values is small enough and that the values themselves are also small enough
(so that the whole set fits in $\O(1)$ machine words). Now we will show how to
lift the restriction on the magnitude of the values and still keep constant time
-complexity. We will describe a~simplified version of the Q-heaps developed by
+complexity. We will describe a~slightly simplified version of the Q-heaps developed by
Fredman and Willard in~\cite{fw:transdich}.
The Q-heap represents a~set of at most~$k$ word-sized integers, where $k\le W^{1/4}$
we are willing to spend~$\O(2^{k^4})$ time on preprocessing.
The exponential-time preprocessing may sound alarming, but a~typical application uses
-Q-heaps of size $k=\log^{1/4} N$, where $N$ is the size of the algorithm's input,
-which guarantees that $k\le W^{1/4}$ and $\O(2^{k^4}) = \O(N)$. Let us however
+Q-heaps of size $k=\log^{1/4} N$, where $N$ is the size of the algorithm's input.
+This guarantees that $k\le W^{1/4}$ and $\O(2^{k^4}) = \O(N)$. Let us however
remark that the whole construction is primarily of theoretical importance
and that the huge constants involved everywhere make these heaps useless
-for practical algorithms. However, many of the tricks used prove themselves
+in practical algorithms. Many of the tricks used however prove themselves
useful even in real-life implementations.
-Preprocessing makes it possible to precompute tables for almost arbitrary functions
-and then assume that they can be evaluated in constant time:
+Spending the time on reprocessing makes it possible to precompute tables for
+almost arbitrary functions and then assume that they can be evaluated in
+constant time:
\lemma\id{qhprecomp}%
When~$f$ is a~function computable in polynomial time, $\O(2^{k^4})$ time is enough
-to precompute a~table of the values of~$f$ for the values of its arguments whose total
-bit size is $\O(k^3)$.
+to precompute a~table of the values of~$f$ for all arguments whose size is $\O(k^3)$ bits.
\proof
There are $2^{\O(k^3)}$ possible combinations of arguments of the given size and for each of
\para
We will first show an~auxiliary construction based on tries and then derive
-the actual definition of the Q-heap from it.
+the real definition of the Q-heap from it.
\nota
Let us introduce some notation first:
\itemize\ibull
\:$W$ --- the word size of the RAM,
\:$k = \O(W^{1/4})$ --- the limit on the size of the heap,
-\:$n\le k$ --- the number of elements in the represented set,
+\:$n\le k$ --- the number of elements stored in the heap,
\:$X=\{x_1, \ldots, x_n\}$ --- the elements themselves: distinct $W$-bit numbers
indexed in a~way that $x_1 < \ldots < x_n$,
-\:$g_i = \<MSB>(x_i \bxor x_{i+1})$ --- the most significant bit of those in which $x_i$ and~$x_{i+1}$ differ,
-\:$R_X(x)$ --- the rank of~$x$ in~$X$, that is the number of elements of~$X$, which are less than~$x$
+\:$g_i = \<MSB>(x_i \bxor x_{i+1})$ --- the position of the most significant bit in which $x_i$ and~$x_{i+1}$ differ,
+\:$R_X(x)$ --- the rank of~$x$ in~$X$, that is the number of elements of~$X$ which are less than~$x$
(where $x$~itself need not be an~element of~$X$).\foot{We will dedicate the whole chapter \ref{rankchap} to the
study of various ranks.}
\endlist
A~\df{trie} for a~set of strings~$S$ over a~finite alphabet~$\Sigma$ is
a~rooted tree whose vertices are the prefixes of the strings in~$S$ and there
is an~edge going from a~prefix~$\alpha$ to a~prefix~$\beta$ iff $\beta$ can be
-obtained from~$\alpha$ by adding a~single symbol of the alphabet. The edge
-will be labeled with the particular symbol. We will also define a~\df{letter depth}
-of a~vertex to be the length of the corresponding prefix and mark the vertices
+obtained from~$\alpha$ by appending a~single symbol of the alphabet. The edge
+will be labeled with the particular symbol. We will also define the~\df{letter depth}
+of a~vertex as the length of the corresponding prefix. We mark the vertices
which match a~string of~$S$.
-A~\df{compressed trie} is obtained from the trie by removing the vertices of outdegree~1.
-Whereever is a~directed path whose internal vertices have outdegree~1, we replace this
-path by a~single edge labeled with the contatenation of the original edge's labels.
+A~\df{compressed trie} is obtained from the trie by removing the vertices of outdegree~1
+except for the root and marked vertices.
+Whereever is a~directed path whose internal vertices have outdegree~1 and they carry
+no mark, we replace this path by a~single edge labeled with the contatenation
+of the original edge's labels.
-In both kinds of tries, we will order the outgoing edges of every vertex by their labels
+In both kinds of tries, we order the outgoing edges of every vertex by their labels
lexicographically.
