progress

book.tex

@@ -2660,8 +2660,8 @@ with $t.1$.
 \node (w) at (3,0)    {$w:-,\{\}$};
 \node (x) at (6,0)    {$x:-,\{\}$};
 \node (y) at (3,-1.5) {$y:-,\{\}$};
-\node (z) at (6,-1.5) {$z:-,\{0\}$};
-\node (t1) at (9,-1.5)   {$t.1:0,\{\}$};
+\node (z) at (6,-1.5) {$z:-,\{\mathbf{0}\}$};
+\node (t1) at (9,-1.5)   {$t.1:\mathbf{0},\{\}$};
 \draw (v) to (w);
 \foreach \i in {w,x,y}
 {
@@ -2680,11 +2680,11 @@ vertex, which in this case is $z$. We color $z$ with $1$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
 \node (v) at (0,0)    {$v:-,\{\}$};
-\node (w) at (3,0)    {$w:-,\{1\}$};
+\node (w) at (3,0)    {$w:-,\{\mathbf{1}\}$};
 \node (x) at (6,0)    {$x:-,\{\}$};
-\node (y) at (3,-1.5) {$y:-,\{1\}$};
-\node (z) at (6,-1.5) {$z:1,\{0\}$};
-\node (t1) at (9,-1.5)   {$t.1:0,\{\}$};
+\node (y) at (3,-1.5) {$y:-,\{\mathbf{1}\}$};
+\node (z) at (6,-1.5) {$z:\mathbf{1},\{0\}$};
+\node (t1) at (9,-1.5)   {$t.1:0,\{\mathbf{1}\}$};
 \draw (t1) to (z);
 \draw (v) to (w);
 \foreach \i in {w,x,y}
@@ -2698,16 +2698,16 @@ vertex, which in this case is $z$. We color $z$ with $1$.
 \draw (z) to (y);
 \end{tikzpicture}
 \]
-The most saturated vertices are now $w$ and $y$. We color $w$ with the
-first available color which is $0$.
+The most saturated vertices are now $w$ and $y$. We color $y$ with the
+first available color, which is $0$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (v) at (0,0)    {$v:-,\{0\}$};
-\node (w) at (3,0)    {$w:0,\{1\}$};
-\node (x) at (6,0)    {$x:-,\{0,\}$};
-\node (y) at (3,-1.5) {$y:-,\{0,1\}$};
-\node (z) at (6,-1.5) {$z:1,\{0\}$};
-\node (t1) at (9,-1.5)   {$t.1:0,\{\}$};
+\node (v) at (0,0)    {$v:-,\{\}$};
+\node (w) at (3,0)    {$w:-,\{\mathbf{0},1\}$};
+\node (x) at (6,0)    {$x:-,\{\mathbf{0},\}$};
+\node (y) at (3,-1.5) {$y:\mathbf{0},\{1\}$};
+\node (z) at (6,-1.5) {$z:1,\{\mathbf{0}\}$};
+\node (t1) at (9,-1.5)   {$t.1:0,\{1\}$};
 \draw (t1) to (z);
 \draw (v) to (w);
 \foreach \i in {w,x,y}
@@ -2721,14 +2721,14 @@ first available color which is $0$.
 \draw (z) to (y);
 \end{tikzpicture}
 \]
-Vertex $y$ is the next most highly saturated, so we color $y$ with $2$.
+Vertex $w$ is now the most highly saturated, so we color $w$ with $2$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (v) at (0,0)   {$v:-,\{0\}$};
-\node (w) at (3,0)   {$w:0,\{1,2\}$};
-\node (x) at (6,0)   {$x:-,\{0,2\}$};
-\node (y) at (3,-1.5)  {$y:2,\{0,1\}$};
-\node (z) at (6,-1.5)  {$z:1,\{0,2\}$};
+\node (v) at (0,0)   {$v:-,\{2\}$};
+\node (w) at (3,0)   {$w:\mathbf{2},\{0,1\}$};
+\node (x) at (6,0)   {$x:-,\{0,\mathbf{2}\}$};
+\node (y) at (3,-1.5)  {$y:0,\{1,\mathbf{2}\}$};
+\node (z) at (6,-1.5)  {$z:1,\{0,\mathbf{2}\}$};
 \node (t1) at (9,-1.5)   {$t.1:0,\{\}$};
 \draw (t1) to (z);
 \draw (v) to (w);
@@ -2746,10 +2746,10 @@ Vertex $y$ is the next most highly saturated, so we color $y$ with $2$.
 Now $x$ has the highest saturation, so we color it $1$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (v) at (0,0)   {$v:-,\{0\}$};
-\node (w) at (3,0)   {$w:0,\{1,2\}$};
-\node (x) at (6,0)   {$x:1,\{0,2\}$};
-\node (y) at (3,-1.5)  {$y:2,\{0,1\}$};
+\node (v) at (0,0)   {$v:-,\{2\}$};
+\node (w) at (3,0)   {$w:2,\{0,\mathbf{1}\}$};
+\node (x) at (6,0)   {$x:\mathbf{1},\{0,2\}$};
+\node (y) at (3,-1.5)  {$y:0,\{\mathbf{1},2\}$};
 \node (z) at (6,-1.5)  {$z:1,\{0,2\}$};
 \node (t1) at (9,-1.5)   {$t.1:0,\{\}$};
 \draw (t1) to (z);
@@ -2765,13 +2765,13 @@ Now $x$ has the highest saturation, so we color it $1$.
 \draw (z) to (y);
 \end{tikzpicture}
 \]
-In the last two steps of the algorithm, we color $v$ with $1$.
+In the last step of the algorithm, we color $v$ with $0$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (v) at (0,0)   {$v:1,\{0\}$};
-\node (w) at (3,0)   {$w:0,\{1,2\}$};
+\node (v) at (0,0)   {$v:\mathbf{0},\{2\}$};
+\node (w) at (3,0)   {$w:2,\{\mathbf{0},1\}$};
 \node (x) at (6,0)   {$x:1,\{0,2\}$};
-\node (y) at (3,-1.5)  {$y:2,\{0,1\}$};
+\node (y) at (3,-1.5)  {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5)  {$z:1,\{0,2\}$};
 \node (t1) at (9,-1.5)   {$t.1:0,\{\}$};
 \draw (t1) to (z);
@@ -2799,18 +2799,16 @@ is the mapping of colors to registers and stack allocations.
 \]
 Putting this mapping together with the above coloring of the variables, we
 arrive at the assignment:
-
-UNDER CONSTRUCTION
 \begin{gather*}
-  \{ v \mapsto \key{-8(\%rbp)}, \,
-  w \mapsto \key{\%rcx},  \,
-  x \mapsto \key{-8(\%rbp)}, \,
-  y \mapsto \key{-16(\%rbp)},  \,
+  \{ v \mapsto \key{\%rcx}, \,
+  w \mapsto \key{-16(\%rbp)},  \,
+  x \mapsto \key{-8(\%rbp)}, \\
+  y \mapsto \key{\%rcx},  \,
   z\mapsto \key{-8(\%rbp)}, 
   t.1\mapsto \key{\%rcx} \}
 \end{gather*}
-Applying this assignment to our running example
-(Figure~\ref{fig:reg-eg}) yields the program on the right.\\
+Applying this assignment to our running example, on the left, yields
+the program on the right.\\
 % why frame size of 32? -JGS
 \begin{minipage}{0.4\textwidth}
 \begin{lstlisting}
@@ -2834,17 +2832,17 @@ $\Rightarrow$
 \begin{minipage}{0.45\textwidth}
 \begin{lstlisting}
 (block ()
-  (movq (int 1) (reg rdx))
-  (movq (int 46) (reg rcx))
-  (movq (reg rdx) (reg rdx))
-  (addq (int 7) (reg rdx))
-  (movq (reg rdx) (deref rbp -8))
-  (addq (int 4) (deref rbp -8))
-  (movq (reg rdx) (reg rdx))
-  (addq (reg rcx) (reg rdx))
+  (movq (int 1) (reg rcx))
+  (movq (int 46) (deref rbp -16))
+  (movq (reg rcx) (deref rbp -8))
+  (addq (int 7) (deref rbp -8))
   (movq (deref rbp -8) (reg rcx))
+  (addq (int 4) (reg rcx))
+  (movq (deref rbp -8) (deref rbp -8))
+  (addq (deref rbp -16) (deref rbp -8))
+  (movq (reg rcx) (reg rcx))
   (negq (reg rcx))
-  (movq (reg rdx) (reg rax))
+  (movq (deref rbp -8) (reg rax))
   (addq (reg rcx) (reg rax))
   (jmp conclusion))
 \end{lstlisting}