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| | '''Strassen algorithm for a symmetric matrix:''' | | '''Strassen algorithm for a symmetric matrix:''' |
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| − | For a [[w:Symmetric matrix|<span class="wikipedia">symmetric matrix</span>]] <math>A \in \mathbb{C}^{n \times n}</math> where <math>n \in \mathbb{N}^*</math>, the [[w:Runtime (program lifecycle phase)|<span class="wikipedia">runtime</span>]] <math>T_s(n)</math> of the [[w:Strassen algorithm|<span class="wikipedia">Strassen algorithm</span>]] for the [[w:Matrix multiplication|<span class="wikipedia">matrix product</span>]] <math>A^2</math> is about half that of the original algorithm in <math>\mathcal{O}(n^{(_2 7)})</math>. | + | For a square [[w:Matrix (mathematics)|<span class="wikipedia">matrix</span>]] <math>A \in \mathbb{C}^{n \times n}</math> where sufficiently big <math>2^k := n, k \in \mathbb{N}^*</math>, the [[w:Runtime (program lifecycle phase)|<span class="wikipedia">runtime</span>]] <math>T_q(n)</math> of the [[w:Strassen algorithm|<span class="wikipedia">Strassen algorithm</span>]] for the [[w:Matrix multiplication|<span class="wikipedia">matrix product</span>]] <math>AA^T</math> roughly halves that of the original algorithm in <math>\mathcal{O}(n^{(_2 7)})</math>. |
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| − | '''Proof:''' For <math>A :=
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| − | \begin{pmatrix}
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| − | A_{11} & A_{12} \\
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| − | A_{12} & A_{22}
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| − | \end{pmatrix}</math>, it holds that <math>A^TA =
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| − | \begin{pmatrix}
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| − | A_{11}^TA_{11}+A_{12}^TA_{12} & A_{11}^TA_{12}+A_{12}^TA_{22} \\
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| − | A_{12}^TA_{11}+A_{22}^TA_{12} & A_{12}^TA_{12}+A_{22}^TA_{22}
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| − | \end{pmatrix}</math> and <math>T_s(2n) = 3T_s(n) + 2n^{(_2 7)}</math>. Thus <math>T_s(n) = 3T_s(n/2) + 2(n/2)^{(_2 7)}</math> and <math>T_s(n/2) = 3T_s(n/4) + 2(n/4)^{(_2 7)}</math>.
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| − | The [[w:Geometric series|<span class="wikipedia">geometric series</span>]] yields because of <math>T_s(1) = 1</math>: <math>T_s(n) = 27T_s(n/8) + 2/7n^{(_2 7)}(1+3/7 + (3/7)^2 + ...) = 3^{(_2n)} + 2/7n^{(_2 7)} (1-(3/7)^{(_2n)})/(1-3/7)</math> <math>= n^{(_2 3)} + \hat{2}(n^{(_2 7)}-n^{(_2 3)}) = \hat{2} (n^{(_2 3)} + n^{(_2 7)})</math>.<math>\square</math>
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| − | '''Strassen algorithm for a square matrix:'''
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| − | For a square [[w:Matrix (mathematics)|<span class="wikipedia">matrix</span>]] <math>A \in \mathbb{C}^{n \times n}</math> where <math>n \in \mathbb{N}^*</math>, the runtime <math>T_q(n)</math> of the Strassen algorithm is for the matrix product <math>A^TA</math> about <math>4/7</math> that of the original algorithm in <math>\mathcal{O}(n^{(_2 7)})</math>.
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| | '''Proof:''' For <math>A := | | '''Proof:''' For <math>A := |
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| | A_{11} & A_{12} \\ | | A_{11} & A_{12} \\ |
| | A_{21} & A_{22} | | A_{21} & A_{22} |
| − | \end{pmatrix}</math>, it holds that <math>A^TA = | + | \end{pmatrix}</math>, it holds that <math>AA^T = |
| | \begin{pmatrix} | | \begin{pmatrix} |
| − | A_{11}^TA_{11}+A_{21}^TA_{21} & A_{11}^TA_{12}+A_{21}^TA_{22} \\ | + | A_{11}A_{11}^T+A_{12}A_{12}^T & A_{11}A_{21}^T+A_{12}A_{22}^T \\ |
| − | A_{12}^TA_{11}+A_{22}^TA_{21} & A_{12}^TA_{12}+A_{22}^TA_{22}
| + | A_{21}A_{11}^T+A_{22}A_{12}^T & A_{21}A_{21}^T+A_{22}A_{22}^T |
| − | \end{pmatrix}</math> such that <math>T_q(2n) = 4T_s(n) + 2n^{(_2 7)}</math> and <math>T_q(n) = 4T_s(n/2) + 2/7n^{(_2 7)} = 2/3n^{(_2 3)} + 4/7n^{(_2 7)}</math>.<math>\square</math> | + | \end{pmatrix}</math>. |
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| − | '''New algorithm for two symmetric matrices:'''
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| − | For two symmetric matrices <math>A, B \in \mathbb{C}^{n \times n}</math> where <math>n \in \mathbb{N}^*</math> the runtime is for the matrix product <math>AB</math> about <math>6/7</math> that of the original algorithm in <math>\mathcal{O}(n^{(_2 7)})</math>.
