add new stackrel-compatible-auto-inserting-padding-environment

ana8.tex

@@ -15,28 +15,28 @@
 
 \begin{proof}
     Sei $x \in \mathbb{K}^{n}$. Dann ist
-    \begin{align*}
-        \Vert (\mathbb{I} + B) x \Vert \qquad
-        &= \qquad \Vert x + B x\Vert \\
-        &\stackrel{\text{Dreiecksungl.}}{\ge } \qquad \Vert x \Vert - \Vert Bx \Vert \\
+    \begin{salign}
+        \Vert (\mathbb{I} + B) x \Vert 
+        &=  \Vert x + B x\Vert \\
+        &\stackrel{\text{Dreiecksungl.}}{\ge }  \Vert x \Vert - \Vert Bx \Vert \\
         &\stackrel{\Vert Bx \Vert \le \Vert B \Vert \Vert x \Vert}{\ge } 
-        \qquad \Vert x \Vert - \Vert B \Vert \cdot \Vert x \Vert \\
-        &= \qquad ( \underbrace{1  - \Vert B \Vert}_{> 0}) \Vert x \Vert
-    .\end{align*}
+         \Vert x \Vert - \Vert B \Vert \cdot \Vert x \Vert \\
+        &=  ( \underbrace{1  - \Vert B \Vert}_{> 0}) \Vert x \Vert
+    .\end{salign}
     Also hat die Gleichung $(\mathbb{I} + B) x = 0$ nur die Lösung $x = 0$, also
     ist $(\mathbb{I} + B)$ injektiv und mit \ref{lemma:linabb} regulär.
 
     Bleibt zu zeigen: $\Vert (\mathbb{I} + B)^{-1} \Vert \le \frac{1}{1 - \Vert B \Vert}$.
     Es gilt
-    \begin{align*}
-        1 \qquad &= \qquad \Vert \mathbb{I}\Vert \\
-        &= \qquad \Vert (\mathbb{I} + B) (\mathbb{I} + B)^{-1} \Vert \\
-        &= \qquad \Vert (\mathbb{I} + B)^{-1} + B (\mathbb{I} + B)^{-1} \Vert \\
-        &\stackrel{\text{Dreicksungl.}}{\ge } \qquad \Vert (\mathbb{I} + B)^{-1} \Vert
+    \begin{salign}
+        1 &=  \Vert \mathbb{I}\Vert \\
+        &=  \Vert (\mathbb{I} + B) (\mathbb{I} + B)^{-1} \Vert \\
+        &=  \Vert (\mathbb{I} + B)^{-1} + B (\mathbb{I} + B)^{-1} \Vert \\
+        &\stackrel{\text{Dreicksungl.}}{\ge }  \Vert (\mathbb{I} + B)^{-1} \Vert
         - \Vert B (\mathbb{I} + B)^{-1} \Vert \\
-        &\ge \qquad \Vert (\mathbb{I} + B)^{-1} \Vert - \Vert B \Vert \cdot \Vert (\mathbb{I} + B)^{-1} \Vert \\
-        &= \qquad (1 - \Vert B \Vert) \Vert (\mathbb{I} + B)^{-1} \Vert
-    .\end{align*}
+        &\ge  \Vert (\mathbb{I} + B)^{-1} \Vert - \Vert B \Vert \cdot \Vert (\mathbb{I} + B)^{-1} \Vert \\
+        &=  (1 - \Vert B \Vert) \Vert (\mathbb{I} + B)^{-1} \Vert
+    .\end{salign}
     Damit folgt die Behauptung.
 \end{proof}
 

lecture.cls

@@ -24,6 +24,7 @@
 \RequirePackage[hidelinks, unicode]{hyperref} %[unicode, hidelinks]{hyperref}
 \RequirePackage{bookmark}
 \RequirePackage{wasysym}
+\RequirePackage{environ}
 
 \usetikzlibrary{quotes, angles}
 
@@ -153,3 +154,66 @@
 
 % people seem to prefer varepsilon over epsilon
 \renewcommand{\epsilon}{\varepsilon}
+
+\ExplSyntaxOn
+
+% S-tackrelcompatible ALIGN environment
+% some might also call it the S-uper ALIGN environment
+% uses regular expressions to calculate the widest stackrel
+% to put additional padding on both sides of relation symbols
+\NewEnviron{salign}
+{
+    \begin{align*}
+        \lec_insert_padding:V \BODY
+    .\end{align*}
+}
+
+% some helper variables
+\tl_new:N \l__lec_text_tl
+\seq_new:N \l_lec_stackrels_seq
+\int_new:N \l_stackrel_count_int
+\int_new:N \l_idx_int
+\box_new:N \l_tmp_box
+\dim_new:N \l_tmp_dim_a
+\dim_new:N \l_tmp_dim_b
+\dim_new:N \l_tmp_dim_needed
+
+% function to insert padding according to widest stackrel
+\cs_new_protected:Nn \lec_insert_padding:n
+ {
+  \tl_set:Nn \l__lec_text_tl { #1 }
+  % get all stackrels in this align environment
+  \regex_extract_all:nnN { \c{stackrel}{(.*?)}{(.*?)} } { #1 } \l_lec_stackrels_seq
+  % get number of stackrels
+  \int_set:Nn \l_stackrel_count_int { \seq_count:N \l_lec_stackrels_seq }
+  \int_set:Nn \l_idx_int { 1 }
+  \dim_set:Nn \l_tmp_dim_needed { 0pt }
+  % iterate over stackrels
+  \int_while_do:nn { \l_idx_int <= \l_stackrel_count_int }
+  {
+      % calculate width of text
+      \hbox_set:Nn \l_tmp_box {$\seq_item:Nn \l_lec_stackrels_seq { \l_idx_int + 1 }$}
+      \dim_set:Nn \l_tmp_dim_a {\box_wd:N \l_tmp_box}
+      % calculate width of relation symbol
+      \hbox_set:Nn \l_tmp_box {$\seq_item:Nn \l_lec_stackrels_seq { \l_idx_int + 2 }$}
+      \dim_set:Nn \l_tmp_dim_b {\box_wd:N \l_tmp_box}
+      % check if 0.5*(a-b) > minimum padding, if yes updated minimum padding
+      \dim_compare:nNnTF
+        { 1pt * \dim_ratio:nn { \l_tmp_dim_a - \l_tmp_dim_b } { 2pt } } > { \l_tmp_dim_needed }
+        { \dim_set:Nn \l_tmp_dim_needed { 1pt * \dim_ratio:nn { \l_tmp_dim_a - \l_tmp_dim_b } { 2pt } } }
+        { }
+      \quad
+      % increment list index by three, as every stackrel produces three list entries
+      \int_incr:N \l_idx_int
+      \int_incr:N \l_idx_int
+      \int_incr:N \l_idx_int
+  }
+  % replace all relations with align characters (&) and add the needed padding
+  \regex_replace_all:nnN
+      { (&=|&\c{le}|&\c{ge}|&\c{stackrel}{.*?}{.*?}|&\c{neq}) }
+      { \c{kern} \u{l_tmp_dim_needed} \1 \c{kern} \u{l_tmp_dim_needed} }
+      \l__lec_text_tl
+  \l__lec_text_tl
+ }
+\cs_generate_variant:Nn \lec_insert_padding:n { V }
+\ExplSyntaxOff