Minor fixes.

[saga.git] / rank.tex

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\chapter{Ranking Combinatorial Structures}

\section{Ranking and unranking}

The techniques for building efficient data structures on the RAM described
in Chapter~\ref{ramchap} can be also used for a~variety of problems related
to ranking of combinatorial structures. Generally, the problems are stated
in the following way:

\defn\id{rankdef}%
Let~$C$ be a~set of objects and~$\prec$ a~linear order on~$C$. The \df{rank}
$R_{C,\prec}(x)$ of an~element $x\in C$ is the number of elements $y\in C$ such that $y\prec x$.
When~$\prec$ is defined on some superset of~$C$, we naturally extend the rank
to elements outside~$C$.
We will call the function $R_{C,\prec}$ the \df{ranking function} for $C$ and~$\prec$
and its inverse $R^{-1}_{C,\prec}$ the \df{unranking function}. When the set
and the order are clear from the context, we will use just~$R(x)$ and $R^{-1}(x)$.

\example
Let us consider the set $C_k=\{\0,\1\}^k$ of all binary strings of length~$k$ ordered
lexicographically. Then $R^{-1}(i)$ is the $i$-th smallest element of this set, that
is the number~$i$ written in binary and padded to~$k$ digits (i.e., $\(i)_k$ in the
notation of Section~\ref{bitsect}). Obviously, $R(x)$ is the integer whose binary
representation is~$x$.

\para
In this chapter, we will investigate how to compute the ranking and unranking
functions on different sets. Usually, we will make use of the fact that the ranks
(and hence the input and output of our algorithm) are large numbers, so we can
use the integers of a~similar magnitude to represent non-trivial data structures.
This will allow us to design efficient algorithms on the RAM.

\FIXME{We will assume RAM in the whole chapter.}

\section{Ranking of permutations}
\id{pranksect}

One of the most common ranking problems is ranking of permutations on the set~$[n]=\{1,2,\ldots,n\}$.
This is frequently used to create arrays indexed by permutations: for example in Ruskey's algorithm
for finding Hamilton cycles in Cayley graphs (see~\cite{ruskey:ham} and \cite{ruskey:hce})
or when exploring state spaces of combinatorial puzzles like the Loyd's Fifteen \cite{ss:fifteen}.
Many other applications are surveyed by Critani et al.~in~\cite{critani:rau} and in
most cases, the time complexity of the whole algorithm is limited by the efficiency
of the (un)ranking functions.

In most cases, the permutations are ranked according to their lexicographic order.
In fact, an~arbitrary order is often sufficient if the ranks are used solely
for indexing of arrays, but the lexicographic order has an~additional advantage
of a~nice structure allowing many additional operations on permutations to be
performed directly on their ranks.

Na\"\i{}ve algorithms for lexicographic ranking require time $\Theta(n^2)$ in the
worst case \cite{reingold:catp} and even on average~\cite{liehe:raulow}.
This can be easily improved to $O(n\log n)$ by using either a binary search
tree to calculate inversions, or by a divide-and-conquer technique, or by clever
use of modular arithmetic (all three algorithms are described in
\cite{knuth:sas}). Myrvold and Ruskey \cite{myrvold:rank} mention further
improvements to $O(n\log n/\log \log n)$ by using the RAM data structures of Dietz
\cite{dietz:oal}.

Linear time complexity was reached by Myrvold and Ruskey \cite{myrvold:rank}
for a~non-lexicographic order, which is defined locally by the history of the
data structure --- in fact, they introduce a linear-time unranking algorithm
first and then they derive an inverse algorithm without describing the order
explicitly. However, they leave the problem of lexicographic ranking open.

We will describe a~general procedure which, when combined with suitable
RAM data structures, yields a~linear-time algorithm for lexicographic
(un)ranking.

\nota\id{brackets}%
We will view permutations on a~ordered set~$A\subseteq {\bb N}$ as ordered $\vert A\vert$-tuples
(in other words, arrays) containing every element of~$A$ exactly once. We will
use square brackets for indexing of these tuples: $\pi=(\pi[1],\ldots,\pi[\vert A\vert])$.
The corresponding lexicographic ranking and unranking functions will be denoted by~$L(\pi,A)$
and $L^{-1}(i,A)$ respectively.

\obs
Let us first observe that permutations have a simple recursive structure.
If we fix the first element $\pi[1]$ of a~permutation~$\pi$ on the set~$[n]$, the
elements $\pi[2], \ldots, \pi[n]$ form a~permutation on $[n]-\{\pi[1]\} = \{1,\ldots,\pi[1]-1,\pi[1]+1,\ldots,n\}$.
The lexicographic order of two permutations $\pi$ and~$\pi'$ on the original set is then determined
by $\pi[1]$ and $\pi'[1]$ and only if these elements are equal, it is decided
by the lexicographic comparison of permutations $(\pi[2],\ldots,\pi[n])$ and
$(\pi'[2],\ldots,\pi'[n])$. Therefore the rank of $\pi$ is $(\pi[1]-1)\cdot
(n-1)!$ plus the rank of $(\pi[2],\ldots,\pi[n])$.

