X-Git-Url: http://mj.ucw.cz/gitweb/?a=blobdiff_plain;f=ram.tex;h=d39d9ba73f727dcafe19cb5c368457090c72c09f;hb=e4ece93696bd08db4ef45ed8b29e76ba6c7848c4;hp=fad390c8e3e5bebe7e915c0af04788b4dafb66a1;hpb=049b863bee2992aed6230ac377dee53c4a08cc85;p=saga.git

diff --git a/ram.tex b/ram.tex
index fad390c..d39d9ba 100644
--- a/ram.tex
+++ b/ram.tex
@@ -3,6 +3,7 @@
 \fi
 
 \chapter{Fine Details of Computation}
+\id{ramchap}
 
 \section{Models and machines}
 
@@ -160,7 +161,7 @@ constant time).
 We can therefore view the whole memory as a~directed graph, whose vertices
 correspond to the cells (the registers are stored in a~single special cell).
 The outgoing edges of each vertex correspond to pointer fields of the cells and they are
-labelled with distinct labels drawn from a~finite set. In addition to that,
+labeled with distinct labels drawn from a~finite set. In addition to that,
 each vertex contains a~fixed amount of symbols. The program can directly access
 vertices within distance~2 from the register vertex.
 
@@ -306,6 +307,7 @@ scanning all~$n$ buckets takes $\O(n+m)$ time.
 %--------------------------------------------------------------------------------
 
 \section{Data structures on the RAM}
+\id{ramdssect}
 
 There is a~lot of data structures designed specifically for the RAM, taking
 advantage of both indexing and arithmetics. In many cases, they surpass the known
@@ -342,7 +344,7 @@ and Willard. It will also form a~basis for the rest of this chapter.
 
 %--------------------------------------------------------------------------------
 
-\section{Bits and vectors}
+\section{Bits and vectors}\id{bitsect}
 
 In this rather technical section, we will show how the RAM can be used as a~vector
 computer to operate in parallel on multiple elements, as long as these elements
@@ -429,7 +431,7 @@ replaced by a~different value in $\O(1)$ time by masking and shifts.
 \medskip
 }
 
-\algn{Operations on vectors with $d$~elements of $b$~bits each}
+\algn{Operations on vectors with $d$~elements of $b$~bits each}\id{vecops}
 
 \itemize\ibull
 
@@ -442,7 +444,7 @@ the result fits in $b$~bits:
 
 \alik{\<Sum>(x) = x \bmod \1^{b+1}. \cr}
 
-This works because when we work modulo~$\1^{b+1}$, the number $2^{b+1}=\1\0^{b+1}$
+This is correct because when we calculate modulo~$\1^{b+1}$, the number $2^{b+1}=\1\0^{b+1}$
 is congruent to~1 and thus $x = \sum_i 2^{(b+1)i}\cdot x_i \equiv \sum_i 1^i\cdot x_i \equiv \sum_i x_i$.
 As the result should fit in $b$~bits, the modulo makes no difference.
 
@@ -490,20 +492,27 @@ carries from propagating, so the fields do not interact with each other:
 It only remains to shift the separator bits to the right positions, negate them
 and mask out all other bits.
 
-\:$\<Rank>(x,\alpha)$ --- return the number of elements of~${\bf x}$ which are less than~$\alpha$,
+\:$\<Rank>(x,\alpha)$ --- returns the number of elements of~${\bf x}$ which are less than~$\alpha$,
 assuming that the result fits in~$b$ bits:
 
 \alik{
 \<Rank>(x,\alpha) = \<Sum>(\<Cmp>(x,\<Replicate>(\alpha))). \cr
 }
 
-\:$\<Insert>(x,\alpha)$ --- insert~$\alpha$ into a~sorted vector $\bf x$:
+\:$\<Insert>(x,\alpha)$ --- inserts~$\alpha$ into a~sorted vector $\bf x$:
 
-Calculate $k = \<Rank>(x,\alpha)$ first, then insert~$\alpha$ as the $k$-th
+We calculate the rank of~$\alpha$ in~$x$ first, then we insert~$\alpha$ as the $k$-th
 field of~$\bf x$ using masking operations and shifts.
 
