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@@ -5771,11 +5771,11 @@ instructions. \racket{The first instruction is \lstinline{movq $1, v},
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so \code{v} interferes with \code{rsp}.}
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%
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\python{The first instruction is \lstinline{movq $1, v}, and the
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- live-after set is $\{\ttm{v}\}$. Rule 1 applies but there is
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+ live-after set is $\{\ttm{v}\}$. Rule 1 applies, but there is
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no interference because $\ttm{v}$ is the destination of the move.}
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%
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\racket{The fourth instruction is \lstinline{addq $7, x}, and the
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- live-after set is $\{\ttm{w},\ttm{x},\ttm{rsp}\}$. Rule 2 applies so
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+ live-after set is $\{\ttm{w},\ttm{x},\ttm{rsp}\}$. Rule 2 applies, so
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$\ttm{x}$ interferes with \ttm{w} and \ttm{rsp}.}
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%
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\python{The fourth instruction is \lstinline{addq $7, x}, and the
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@@ -6309,16 +6309,17 @@ So, we obtain the following coloring:
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{\if\edition\pythonEd\pythonColor
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%
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With the DSATUR algorithm in hand, let us return to the running
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-example and consider how to color the interference graph in
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-figure~\ref{fig:interfere}. We annotate each variable node with a dash
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-to indicate that it has not yet been assigned a color. Each register
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-node (not shown) should be assigned the number that the register
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-corresponds to, for example, color \code{rcx} with the number \code{0}
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-and \code{rdx} with \code{1}. The saturation sets are also shown for
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-each node; all of them start as the empty set. We do not show the
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-register nodes in the graph below because there were no interference
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-edges involving registers in this program, but in general there can
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-be.
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+example and consider how to color the interference graph shown in
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+figure~\ref{fig:interfere}, again mapping 1 to blank, 2 to white, and
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+3 to gray. We annotate each variable node with a dash to indicate that
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+it has not yet been assigned a color. Each register node (not shown)
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+should be assigned the number that the register corresponds to, for
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+example, color \code{rcx} with the number \code{0} and \code{rdx} with
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+\code{1}. The saturation sets are also shown for each node; all of
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+them start as the empty set. We do not show the register nodes in the
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+following graph because there were no interference edges involving
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+registers in this program; however, in general there can be inference
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+edges that involve registers.
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%
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\[
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\begin{tikzpicture}[baseline=(current bounding box.center)]
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@@ -6341,7 +6342,7 @@ be.
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\]
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The algorithm says to select a maximally saturated vertex, but they
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are all equally saturated. So we flip a coin and pick $\ttm{tmp\_0}$
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-then color it with the first available integer, which is $0$. We mark
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+and then we color it with the first available integer, which is $0$. We mark
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$0$ as no longer available for $\ttm{tmp\_1}$ and $\ttm{z}$ because
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they interfere with $\ttm{tmp\_0}$.
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\[
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@@ -6859,7 +6860,7 @@ allocated.
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\code{rbx}, and variables \code{w} and \code{t} was assigned to \code{rcx}.}
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%
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\python{Variables \code{v}, \code{x}, \code{y}, and \code{tmp\_0}
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- were assigned to \code{rcx} and variables \code{w} and \code{tmp\_1}
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+ were assigned to \code{rcx}, and variables \code{w} and \code{tmp\_1}
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were assigned to \code{rbx}.}
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%
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Variable \racket{\code{y}}\python{\code{z}} was spilled to the stack
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@@ -7245,7 +7246,7 @@ were \code{tmp\_1}, \code{w}, and \code{y}.
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\end{tikzpicture}
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\]
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We have arbitrarily chosen to color \code{w} instead of \code{tmp\_1}
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-or \code{y}, but note that \code{w} is not move related to any
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+or \code{y}. Note, however, that \code{w} is not move related to any
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variables, whereas \code{y} and \code{tmp\_1} are move related to
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\code{tmp\_0} and \code{z}, respectively. If we instead choose
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\code{y} and color it $0$, we can delete another move instruction.
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Jeremy G. Siek