# 施特拉森演算法

## 算法

### 定義

${\displaystyle A}$${\displaystyle B}$${\displaystyle F}$上的方矩陣。求兩者的積${\displaystyle C}$。一般矩陣可以填0的方法計算令它成為${\displaystyle 2^{n}\times 2^{n}}$矩陣

${\displaystyle \mathbf {C} =\mathbf {A} \mathbf {B} \qquad \mathbf {A} ,\mathbf {B} ,\mathbf {C} \in F^{2^{n}\times 2^{n}}}$

### 計算

A, B, C分成相等大小的方塊矩陣：

${\displaystyle \mathbf {A} ={\begin{bmatrix}\mathbf {A} _{1,1}&\mathbf {A} _{1,2}\\\mathbf {A} _{2,1}&\mathbf {A} _{2,2}\end{bmatrix}}{\mbox{ , }}\mathbf {B} ={\begin{bmatrix}\mathbf {B} _{1,1}&\mathbf {B} _{1,2}\\\mathbf {B} _{2,1}&\mathbf {B} _{2,2}\end{bmatrix}}{\mbox{ , }}\mathbf {C} ={\begin{bmatrix}\mathbf {C} _{1,1}&\mathbf {C} _{1,2}\\\mathbf {C} _{2,1}&\mathbf {C} _{2,2}\end{bmatrix}}}$

${\displaystyle \mathbf {A} _{i,j},\mathbf {B} _{i,j},\mathbf {C} _{i,j}\in F^{2^{n-1}\times 2^{n-1}}}$

${\displaystyle \mathbf {C} _{1,1}=\mathbf {A} _{1,1}\mathbf {B} _{1,1}+\mathbf {A} _{1,2}\mathbf {B} _{2,1}}$
${\displaystyle \mathbf {C} _{1,2}=\mathbf {A} _{1,1}\mathbf {B} _{1,2}+\mathbf {A} _{1,2}\mathbf {B} _{2,2}}$
${\displaystyle \mathbf {C} _{2,1}=\mathbf {A} _{2,1}\mathbf {B} _{1,1}+\mathbf {A} _{2,2}\mathbf {B} _{2,1}}$
${\displaystyle \mathbf {C} _{2,2}=\mathbf {A} _{2,1}\mathbf {B} _{1,2}+\mathbf {A} _{2,2}\mathbf {B} _{2,2}}$

${\displaystyle \mathbf {M} _{1}:=(\mathbf {A} _{1,1}+\mathbf {A} _{2,2})(\mathbf {B} _{1,1}+\mathbf {B} _{2,2})}$
${\displaystyle \mathbf {M} _{2}:=(\mathbf {A} _{2,1}+\mathbf {A} _{2,2})\mathbf {B} _{1,1}}$
${\displaystyle \mathbf {M} _{3}:=\mathbf {A} _{1,1}(\mathbf {B} _{1,2}-\mathbf {B} _{2,2})}$
${\displaystyle \mathbf {M} _{4}:=\mathbf {A} _{2,2}(\mathbf {B} _{2,1}-\mathbf {B} _{1,1})}$
${\displaystyle \mathbf {M} _{5}:=(\mathbf {A} _{1,1}+\mathbf {A} _{1,2})\mathbf {B} _{2,2}}$
${\displaystyle \mathbf {M} _{6}:=(\mathbf {A} _{2,1}-\mathbf {A} _{1,1})(\mathbf {B} _{1,1}+\mathbf {B} _{1,2})}$
${\displaystyle \mathbf {M} _{7}:=(\mathbf {A} _{1,2}-\mathbf {A} _{2,2})(\mathbf {B} _{2,1}+\mathbf {B} _{2,2})}$

${\displaystyle \mathbf {C} _{1,1}=\mathbf {M} _{1}+\mathbf {M} _{4}-\mathbf {M} _{5}+\mathbf {M} _{7}}$
${\displaystyle \mathbf {C} _{1,2}=\mathbf {M} _{3}+\mathbf {M} _{5}}$
${\displaystyle \mathbf {C} _{2,1}=\mathbf {M} _{2}+\mathbf {M} _{4}}$
${\displaystyle \mathbf {C} _{2,2}=\mathbf {M} _{1}-\mathbf {M} _{2}+\mathbf {M} _{3}+\mathbf {M} _{6}}$

## 参考来源

1. ^ Virginia Vassilevska Williams. Multiplying matrices faster than Coppersmith-Winograd (PDF). [2014-01-14]. （原始内容存档 (PDF)于2013-10-08）.而1990年Coppersmith-Winograd方法提出时的算法复杂度为${\displaystyle O(n^{2.3727})}$