Commit 7358e5e9 authored by Filip Stedronsky's avatar Filip Stedronsky
Browse files

Succinct: progress

parent 328470af
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 $r(f_2)$:
$$\eqalign{r(f_2) &= M - \log(Y\cdot C)= M - \log(\overbrace{Y\cdot \lfloor 2^M / Y \rfloor}^{\ge 2^M - Y})\cr
r(f_2) &\le M - \log(2^M-Y)= \log{2^M\over 2^M-Y} = \log{1 \over 1-{Y \over 2^M}}}$$
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