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Describe the standard O(n^3) time dynamic
Dynamic programming: Matrix Chain Multiplication
Description In this assignment you are asked to implement a dynamic programming algorithm for the matrix chain multiplication problem , the goal is to find the most computationally efficient matrix order when multiplying an arbitrary number of matrices. in a row. you can assume that the entire input will be given as integers that can be stored using the standard C++ int type and that matrix sizes will be at least 1.
Input The input has the following format. The first number, n≥1, in the test case will tell you how many matrices are in the sequence. The first number will be then followed by n+1 is enough to fully specify the dimensions of the multiplied.
Output First you need to output the minimum number of scalar multiplication needed to multiply the given matrices. Then, print the matrix multiplication sequence, via parentheses, that minimizes the total number of number multiplication.
Each matrix should be named A#, where # is the matrix number starting at 0 (zero) and ending at n-1. see the examples below.
Examples of input and output:
2
2 3 5
30
(A0A1)
3
10 10)A2)
6
30 35 10 5 50
7500
((A0A1)A2)
3
10 30 5 60
4500
((A0A15 5 10 20 25
15125
((A0(A1A2))((A3A4)A5))
Recurrence relation:
A recurrence relation is an equation that uses recursion to relate terms in a sequence or elements in an array. It is a way to define a sequence or array in terms of itself. Recurrence relation have application in many areas of mathematics:
Sometimes, a recurrence relation can be "solved" by defining the terms of a sequence in terms of its index rather than previous terms in the sequence. This given a closed form expression for each term in the sequence and eliminates the need for an iterative process to solve for terms in the sequence. there are several ways to accomplish this:
Solving linear recurrence relations
solving recurrence relations with generating functions
solving recurrence relation with the substitution method
solving recurrence relation with the method of summation factors.
Even if the solution of this form is not possible, a recurrence relation is still useful, as it can use to develop computer algorithms. When terms in a sequence are stored, dynamic programming allows one to compute new terms in a sequence efficiently. Recurrence relation is also applicable for recursive backtracking, in which recursion is used to optimize algorithms.
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- Python programming language please help me understand this question with as much explanation as possible please.arrow_forwarda. Correctness of dynamic programming algorithm: Usually, a dynamic programming algorithm can be seen as a recursion and proof by induction is one of the easiest way to show its correctness. The structure of a proof by strong induction for one variable, say n, contains three parts. First, we define the Proposition P(n) that we want to prove for the variable n. Next, we show that the proposition holds for Base case(s), such as n = 0, 1, . . . etc. Finally, in the Inductive step, we assume that P(n) holds for any value of n strictly smaller than n' , then we prove that P(n') also holds. Use the proof by strong induction properly to show that the algorithm of the Knapsack problem above is correct. b. Bounded Knapsack Problem: Let us consider a similar problem, in which each item i has ci > 0 copies (ci is an integer). Thus, xi is no longer a binary value, but a non-negative integer at most equal to ci , 0 ≤ xi ≤ ci . Modify the dynamic programming algorithm seen at class for this…arrow_forwardA. Construct a RECURSIVE solution for following iterative solution to find a value in a circular-linked list B. Analyze the runtime complexity (correctly explain the scenario, show how much work would be done, and represent the work using asymptotic notations, i.e. big O), of the given iterative solution and your recursive solution for the best case scenario and the worse case scenario. C. prove the correctness of your recursive solution by induction /**@param value - a value to search for@return true if the value is in the list and set the current reference to itotherwise return false and not updating the current reference*/public boolean find(T value){ if(this.cur == null) return false; //get out, nothing is in here Node<T> tmp = this.cur; //start at the current position if(tmp.data == value) return true; // found it at the starting location tmp = tmp.next; //adv. to next node, if there is one while(tmp != cur) { if(tmp.data == value){ this.cur = tmp;…arrow_forward
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