although the authors rarely mention the details. In some cases, a~non-uniform variant
of the Word-RAM is considered as well (e.g., in~\cite{hagerup:dd}):
-\defn
+\defn\id{nonuniform}%
A~Word-RAM is called \df{weakly non-uniform,} if it is equipped with $\O(1)$-time
access to a~constant number of word-sized constants, which depend only on the word
size. These are called \df{native constants} and they are available in fixed memory
will be denoted as $x_0,\ldots,x_{d-1}$.
\notan{Bit strings}\id{bitnota}%
-We will work with binary representations of natural numbers as strings over the
+We will work with binary representations of natural numbers by strings over the
alphabet $\{\0,\1\}$: we will use $\(x)$ for the number~$x$ written in binary,
-$\(x)_b$ for the same padded to exactly $b$ bits by adding zeroes at the beginning.
-Strings will be denoted by Greek letters and the usual conventions will be utilized
-for string operations: $\alpha\beta$ for concatenation and $\alpha^k$ for the
-string~$\alpha$ repeated $k$~times. When the meaning is clear from the context,
+$\(x)_b$ for the same padded to exactly $b$ bits by adding zeroes at the beginning
+and $x[k]$ for the value of the $k$-th bit of~$x$.
+The usual conventions for operations on strings will be utilized: When $s$
+and~$t$ are strings, we write $st$ for their concatenation and
+$s^k$ for the string~$s$ repeated $k$~times.
+When the meaning is clear from the context,
we will use $x$ and $\(x)$ interchangeably to avoid outbreak of symbols.
\defn
The \df{bitwise encoding} of a~vector ${\bf x}=(x_0,\ldots,x_{d-1})$ of~$b$-bit numbers
-is an~integer $\0\(x_{d-1})_b\0\(x_{d-2})_b\0\ldots\0\(x_0)_b = \sum_i 2^{(b+1)i}\cdot x_i$.
+is an~integer~$x$ such that $\(x)=\0\(x_{d-1})_b\0\(x_{d-2})_b\0\ldots\0\(x_0)_b = \sum_i 2^{(b+1)i}\cdot x_i$.
+(We have interspersed the elements with \df{separator bits.})
\para
If we wish for the whole vector to fit in a~single word, we need $(b+1)d\le W$.
By using multiple-precision arithmetics, we can encode all vectors satisfying $bd=\O(W)$.
We will now describe how to translate simple vector manipulations to $\O(1)$ RAM operations
-on their codes.
+on their codes. As we are interested in asymptotic complexity only, we prefer clarity
+of the algorithms over saving instructions. Among other things, we freely use calculations
+on words of size $\O(bd)$, assuming that the Multiple-precision lemma comes to save us
+when necessary.
-\FIXME{Mention multi-prec. intermediates}
-\FIXME{Mention isolation of fields?}
-\FIXME{Fix notation and add it to the table}
-\FIXME{Masking}
-\FIXME{Optimality}
+\para
+First of all, let us observe that we can use $\band$ and $\bor$ with suitable constants
+to write zeroes or ones to an~arbitrary set of bit positions in constant time. These operations
+are usually called \df{bit masking}. Also, an~element of a~vector can be extracted or replaced
+using masking and shifts.
\newdimen\slotwd
\def\setslot#1{\setbox0=#1\slotwd=\wd0}
\def\alik#1{%
\medskip
-\halign{\hskip 0.2\hsize\hfil $ ##$&\hbox to 0.6\hsize{${}##$ \hss}\cr
+\halign{\hskip 0.15\hsize\hfil $ ##$&\hbox to 0.6\hsize{${}##$ \hss}\cr
#1}
\medskip
}
\>$\<Replicate>(\alpha)$ --- creates a~vector $(\alpha,\ldots,\alpha)$:
-\alik{\alpha*(\0^b\1)^d \cr}
+\alik{\<Replicate>(\alpha)=\alpha\cdot(\0^b\1)^d. \cr}
\>$\<Sum>(x)$ --- calculates the sum of the elements of~${\bf x}$, assuming that
it fits in~$b$ bits:
-\alik{x \bmod \1^{b+1} \cr}
+\alik{\<Sum>(x) = x \bmod \1^{b+1}. \cr}
-This works because when taken modulo~$\1^{b+1}$, the number $2^{b+1}=\1\0^{b+1}$
+This works because when we work 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 cannot change it.
-~ \0 \[y_{d-1}] \0 \[y_{d-2}] \[\cdots] \0 \[y_1] \0 \[y_0] \cr
}
-We replace the separator zeroes in~$x$ by ones (this is just an~$\bor$ with
-a~suitable constant) and subtract~$y$. These ones change back to zeroes exactly at
+We replace the separator zeroes in~$x$ by ones and subtract~$y$. These ones change back to zeroes exactly at
the positions where $x_i<y_i$ and they stop 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 zero out all other bits.
\<Sum>(\<Cmp>(x,\<Replicate>(\alpha))) \cr
}
-\>$\<Insert>(x,\alpha)$ --- insert~$\alpha$ to a~sorted vector $\bf x$:
+\>$\<Insert>(x,\alpha)$ --- insert~$\alpha$ into a~sorted vector $\bf x$:
Calculate $k = \<Rank>(x,\alpha)$ first, then insert~$\alpha$ as the $k$-th
field of~$\bf x$ using masking operations.
\>$\<Unpack>_\varphi(\alpha)$ --- like \<Unpack>, but changes the order of the
bits according to a~fixed permutation~$\varphi$: The $i$-th element of the
-resulting vector is equal to the $\pi(i)$-th bit of~$\alpha$.
+resulting vector is equal to~$\alpha[\pi(i)]$.
Implemented as above, but with mask~$y=(2^{\pi(b-1)},\ldots,2^{\pi(0)})$.
the separators. (Alternatively, we can observe that $\1^b$ is the only case
affected, so we can handle it separately.)
-\FIXME{Note constants}
-
\para
We can use the above tricks to perform interesting operations on individual
numbers in constant time, too. Let us assume for a~while that we are
Perform $\<Unpack>_\pi$ and \<Pack> back.
\>$\<LSB>(\alpha)$ --- find the least significant bit of~$\alpha$,
-i.e., the minimum~$i$ such that the $i$-th bit of~$\alpha$ is one.
+i.e., the minimum~$i$ such that $\alpha[i]=1$.
By a~combination of subtraction with $\bxor$, we create a~number
which contain ones exactly at positions below $\<LSB>(\alpha)$:
\algn{LSB in linear workspace}
-\algo
+\algo\id{lsb}
\algin A~$w$-bit number~$\alpha$.
\:$b\=\lceil\sqrt{w}\rceil$. \cmt{size of a~block}
\:$x\=(\alpha \band (\0\1^b)^b) \bor (\alpha \band (\1\0^b)^b)$.
\algout $\<LSB>(\alpha) = bp+q$.
\endalgo
+\rem
+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
+are the~$w$ and~$b$ in the LSB algorithm \ref{lsb}, which we are unable to produce
+in constant time.
+
+
\endpart
projects / saga.git / blobdiff