Introduction to Algorithms
3rd Edition
ISBN: 9780262033848
Author: Thomas H. Cormen, Ronald L. Rivest, Charles E. Leiserson, Clifford Stein
Publisher: MIT Press
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Chapter 27.1, Problem 7E
Program Plan Intro
The following is the analysis of work, span, and parallelism of the multithreaded
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## Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)#
def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)]
# Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1
for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…
Suppose that two matrices A and B with dimension n x n are stored on a 2D mesh of n2 processors such that P_(i,j) holds A[I,j] and B[j,i] . Write C-style pseudocode for an asynchronous algorithm that computes the product of A and B in O(n) time. (10 points) (WRITE THE PSEUDOCODE! IF YOU DON'T WRITE PSEUDOCODE, YOU WILL BE WRONG!! C-STYLE ONLY!!!)
Suppose that two n x n matrices A and B are stored on a mesh of n2 processors such that Pi;j holds A[i; j] and B[j; i]. Write pseudocode for an asynchronous algorithm that computes the product of A and B in O(n).
Chapter 27 Solutions
Introduction to Algorithms
Ch. 27.1 - Prob. 1ECh. 27.1 - Prob. 2ECh. 27.1 - Prob. 3ECh. 27.1 - Prob. 4ECh. 27.1 - Prob. 5ECh. 27.1 - Prob. 6ECh. 27.1 - Prob. 7ECh. 27.1 - Prob. 8ECh. 27.1 - Prob. 9ECh. 27.2 - Prob. 1E
Ch. 27.2 - Prob. 2ECh. 27.2 - Prob. 3ECh. 27.2 - Prob. 4ECh. 27.2 - Prob. 5ECh. 27.2 - Prob. 6ECh. 27.3 - Prob. 1ECh. 27.3 - Prob. 2ECh. 27.3 - Prob. 3ECh. 27.3 - Prob. 4ECh. 27.3 - Prob. 5ECh. 27.3 - Prob. 6ECh. 27 - Prob. 1PCh. 27 - Prob. 2PCh. 27 - Prob. 3PCh. 27 - Prob. 4PCh. 27 - Prob. 5PCh. 27 - Prob. 6P
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- #plea# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…arrow_forwardConsider the following multithreaded algorithm for performing pairwise multiplication on n-element arrays A[1..n] and B[1..n], storing the multiplications in C[1..n]:Analyze the work, span and parallelism of this algorithm.arrow_forwardGive a MPI program segment to convert a n-by-n matrix distributed on a n-by-n 2D mesh such that allits rows and columns get sorted in ascending order. Show only the iterative loop. Only communicationsallowed are to the four direct neighbors. (Hint: You may employ a variant of odd-even transpositionsort, but the whole matrix does not need to be sorted.)arrow_forward
- Write a MPI program in C to multiply two n-by-n matrices using p processors with 1 <= p <= 8. Fill up the matrices with some constant values so that it would be easier for you to verify the resulting matrix for correctness. You may use collective operations or send/receive primitives.arrow_forwardOne of the methods used to keep graphy in memory is the use of two-array. In this method, connection information is stored in one of the directories, and the index values of the nodes in the other. a) Show the retention of a costly graphene on a two-array by taking the following neighborhood matrix as an example. b) Report the matrices in C language by giving the initial values as well NOT:It should be in C Language, Not Another Languagearrow_forwardUsing Java Modify the program 1 in guided activity 3 to compute the multiplication of two vectors with the sizes 1xn and nx1. The result is a number (one element matrix) computed as the sum of the products of the corresponding elements from the first vector with the ones from the second vector: m1[0][0]m2[0][0]+m1[0][1]*m2[1][0]+...+m1[0][n-1]m2[n-1][0arrow_forward
- Lecture: Analysis of Algorithm >> Matrix Chain Multiplication << Please write C code of the matrix chain multiplication (A1.A2.A3) by using dynamic programming as given below. Then calculate the algorithmic complexity. A1: 3x4 P0=3 P1=4 P2=2 P3=5A2: 4x2A3: 2x5arrow_forward#com# Count the number of unique paths from a[0][0] to a[m-1][n-1]# We are allowed to move either right or down from a cell in the matrix.# Approaches-# (i) Recursion- Recurse starting from a[m-1][n-1], upwards and leftwards,# add the path count of both recursions and return count.# (ii) Dynamic Programming- Start from a[0][0].Store the count in a count# matrix. Return count[m-1][n-1]# T(n)- O(mn), S(n)- O(mn)# def count_paths(m, n): if m < 1 or n < 1: return -1 count = [[None for j in range(n)] for i in range(m)] # Taking care of the edge cases- matrix of size 1xn or mx1 for i in range(n): count[0][i] = 1 for j in range(m): count[j][0] = 1 for i in range(1, m): for j in range(1, n): # Number of ways to reach a[i][j] = number of ways to reach # a[i-1][j] + a[i][j-1] count[i][j] = count[i - 1][j] + count[i][j - 1]…arrow_forwardGiven an n x n matrix where each of the rows and columns is sorted in ascending order, return the kth smallest element in the matrix. Note that it is the kth smallest element in the sorted order, not the kth distinct element. You must find a solution with a memory complexity better than O(n2). Example 1: Input: matrix = [[1,5,9],[10,11,13],[12,13,15]], k = 8 Output: 13 Explanation: The elements in the matrix are [1,5,9,10,11,12,13,13,15], and the 8th smallest number is 13 Example 2: Input: matrix = [[-5]], k = 1 Output: -5arrow_forward
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