# 0.4 Universal compression for context tree sources

 Page 1 / 3

## Semi-predictive approach

Recall that a context tree source is similar to a Markov source, where the number of states is greatly reduced. Let $T$ be the set of leaves of a context tree source, then the redundancy is

$r\lesssim \frac{|T|\left(r-1\right)}{2}\left(log,\left(\frac{n}{|T|}\right),+,O,\left(1\right)\right),$

where $|T|$ is the number of leaves, and we have $log\left(\frac{n}{|T|}\right)$ instead of $log\left(n\right)$ , because each state generated $\frac{n}{|T|}$ symbols, on average. In contrast, the redundancy for a Markov representation of the tree source $T$ is much larger. Therefore, tree sources are greatly preferable in practice, they offer a significant reductionin redundancy.

How can we compress universally over the parametric class of tree sources? Suppose that we knew $T$ , that is we knew the set of leaves. Then we could process $x$ sequentially, where for each ${x}_{i}$ we can determine what state its context is in, that is the unique suffix of ${x}_{1}^{i-1}$ that belongs to the set of leaf labels in $T$ . Having determined that we are in some state $s$ , $Pr\left({x}_{i}=0|s,{x}_{1}^{i-1}\right)$ can be computed by examining all previous times that we were in state $s$ and computing the probability with the Krichevsky-Trofimov approach based on the number of times that the following symbol(after $s$ ) was 0 or 1. In fact, we can store symbol counts ${n}_{x}\left(s,0\right)$ and ${n}_{x}\left(s,1\right)$ for all $s\in T$ , update them sequentially as we process $x$ , and compute $Pr\left({x}_{i}=0|s,{x}_{1}^{i-1}\right)$ efficiently. (The actual translation to bits is performed with an arithmetic encoder.)

While promising, this approach above requires to know $T$ . How do we compute the optimal ${T}^{*}$ from the data?

Semi-predictive coding : The semi-predictive approach to encoding for context tree sources  [link] is to scan the data twice, where in the first scan we estimate ${T}^{*}$ and in the second scan we encode $x$ from ${T}^{*}$ , as described above. Let us describe a procedure for computing the optimal ${T}^{*}$ among tree sources whose depth is bounded by $D$ . This procedure is visualized in [link] . As suggested above, we count ${n}_{x}\left(s,a\right)$ , the number of times that each possible symbol appeared in context $s$ , for all $s\in {\alpha }^{D},a\in \alpha$ . Having computed all the symbol counts, we process the depth- $D$ tree in a bottom-top fashion, from the leaves to the root, where for each internal node $s$ of the tree (that is, $s\in {\alpha }^{d}$ where $d ), we track ${T}_{s}^{*}$ , the optimal tree structure rooted at $s$ to encode symbols whose context ends with $s$ , and $\text{MDL}\left(s\right)$ the minimum description lengths (MDL) required for encoding these symbols.

Without loss of generality, consider the simple case of a binary alphabet $\alpha =\left\{0,1\right\}$ . When processing $s$ we have already computed the symbol counts ${n}_{x}\left(0s,0\right)$ and ${n}_{x}\left(0s,1\right)$ , ${n}_{x}\left(1s,0\right)$ , ${n}_{x}\left(1s,1\right)$ , the optimal trees ${T}_{0s}^{*}$ and ${T}_{1s}^{*}$ , and the minimum description lengths (MDL) $\text{MDL}\left(0s\right)$ and $\text{MDL}\left(1S\right)$ . We have two options for state $s$ .

1. Keep ${T}_{0S}^{*}$ and ${T}_{1S}^{*}$ . The coding length required to do so is $\text{MDL}\left(0S\right)+\text{MDL}\left(1S\right)+1$ , where the extra bit is spent to describe the structure of the maximizing tree.
2. Merge both states (this is also called tree pruning ). The symbol counts will be ${n}_{x}\left(s,\alpha \right)={n}_{x}\left(0s,\alpha \right)+{n}_{x}\left(1s,\alpha \right),\alpha \in \left\{0,1\right\}$ , and the coding length will be
$\text{KT}\left({n}_{x}\left(s,0\right),{n}_{x}\left(s,1\right)\right)+1,$
where $\text{KT}\left(·,·\right)$ is the Krichevsky-Trofimov length  [link] , and we again included an extra bit for the structure of the tree.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!