-\:$\<Unpack>(\alpha)$ --- create a~vector whose elements are the bits of~$\(\alpha)_d$.
-In other words, insert blocks~$\0^b$ between the bits of~$\alpha$. Assuming that $b\ge d$,
+\algo
+\:$k\=\<Rank>(x,\alpha)$.
+\:$\ell\=x \band \1^{(b+1)(n-k-1)}\0^{(b+1)(k+1)}$. \cmt{``left'' part of the vector}
+\:$r=x \band \1^{(b+1)k}$. \cmt{``right'' part}
+\:Return $(\ell\shl (b+1)) \bor (\alpha\shl ((b+1)k)) \bor r$.
+\endalgo
+
+\:$\<Unpack>(\alpha)$ --- creates a~vector whose elements are the bits of~$\(\alpha)_d$.
+In other words, inserts blocks~$\0^b$ between the bits of~$\alpha$. Assuming that $b\ge d$,
 we can do it as follows:
 
 \algo
@@ -550,7 +559,7 @@ affected, so we can handle it separately.)
 \endlist
 
 \para
-We can use the above tricks to perform interesting operations on individual
+We can use the aforementioned tricks to perform interesting operations on individual
 numbers in constant time, too. Let us assume for a~while that we are
 operating on $b$-bit numbers and the word size is at least~$b^2$.
 This enables us to make use of intermediate vectors with $b$~elements
@@ -560,16 +569,16 @@ of $b$~bits each.
 
 \itemize\ibull
 
-\:$\<Weight>(\alpha)$ --- compute the Hamming weight of~$\alpha$, i.e., the number of ones in~$\(\alpha)$.
+\:$\<Weight>(\alpha)$ --- computes the Hamming weight of~$\alpha$, i.e., the number of ones in~$\(\alpha)$.
 
-Perform \<Unpack> and then \<Sum>.
+We perform \<Unpack> and then \<Sum>.
 
-\:$\<Permute>_\pi(\alpha)$ --- shuffle the bits of~$\alpha$ according
+\:$\<Permute>_\pi(\alpha)$ --- shuffles the bits of~$\alpha$ according
 to a~fixed permutation~$\pi$.
 
-Perform $\<Unpack>_\pi$ and \<Pack> back.
+We perform $\<Unpack>_\pi$ and \<Pack> back.
 
-\:$\<LSB>(\alpha)$ --- find the least significant bit of~$\alpha$,
+\:$\<LSB>(\alpha)$ --- finds the least significant bit of~$\alpha$,
 i.e., the smallest~$i$ such that $\alpha[i]=1$.
 
 By a~combination of subtraction with $\bxor$, we create a~number
@@ -583,7 +592,7 @@ which contains ones exactly at the position of $\<LSB>(\alpha)$ and below:
 
 Then we calculate the \<Weight> of the result and subtract~1.
 
-\:$\<MSB>(\alpha)$ --- find the most significant bit of~$\alpha$ (the position
+\:$\<MSB>(\alpha)$ --- finds the most significant bit of~$\alpha$ (the position
 of the highest bit set).
 