\obs
-In both tries, the root of the tree is the empty word and for every vertex, the
+In both tries, the root of the tree is the empty word and for every other vertex, the
corresponding prefix is equal to the concatenation of edge labels on the path
-leading from the root to the vertex. The letter depth of the vertex is equal to
+leading from the root to that vertex. The letter depth of the vertex is equal to
the total size of these labels. All leaves correspond to strings in~$S$, but so can
some internal vertices if there are two strings in~$S$ such that one is a~prefix
of the other.
symbols. This allows us to search in the trie efficiently: when looking for
a~string~$x$, we follow the path from the root and whenever we visit
an~internal vertex of letter depth~$d$, we test the $d$-th character of~$x$,
-follow the edge whose label starts with this character, and check that the
+we follow the edge whose label starts with this character, and we check that the
rest of the label matches.
The compressed trie is also efficient in terms of space consumption --- it has
lengths of the strings in~$S$.
\defn
-For our set~$X$, we will define~$T$ as a~compressed trie for the set of binary
+For our set~$X$, we define~$T$ as a~compressed trie for the set of binary
encodings of the numbers~$x_i$, padded to exactly $W$~bits, i.e., for $S = \{ \(x)_W ; x\in X \}$.
\obs
\para
Let us now modify the algorithm for searching in the trie and make it compare
only the first symbols of the edges. In other words, we will test only the bits~$g_i$
-which will be called \df{guides,} as they guide us through the tree. For $x\in
+which will be called \df{guides} (as they guide us through the tree). For $x\in
X$, the modified algorithm will still return the correct leaf. For all~$x$ outside~$X$
it will no longer fail and instead it will land on some leaf~$x_i$. At the
-first sight this vertex may seem unrelated, but we will show that it can be
+first sight the number~$x_i$ may seem unrelated, but we will show that it can be
used to determine the rank of~$x$ in~$X$, which will later form a~basis for all
-other Q-heap operations:
+Q-heap operations:
\lemma\id{qhdeterm}%
The rank $R_X(x)$ is uniquely determined by a~combination of:
\itemize\ibull
\:the trie~$T$,
-\:the index~$i$ of the leaf visited when searching for~$x$ in~$T$,
+\:the index~$i$ of the leaf found when searching for~$x$ in~$T$,
\:the relation ($<$, $=$, $>$) between $x$ and $x_i$,
\:the bit position $b=\<MSB>(x\bxor x_i)$ of the first disagreement between~$x$ and~$x_i$.
\endlist
\proof
-If $x\in X$, we detect that from $x_i=x$ and the rank is obviously~$i$ itself.
+If $x\in X$, we detect that from $x_i=x$ and the rank is obviously~$i-1$.
Let us assume that $x\not\in X$ and imagine that we follow the same path as when
searching for~$x$,
but this time we check the full edge labels. The position~$b$ is the first position
the search were leading either to subtrees containing elements all smaller than~$x$
or all larger than~$x$ and the only values not known yet are those in the subtree
below the edge which we currently consider. Now if $x[b]=0$ (and therefore $x<x_i$),
-all values in the subtree have $x_j[b]=1$ and thus they are larger. In the other
+all values in that subtree have $x_j[b]=1$ and thus they are larger than~$x$. In the other
case, $x[b]=1$ and $x_j[b]=0$, so they are smaller.
\qed
\para
The preceding lemma shows that the rank can be computed in polynomial time, but
-unfortunately the variables on which it depends are too large for the table to
-be efficiently precomputed. We will therefore carefully choose a~representation
+unfortunately the variables on which it depends are too large for a~table to
+be efficiently precomputed. We will carefully choose an~equivalent representation
of the trie which is compact enough.
\lemma\id{citree}%
\qed
\para
-However, the vector of the $g_i$'s is also too long (is has $k\log W$ bits
+Unfortunately, the vector of the $g_i$'s is also too long (is has $k\log W$ bits
and we have no upper bound on~$W$ in terms of~$k$), so we will compress it even
further:
\nota
\itemize\ibull
\:$B = \{g_1,\ldots,g_n\}$ --- the set of bit positions of all the guides, stored as a~sorted array,
-\:$G : \{1,\ldots,n\} \rightarrow \{1,\ldots,n\}$ --- a~function such that
-$g_i = B[G(i)]$, which maps guides to their bit positions,
+\:$G : \{1,\ldots,n\} \rightarrow \{1,\ldots,n\}$ --- a~function mapping
+the guides to their bit positions in~$B$: $g_i = B[G(i)]$,
\:$x[B]$ --- a~bit string containing the bits of~$x$ originally located
at the positions given by~$B$, i.e., the concatenation of bits $x[B[1]],
x[B[2]],\ldots, x[B[n]]$.
