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| | \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> 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> |
| | | | |
| − | '''Theorem (decomposition method):'''
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| − |
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| − | For <math>A, B, D, E \in \mathbb{C}^{n \times n}</math> where <math>2^k := n, k \in \mathbb{N}^*</math>, the runtime is <math>T_z(n) = \mathcal{O}(n^{(2^6)})</math> for the matrix product <math>QR</math> if <math>A</math> has also the following form of <math>B</math>:
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| − |
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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>AB =
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| − | \begin{pmatrix}
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| − | A_{11}B_{11}+A_{12}B_{12} & A_{11}B_{12}+A_{12}B_{22} \\
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| − | A_{12}B_{11}+A_{22}B_{12} & A_{12}B_{12}+A_{22}B_{22}
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| − | \end{pmatrix} =: C</math>. Putting
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| − |
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| − | : <math>M_{1} := A_{12} \cdot (B_{11} + B_{12})</math>
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| − | : <math>M_{2} := A_{12} \cdot (B_{12} + B_{22})</math>
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| − | : <math>M_{3} := A_{22} \cdot (B_{12} + B_{22})</math>
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| − | : <math>M_{4} := (A_{11} - A_{12})\cdot B_{11}</math>
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| − | : <math>M_{5} := (A_{11} - A_{12})\cdot B_{12}</math>
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| − | : <math>M_{6} := (A_{12} - A_{22})\cdot B_{22}</math>
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| − |
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| − | implies
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| − |
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| − | : <math>C_{11} = M_{1} + M_{4}</math>
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| − | : <math>C_{12} = M_{2} + M_{5}</math>
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| − | : <math>C_{21} = M_{1} - C_{22} + M_{3}</math>
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| − | : <math>C_{22} = M_{2} - M_{6} .</math>
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| − |
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| − | Let <math>D</math> have the following form of <math>E</math>:
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| − |
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| − | For <math>E :=
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| − | \begin{pmatrix}
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| − | E_{11} & E_{12} \\
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| − | E_{21} & E_{11}
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| − | \end{pmatrix}</math>, it holds that <math>DE =
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| − | \begin{pmatrix}
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| − | D_{11}E_{11}+D_{12}E_{21} & D_{11}E_{12}+D_{12}E_{11} \\
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| − | D_{21}E_{11}+D_{11}E_{21} & D_{21}E_{12}+D_{11}E_{11}
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| − | \end{pmatrix} =: F</math>. Putting
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| − |
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| − | : <math>N_{1} := D_{11} \cdot (E_{11} + E_{12})</math>
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| − | : <math>N_{2} := D_{21} \cdot (E_{11} + E_{12})</math>
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| − | : <math>N_{3} := D_{12} \cdot (E_{11} - E_{21})</math>
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| − | : <math>N_{4} := (D_{11} - D_{12})\cdot E_{11}</math>
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| − | : <math>N_{5} := (D_{11} - D_{21})\cdot E_{11}</math>
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| − | : <math>N_{6} := (D_{11} + D_{12})\cdot E_{21}</math>
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| − |
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| − | implies
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| − |
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| − | : <math>F_{11} = N_{3} + N_{4}</math>
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| − | : <math>F_{12} = N_{1} - N_{4}</math>
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| − | : <math>F_{21} = F_{11} - N_{5} + N_{6}</math>
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| − | : <math>F_{22} = N_{2} + N_{5} .</math>
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| − |
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| − | Then it holds for <math>D</math> and <math>B</math>:
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| − |
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| − | <math>DB =
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| − | \begin{pmatrix}
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| − | D_{11}B_{11}+D_{12}B_{12} & D_{11}B_{12}+D_{12}B_{22} \\
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| − | D_{21}B_{11}+D_{11}B_{12} & D_{21}B_{12}+D_{11}B_{22}
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| − | \end{pmatrix} =: G</math>. Putting
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| − |
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| − | : <math>O_{1} := D_{12} \cdot (B_{12} + B_{22})</math>
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| − | : <math>O_{2} := D_{11} \cdot (B_{12} + B_{22})</math>
