edits to reg. alloc

book.tex

@@ -5493,6 +5493,8 @@ locations before instruction $I_k$.  \racket{We recommend representing
 \end{figure}
 \fi}
 
+% TODO: add a python version of the reference box for sets. -Jeremy
+
 The locations that are live after an instruction are its
 \emph{live-after}\index{subject}{live-after} set, and the locations
 that are live before an instruction are its
@@ -5905,7 +5907,7 @@ registers in the interference graph.
 \fi}
 {\if\edition\pythonEd\pythonColor
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}$};
 \node (z) at (3,2)  {$\ttm{z}$};
@@ -6103,7 +6105,7 @@ register, obtaining the following annotated graph. For example, the
 saturation for \code{t} is $\{-1,-2\}$ because it interferes with both
 \code{rax} and \code{rsp}.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1\}$};
 \node (t1) at (0,2) {$\ttm{t}:-,\{-1,-2\}$};
@@ -6135,7 +6137,7 @@ $\ttm{t}$ and color it with the first available integer, which is
 $0$. We mark $0$ as no longer available for $\ttm{z}$, $\ttm{rax}$,
 and \ttm{rsp} because they interfere with $\ttm{t}$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1,0\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{-1,-2\}$};
@@ -6167,7 +6169,7 @@ choosing \code{z}, and coloring it with the first available number, which
 is $1$. We add $1$ to the saturation for the neighboring vertices
 \code{t}, \code{y}, \code{w}, and \code{rsp}.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1,0,1\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{-1,1,-2\}$};
@@ -6197,7 +6199,7 @@ is $1$. We add $1$ to the saturation for the neighboring vertices
 The most saturated vertices are now \code{w} and \code{y}. We color
 \code{w} with the first available color, which is $0$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1,0,1\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{-1,1,-2\}$};
@@ -6229,7 +6231,7 @@ with $2$.  We cannot choose $0$ or $1$ because those numbers are in
 \code{y}'s saturation set. Indeed, \code{y} interferes with \code{w}
 and \code{z}, whose colors are $0$ and $1$ respectively.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{-1,1,-2\}$};
@@ -6258,7 +6260,7 @@ and \code{z}, whose colors are $0$ and $1$ respectively.
 \]
 Now \code{x} and \code{v} are the most saturated, so we color \code{v} with $1$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{-1,1,-2\}$};
@@ -6287,7 +6289,7 @@ Now \code{x} and \code{v} are the most saturated, so we color \code{v} with $1$.
 \]
 In the last step of the algorithm, we color \code{x} with $1$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (10,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{-1,1,-2\}$};
@@ -6333,19 +6335,15 @@ So, we obtain the following coloring:
 %
 With the DSATUR algorithm in hand, let us return to the running
 example and consider how to color the interference graph shown in
-figure~\ref{fig:interfere}, again mapping 1 to blank, 2 to white, and
-3 to gray. We annotate each variable node with a dash to indicate that
-it has not yet been assigned a color.  Each register node (not shown)
-should be assigned the number that the register corresponds to, for
-example, color \code{rcx} with the number \code{0} and \code{rdx} with
-\code{1}. The saturation sets are also shown for each node; all of
-them start as the empty set.  We do not show the register nodes in the
-following graph because there were no interference edges involving
-registers in this program; however, in general there can be inference
-edges that involve registers.
+figure~\ref{fig:interfere}. We annotate each variable node with a dash
+to indicate that it has not yet been assigned a color.  Each register
+node (not shown) should be assigned the number that the register
+corresponds to, for example, color \code{rcx} with the number \code{0}
+and \code{rdx} with \code{1}. The saturation sets are also shown for
+each node; all of them start as the empty set. 
 %
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: -, \{\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{\}$};
 \node (z) at (3,2)  {$\ttm{z}: -, \{\}$};
@@ -6369,7 +6367,7 @@ and then we color it with the first available integer, which is $0$. We mark
 $0$ as no longer available for $\ttm{tmp\_1}$ and $\ttm{z}$ because
 they interfere with $\ttm{tmp\_0}$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: -, \{0\}$};
@@ -6392,7 +6390,7 @@ We repeat the process. The most saturated vertices are \code{z} and
 available number, which is $1$. We add $1$ to the saturation for the
 neighboring vertices \code{tmp\_0}, \code{y}, and \code{w}.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0\}$};
@@ -6414,7 +6412,7 @@ The most saturated vertices are now \code{tmp\_1}, \code{w}, and
 \code{y}. We color \code{w} with the first available color, which
 is $0$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0\}$};
@@ -6434,7 +6432,7 @@ is $0$.
 \]
 Now \code{y} is the most saturated, so we color it with $2$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0,2\}$};
@@ -6455,7 +6453,7 @@ Now \code{y} is the most saturated, so we color it with $2$.
 The most saturated vertices are \code{tmp\_1}, \code{x}, and \code{v}.
 We choose to color \code{v} with $1$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0,2\}$};
@@ -6475,7 +6473,7 @@ We choose to color \code{v} with $1$.
 \]
 We color the remaining two variables, \code{tmp\_1} and \code{x}, with $1$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: 1, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0,2\}$};
@@ -6556,8 +6554,7 @@ With the coloring complete, we finalize the assignment of variables to
 registers and stack locations. We map the first $k$ colors to the $k$
 registers and the rest of the colors to stack locations.  Suppose for
 the moment that we have just one register to use for register
-allocation, \key{rcx}. Then we have the following map from colors to
-locations.
+allocation, \key{rcx}. Then we have the following assignment.
 \[
   \{ 0 \mapsto \key{\%rcx}, \; 1 \mapsto \key{-8(\%rbp)}, \; 2 \mapsto \key{-16(\%rbp)} \}
 \]
@@ -6589,7 +6586,6 @@ Adapt the code from the \code{assign\_homes} pass
 (section~\ref{sec:assign-Lvar}) to replace the variables with their
 assigned location. Applying this assignment to our running
 example shown next, on the left, yields the program on the right.
-% why frame size of 32? -JGS
 \begin{center}
 {\if\edition\racketEd      
 \begin{minipage}{0.35\textwidth}
@@ -6834,7 +6830,7 @@ and \code{pushq} subtracts $8$ from the \code{rsp}.
 {\if\edition\racketEd      
 \begin{figure}[tbp]
 \begin{tcolorbox}[colback=white]
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (Lvar) at (0,2)  {\large \LangVar{}};
 \node (Lvar-2) at (3,2)  {\large \LangVar{}};
 \node (Lvar-3) at (7,2)  {\large \LangVarANF{}};
@@ -7077,10 +7073,10 @@ two stack locations.
 
