Commit 7358e5e9 by Filip Stedronsky

### Succinct: progress

parent 328470af
 TOP=.. PICS=sole sole_boxes sole_hilevel mixer PICS=sole sole_boxes sole_hilevel mixer composition include ../Makerules ... ...
 //import ads; //import flowchart; //draw(roundrectangle("f", (0,0))); //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); // XXX this does not work when setting unitsize draw(point(f1, SE) -- point(f2, NW), Arrow); draw(point(f1, SW) -- point(f3, NE), Arrow); draw(point(f2, W) -- point(f3, E), Arrow); draw(roundbox(bbox(), xmargin=0.35cm)); 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));
 ... ... @@ -284,6 +284,23 @@ want to set $C$ so that it satisfies the inequality $C \cdot Y \le 2^M$. Basical 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 $C := \lfloor 2^M / Y \rfloor$. 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 &= 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} 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}}} \endchapter
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