From c919d7e1f0da6c9558b5feb995ad66f065498a69 Mon Sep 17 00:00:00 2001
From: =?UTF-8?q?Andreas=20M=C3=BCller?= <andreas.mueller@ost.ch>
Date: Fri, 10 Sep 2021 18:06:24 +0200
Subject: more formatting perfection

---
 buch/papers/reedsolomon/codebsp.tex | 25 +++++++++++++------------
 1 file changed, 13 insertions(+), 12 deletions(-)

(limited to 'buch/papers/reedsolomon/codebsp.tex')

diff --git a/buch/papers/reedsolomon/codebsp.tex b/buch/papers/reedsolomon/codebsp.tex
index 5d3daa5..037fba7 100644
--- a/buch/papers/reedsolomon/codebsp.tex
+++ b/buch/papers/reedsolomon/codebsp.tex
@@ -119,17 +119,18 @@ in die von $a$ abhängige Schreibweise
 \index{Einheitswurzel, primitiv}%
 Wenn wir jetzt Zahlen von $\mathbb{F}_{11}$ an Stelle von $a$ einsetzen, erhalten wir
 \begin{center}
-\begin{tabular}{c c c c c c c}
-$a = 1$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 1, 1, 1, 1, 1, 1, 1, 1, 1\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 2$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 2, 4, 8, 5, 10, 9, 7, 3, 6\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 3$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 3, 9, 5, 4, 1, 3, 9, 5, 4\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 4$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 4, 5, 9, 3, 1, 4, 5, 9, 3\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 5$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 5, 3, 4, 9, 1, 5, 3, 4, 9\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 6$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 6, 3, 7, 9, 10, 5, 8, 4, 2\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 7$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 7, 5, 2, 3, 10, 4, 6, 9, 8\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 8$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 8, 9, 6, 4, 10, 3, 2, 5, 7\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 9$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 9, 4, 3, 5, 1, 9, 4, 3, 5\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
-$a = 10$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 10, 1, 10, 1, 10, 1, 10, 1, 10\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$. \\
+\def\s{\phantom{0}}
+\begin{tabular}{c c c c c c l}
+$a = \s1$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s1,\s1,\s1,\s1,\s1,\s1,\s1,\s1,\s1\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s2$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s2,\s4,\s8,\s5,10,\s9,\s7,\s3,\s6\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s3$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s3,\s9,\s5,\s4,\s1,\s3,\s9,\s5,\s4\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s4$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s4,\s5,\s9,\s3,\s1,\s4,\s5,\s9,\s3\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s5$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s5,\s3,\s4,\s9,\s1,\s5,\s3,\s4,\s9\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s6$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s6,\s3,\s7,\s9, 10,\s5,\s8,\s4,\s2\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s7$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s7,\s5,\s2,\s3,10,\s4,\s6,\s9,\s8\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s8$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s8,\s9,\s6,\s4, 10,\s3,\s2,\s5,\s7\}$ & $ = $ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = \s9$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1,\s9,\s4,\s3,\s5,\s1,\s9,\s4,\s3,\s5\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$ \\
+$a = 10$ & $\Rightarrow$ & $\{a^i | 0 \le i \le 10\}$ & $=$ & $\{1, 10,\s1,10,\s1, 10,\s1, 10,\s1, 10\}$ & $\neq$ & $\mathbb{F}_{11}\setminus\{0\}$. \\
 \end{tabular}
 \end{center}
 %\begin{center}
@@ -164,7 +165,7 @@ Für die Codierung setzen wir alle Zahlen in $\mathbb{F}_{11}\setminus\{0\}$ nac
 	\begin{tabular}{c}
 		$m(8^0) = 4 \cdot 1^5 + 7 \cdot 1^4 + 2 \cdot 1^3 + 5 \cdot 1^2 + 8 \cdot 1^1 + 1 = 5$ \\
 		$m(8^1) = 4 \cdot 8^5 + 7 \cdot 8^4 + 2 \cdot 8^3 + 5 \cdot 8^2 + 8 \cdot 8^1 + 1 = 3$ \\
-		\vdots \\
+		\vdots \\[5pt]
 		$m(8^9) = 4 \cdot 7^5 + 7 \cdot 7^4 + 2 \cdot 7^3 + 5 \cdot 7^2 + 8 \cdot 7^1 + 1 = 4$
 	\end{tabular}
 \end{center}
-- 
cgit v1.2.1