 Reverse the bits of the number~$\alpha$ first by calling \<Permute>, then apply \<LSB>
@@ -627,9 +636,381 @@ relocate the bits we have overwritten.}
 We have used a~plenty of constants which depend on the format of the vectors.
 Either we can write non-uniform programs (see \ref{nonuniform}) and use native constants,
 or we can observe that all such constants can be easily manufactured. For example,
-$(\0^b\1)^d = \1^{bd} / \1^{b+1} = (2^{bd}-1)/(2^{b+1}-1)$. The only exceptions
+$(\0^b\1)^d = \1^{(b+1)d} / \1^{b+1} = (2^{(b+1)d}-1)/(2^{b+1}-1)$. The only exceptions
 are the~$w$ and~$b$ in the LSB algorithm \ref{lsb}, which we are unable to produce
-in constant time.
+in constant time. In practice we use the ``bit tricks'' as frequently called subroutines
+in an~encompassing algorithm, so we usually can spend a~lot of time on the precalculation
+of constants performed once during algorithm startup.
+
+%--------------------------------------------------------------------------------
+
+\section{Q-Heaps}\id{qheaps}%
+
+We have shown how to perform non-trivial operations on a~set of values
+in constant time, but so far only under the assumption that the number of these
+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~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 $k\le W^{1/4}$
+and $W$ is the word size of the machine. It will support insertion, deletion, finding
+of minimum and maximum, and other operations described below, in constant time, provided that
+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.
+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
+in practical algorithms. Many of the tricks used however prove themselves
+useful even in real-life implementations.
+
+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 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
+them we spend $\poly(k)$ time on calculating the function. It remains
+to observe that $2^{\O(k^3)}\cdot \poly(k) = \O(2^{k^4})$.
+\qed
+
+\para
+We will first show an~auxiliary construction based on tries and then derive
+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 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 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
+
+\defn
+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 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
+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 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 other vertex, the
+corresponding prefix is equal to the concatenation of edge labels on the path
+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.
+
+Furthermore, the labels of all edges leaving a~common vertex are always
+distinct and when we compress the trie, no two such labels have share their initial
+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~$x$,
+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
+$\O(\vert S\vert)$ vertices (this can be easily shown by induction on~$\vert S\vert$)
+and all edge labels can be represented in space linear in the sum of the
+lengths of the strings in~$S$.
+
+\defn
+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 \}$.
+
+\obs
+The trie~$T$ has several interesting properties. Since all words in~$S$ have the same
+length, the leaves of the trie correspond to these exact words, that is to the numbers~$x_i$.
+The inorder traversal of the trie enumerates the words of~$S$ in lexicographic order
+and therefore also the~$x_i$'s in the order of their values. Between each
+pair of leaves $x_i$ and~$x_{i+1}$ it visits an~internal vertex whose letter depth
+is exactly~$W-1-g_i$.
+
+\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 $x\in
+X$, 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 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
+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 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~$i-1$.
+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
+where~$\(x)$ disagrees with a~label. Before this point, all edges not taken by
+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 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 a~table to
+be efficiently precomputed. We will carefully choose an~equivalent representation
+of the trie which is compact enough.
+
+\lemma\id{citree}%
+The trie is uniquely determined by the order of the guides~$g_1,\ldots,g_{n-1}$.
+
+\proof
+We already know that the letter depths of the trie vertices are exactly
+the numbers~$W-1-g_i$. The root of the trie must have the smallest of these
+letter depths, i.e., it must correspond to the highest numbered bit. Let
+us call this bit~$g_i$. This implies that the values $x_1,\ldots,x_i$
+must lie in the left subtree of the root and $x_{i+1},\ldots,x_n$ in its
+right subtree. Both subtrees can be then constructed recursively.\foot{This
+construction is also known as the \df{cartesian tree} for the sequence
+$g_1,\ldots,g_n$ and it is useful in many other algorithms as it can be
+built in $\O(n)$ time. A~nice application on the Lowest Common Ancestor
+and Range Minimum problems has been described by Bender et al.~in \cite{bender:lca}.}
+\qed
+
+\para
+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\id{qhnota}%
+\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 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 $x[B[1]],
+x[B[2]],\ldots, x[B[n]]$.
+\endlist
 
