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	Add exercise: [▶M17] Factorials inverted - libzahl - big integer library
	git clone git://git.suckless.org/libzahl
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	commit dbb82e8a1184eaa7f6fa4b04e0560589cc6092e9
	parent 58cbdbd892c9a83369e3e46aa9700cc7df98a17b
	Author: Mattias Andrée <[email protected]>
	Date: Sun, 24 Jul 2016 17:39:27 +0200

	Add exercise: [▶M17] Factorials inverted

	Signed-off-by: Mattias Andrée <[email protected]>

	Diffstat:
	M doc/exercises.tex \| 91 +++++++++++++++++++++++++++++…
	M 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}}…
	-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 $…
	+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}