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| author | Mattias Andrée <maandree@kth.se> | 2016-07-25 15:58:29 +0200 |
|---|---|---|
| committer | Mattias Andrée <maandree@kth.se> | 2016-07-25 15:58:29 +0200 |
| commit | 4d2e79e7eec793a557c26d1253bcfc13f6b555d6 (patch) | |
| tree | f8b89ad4245c3e9657cac43577ea29d4847d415f | |
| parent | Manual: how to calculate the legendre symbol (diff) | |
| download | libzahl-4d2e79e7eec793a557c26d1253bcfc13f6b555d6.tar.gz libzahl-4d2e79e7eec793a557c26d1253bcfc13f6b555d6.tar.bz2 libzahl-4d2e79e7eec793a557c26d1253bcfc13f6b555d6.tar.xz | |
Style fix
Signed-off-by: Mattias Andrée <maandree@kth.se>
| -rw-r--r-- | doc/exercises.tex | 8 |
1 files changed, 4 insertions, 4 deletions
diff --git a/doc/exercises.tex b/doc/exercises.tex index 23b8ef4..e89c2b3 100644 --- a/doc/exercises.tex +++ b/doc/exercises.tex @@ -168,13 +168,13 @@ For improved performance, instead of using \texttt{zmod}, you can use the recursive function % \( \displaystyle{ - k ~\mbox{Mod}~ 2^n - 1 = + k \mod (2^n - 1) = \left ( - (k ~\mbox{Mod}~ 2^n) + \lfloor k \div 2^n \rfloor - \right ) ~\mbox{Mod}~ 2^n - 1, + (k \mod 2^n) + \lfloor k \div 2^n \rfloor + \right ) \mod (2^n - 1), }\) % -where $k ~\mbox{Mod}~ 2^n$ is efficiently calculated +where $k \mod 2^n$ is efficiently calculated using \texttt{zand($k$, $2^n - 1$)}. (This optimisation is not part of the difficulty rating of this problem.) |