+\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 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~$B$, 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 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:
+\itemize\ibull
+\:the function~$G$,
+\:the values $x_1,\ldots,x_n$,
+\:the bit string~$x[B]$,
+\:$x$ itself.
+\endlist
+
+\proof
+Let us prove that all ingredients of Lemma~\ref{qhdeterm} are either small
+enough or computable in constant time.
+
+We know that the shape of the trie~$T$ is uniquely determined by the order of the $g_i$'s
+and therefore by the function~$G$ since the array~$B$ is sorted. The shape of
+the trie together with the bits in $x[B]$ determine the leaf~$x_i$ found when searching
+for~$x$ using only the guides. This can be computed in polynomial time and it
+depends on $\O(k\log k)$ bits of input, so according to Lemma~\ref{qhprecomp}
+we can look it up in a~precomputed table.
+
+The relation between $x$ and~$x_i$ can be obtained directly as we know the~$x_i$.
+The bit position of the first disagreement can be calculated in constant time
+using the LSB/MSB algorithm (\ref{lsb}).
+
+All these ingredients can be stored in $\O(k\log k)$ bits, so we may assume
+that the rank can be looked up in constant time as well.
+\qed
+
+\para
+In the Q-heap we would like to store the set~$X$ as a~sorted array together
+with the corresponding trie, which will allow us to determine the position
+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 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 for the description
+of the basic operations on it.
+
+\defn
+A~\df{Q-heap} consists of:
+\itemize\ibull
+\:$k$, $n$ --- 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 $X[\varrho(i)]$),
+\:$B$ --- 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~$B$
+(a~vector of~$\O(n\log k)$ bits),
+\:precomputed tables of various functions.
+\endlist
+
+\algn{Search in the Q-heap}\id{qhfirst}%
+\algo
+\algin A~Q-heap and an~integer~$x$ to search for.
+\:$i\=R_X(x)+1$, using Lemma~\ref{qhrank} to calculate the rank.
+\:If $i\le n$ return $x_i$, otherwise return {\sc undefined.}
+\algout The smallest element of the heap which is greater or equal to~$x$.
+\endalgo
+
+\algn{Insertion to the Q-heap}
+\algo
+\algin A~Q-heap and an~integer~$x$ to insert.
+\:$i\=R_X(x)+1$, using Lemma~\ref{qhrank} to calculate the rank.
+\:If $x=x_i$, return immediately (the value is already present).
+\:Insert the new value to~$X$:
+\::$n\=n+1$.
+\::$X[n]\=x$.
+\::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~$G$.
+\::Recalculate $g_{i-1}$ and~$g_i$ according to the definition.
+   \hfil\break Update~$B$ and~$G$ as described in~\ref{qhsetb}.
+\algout The updated Q-heap.
+\endalgo
+
+\algn{Deletion from the Q-heap}
+\algo
+\algin A~Q-heap and an~integer~$x$ to be deleted from it.
+\:$i\=R_X(x)+1$, using Lemma~\ref{qhrank} to calculate the rank.
+\:If $i>n$ or $x_i\ne x$, return immediately (the value is not in the heap).
+\:Delete the value from~$X$:
+\::$X[\varrho(i)]\=X[n]$.
+\::Find $j$ such that~$\varrho(j)=n$ and set $\varrho(j)\=\varrho(i)$.
+\::$n\=n-1$.
+\:Update the $g_j$'s like in the previous algorithm.
+\algout The updated Q-heap.
+\endalgo
+
+\algn{Finding the $i$-th smallest element in the Q-heap}\id{qhlast}%
+\algo
+\algin A~Q-heap and an~index~$i$.
+\:If $i<1$ or $i>n$, return {\sc undefined.}
+\:Return~$x_i$.
+\algout The $i$-th smallest element in the heap.
+\endalgo
+
+\para
+The heap algorithms we have just described have been built from primitives
+operating in constant time, with one notable exception: the extraction
+$x[B]$ of all bits of~$x$ at positions specified by the set~$B$. This cannot be done
+in~$\O(1)$ time on the Word-RAM, but we can implement it with ${\rm AC}^0$
+instructions as suggested by Andersson in \cite{andersson:fusion} or even
+with those ${\rm AC}^0$ instructions present on real processors (see Thorup
+\cite{thorup:aczero}). On the Word-RAM, we need to make use of the fact
+that the set~$B$ is not changing too much --- there are $\O(1)$ changes
+per Q-heap operation. As Fredman and Willard have shown, it is possible
+to maintain a~``decoder'', whose state is stored in $\O(1)$ machine words,
+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,\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
+can be updated in constant time, as long as $\vert B\vert \le k$.
+
+\proof
+See Fredman and Willard \cite{fw:transdich}.
+\qed
+
+\para
+This was the last missing bit of the mechanics of the Q-heaps. We are
+therefore ready to conclude this section by the following theorem
+and its consequences:
+
+\thmn{Q-heaps, Fredman and Willard \cite{fw:transdich}}\id{qh}%
+Let $W$ and~$k$ be positive integers such that $k=\O(W^{1/4})$. Let~$Q$
+be a~Q-heap of at most $k$-elements of $W$~bits each. Then the Q-heap
+operations \ref{qhfirst} to \ref{qhlast} on~$Q$ (insertion, deletion,
+search for a~given value and search for the $i$-th smallest element)
+run in constant time on a~Word-RAM with word size~$W$, after spending
+time $\O(2^{k^4})$ on the same RAM on precomputing of tables.
+
+\proof
+Every operation on the Q-heap can be performed in a~constant number of
+vector operations and calculations of ranks. The ranks are computed
+in $\O(1)$ steps involving again $\O(1)$ vector operations, binary
+logarithms and bit extraction. All these can be calculated in constant
+time using the results of section \ref{bitsect} and Lemma \ref{qhxtract}.
+\qed
+
+\rem
+We can also use the Q-heaps as building blocks of more complex structures
+like Atomic heaps and AF-heaps (see once again \cite{fw:transdich}). We will
+show a~simpler, but useful construction, sometimes called the \df{Q-heap tree.}
+Suppose we have a~Q-heap of capacity~$k$ and a~parameter $d\in{\bb N}^+$. We
+can build a~balanced $k$-ary tree of depth~$d$ such that its leaves contain
+a~given set and every internal vertex keeps the minimum value in the subtree
+rooted in it, together with a~Q-heap containing the values in all its sons.
+This allows minimum to be extracted in constant time (it is placed in the root)
+and when any element is changed, it is sufficient to recalculate the values
+from the path from this element to the root, which takes $\O(d)$ Q-heap
+operations.
+
+\corn{Q-heap trees}\id{qhtree}%
+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 $W=\Omega(r^{\delta})$, after spending time
+$\O(2^{r^\delta})$ on precomputing of tables.
+
+\proof
+Choose $\delta' \le \delta$ such that $r^{\delta'} = \O(W^{1/4})$. 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~$\log N$
+and we can spend time $\O(N)$ on preprocessing, so we can choose $r=\log N$ and
+$\delta=1$ in the above corollary and get a~heap of size $\log N$ working in
+constant time per operation.
 
 \endpart