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| author | Mattias Andrée <maandree@kth.se> | 2016-07-28 15:05:18 +0200 |
|---|---|---|
| committer | Mattias Andrée <maandree@kth.se> | 2016-07-28 15:05:18 +0200 |
| commit | 2864613b162ed4c7a28a79c8c4dcd24893cc2128 (patch) | |
| tree | f1e65750427df41396a53b1d071040eaa78aab48 | |
| parent | Fix difficulty ratings on exercises (diff) | |
| download | libzahl-2864613b162ed4c7a28a79c8c4dcd24893cc2128.tar.gz libzahl-2864613b162ed4c7a28a79c8c4dcd24893cc2128.tar.bz2 libzahl-2864613b162ed4c7a28a79c8c4dcd24893cc2128.tar.xz | |
Fix spacing
Signed-off-by: Mattias Andrée <maandree@kth.se>
Diffstat (limited to '')
| -rw-r--r-- | doc/exercises.tex | 6 |
1 files changed, 3 insertions, 3 deletions
diff --git a/doc/exercises.tex b/doc/exercises.tex index 757788b..94d780a 100644 --- a/doc/exercises.tex +++ b/doc/exercises.tex @@ -179,10 +179,10 @@ For improved performance, instead of using \texttt{zmod}, you can use the recursive function % \( \displaystyle{ - k \mod (2^n - 1) = + k \text{ mod } (2^n - 1) = \left ( - (k \mod 2^n) + \lfloor k \div 2^n \rfloor - \right ) \mod (2^n - 1), + (k \text{ mod } 2^n) + \lfloor k \div 2^n \rfloor + \right ) \text{ mod } (2^n - 1), }\) % where $k \mod 2^n$ is efficiently calculated |