## Tuesday, July 29, 2014

### Fractals: parameters estimation (part IV)

Introduction
In the former posts we discussed about the following points:
1. There are special points in the contour of the fractal that can be used to derive its contour.
2. Such points can be used to describe the fractal thru an iterative and deterministic algorithm.
It's still open the main issue: can we leverage the two above findings to determine the fractals parameters?
The answer is yes, provided that we use a good technique to extract the contour points of the fractal image.

The algorithm (practical explanation)
We already noticed that points in the contour of the fractal are in tight relationship each other.
Since each linear transformation is described by exactly six parameters, for their determination we need at least 6 points for each transformation function.
The below image makes the concept clear:
 Image-1: Relationship between points of the fractal contour

The steps to obtain the estimation of the params are the following:
1. Extraction of the contour points thru convex hull algorithm;
2. Build relationships among the points.

 Left Image: Extraction of contour points thru Convex Hull. Right Image: Relationship among the points extracted.
How to build the relationships among the points?
As showed both in the deterministic algorithm presented in the last post and in the above image, the points preserve an ordering that can be leveraged to simplify the complexity of the problem.
Consider for instance the fern case:
• the transformation that leads to the fixed point lying on the contour (highlighted in image-1) can be obtained (as explained in the points 1,2 of the deterministic algorithm) just creating the relationship among the consecutive points of the set $l_1$ described in Image-1 and in the below image:

 Set of points to determine the Transformations
At this point we obtained the parameters of the first transformation, but still, we don't know neither the number of transformations we really need to describe the fractal image,  nor the probability associated to them.
For the first problem, we have to strategies: the brute force, that is, try all the possible combinations, or more wisely try the soft computing approach described below.

The fuzzy clustering approach (...unfortunately here some formulas are required)
The fuzzy clustering is a very well known algorithm that assigns the membership of a point to a cluster with a certain "degree", so the point $p_i$ might belong to the cluster $A$ with degree $x$ and to the cluster $B$ with degree $y$.

In this problem I modified the algorithm in this way:
• the clusters are the mapping functions that we want to discover.
• the probability associated to the fractal maps can be easily derived by the "size" of each cluster.
• The fuzzy-membership function is based on the minimisation of the distance between the point obtained applying to $x_i$ the contractive map $x_i^{'} = \tau_j(x_i)$  and all the other points falling in a reasonable neighbour. The animation below describes it.

• The update step of the centroids is aimed to minimise the distance between $d(\tau (x_i)),\phi(x_i,\tau_j))$ computed over each $\tau$ and each $x_i$, where
• $\phi(x_i,\tau_j) = \arg\min_{x_s} d(\tau_j(x_i),x_s)$
• To minimise the above function,  I used the gradient technique, working on the the following cost function:
• $E(\theta)= \sum_{i=1}^{n}{(\mu_{x_i}(\tau_j)\cdot[d(\tau_j(x_i),\phi(x_i,\tau_j))]^2)}$
• For each mapping function,  the correction to each param is the following:
• $\theta_i= \theta_i - \eta \frac{\partial E(\theta)}{\partial \theta_i}$

Results
Starting from the image of the fern (a jpg image containing 100.000 points) the application of the algorithm for the determination of the contractive maps gives the following result:
 Contractive maps detection algorithm: in black the points of the original fern, in red,green, blue the three estimated maps.
• The results are not bad, the weakness of the algorithm lays on the extraction of the points of the Convex hull of the fractal.
• The application of smarter contour extraction algorithm should improve further the accuracy of the algorithm.
Disclaimer
In the last posts about fractals I often abused the term contour of fractal. ...Theoretically speaking, a fractal cannot have a contour, but to make clear the pragmatical aspects I decided voluntarily such terminology.

## Saturday, June 7, 2014

### Fractals: a deterministic and recursive algorithm to reproduce it (Part II)

Abstract:
 Fern fractal estimation thru recursive and deterministic algorithm.

A fractal can be described as fixed point of a Markov process: given the proper contractive maps it's easy to generate it.
The question is: given the contractive maps, is it possible to generate the fractal using a pure deterministic approach?

Problem
We already observed that the iteration of the contractive maps starting from a fixed point that lyes on the border of the fractal returns points of the convex hull of the fractal.

• What's happen if we recursively apply such schema to the points obtained from the the above procedure?
• Is it possible recreate the markov process (or at least approximate it) removing any probabilistic aspects?
The below algorithm, given the contractive maps of the fractal, returns at each iteration better approximation of it.

At each iteration it considers the fractal as better approximation of the contour of the original image.

The Algorithm (by practical means)
1. Consider a fixed point $P_0$ of a contractive map $\tau_1$ that lyes on the convex hull of the fractal.
2. Choose randomly a point $P_i$ of the fractal and apply the above contractive map until the distance to $P_0$ is negligible.
3. Map the point calculated at step 2 using sequentially all the contractive maps.
4. Map each point obtained from point 3 the former step with $\tau_1$ till the distance to $P_0$ is negligible.
5. If[ITERATIONS< K]:
1. K--;
2. For each point obtained from point 4 go to point 3.
To explain how it works I plotted the list of points got using $K=1$ iterations of the algorithm:

Bigger is $K$, more accurate will be the result.
The above procedure works only if the contractive map $\tau_1$ has a fixed point that lyes on the convex hull of the fractal!!

Results:

I iterated the algorithm with $K=4$ times. At each iteration the algorithm returns a deeper description of the contour of the fractal (...even though definition of contour for a fractal doesn't make any sense, it gives at least a practical meaning):
 Results of the Recursive Algorithm iterated with K=1,2,3,4.

If we try to overlap the results obtained with the original fern we get:
 Original Fern, and the overlap with the results obtained using the recursive algorithm (K=4).
Conclusion and next steps
I showed how to depict the IFS generated thru the markovian process as a recursive and completely deterministic process.
We noticed also, that in the fractal there are special points (as for instance $P_0$) that play a crucial role to describe the IFS.

The question now is the following:
is it possible leverage such special points and the above recursive algorithm to estimate the parameters of the contractive maps?
... I will show you a procedure that partially answer the question and some other example of the algorithm above described!
Stay Tuned.
Cristian

## Saturday, May 24, 2014

### Fractal parameters estimation thru contour analysis - Part I

Introduction
I really love fractals theory. