updated the rest of the examples in chap 3

book.tex

@@ -1859,7 +1859,7 @@ Source program:
   (let ([x (+ v 7)])
   (let ([y (+ 4 x)])
   (let ([z (+ x w)])
-       (- z y)))))))
+       (+ z (- y))))))))
 \end{lstlisting}
 \end{minipage}
 \begin{minipage}{0.45\textwidth}
@@ -2262,6 +2262,9 @@ with a dash for their color and an empty set for the saturation.
 \node (x) at (6,0)    {$x:-,\{\}$};
 \node (y) at (3,-1.5) {$y:-,\{\}$};
 \node (z) at (6,-1.5) {$z:-,\{\}$};
+\node (t1) at (9,0)   {$t.1:-,\{\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2272,10 +2275,12 @@ with a dash for their color and an empty set for the saturation.
 }
 \draw (z) to (w);
 \draw (z) to (y);
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \end{tikzpicture}
 \]
 We select a maximally saturated node and color it $0$. In this case we
-have a 5-way tie, so we arbitrarily pick $y$. The then mark color $0$
+have a 7-way tie, so we arbitrarily pick $y$. The then mark color $0$
 as no longer available for $w$, $x$, and $z$ because they interfere
 with $y$.
 \[
@@ -2285,6 +2290,8 @@ with $y$.
 \node (x) at (6,0)    {$x:-,\{0\}$};
 \node (y) at (3,-1.5) {$y:0,\{\}$};
 \node (z) at (6,-1.5) {$z:-,\{0\}$};
+\node (t1) at (9,0)   {$t.1:-,\{\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2295,6 +2302,8 @@ with $y$.
 }
 \draw (z) to (w);
 \draw (z) to (y);
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \end{tikzpicture}
 \]
 Now we repeat the process, selecting another maximally saturated node.
@@ -2307,6 +2316,10 @@ $w$ with $1$.
 \node (x) at (6,0)    {$x:-,\{0,1\}$};
 \node (y) at (3,-1.5) {$y:0,\{1\}$};
 \node (z) at (6,-1.5) {$z:-,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2328,6 +2341,10 @@ next available color which is $2$.
 \node (x) at (6,0)    {$x:2,\{0,1\}$};
 \node (y) at (3,-1.5) {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5) {$z:-,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2340,8 +2357,7 @@ next available color which is $2$.
 \draw (z) to (y);
 \end{tikzpicture}
 \]
-We have only two nodes left to color, $v$ and $z$, but $z$ is
-more highly saturated, so we color $z$ with $2$.
+Node $z$ is the next most highly saturated, so we color $z$ with $2$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
 \node (v) at (0,0)   {$v:-,\{1\}$};
@@ -2349,6 +2365,10 @@ more highly saturated, so we color $z$ with $2$.
 \node (x) at (6,0)   {$x:2,\{0,1\}$};
 \node (y) at (3,-1.5)  {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5)  {$z:2,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{2\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2361,7 +2381,8 @@ more highly saturated, so we color $z$ with $2$.
 \draw (z) to (y);
 \end{tikzpicture}
 \]
-The last iteration of the coloring algorithm assigns color $0$ to $v$.
+We have a 2-way tie between $v$ and $t.1$. We choose to color $v$ with
+$0$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
 \node (v) at (0,0)   {$v:0,\{1\}$};
@@ -2369,6 +2390,10 @@ The last iteration of the coloring algorithm assigns color $0$ to $v$.
 \node (x) at (6,0)   {$x:2,\{0,1\}$};
 \node (y) at (3,-1.5)  {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5)  {$z:2,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{2\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2381,6 +2406,33 @@ The last iteration of the coloring algorithm assigns color $0$ to $v$.
 \draw (z) to (y);
 \end{tikzpicture}
 \]
+In the last two steps of the algorithm, we color $t.1$ with $0$
+then $t.2$ with $1$.
+\[
+\begin{tikzpicture}[baseline=(current  bounding  box.center)]
+\node (v) at (0,0)   {$v:0,\{1\}$};
+\node (w) at (3,0)   {$w:1,\{0,2\}$};
+\node (x) at (6,0)   {$x:2,\{0,1\}$};
+\node (y) at (3,-1.5)  {$y:0,\{1,2\}$};
+\node (z) at (6,-1.5)  {$z:2,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:0,\{2,1\}$};
+\node (t2) at (9,-1.5) {$t.2:1,\{0\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
+\draw (v) to (w);
+\foreach \i in {w,x,y} 
+{
+  \foreach \j in {w,x,y}
+  { 
+    \draw (\i) to (\j);
+  }
+}
+\draw (z) to (w);
+\draw (z) to (y);
+\end{tikzpicture}
+\]
+
+
 With the coloring complete, we can finalize the assignment of
 variables to registers and stack locations. Recall that if we have $k$
 registers, we map the first $k$ colors to registers and the rest to
@@ -2392,13 +2444,15 @@ following 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:
-\[
-  \{ v \mapsto \key{\%rbx}, \;
-  w \mapsto \key{-8(\%rbp)},  \;
-  x \mapsto \key{-16(\%rbp)}, \;
-  y \mapsto \key{\%rbx},  \;
-  z\mapsto \key{-16(\%rbp)} \}
-\]
+\begin{gather*}
+  \{ v \mapsto \key{\%rbx}, \,
+  w \mapsto \key{-8(\%rbp)},  \,
+  x \mapsto \key{-16(\%rbp)}, \,
+  y \mapsto \key{\%rbx},  \,
+  z\mapsto \key{-16(\%rbp)}, \\
+  t.1\mapsto \key{\%rbx} ,\,
+  t.2\mapsto \key{-8(\%rbp)} \}
+\end{gather*}
 Applying this assignment to our running example
 (Figure~\ref{fig:reg-eg}) yields the program on the right.
 
@@ -2414,14 +2468,17 @@ Applying this assignment to our running example
     (addq (int 4) (var y))
     (movq (var x) (var z))
     (addq (var w) (var z))
-    (movq (var z) (reg rax))
-    (subq (var y) (reg rax)))
+    (movq (var y) (var t.1))
+    (negq (var t.1))
+    (movq (var z) (var t.2))
+    (addq (var t.1) (var t.2))
+    (movq (var t.2) (reg rax)))
 \end{lstlisting}
 \end{minipage}
 $\Rightarrow$
 \begin{minipage}{0.45\textwidth}
 \begin{lstlisting}
-(program 32
+(program 16
   (movq (int 1) (reg rbx))
   (movq (int 46) (stack -8))
   (movq (reg rbx) (stack -16))
@@ -2430,8 +2487,11 @@ $\Rightarrow$
   (addq (int 4) (reg rbx))
   (movq (stack -16) (stack -16))
   (addq (stack -8) (stack -16))
-  (movq (stack -16) (reg rax))
-  (subq (reg rbx) (reg rax)))
+  (movq (reg rbx) (reg rbx))
+  (negq (reg rbx))
+  (movq (stack -16) (stack -8))
+  (addq (reg rbx) (stack -8))
+  (movq (stack -8) (reg rax)))
 \end{lstlisting}
 \end{minipage}
 
@@ -2566,24 +2626,30 @@ described in the last section, we get the following program.
     (addq (int 4) (var y))
     (movq (var x) (var z))
     (addq (var w) (var z))
-    (movq (var z) (reg rax))
-    (subq (var y) (reg rax)))
+    (movq (var y) (var t.1))
+    (negq (var t.1))
+    (movq (var z) (var t.2))
+    (addq (var t.1) (var t.2))
+    (movq (var t.2) (reg rax)))
 \end{lstlisting}
 \end{minipage}
 $\Rightarrow$
 \begin{minipage}{0.45\textwidth}
 \begin{lstlisting}
-  (program 0
-    (movq (int 1) (reg rbx))
-    (movq (int 46) (reg rcx))
-    (movq (reg rbx) (reg rdx))
-    (addq (int 7) (reg rdx))
-    (movq (reg rdx) (reg rbx))
-    (addq (int 4) (reg rbx))
-    (movq (reg rdx) (reg rdx))
-    (addq (reg rcx) (reg rdx))
-    (movq (reg rdx) (reg rax))
-    (subq (reg rbx) (reg rax)))
+(program 0
+  (movq (int 1) (reg rbx))
+  (movq (int 46) (reg rcx))
+  (movq (reg rbx) (reg rdx))
+  (addq (int 7) (reg rdx))
+  (movq (reg rdx) (reg rbx))
+  (addq (int 4) (reg rbx))
+  (movq (reg rdx) (reg rdx))
+  (addq (reg rcx) (reg rdx))
+  (movq (reg rbx) (reg rbx))
+  (negq (reg rbx))
+  (movq (reg rdx) (reg rcx))
+  (addq (reg rbx) (reg rcx))
+  (movq (reg rcx) (reg rax)))
 \end{lstlisting}
 \end{minipage}
 
@@ -2612,6 +2678,10 @@ similar to how we represented interference.  The following is the
 \node (x) at (6,0)    {$x$};
 \node (y) at (3,-1.5) {$y$};
 \node (z) at (6,-1.5) {$z$};
+\node (t1) at (9,0)   {$t.1$};
+\node (t2) at (9,-1.5) {$t.2$};
+\draw (t1) to (y);
+\draw (t2) to (z);
 \draw[bend left=20] (v) to (x);
 \draw (x) to (y);
 \draw (x) to (z);
@@ -2628,6 +2698,10 @@ saturated vertex is $z$.
 \node (x) at (6,0)    {$x:2,\{0,1\}$};
 \node (y) at (3,-1.5) {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5) {$z:-,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2654,6 +2728,10 @@ case choosing $2$ because that's the color of $x$.
 \node (x) at (6,0)    {$x:2,\{0,1\}$};
 \node (y) at (3,-1.5) {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5) {$z:2,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{2\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2667,9 +2745,9 @@ case choosing $2$ because that's the color of $x$.
 \end{tikzpicture}
 \]
 
-The last variable to color is $v$, and we just need to avoid choosing
-$1$ because of the interference with $w$. Last time we choose the
-color $0$, simply because it was the lowest, but this time we know
+Next we consider coloring the variable $v$, and we just need to avoid
+choosing $1$ because of the interference with $w$. Last time we choose
+the color $0$, simply because it was the lowest, but this time we know
 that $v$ is move related to $x$, so we choose the color $2$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
@@ -2678,6 +2756,10 @@ that $v$ is move related to $x$, so we choose the color $2$.
 \node (x) at (6,0)    {$x:2,\{0,1\}$};
 \node (y) at (3,-1.5) {$y:0,\{1,2\}$};
 \node (z) at (6,-1.5) {$z:2,\{0,1\}$};
+\node (t1) at (9,0)   {$t.1:-,\{2\}$};
+\node (t2) at (9,-1.5) {$t.2:-,\{\}$};
+\draw (t1) to (z);
+\draw (t2) to (t1);
 \draw (v) to (w);
 \foreach \i in {w,x,y} 
 {
@@ -2705,40 +2787,48 @@ to obtain the code on right.
     (addq (int 4) (var y))
     (movq (var x) (var z))
     (addq (var w) (var z))
-    (movq (var z) (reg rax))
-    (subq (var y) (reg rax)))
+    (movq (var y) (var t.1))
+    (negq (var t.1))
+    (movq (var z) (var t.2))
+    (addq (var t.1) (var t.2))
+    (movq (var t.2) (reg rax)))
 \end{lstlisting}
 \end{minipage}
 $\Rightarrow$
 \begin{minipage}{0.45\textwidth}
 \begin{lstlisting}
-  (program 0
-    (movq (int 1) (reg rdx))
-    (movq (int 46) (reg rcx))
-    (movq (reg rdx) (reg rdx))
-    (addq (int 7) (reg rdx))
-    (movq (reg rdx) (reg rbx))
-    (addq (int 4) (reg rbx))
-    (movq (reg rdx) (reg rdx))
-    (addq (reg rcx) (reg rdx))
-    (movq (reg rdx) (reg rax))
-    (subq (reg rbx) (reg rax)))
+(program 0
+  (movq (int 1) (reg rdx))
+  (movq (int 46) (reg rcx))
+  (movq (reg rdx) (reg rdx))
+  (addq (int 7) (reg rdx))
+  (movq (reg rdx) (reg rbx))
+  (addq (int 4) (reg rbx))
+  (movq (reg rdx) (reg rdx))
+  (addq (reg rcx) (reg rdx))
+  (movq (reg rbx) (reg rbx))
+  (negq (reg rbx))
+  (movq (reg rdx) (reg rcx))
+  (addq (reg rbx) (reg rcx))
+  (movq (reg rcx) (reg rax)))
 \end{lstlisting}
 \end{minipage}
 
 The \code{patch-instructions} then removes the trivial moves from
-\key{v} to \key{x} and from \key{x} to \key{z} to obtain the following
-result.
+\key{v} to \key{x}, from \key{x} to \key{z}, and from \key{y} to
+\key{t.1}, to obtain the following result.
 \begin{lstlisting}
-  (program 0
-    (movq (int 1) (reg rdx))
-    (movq (int 46) (reg rcx))
-    (addq (int 7) (reg rdx))
-    (movq (reg rdx) (reg rbx))
-    (addq (int 4) (reg rbx))
-    (addq (reg rcx) (reg rdx))
-    (movq (reg rdx) (reg rax))
-    (subq (reg rbx) (reg rax)))
+(program 0
+  (movq (int 1) (reg rdx))
+  (movq (int 46) (reg rcx))
+  (addq (int 7) (reg rdx))
+  (movq (reg rdx) (reg rbx))
+  (addq (int 4) (reg rbx))
+  (addq (reg rcx) (reg rdx))
+  (negq (reg rbx))
+  (movq (reg rdx) (reg rcx))
+  (addq (reg rbx) (reg rcx))
+  (movq (reg rcx) (reg rax)))
 \end{lstlisting}
 
 \begin{exercise}\normalfont
@@ -3678,8 +3768,8 @@ short-circuiting behavior in the order of evaluation of its arguments.
 \begin{array}{lcl}
   \Type &::=& \ldots \mid (\key{Vector}\;\Type^{+}) \\
   \Exp &::=& \ldots \mid (\key{vector}\;\Exp^{+}) \mid 
-    (\key{vector-ref}\;\Exp\;\Exp) \\
-  &\mid& (\key{vector-set!}\;\Exp\;\Exp\;\Exp)\\
+    (\key{vector-ref}\;\Exp\;\Int) \\
+  &\mid& (\key{vector-set!}\;\Exp\;\Int\;\Exp)\\
   R_3 &::=& (\key{program} \; \Exp)
 \end{array}
 \]