线程并行高效 [英] Thread Parallel Efficient

问题描述

编写一个并行函数，该函数将用可变线程数将两个n x n矩阵相乘.
通过选择线程数和每个线程的块数来提高效率.

在此先感谢！

Write a parallel function that will multiply two n by n matrices with a variable number of threads.
This should be an efficient by choosing the number of threads and number of blocks for each thread.

Thanks in advance!

推荐答案

1. 计算出乘积兼容性，(m x n)可以乘以(n x o)个矩阵，这将产生一个(m x o)个矩阵.此处并不需要，因为我们只查看nxn的平方矩阵，因此具有正确的乘法尺寸.
2. 选择块大小: b
3. 创建尺寸为(mxo)的新矩阵.
4. 按行(或逐列)方式遍历新矩阵，以批量创建 b 的元组.
   b = 2且m = n = o = 3且基于一个的数组索引的示例:
   1.(1,1)(1,2)
   2.(1,3)(2,1)
   3.(2,2)(2,3)
   4.(3,1)(3,2)
   5.(3,3)
   
5. 为每个批次启动一个线程.使用Thread.Start，您将把元组列表传递给线程，并以左手，右手的顺序传递对新创建的mxo矩阵以及两个原始矩阵的引用.
6. 在线程工作方法中，您将对传递给线程的列表的每个元组(x，y)，将x行乘以y列.生成的标量将放置在新矩阵中的位置(x，y).
7. 在主程序中，等待所有线程完成(Join).乘法结果现在位于第3步中新创建的矩阵中.

1. Work out multiplication compatibility, (m x n) can multiply (n x o) matrices which will yield a (m x o) matrix. That is not really needed here since we are only looking at square matrices which are n x n and thus have the correct dimensions for multiplication.
2. Choose block size: b
3. Create a new matrix with the dimensions (m x o).
4. Traversing the new matrix in a row by row (or column by column) fashion create tuples in batches size b.
   Example with b = 2 and m = n = o = 3 and array indices based on one:
   1. (1,1) (1,2)
   2. (1,3) (2,1)
   3. (2,2) (2,3)
   4. (3,1) (3,2)
   5. (3,3)
   
5. Start a thread for each of the batches. Using Thread.Start you will pass the list of tuples to the thread as well as a reference to the the newly created m x o matrix as well as both original matrices in the order left hand side, right hand side.
6. In the threads work method you''ll do the multiplication of row x by column y for each tuple (x, y) of the list you passed to the thread. The resulting scalar is placed into the new matrix at position (x,y).
7. In your main program wait for all threads to finish (Join). The multiplication result is now in the newly created matrix from step 3.

这就是全部.行与列或矩阵乘法 [

That is all there is to it really. The multiplication of a row by a column or matrix multiplication[^] is straight forward enough so I won''t go into that.

Regards,

Manfred

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