Commit 328470af by Filip Stedronsky

### Succinct: progress

parent b662898c
 ... ... @@ -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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