libzahl

big integer library
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commit dbb82e8a1184eaa7f6fa4b04e0560589cc6092e9
parent 58cbdbd892c9a83369e3e46aa9700cc7df98a17b
Author: Mattias Andrée <maandree@kth.se>
Date:   Sun, 24 Jul 2016 17:39:27 +0200

Add exercise: [▶M17] Factorials inverted

Signed-off-by: Mattias Andrée <maandree@kth.se>

Diffstat:
doc/exercises.tex | 91+++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++------
doc/libzahl.tex | 2+-
2 files changed, 86 insertions(+), 7 deletions(-)

diff --git a/doc/exercises.tex b/doc/exercises.tex @@ -24,8 +24,10 @@ \item {[\textit{M10}]} \textbf{Convergence of the Lucas Number ratios} -Find an approximation for $\displaystyle{ \lim_{n \to \infty} \frac{L_{n + 1}}{L_n}}$, -where $L_n$ is the $n^{\text{th}}$ Lucas Number \psecref{sec:Lucas numbers}. +Find an approximation for +$\displaystyle{ \lim_{n \to \infty} \frac{L_{n + 1}}{L_n}}$, +where $L_n$ is the $n^{\text{th}}$ +Lucas Number \psecref{sec:Lucas numbers}. \( \displaystyle{ L_n \stackrel{\text{\tiny{def}}}{\text{=}} \left \{ \begin{array}{ll} @@ -48,9 +50,11 @@ Implement the function \vspace{-1em} \noindent -which prints the prime factorisation of $n!$ (the $n^{\text{th}}$ factorial). -The function shall be efficient for all $n$ where all primes $p \le n$ can -be found efficiently. You can assume that $n \ge 2$. You should not evaluate $n!$. +which prints the prime factorisation of $n!$ +(the $n^{\text{th}}$ factorial). The function shall +be efficient for all $n$ where all primes $p \le n$ +can be found efficiently. You can assume that +$n \ge 2$. You should not evaluate $n!$. @@ -76,6 +80,30 @@ $P_i$ is \texttt{P[i - 1]} and $K_i$ is \texttt{K[i - 1]}. +\item {[$\RHD$\textit{M17}]} \textbf{Factorials inverted} + +Implement the function + +\vspace{-1em} +\begin{alltt} + void unfact(z_t x, z_t n); +\end{alltt} +\vspace{-1em} + +\noindent +which given a factorial number $n$, i.e. on the form +$x! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot x$, +calculates $x = n!^{-1}$. You can assume that +$n$ is a perfect factorial number and that $x \ge 1$. +Extra credit if you can detect when the input, $n$, +is not a factorial number. Such function would of +course return an \texttt{int} 1 if the input is a +factorial and 0 otherwise, or alternatively 0 +on success and $-1$ with \texttt{errno} set to +\texttt{EDOM} if the input is not a factorial. + + + \item {[\textit{05}]} \textbf{Fast primality test} $(x + y)^p \equiv x^p + y^p ~(\text{Mod}~p)$ @@ -153,9 +181,60 @@ of $x!$. $f(p, k)$ is defined as: +\item \textbf{Factorials inverted} + +Use \texttt{zlsb} to get the power of 2 in the +prime factorisation of $n$, that is, the number +of times $n$ is divisible by 2. If we write $n$ on +the form $1 \cdot 2 \cdot 3 \cdot \ldots \cdot x$, +every $2^\text{nd}$ factor is divisible by 2, every +$4^\text{th}$ factor is divisible by $2^2$, and so on. +From call \texttt{zlsb} we know how many times, +$n$ is divisible by 2, but know how many of the factors +are divisible by 2, but this can be calculated with +the following algorithm, where $k$ is the number +times $n$ is divisible by 2: + +\vspace{1em} +\hspace{-2.8ex} +\begin{minipage}{\linewidth} +\begin{algorithmic} + \STATE $k^\prime \gets 0$ + \WHILE{$k > 0$} + \STATE $a \gets 0$ + \WHILE{$2^a \le k$} + \STATE $k \gets k - 2^a$ + \STATE $a \gets a + 1$ + \ENDWHILE + \STATE $k^\prime \gets k^\prime + 2^{a - 1}$ + \ENDWHILE + \RETURN $k^\prime$ +\end{algorithmic} +\end{minipage} +\vspace{1em} + +\noindent +Note that $2^a$ is efficiently calculated with, +\texttt{zlsh}, but it is more efficient to use +\texttt{zbset}. + +Now that we know $k^\prime$, the number of +factors ni $1 \cdot \ldots \cdot x$ that are +divisible by 2, we have two choices for $x$: +$k^\prime$ and $k^\prime + 1$. To check which, we +calculate $(k^\prime - 1)!!$ (the semifactoral, i.e. +$1 \cdot 3 \cdot 5 \cdot \ldots \cdot (k^\prime - 1)$) +naïvely and shift the result $k$ steps to the left. +This gives us $k^\prime!$. If $x! \neq k^\prime!$, then +$x = k^\prime + 1$ unless $n$ is not factorial number. +Of course, if $x! = k^\prime!$, then $x = k^\prime$. + + + \item \textbf{Fast primality test} -If we select $x = y = 1$ we get $2^p \equiv 2 ~(\text{Mod}~p)$. This gives us +If we select $x = y = 1$ we get +$2^p \equiv 2 ~(\text{Mod}~p)$. This gives us \vspace{-1em} \begin{alltt} diff --git a/doc/libzahl.tex b/doc/libzahl.tex @@ -3,7 +3,7 @@ \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{algorithmic, algorithm, colonequals, alltt} -\usepackage{amsmath, amssymb, mathtools, MnSymbol, mathrsfs, esvect} +\usepackage{amsmath, amssymb, mathtools, MnSymbol, mathrsfs, esvect, wasysym} \usepackage{tipa, color, graphicx} \usepackage{shorttoc, minitoc} \usepackage{enumitem}