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Commit 328470af authored by Filip Stedronsky's avatar Filip Stedronsky
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Succinct: progress

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......@@ -12,4 +12,4 @@ label((0, dist), "\vbox{\hbox{$x\in[X]$}\hbox{\eightrm (input)}}", N);
label((-dist, 0), "\vbox{\hbox{$y\in[Y]$}\hbox{\eightrm (carry in)}}", W);
label((dist, 0), "\vbox{\hbox{$s\in[S]$}\hbox{\eightrm (carry out)}}", E);
label((0, -dist), "\vbox{\hbox{$m\in[2^M]$}\hbox{\eightrm (output)}}", S);
label((0, 0), "$t\in [T]$");
label((0, 0), "$c\in [C]$");
......@@ -153,7 +153,7 @@ into a pair from alphabets $[C]$ and $[D]$ as long as $C\cdot D
\ge A \cdot B$ (we need an output universe large enough to hold all
possible input combinations).
We will use this kind of alphabet re-encoding by pair heavily in the SOLE
We will use this kind of alphabet re-encoding by pairs heavily in the SOLE
encoding. The best way to explain the exact scheme is with a diagram (fig. \figref{sole}).
\figure[sole]{sole.pdf}{}{SOLE alphabet re-encoding scheme}
......@@ -236,30 +236,54 @@ into the output. The final carry is then used to output some extra blocks at the
\figure[mixer]{mixer.pdf}{}{General structure of a mixer}
At a high level, a mixer can be thought of as a mapping $f: [X]\times[Y] \rightarrow [2^M]\times[S]$
with the property that when $(m,s) = f(x,y)$, $s$ depends only on $x$.
with the property that when $(m,s) = f(x,y)$, $s$ depends only on $x$. This is the key property
that allows local decoding and modification because carry does not cascade.
Internally, the a mixer is
always implemented as a composition of two mappings, $f_1$ that transforms $x \rightarrow (t,s)$ and $f2$
that transforms $(y,t) \rightarrow m$. See fig. \figref{mixer}. Both $f_1$ and $f_2$ must be injective
always implemented as a composition of two mappings, $f_1$ that transforms $x \rightarrow (c,s)$ and $f2$
that transforms $(y,c) \rightarrow m$. See fig. \figref{mixer}. Both $f_1$ and $f_2$ must be injective
so that the encoding is reversible.
The mappings $f_1$ and $f_2$ themselves are trivial alphabet translations similar to what we
used in the SOLE encoding. You can for example use $f_1(x) = (\lceil x/S \rceil, x \bmod S)$
and $f_2(y,t) = t\cdot Y + y$.
and $f_2(y,c) = c\cdot Y + y$.
Thus implementing the mixer is simple as long as the parameters allow its existence. A mixer
with parameters $X$, $Y$, $S$, $M$ can exist if and only if there exists $T$ such that
$S\cdot T \le X$ and $2^M \le T\cdot Y$ (once again, the alphabet translations need their
with parameters $X$, $Y$, $S$, $M$ can exist if and only if there exists $C$ such that
$S\cdot C \ge X$ and $C\cdot Y \le 2^M$ (once again, the alphabet translations need their
range to be as large as their domain in order to work).
\lemma{
A mixer $f$ has the following properties (as long as all inputs and outputs fit into a constant
number of words):
\tightlist{o}
\: $f$ can be computed on a RAM in constant time
\: $s$ depends only on $x$, not $y$
\: $x$ can be decoded given $m$, $s$ in constant time
\: $y$ can be decoded given $m$ in constant time
\endlist
}
All these properties should be evident from the construction.
\defn{The redundancy of a mixer is $$r(f) := \underbrace{M + \log S}_{\hbox{output entropy}} - \quad \underbrace{(\log X + \log Y)}_{\hbox{input entropy}}.$$}
\subsection{On the existence of certain kinds of mixers}
Now we would like to show that mixers with certain parameters do exist.
\lemma{For $X,Y \le 2^w$ there exists a mixer $f: [X]\times[Y] \rightarrow [2^M]\times[S]$ such that:
\lemma{For $X,Y$ there exists a mixer $f: [X]\times[Y] \rightarrow [2^M]\times[S]$
such that:
\tightlist{o}
\: $S = \O(\sqrt{X})$
\: $S = \O(\sqrt{X})$, $2^M = \O(Y\cdot\sqrt{X})$
\: $r(f) = \O(1/\sqrt{X})$
\endlist
}
\proof{
First, let's assume we have chosen an $M$ (which we shall do later). Then we
want to set $C$ so that it satisfies the inequality $C \cdot Y \le 2^M$. Basically
we are asking the question how much information can we fit in $m$ in addition to
the whole of $y$. Clearly we want $C$ to be as high as possibly, thus we set
$C := \lfloor 2^M / Y \rfloor$.
}
\endchapter
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