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fixed typo
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2 changed files with 30 additions and 18 deletions
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@ -8,28 +8,40 @@ wobei $L$ eine invertierbare, untere Dreiecksmatrix ist.
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Geben Sie die Formel zur Berechnung von $y_i$ an.
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Geben Sie die Formel zur Berechnung von $y_i$ an.
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\textbf{Lösung:}
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\textbf{Lösung:}
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\[y_i = \frac{b_i - \sum_{k=1}^{i-1} l_{ik} \cdot y_k}{l_{ii}}\]
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\[y_i = \frac{b_i - \sum_{k=i}^{i-1} l_{ik} \cdot y_k}{l_{ii}}\]
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\begin{algorithm}[H]
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\begin{algorithmic}
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\Require Lower, invertable, triangular Matrix $L \in \mathbb{R}^{n \times n}$, Vektor $b$
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\Procedure{solve}{$L$, $b$}
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\For{$i \in \Set{1, \dots n}$}
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\State $y_i \gets b_i$
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\For{$k \in \Set{1, \dots, i-1}$}
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\State $y_i \gets y_i - l_{ik} \cdot y_k$
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\EndFor
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\State $y_i \gets \frac{y_i}{l_{ii}}$
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\EndFor
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\EndProcedure
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\end{algorithmic}
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\caption{Calculate $y$ in $Ly = b$}
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\end{algorithm}
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\subsection*{Teilaufgabe b}
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\subsection*{Teilaufgabe b}
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\[Ax = b ? PAx = Pb ? LRx = Pb \]
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\[Ax = b \Leftrightarrow PAx = Pb \Leftrightarrow LRx = Pb \]
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Pseudocode:
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\begin{algorithm}[H]
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\begin{algorithmic}
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\begin{algorithm}[H]
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\Require Matrix $A$, Vektor $b$
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\begin{algorithmic}
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\Procedure{LoeseLGS}{$A$, $b$}
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\Require Matrix $A$, Vektor $b$
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\State $P, L, R \gets \Call{LRZer}{A}$
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\Procedure{CalculateLegendre}{$A$, $b$}
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\State $b^* \gets Pb$
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\State $P, L, R \gets \Call{LRZerlegung}{A}$
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\State $c \gets \Call{VorSub}{L, b^*}$
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\State $b^* \gets Pb$
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\State $x \gets \Call{RueckSub}{R, c}$
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\State $c \gets \Call{VorwärtsSubstitution}{L, b^*}$
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\State \Return $x$
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\State $x \gets \Call{RückwärtsSubstitution}{R, c}$
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\EndProcedure
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\State \Return $x$
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\end{algorithmic}
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\EndProcedure
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\caption{Löse ein LGS $Ax = b$}
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\end{algorithmic}
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\end{algorithm}
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\caption{Calculate TODO}
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\label{alg:TODO}
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\end{algorithm}
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\subsection*{Teilaufgabe c}
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\subsection*{Teilaufgabe c}
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Der Gesamtaufwand ist:
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Der Gesamtaufwand ist:
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