 We recommend representing the move relationships in a graph, similarly
 to how we represented interference.  The following is the \emph{move
-  graph} for our running example.
+  graph} for our example.
 {\if\edition\racketEd      
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}$};
 \node (rsp) at (9,2) {$\ttm{rsp}$};
 \node (t) at (0,2) {$\ttm{t}$};
@@ -7100,7 +7096,7 @@ to how we represented interference.  The following is the \emph{move
 %
 {\if\edition\pythonEd\pythonColor
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}$};
 \node (z) at (3,2)  {$\ttm{z}$};
@@ -7123,7 +7119,7 @@ Now we replay the graph coloring, pausing to see the coloring of
 \code{y}. Recall the following configuration. The most saturated vertices
 were \code{w} and \code{y}.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (9,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{1,-2\}$};
@@ -7156,7 +7152,7 @@ that \code{w} is not move-related to any vertex, but \code{y} is
 move-related to \code{t}.  So we choose to color \code{y} with $0$,
 the same color as \code{t}.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (9,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{1,-2\}$};
@@ -7185,7 +7181,7 @@ the same color as \code{t}.
 \]
 Now \code{w} is the most saturated, so we color it $2$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (9,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t1) at (0,2) {$\ttm{t}:0,\{1,-2\}$};
@@ -7216,7 +7212,7 @@ At this point, vertices \code{x} and \code{v} are most saturated, but
 \code{x} is move related to \code{y} and \code{z}, so we color
 \code{x} to $0$ to match \code{y}. Finally, we color \code{v} to $0$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (rax) at (0,0) {$\ttm{rax}:-1,\{0,-2\}$};
 \node (rsp) at (9,2) {$\ttm{rsp}:-2,\{-1,0,1,2\}$};
 \node (t) at (0,2) {$\ttm{t}:0,\{1,-2\}$};
@@ -7250,7 +7246,7 @@ Now we replay the graph coloring, pausing before the coloring of
 \code{w}. Recall the following configuration. The most saturated vertices
 were \code{tmp\_1}, \code{w}, and \code{y}.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0\}$};
@@ -7274,7 +7270,7 @@ variables, whereas \code{y} and \code{tmp\_1} are move related to
 \code{tmp\_0} and \code{z}, respectively. If we instead choose
 \code{y} and color it $0$, we can delete another move instruction.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0\}$};
@@ -7294,7 +7290,7 @@ variables, whereas \code{y} and \code{tmp\_1} are move related to
 \]
 Now \code{w} is the most saturated, so we color it $2$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: -, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0\}$};
@@ -7315,7 +7311,7 @@ Now \code{w} is the most saturated, so we color it $2$.
 To finish the coloring, \code{x} and \code{v} get $0$ and
 \code{tmp\_1} gets $1$.
 \[
-\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\begin{tikzpicture}[baseline=(current  bounding  box.center),scale=0.9]
 \node (t0) at (0,2) {$\ttm{tmp\_0}: 0, \{1\}$};
 \node (t1) at (0,0) {$\ttm{tmp\_1}: 1, \{0\}$};
 \node (z) at (3,2)  {$\ttm{z}: 1, \{0\}$};