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| − | '''Proof:''' For <math>B :=
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| − | \begin{pmatrix}
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| − | B_{11} & B_{12} \\
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| − | B_{12} & B_{22}
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| − | \end{pmatrix}</math>, it holds that <math>A^TB =
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| − | \begin{pmatrix}
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| − | A_{11}^TB_{11}+A_{12}^TB_{12} & A_{11}^TB_{12}+A_{12}^TB_{22} \\ | |
| − | A_{12}^TB_{11}+A_{22}^TB_{12} & A_{12}^TB_{12}+A_{22}^TB_{22}
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| − | \end{pmatrix} =: C</math>. Putting | |
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| − | : <math>M_{1} := A_{12} \cdot (B_{11} + B_{12})</math>
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| − | : <math>M_{2} := A_{22} \cdot (B_{12} + B_{22})</math>
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| − | : <math>M_{3} := A_{12} \cdot (B_{12} + B_{22})</math>
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| − | : <math>M_{4} := (A_{12} - A_{11})\cdot B_{12}</math>
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| − | : <math>M_{5} := (A_{12} - A_{22})\cdot B_{22}</math>
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| − | : <math>M_{6} := (A_{12} - A_{11})\cdot B_{11}</math>
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| − | gesetzt, so gilt
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| − | : <math>C_{11} = M_{1} - M_{6}</math>
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| − | : <math>C_{12} = M_{3} - M_{4}</math>
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| − | : <math>C_{21} = M_{1} + M_{2} - M_{3} + M_{5}</math>
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| − | : <math>C_{22} = M_{3} - M_{5} .\square</math>
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| − | '''New algorithm for a symmetric matrix:'''
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| − | For a symmetric matrix <math>A \in \mathbb{C}^{n \times n}</math> where <math>n \in \mathbb{N}^*</math> the runtime <math>T_p(n)</math> of the new algorithm is for the matrix product <math>A^2</math> by computing analogously as above about <math>3/7</math> that of the original algorithm in <math>\mathcal{O}(n^{(_2 7)}).\square</math>
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| − | '''New algorithm for a square matrix:'''
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| − | For a square matrix <math>A \in \mathbb{C}^{n \times n}</math> where <math>n \in \mathbb{N}^*</math> the runtime <math>T_r(n)</math> of the new algorithm is for the matrix product <math>A^TA</math> about <math>24/49</math> that of the original algorithm (see above) in <math>\mathcal{O}(n^{(_2 7)}).\square</math>
| + | The [[w:Geometric series|<span class="wikipedia">geometric series</span>]] yields the claim because of the halvings above (see top right).<math>\square</math> |
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| | == See also == | | == See also == |
Strassen algorithm for a symmetric matrix:
For a square matrix [math]\displaystyle{ A \in \mathbb{C}^{n \times n} }[/math] where sufficiently big [math]\displaystyle{ 2^k := n, k \in \mathbb{N}^* }[/math], the runtime [math]\displaystyle{ T_q(n) }[/math] of the Strassen algorithm for the matrix product [math]\displaystyle{ AA^T }[/math] roughly halves that of the original algorithm in [math]\displaystyle{ \mathcal{O}(n^{(_2 7)}) }[/math].
Proof: For [math]\displaystyle{ A :=
\begin{pmatrix}
A_{11} & A_{12} \\
A_{21} & A_{22}
\end{pmatrix} }[/math], it holds that [math]\displaystyle{ AA^T =
\begin{pmatrix}
A_{11}A_{11}^T+A_{12}A_{12}^T & A_{11}A_{21}^T+A_{12}A_{22}^T \\
A_{21}A_{11}^T+A_{22}A_{12}^T & A_{21}A_{21}^T+A_{22}A_{22}^T
\end{pmatrix} }[/math].
The geometric series yields the claim because of the halvings above (see top right).[math]\displaystyle{ \square }[/math]
See also
List of mathematical symbols