\obs\id{qhsetb}%
The set~$B$ has $\O(k\log W)=\O(W)$ bits, so it can be stored in a~constant number
-of machine words as a~sorted vector. The function~$G$ can be also stored as a~vector
-of $k\log k$ bits. We can change a~single~$g_i$ in constant time using
+of machine words in form of a~sorted vector. The function~$G$ can be also stored as a~vector
+of $\O(k\log k)$ bits. We can change a~single~$g_i$ in constant time using
vector operations: First we delete the original value of~$g_i$ from~$B$ if it
is not used anywhere else. Then we add the new value to~$B$ if it was not
there yet and we write its position in~$B$ to~$G(i)$. Whenever we insert
-or delete a~value in~$B$, values at the higher positions shift one position
+or delete a~value in~$B$, the values at the higher positions shift one position
up or down and we have to update the pointers in~$G$. This can be fortunately
-accomplished by adding or subtracting a~result of parallel comparison.
+accomplished by adding or subtracting a~result of vector comparison.
+
+In this representation, we can reformulate our lemma on ranks as follows:
\lemma\id{qhrank}%
The rank $R_X(x)$ can be computed in constant time from:
for a~newly inserted element in constant time. However, the set is too large
to fit in a~vector and we cannot perform insertion on an~ordinary array in
constant time. This can be worked around by keeping the set in an~unsorted
-array together with a~vector containing a~permutation which sorts the array.
+array together with a~vector containing the permutation which sorts the array.
We can then insert a~new element at an~arbitrary place in the array and just
update the permutation to reflect the correct order.
-We are now ready for the real definition of the Q-heap and the description
+We are now ready for the real definition of the Q-heap and for the description
of the basic operations on it.
\defn
\:$k$, $n$ --- the capacity of the heap and the current number of elements (word-sized integers),
\:$X$ --- the set of word-sized elements stored in the heap (an~array of words in an~arbitrary order),
\:$\varrho$ --- a~permutation on~$\{1,\ldots,n\}$ such that $X[\varrho(1)] < \ldots < X[\varrho(n)]$
-(a~vector of $\O(n\log k)$ bits; we will write $x_i$ for $X[\varrho[i]]$),
+(a~vector of $\O(n\log k)$ bits; we will write $x_i$ for $X[\varrho(i)]$),
\:$B$ --- a~set of ``interesting'' bit positions
(a~sorted vector of~$\O(n\log W)$ bits),
\:$G$ --- the function which maps the guides to the bit positions in~$B$
\:$i\=R_X(x)+1$, using Lemma~\ref{qhrank} to calculate the rank.
\:If $x=x_i$, return immediately (the value is already present).
\:Insert the new value to~$X$:
-\::$n\=n+1$
-\::$X[n]\=x$ and insert~$n$ at the $i$-th position in the permutation~$\varrho$.
+\::$n\=n+1$.
+\::$X[n]\=x$.
+\::Insert~$n$ at the $i$-th position in the permutation~$\varrho$.
\:Update the $g_j$'s:
\::Move all~$g_j$ for $j\ge i$ one position up. \hfil\break
This translates to insertion in the vector representing~$G$.
\::$X[\varrho(i)]\=X[n]$.
\::Find $j$ such that~$\varrho(j)=n$ and set $\varrho(j)\=\varrho(i)$.
\::$n\=n-1$.
-\:Update the $g_i$'s like in the previous algorithm.
+\:Update the $g_j$'s like in the previous algorithm.
\algout The updated Q-heap.
\endalgo
and which helps us to extract $x[B]$ in a~constant number of operations:
\lemman{Extraction of bits}\id{qhxtract}%
-Under the assumptions on~$k$, $W$ and the preprocessing time as in the Q-heaps,
+Under the assumptions on~$k$, $W$ and the preprocessing time as in the Q-heaps,\foot{%
+Actually, this is the only place where we need~$k$ to be as low as $W^{1/4}$.
+In the ${\rm AC}^0$ implementation, it is enough to ensure $k\log k\le W$.
+On the other hand, we need not care about the exponent because it can
+be arbitrarily increased using the Q-heap trees described below.}
it is possible to maintain a~data structure for a~set~$B$ of bit positions,
which allows~$x[B]$ to be extracted in $\O(1)$ time for an~arbitrary~$x$.
When a~single element is inserted to~$B$ or deleted from~$B$, the structure
For every positive integer~$r$ and $\delta>0$ there exists a~data structure
capable of maintaining the minimum of a~set of at most~$r$ word-sized numbers
under insertions and deletions. Each operation takes $\O(1)$ time on a~Word-RAM
-with word size at least $\O(r^{\delta})$, after spending time
+with word size $W=\Omega(r^{\delta})$, after spending time
$\O(2^{r^\delta})$ on precomputing of tables.
\proof
-Choose $\delta' \le \delta$ such that $r^{\delta'} = \O(W^{1/4})$, where $W$
-is the word size of the RAM. Build a~Q-heap tree of depth $d=\lceil \delta/\delta'\rceil$
-containing Q-heaps of size $k=r^{\delta'}$.
-\qed
+Choose $\delta' \le \delta$ such that $r^{\delta'} = \O(W^{1/4})$. Build
+a~Q-heap tree of depth $d=\lceil \delta/\delta'\rceil$ containing Q-heaps of
+size $k=r^{\delta'}$. \qed
\rem\id{qhtreerem}%
When we have an~algorithm with input of size~$N$, the word size is at least~$\log N$
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