Minor fixes

manuscrit/60_staticdeps/10_types_of_deps.tex

@@ -37,7 +37,9 @@ physical registers than what is exposed to the user through the ISA; a renaming
 phrase will allocate those registers to avoid the effects of WaR and WaW
 dependencies as much as possible.
 
-\smallskip{} For the present chapter, \textit{we only consider read-after-write
+\smallskip{}
+
+For the present chapter, \textit{we only consider read-after-write
 dependencies}; however, all the techniques we present are applicable to other
 types of dependencies if the considered architecture requires to take them into
 account.

manuscrit/60_staticdeps/30_static_principle.tex

@@ -11,7 +11,7 @@ detected by keeping track of which instruction last wrote each register in a
 Loop-carried dependencies can, to some extent, be detected the same way. As the
 basic block is always assumed to be the body of an infinite loop, a
 straight-line analysis can be performed on a duplicated kernel. This strategy
-is \eg{} adopted by \osaca{}~\cite{osaca2} (§II.D). When dealing only with
+is \eg{} adopted by \osaca{}~\cite[§II.D]{osaca2}. When dealing only with
 register accesses, this strategy is always sufficient: as each iteration always
 executes the same basic block, it is not possible for an instruction to depend
 on another instruction two iterations earlier or more.

manuscrit/60_staticdeps/35_rob_proof.tex

@@ -20,10 +20,11 @@ performance}\label{ssec:staticdeps:rob_proof}
     instructions $\left(I_{p+kn}, I_{q+kn}\right)_{k\in\nat}$.
 \end{theorem}
 
-To prove this assertion, we require a few postulates describing the functioning
-of a CPU and, in particular, how \uops{} transit in (decoded) and out (retired)
-the reorder buffer.
+To prove this assertion, we require a few postulates to model the functioning
+of a CPU and, in particular, how \uops{} transit in (decoded) and out
+(retired) the reorder buffer.
 
+\needspace{10em}
 \begin{postulate}[Reorder buffer as a circular buffer]
     The reorder buffer is a circular buffer of size $R \in \nat^+$.
     It contains only decoded \uops{}.
@@ -90,6 +91,7 @@ Consequently, if both $i_q$  and $i_{q'}$ are in-flight then
 $|q'-q|<R$, and thus $\distance{I_p}{I_{p'}} < R - 1$.
 \end{proof}
 
+\needspace{15em}
 \begin{postulate}[Issue delay]
     Reasons why the issue of a \uop{} $i$ is delayed can be:
     \begin{enumerate}