施特拉森演算法

設${\displaystyle A}$、${\displaystyle B}$為域${\displaystyle F}$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。

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

將A, B, C分成相等大小的方塊矩陣：

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

即

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

於是

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

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

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

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

引入新矩陣

${\displaystyle \mathbf {M} _{1}:=(\mathbf {A} _{1,1}+\mathbf {A} _{2,2})(\mathbf {B} _{1,1}+\mathbf {B} _{2,2})}$

${\displaystyle \mathbf {M} _{2}:=(\mathbf {A} _{2,1}+\mathbf {A} _{2,2})\mathbf {B} _{1,1}}$

${\displaystyle \mathbf {M} _{3}:=\mathbf {A} _{1,1}(\mathbf {B} _{1,2}-\mathbf {B} _{2,2})}$

${\displaystyle \mathbf {M} _{4}:=\mathbf {A} _{2,2}(\mathbf {B} _{2,1}-\mathbf {B} _{1,1})}$

${\displaystyle \mathbf {M} _{5}:=(\mathbf {A} _{1,1}+\mathbf {A} _{1,2})\mathbf {B} _{2,2}}$

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

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

可得：

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

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

${\displaystyle \mathbf {C} _{2,1}=\mathbf {M} _{2}+\mathbf {M} _{4}}$

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

其中${\displaystyle M_{i,j}}$的計算也是使用施特拉森演算法求得。