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| − | : <math>O_{3} := D_{21} \cdot (B_{11} + B_{12})</math>
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| − | : <math>O_{4} := (D_{11} - D_{12})\cdot B_{22}</math>
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| − | : <math>O_{5} := (D_{11} - D_{21})\cdot B_{12}</math>
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| − | : <math>O_{6} := (D_{11} - D_{21})\cdot B_{11}</math>
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| − |
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| − | implies
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| − |
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| − | : <math>G_{11} = O_{1} - G_{12} + G_{21} + O_{6}</math>
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| − | : <math>G_{12} = O_{2} - O_{4}</math>
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| − | : <math>G_{21} = O_{3} + O_{5}</math>
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| − | : <math>G_{22} = O_{2} - O_{5} .</math>
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| − |
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| − | Then it holds for <math>A</math> and <math>E</math>:
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| − |
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| − | <math>AE =
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| − | \begin{pmatrix}
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| − | A_{11}E_{11}+A_{12}E_{21} & A_{11}E_{12}+A_{12}E_{11} \\
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| − | A_{12}E_{11}+A_{22}E_{21} & A_{12}E_{12}+A_{22}E_{11}
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| − | \end{pmatrix} =: H</math>. Putting
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| − |
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| − | : <math>P_{1} := A_{12} \cdot (E_{11} - E_{21})</math>
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| − | : <math>P_{2} := A_{11} \cdot (E_{11} - E_{12})</math>
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| − | : <math>P_{3} := A_{22} \cdot (E_{11} - E_{21})</math>
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| − | : <math>P_{4} := (A_{11} + A_{12})\cdot E_{11}</math>
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| − | : <math>P_{5} := (A_{11} + A_{12})\cdot E_{12}</math>
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| − | : <math>P_{6} := (A_{12} + A_{22})\cdot E_{21}</math>
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| − |
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| − | implies
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| − |
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| − | : <math>H_{11} = P_{4} - P_{1}</math>
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| − | : <math>H_{12} = P_{4} - P_{2}</math>
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| − | : <math>H_{21} = P_{1} + P_{6}</math>
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| − | : <math>H_{22} = H_{21} - H_{12} + P_{3} + P_{5} .</math>
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| − |
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| − | Since <math>QR</math> may be decomposed as follows, the claim is true:
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| − |
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| − | : <math>QR=\left(\left(\begin{matrix}Q_{11}&Q_{12}\\Q_{21}&Q_{11}\\\end{matrix}\right)-\left(\begin{matrix}0&0\\0&{Q_{11}-Q}_{22}\\\end{matrix}\right)\right)\left(\left(\begin{matrix}R_{22}&R_{12}\\R_{21}&R_{22}\\\end{matrix}\right)-\left(\begin{matrix}{R_{22}-R}_{11}&0\\0&0\\\end{matrix}\right)\right).\square</math>
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| | == See also == | | == See also == |
| | [[List of mathematical symbols]] | | [[List of mathematical symbols]] |
Strassen algorithm for a symmetric matrix:
For a symmetric matrix [math]\displaystyle{ A \in \mathbb{C}^{n \times n} }[/math] where [math]\displaystyle{ 2^k := n, k \in \mathbb{N}^* }[/math], the runtime [math]\displaystyle{ T_s(n) }[/math] of the Strassen algorithm for the matrix product [math]\displaystyle{ A^2 }[/math] is about half 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_{12}^T & A_{22}
\end{pmatrix} }[/math], it holds that [math]\displaystyle{ A^TA =
\begin{pmatrix}
A_{11}A_{11}+A_{12}A_{12}^T & A_{11}A_{12}+A_{12}A_{22} \\
A_{12}^TA_{11}+A_{22}A_{12}^T & A_{12}^TA_{12}+A_{22}A_{22}
\end{pmatrix} }[/math] and [math]\displaystyle{ T_s(2n) = 3T_s(n) + 2n^{(_2 7)} }[/math]. Thus [math]\displaystyle{ T_s(n) = 3T_s(n/2) + 2(n/2)^{(_2 7)} }[/math] and [math]\displaystyle{ T_s(n/2) = 3T_s(n/4) + 2(n/4)^{(_2 7)} }[/math].
The geometric series yields because of [math]\displaystyle{ T_s(1) = 1 }[/math]: [math]\displaystyle{ 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]\displaystyle{ = n^{(_2 3)} + \hat{2}(n^{(_2 7)}-n^{(_2 3)}) = \hat{2} (n^{(_2 3)} + n^{(_2 7)}) }[/math].[math]\displaystyle{ \square }[/math]
Strassen algorithm for a square matrix:
For a square matrix [math]\displaystyle{ A \in \mathbb{C}^{n \times n} }[/math] where [math]\displaystyle{ 2^k := n, k \in \mathbb{N}^* }[/math], the runtime [math]\displaystyle{ T_q(n) }[/math] of the Strassen algorithm is for the matrix product [math]\displaystyle{ A^TA }[/math] about [math]\displaystyle{ 4/7 }[/math] 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{ A^TA =
\begin{pmatrix}
A_{11}^TA_{11}+A_{21}^TA_{21} & A_{11}^TA_{12}+A_{21}^TA_{22} \\
A_{12}^TA_{11}+A_{22}^TA_{21} & A_{12}^TA_{12}+A_{22}^TA_{22}
\end{pmatrix} }[/math] such that [math]\displaystyle{ T_q(2n) = 4T_s(n) + 2n^{(_2 7)} }[/math] and [math]\displaystyle{ T_q(n) = 4T_s(n/2) + 2/7n^{(_2 7)} = 2/3n^{(_2 3)} + 4/7n^{(_2 7)} }[/math].[math]\displaystyle{ \square }[/math]
See also
List of mathematical symbols