From 6689b5285c16a5462500f2e1a90292ed29de4311 Mon Sep 17 00:00:00 2001
From: Thibaut Horel <thibaut.horel@gmail.com>
Date: Thu, 25 Sep 2014 00:02:43 -0400
Subject: [ps1] trailing paragraph

---
 ps1/main.tex | 13 ++++++-------
 1 file changed, 6 insertions(+), 7 deletions(-)

(limited to 'ps1')

diff --git a/ps1/main.tex b/ps1/main.tex
index b04d730..b773275 100644
--- a/ps1/main.tex
+++ b/ps1/main.tex
@@ -155,13 +155,6 @@ t = w}
 
 \section*{Problem 3}
 
-\paragraph{First solution \emph{(too easy, feels like cheating).}} We saw in
-class how to compute the most signigicant bit set to one. But observe that
-\texttt{x \& $\sim$(x-1)} will clear all the bits of $x$ except for the least
-significant bit set to one.  Assuming access to a procedure \texttt{msb} to
-compute the most significant bit set to one, we can simply define
-\texttt{lsb(x) $:=$ msb(x \& $\sim$(x-1))}.
-
 \paragraph{Second solution.}
 Using Problem 2 and what we did in class, we assume that we have access to the
 following procedures:
@@ -242,5 +235,11 @@ Note that the above code only requires a constant number of multiplications and
 bitwise operations once the constants $m$, $a$, $b$ and $c$ have been
 pre-computed.
 
+\paragraph{Alternative solution \emph{(too easy, felt like cheating).}} We saw in
+class how to compute the most signigicant bit set to one. But observe that
+\texttt{x \& $\sim$(x-1)} will clear all the bits of $x$ except for the least
+significant bit set to one.  Assuming access to a procedure \texttt{msb} to
+compute the most significant bit set to one, we can simply define
+\texttt{lsb(x) $:=$ msb(x \& $\sim$(x-1))}.
 
 \end{document}
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