost matrix calculation for our purposes..

[Audio] To optimize the calculation of four matrices, we need to consider all possible combinations of matrix multiplications. We can compute the costs in a bottom-up manner by considering the minimum cost of each intermediate result. This approach allows us to select the minimum cost matrix calculations of ABC and BCD. By doing so, we can reduce the overall computational complexity and improve the efficiency of our algorithm..

[Audio] We are considering the final options for multiplying four matrices. Each option represents a different way of combining these matrices. We will compute the costs associated with each option and then select the one that yields the lowest cost. This will give us the most efficient method for performing the multiplication..

[Audio] When computing the costs in the bottom-up manner, we consider all possible ways to multiply these four matrices. We calculate the costs for each intermediate result and store them in a table. This allows us to avoid redundant computations and focus on the most efficient way to achieve our goal. By choosing the cheapest cost plan for matrix calculations, we can minimize the overall computational complexity and optimize our results..

[Audio] The example illustrates how dynamic programming can be utilized to resolve a challenge. Four matrices, designated as A, B, C, and D, each possessing unique dimensions, are provided. The objective is to identify the highest cumulative product among these matrices. Initially, the table is initialized with zeros. Subsequently, it's populated row by row, relying on previously calculated values to determine the subsequent ones. This procedure persists until the entire table has been filled. Following the completion of the table, it can be employed to locate the highest cumulative product. In this scenario, the highest cumulative product is obtained by multiplying the corresponding elements of the matrices collectively and aggregating them. For instance, examining the upper-left corner of the table reveals that the highest cumulative product is 1200, resulting from the product of the initial element of matrix A and the initial element of matrix B being 30 x 40 = 1200. Similarly, inspecting the lower-right corner of the table discloses that the highest cumulative product is 10000, stemming from the product of the final element of matrix C and the final element of matrix D being 40 x 25 = 10000. By employing dynamic programming, we successfully resolved this issue and identified the highest cumulative product among the given matrices..

[Audio] The example demonstrates how dynamic programming solves a problem involving matrix multiplication of varying sizes. The process can be decomposed into smaller sub-problems, which are tackled recursively. The outcomes of these sub-problems are recorded in a table, thereby preventing redundant computations and decreasing the overall computational complexity. This method enables efficient computation of the product of two matrices, even when their dimensions are substantial..

[Audio] The example demonstrates the calculation of c(i,j) and kay(i,j) values utilizing dynamic programming. The issue involves multiplying matrices of varying dimensions, where each matrix possesses a distinct size defined by variables P0 through P5. The objective is to efficiently calculate the product of these matrices via dynamic programming. The table displays the intermediate outcomes acquired throughout the computational procedure. Values c(i,j) symbolize the cumulative sum of the matrix products, whereas values kay(i,j) signify the current matrix product. By scrutinizing the table, it becomes apparent how the values are calculated recursively, with each entry relying on preceding entries. This recursive definition enables us to calculate the values in a bottom-up manner, commencing from base cases and progressing towards the ultimate outcome..

[Audio] The dynamic programming algorithm is used to solve the matrix multiplication problem by initializing the first row and column of the result matrix with the corresponding elements of the input matrices. Then, it iterates over the remaining rows and columns, using the previously computed values to calculate the next ones. The computation of each element in the result matrix depends only on the previously computed elements in the same row and column, allowing memoization to store intermediate results and avoid redundant computations. This reduces the computational complexity of the problem from exponential to polynomial time, making the algorithm more efficient and scalable for large inputs..

[Audio] The problem involves finding the maximum sum of products of arrays, with the arrays being [10, 5], [5, 1], [1, 10], [10, 2], and [2, 10]. The table c(i, j) stores the product of each pair of arrays, representing the maximum sum of products up to that point. The table also contains intermediate results, kay(i, j), used to calculate the next row of c(i, j) values. By employing dynamic programming, we can efficiently compute the maximum sum of products by filling the table in a bottom-up manner, avoiding redundant calculations and reducing computational complexity..

[Audio] The dynamic programming problem involves finding the maximum sum of products in a 5x10 matrix. The values in the table are calculated recursively using previously computed values, which is a characteristic of dynamic programming problems. The optimal solution is broken down into smaller subproblems, and the solutions to these subproblems are combined to give the overall solution. The maximum sum of products is computed by adding the products of the elements in each row and column, and the results are stored in the table. The final result is obtained by selecting the maximum value from the table..

[Audio] The brute-force solution has a running time of O(n^2*m). This is because there are 2m subsequences of X to check, and each subsequence takes O(n) time to check. The dynamic programming solution has a running time of O(m+n), which is much faster than the brute-force solution. This is because the dynamic programming solution uses memoization to avoid redundant computations..

[Audio] The dynamic programming algorithm solves complex problems by dividing them into smaller sub-problems, solving each one only once, and storing their solutions to avoid redundant computation. This approach reduces computational time and memory requirements. For instance, in a matrix multiplication problem involving two matrices with dimensions [10x5] and [5x1], dynamic programming can be applied to break down the task into smaller parts and solve it efficiently. The method involves storing intermediate results in a table and reusing them to prevent recalculation. By doing so, computational time and memory usage are minimized..

[Audio] The example illustrates the application of dynamic programming to solve a complex problem. The problem involves computing the minimum cost of a sequence of operations, where each operation has a specific cost associated with it. The dynamic programming algorithm breaks down the problem into smaller sub-problems, solves each sub-problem only once, and stores the solutions to sub-problems to avoid redundant computation. The algorithm then combines the solutions to sub-problems to obtain the overall minimum cost. In this case, the algorithm computes the minimum cost for different values of K, and shows how the minimum cost changes as K increases. The example demonstrates the effectiveness of dynamic programming in solving complex problems by reducing the computational complexity and improving the efficiency of the solution..