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Chapter 10: Problem 9

Write a CRCW PRAM algorithm that uses \(n^{2}\) processors to multiply two \(n \times n\) matrices. Your algorithm should perform better than the standard \(\Theta\left(n^{3}\right)\) -time serial algorithm.

Short Answer

Expert verified
The algorithm divides the task among different processors to each multiply corresponding elements in the row of the first matrix and column of the second matrix. Multiplications are done in parallel, which is far quicker than the serial algorithm. The time complexity of this algorithm is \(\Theta\left(n^{2}\right)\), assuming add and multiply operations can be done in constant time.

Step by step solution

01

Initialization

Firstly, initialize \(n^{2}\) processors and two matrices A and B of size \(n \times n\) for multiplication.
02

Element Assignment

Assign each element of matrices A and B to a single processor. Given there are \(n^{2}\) processors, each cell in the \(n \times n\) matrix will be assigned to one processor, resulting in distributing the matrices among the processors.
03

Parallel Processing

In parallel, each processor performs the multiplication of the corresponding cell values of matrices A and B. The calculation will be the sum of multiplication of corresponding elements in a row of the first matrix and a column of the second matrix. All processors perform this step simultaneously.
04

Calculating Result

Finally, the partial results calculated by each processor are summed up to compute the value each cell in the resultant matrix.

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Most popular questions from this chapter

Write a scquential algorithm that implements the Tournament Method to find the largest key in an array of \(n\) keys. Show that this algorithm is no more efficient than the standard sequential algorithm.

Consider the problem of adding the numbers in a list of \(n\) numbers, If it takes \(t_{d}(n-1)\) time for one person to add all \(n\) numbers, is it possible for \(m\) people to compute the sum in less than \(\left[t_{e}(n-1)\right] / m\) time? Justify your answer.

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