递归在施特拉森的算法 [英] Recursion in Strassen's Algorithm

问题描述

我想知道你将如何使施特拉森算法的递归调用，并在那里正是他们需要的。

I'm wondering how you would make the recursive calls in Strassen's algorithm, and where exactly they're required.

据我了解，7乘法器更有效率比8，我们就必须以其他方式，但我很困惑，这些乘数是如何递归计算。特别是，如果我们按照分而治之的模式，到底是哪矩阵的一部分，是我们分化和我们如何这样做的，直到我们得到一个基本案例中，我们可以分别征服递归部分？

I understand that the 7 multipliers is more efficient than the 8 we would have otherwise, but I'm confused as to how these multipliers are calculated recursively. In particular, if we are following the divide and conquer paradigm, exactly which part of the matrices are we "dividing" and how are we going about doing that until we get to a base case in which we can conquer the recursive parts separately?

感谢您！

推荐答案

我们进行递归调用，同时计算这7个乘法器。 首先，扩展矩阵的大小为2的幂，然后在每一个步骤中，我们划分每个矩阵成4片

We make recursive calls while calculating these 7 multipliers. At first we extend the size of the matrices to the power of 2 and then on each step we divide the each matrix into 4 pieces.

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