Christoph Ruegg
13 years ago
3 changed files with 133 additions and 1 deletions
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(*** hide ***) |
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#I "../../out/lib/net40" |
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#r "MathNet.Numerics.dll" |
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#r "MathNet.Numerics.FSharp.dll" |
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open MathNet.Numerics.LinearAlgebra |
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(** |
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Linear Equation Systems |
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======================= |
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A system of linear equations is a collection of linear equations involving the same set of variables: |
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$$$ |
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\begin{alignat}{7} |
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3x &\; + \;& 2y &\; - \;& z &\; = \;& 1 & \\ |
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2x &\; - \;& 2y &\; + \;& 4z &\; = \;& -2 & \\ |
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-x &\; + \;& \tfrac{1}{2} y &\; - \;& z &\; = \;& 0 & |
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\end{alignat} |
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More generally, we can write |
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$$$ |
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\begin{alignat}{7} |
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a_{11} x_1 &&\; + \;&& a_{12} x_2 &&\; + \cdots + \;&& a_{1n} x_n &&\; = \;&&& b_1 \\ |
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a_{21} x_1 &&\; + \;&& a_{22} x_2 &&\; + \cdots + \;&& a_{2n} x_n &&\; = \;&&& b_2 \\ |
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\vdots\;\;\; && && \vdots\;\;\; && && \vdots\;\;\; && &&& \;\vdots \\ |
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a_{m1} x_1 &&\; + \;&& a_{m2} x_2 &&\; + \cdots + \;&& a_{mn} x_n &&\; = \;&&& b_m \\ |
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\end{alignat} |
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where we all parameters $a_{ij}$ and $b_i$ are known and we would like to find $x_j$ that satisfy |
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all these equations. If we have the same number $n$ of unknown variables $x_j$ as number of |
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equations $m$, and all these equations are independent, then there is a unique solution. |
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This is a fundamental problem in the domain of linear algebra, and we can use its power to find the solution. |
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Accordingly we can write the equivalent problem with matrices and vectors: |
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$$$ |
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\mathbf{A}= |
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\begin{bmatrix} |
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a_{11} & a_{12} & \cdots & a_{1n} \\ |
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a_{21} & a_{22} & \cdots & a_{2n} \\ |
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\vdots & \vdots & \ddots & \vdots \\ |
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a_{m1} & a_{m2} & \cdots & a_{mn} |
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\end{bmatrix},\quad |
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\mathbf{x}=\begin{bmatrix}x_1\\x_2\\ \vdots \\x_n\end{bmatrix},\quad |
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\mathbf{b}=\begin{bmatrix}b_1\\b_2\\ \vdots \\b_m\end{bmatrix} |
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such that |
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$$$ |
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\mathbf{A}\mathbf{x}=\mathbf{b} |
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The initial example system would then look like this: |
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$$$ |
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\begin{bmatrix}3 & 2 & -1 \\2 & -2 & 4 \\-1 & \tfrac{1}{2} & -1\end{bmatrix} |
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\begin{bmatrix}x\\y\\z\end{bmatrix} |
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\;=\; |
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\begin{bmatrix}1\\-2\\0\end{bmatrix} |
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Which we can solve explicitly with the LU-decomposition, or simply by using the Solve method: |
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[lang=csharp] |
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var A = Matrix<double>.Build.DenseOfArray(new double[,] { |
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{ 3, 2, -1 }, |
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{ 2, -2, 4 }, |
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{ -1, 0.5, -1 } |
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}); |
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var b = Vector<double>.Build.Dense(new double[] { 1, -2, 0 }); |
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var x = A.Solve(b); |
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The resulting $\mathbf{x}$ is $[1,\;-2,\;-2]$, hence the solution $x=1,\;y=-2,\;z=-2$. |
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In F# the syntax is a bit lighter: |
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*) |
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let A = matrix [[ 3.0; 2.0; -1.0 ] |
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[ 2.0; -2.0; 4.0 ] |
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[ -1.0; 0.5; -1.0 ]] |
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let b = vector [ 1.0; -2.0; 0.0 ] |
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let x = A.Solve(b) // 1;-2;-2 |
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(** |
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Normalizing Equation Systems |
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---------------------------- |
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In practice, a linear equation system to be solved is often not in the standard form required |
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to use the linear algebra approach. For example, let's have a look at the following system: |
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$$$ |
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\begin{bmatrix}1 & 2 & 3 & 4\\2 & 3 & 4 & 5\\3 & 4 & 5 & 6\\4 & 5 & 6 & 7\end{bmatrix} |
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\begin{bmatrix}0\\0\\V\\T\end{bmatrix} |
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\;=\; |
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\begin{bmatrix}F\\M\\20\\0\end{bmatrix} |
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The first two values of the solution vector $[0,\;0,\;V,\;T]$ are constant zero, so we can simplify |
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the system to: |
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$$$ |
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\begin{bmatrix}3 & 4\\4 & 5\\5 & 6\\6 & 7\end{bmatrix} |
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\begin{bmatrix}V\\T\end{bmatrix} |
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\;=\; |
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\begin{bmatrix}F\\M\\20\\0\end{bmatrix} |
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Then we need to subtract the two unknowns from the right side back from the left (so that they |
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become zero on the right side), by introducing a new column each. First we subtract |
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$[F,\;0,\;0,\;0]^T$ from both sides: |
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$$$ |
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\begin{bmatrix}3 & 4 & -1\\4 & 5 & 0\\5 & 6 & 0\\6 & 7 & 0\end{bmatrix} |
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\begin{bmatrix}V\\T\\F\end{bmatrix} |
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\;=\; |
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\begin{bmatrix}0\\M\\20\\0\end{bmatrix} |
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Then we subtract $[0,\;M,\;0,\;0]^T$ from both sides the same way: |
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$$$ |
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\begin{bmatrix}3 & 4 & -1 & 0\\4 & 5 & 0 & -1\\5 & 6 & 0 & 0\\6 & 7 & 0 & 0\end{bmatrix} |
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\begin{bmatrix}V\\T\\F\\M\end{bmatrix} |
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\;=\; |
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\begin{bmatrix}0\\0\\20\\0\end{bmatrix} |
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Which is in standard from, so we can solve normally: |
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*) |
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let A' = matrix [[ 3.0; 4.0; -1.0; 0.0 ] |
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[ 4.0; 5.0; 0.0; -1.0 ] |
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[ 5.0; 6.0; 0.0; 0.0; ] |
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[ 6.0; 7.0; 0.0; 0.0 ]] |
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let b' = vector [ 0.0; 0.0; 20.0; 0.0 ] |
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let x' = A'.Solve(b') // -140; 120; 60; 40 |
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