### Succinct: Mixer parameter selecion, lemma proof completed

parent 7358e5e9
 ... ... @@ -5,9 +5,9 @@ //draw(roundrectangle("g", (1,-1))); //draw(roundrectangle("h", (-1,-2))); object f1 = draw("$f_1$", roundbox, (0,0), xmargin=0.5, ymargin=0.5); object f2 = draw("$f_2$", roundbox, (1cm,-1cm), xmargin=0.5, ymargin=0.5); object f3 = draw("$f_3$", roundbox, (-1cm,-1.5cm), xmargin=0.5, ymargin=0.5); object f1 = draw("$g_1$", roundbox, (0,0), xmargin=0.5, ymargin=0.5); object f2 = draw("$g_2$", roundbox, (1cm,-1cm), xmargin=0.5, ymargin=0.5); object f3 = draw("$g_3$", roundbox, (-1cm,-1.5cm), xmargin=0.5, ymargin=0.5); // XXX this does not work when setting unitsize draw(point(f1, SE) -- point(f2, NW), Arrow); ... ... @@ -20,5 +20,5 @@ draw(point(f2, S) -- (xpart(point(f2, S)), -2.5cm), Arrow); draw(point(f3, S) -- (xpart(point(f3, S)), -2.5cm), Arrow); draw((xpart(point(f1, N)), 1cm) -- point(f1, N), Arrow); label("$f$", (xpart(min(currentpicture)), ypart(max(currentpicture))) + (0.25cm, -0.25cm)); label("$g$", (xpart(min(currentpicture)), ypart(max(currentpicture))) + (0.25cm, -0.25cm));
 ... ... @@ -265,7 +265,11 @@ number of words): } All these properties should be evident from the construction. \defn{The redundancy of a mixer is $$r(f) := \underbrace{M + \log S}_{\hbox{output entropy}} - \quad \underbrace{(\log X + \log Y)}_{\hbox{input entropy}}.$$} \defn{The redundancy of a mixer is $$r(f) := \underbrace{M + \log S}_{\hbox{output entropy}} - \quad \underbrace{(\log X + \log Y)}_{\hbox{input entropy}}.$$ In general, the redundancy of a mapping (with possibly multiple inputs and multiple outputs) is the sum of the logs of the output alphabet size, minus the sum of the logs of the input alphabet sizes. Note that there is no rounding (because the inputs and outputs can be from arbitrary alphabets, not necessarily binary) and the redundancy can be non-integer. Compare this to the concept of redundancy for space-efficient datastructures defined above.} \subsection{On the existence of certain kinds of mixers} ... ... @@ -287,20 +291,60 @@ $C := \lfloor 2^M / Y \rfloor$. Now let us calculate the redundancy. First we shall note that we can compute redundancy for $f_1$ and $f_2$ separately and add them up: \eqalign{r(f) &= M + \lceil\log S\rceil - \lceil\log X\rceil - \lceil\log Y\rceil \cr &= \left(M - \lceil\log C\rceil - \lceil\log Y\rceil\right) + \left(\lceil\log C\rceil + \lceil\log S\rceil - \lceil\log X\rceil\right)\cr\eqalign{r(f) &= M + \log S - \log X - \log Y \cr &= \left(M - \log C - \log Y\right) + \left(\log C + \log S - \log X\right)\cr &= r(f_2) + r(f_1)}$$} This is just a telescopic sum. It works similarly for more complex mapping compositions: as long as each intermediate result is used only once as an input to another mapping, you can just sum the redundancies of all the mappings involved. For example, if you have a mapping composition as in fig. \figref{composition}, you can easily see r(f) = r(f_1) + r(f_2) + r(f_3). \figure[composition]{composition.pdf}{}{Mapping composition} For example, if you have a mapping composition as in fig. \figref{composition}, you can easily see r(g) = r(g_1) + r(g_2) + r(g_3). For every edge fully inside the composition, the same number is added once and subtracted once. First, we shall estimate r(f_2):$$\eqalign{r(f_2) &= M - \log(Y\cdot C)= M - \log(\overbrace{Y\cdot \lfloor 2^M / Y \rfloor}^{\ge 2^M - Y})\cr r(f_2) &\le M - \log(2^M-Y)= \log{2^M\over 2^M-Y} = \log{1 \over 1-{Y \over 2^M}}}$$Now we shall use a well-known inequality form analysis:$$\eqalign{ e^x &\ge 1+x\cr x &\ge \log(1+x)\cr -x &\le \log{1 \over 1+x}}$$By substituting x \rightarrow -x we get:$$x \ge \log{1 \over 1-x}$$Thus$$r(f_2) \le {Y\over 2^M} = \O\left({1 \over C}\right)$$Now to r(f_1):$$\eqalign{ r(f_1) &= \log C + \log S - \log X = \log C + \log \left\lceil {X\over C}\right\rceil - \log X = \log\left({C\left\lceil{X \over C}\right\rceil \over X}\right)\cr r(f_1) &\le \log\left({X+C \over X}\right) = \log\left(1 + {C\over X}\right) \le {C \over X}\qquad\hbox{(because $\log(x) \le x-1$)} }$$Putting this together:$$r(f) = r(f_1) + r(f_2) \le \O\left({1 \over C} + {C \over X}\right) In order to minimize this sum, we should set $C = \Theta\left(\sqrt{X}\right)$. Then $r(f) = \O\left({1/\sqrt{X}}\right)$ and $S = \left\lceil{X \over \Theta(\sqrt{X})}\right\rceil = \Theta\left(\sqrt{X}\right)$ as promised. Note that this holds for any value of $Y$. However, we cannot freely set $C$, as we have already decided that $C := \lfloor 2^M / Y \rfloor$. Instead, we need to set a value for $M$ that gives us the right $C$. Now we are almost done. The whole mixer parameter selection process could be as follows (it may be useful to refer back to fig. \figref{mixer}): \tightlist{n.} \: We are given $X$, $Y$ as parameters. \: Set $M := \left\lceil\log\left(Y\sqrt{X}\right)\right\rceil$. \: Set $C := \left\lfloor 2^M / Y \right\rfloor$. This ensures that $2^M \ge C\cdot Y$ and gives us $C = \Theta\left(\sqrt{X}\right)$. \: Set $S := \left\lceil X / C \right\rceil$. This ensures that $C\cdot S \ge X$ and gives us $S = \Theta\left(\sqrt{X}\right)$. \endlist All the inequalities required for mixer existence are satisfied and based on the analysis above the parameters satisfy what our lemma promised. \qed \endchapter
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