This gives us a~reduction from (un)ranking of permutations on $[n]$ to (un)ranking
of permutations on a $(n-1)$-element set, which suggests a straightforward
algorithm, but unfortunately this set is different from $[n-1]$ and it even
depends on the value of~$\pi[1]$. We could renumber the elements to get $[n-1]$,
but it would require linear time per iteration. Instead, we generalize the
problem to permutations on subsets of $[n]$. For a permutation $\pi$ on a~set
$A\subseteq [n]$ of size~$m$, similar reasoning gives a~simple formula:
$$
L((\pi[1],\ldots,\pi[m]),A) = R_A(\pi[1]) \cdot (m-1)! +
L((\pi[2],\ldots,\pi[m]), A\setminus\{\pi[1]\}),
$$
which uses the ranking function~$R_A$ for~$A$. This formula leads to the
following algorithms (a~similar formulation can be found for example in~\cite{knuth:sas}):

\alg $\<Rank>(\pi,i,n,A)$: compute the rank of a~permutation $\pi[i\ldots n]$ on~$A$.
\algo
\:If $i\ge n$, return~0.
\:$a\=R_A(\pi[i])$.
\:$b\=\<Rank>(\pi,i+1,n,A \setminus \{\pi[i]\})$.
\:Return $a\cdot(n-i)! + b$.
\endalgo

\>We can call $\<Rank>(\pi,1,n,[n])$ for ranking on~$[n]$, i.e., to calculate
$L(\pi,[n])$.

\alg $\<Unrank>(j,i,n,A)$: Fill $\pi[i,\ldots,n]$ with the $j$-th permutation on~$A$.
\algo
\:If $i>n$, return immediately.
\:$a\=R^{-1}_A(\lfloor j/(n-i)! \rfloor)$.
\:$\pi\=\<Unrank>(j\bmod (n-i)!,i+1,n,A\setminus \{a\})$.
\:$\pi[i]\=a$.
\endalgo

\>We can call $\<Unrank>(j,1,n,[n])$ for the unranking problem on~$[n]$, i.e., to get $L^{-1}(j,[n])$.

\para
The most time-consuming parts of the above algorithms are of course operations
on the set~$A$. If we use a~data structure of a~known time complexity, the complexity
of the whole algorithm is easy to calculate:

\lemma\id{ranklemma}%
Suppose that there is a~data structure maintaining a~subset of~$[n]$ under insertion
and deletion of single elements and ranking and unranking of elements, all in time
at most $t(n)$ per operation. Then lexicographic ranking and unranking of permutations
can be performed in time $\O(n\cdot t(n))$.

\proof
Let us analyse the above algorithms. The depth of the recursion is~$n$ and in each
nested invokation of the recursive procedure we perform a~constant number of operations.
All of them are either trivial or factorials (which can be precomputed in~$\O(n)$ time)
or operations on the data structure.
\qed

\example
If we store~$A$ in an~ordinary array, we have insertion and deletion in constant time,
but ranking and unranking in~$\O(n)$, so $t(n)=\O(n)$ and the algorithm is quadratic.
Binary search trees give $t(n)=\O(\log n)$. The data structure of Dietz \cite{dietz:oal}
improves it to $t(n)=O(\log n/\log \log n)$. In fact, all these variants are equivalent
to the classical algorithms based on inversion vectors, because at the time of processing~$\pi[i]$,
the value of $R_A(\pi[i])$ is exactly the number of elements forming inversions with~$\pi[i]$.

If we relax the requirements on the data structure to allow ordering of elements
dependent on the history of the structure (i.e., on the sequence of deletes performed
so far), we can implement all three operations in time $\O(1)$. We store
the values in an array and we also keep an inverse permutation. Ranking is done by direct
indexing, unranking by indexing of the inverse permutation and deleting by swapping
the element with the last element. We can observe that although it no longer
gives the lexicographic order, the unranking function is still the inverse of the
ranking function (the sequence of deletes from~$A$ is the same in both functions),
so this leads to $\O(n)$-time ranking and unranking in a~non-lexicographic order.
This order is the same as the one used by Myrvold and Ruskey in~\cite{myrvold:rank}.

However, for our purposes we need to keep the same order on~$A$ over the whole
course of the algorithm. We will make use of the representation of vectors
by integers on the RAM as developed in Section~\ref{bitsect}, but first of
all, we will note that the ranks are large numbers, so the word size of the machine
has to be large as well for them to fit:

\obs
$\log n! = \Theta(n\log n)$, therefore the word size~$W$ must be~$\Omega(n\log n)$.

\proof
We have $n^n \ge n! \ge \lfloor n/2\rfloor^{\lfloor n/2\rfloor}$, so $n\log n \ge \log n! \ge \lfloor n/2\rfloor\cdot\log \lfloor n/2\rfloor$.
\qed

\thmn{Ranking of permutations \cite{mm:rank}} When we order the permutations
on the set~$[n]$ lexicographically, both ranking and unranking can be performed on the
RAM in time~$\O(n)$.

\proof
We will store the elements of the set~$A$ as a~sorted vector. Each element has
$\O(\log n)$ bits, so the whole vector requires $\O(n\log n)$ bits, which by the
above observation fits in a~constant number of machine words. We know from
Algorithm~\ref{vecops} that ranks can be calculated in constant time in such
vectors and that insertions and deletions can be translated to ranks and
masking. Unranking, that is indexing of the vector, is masking alone.
So we can apply the previous Lemma~\ref{ranklemma} with $t(n)=\O(1)$.
\qed

\endpart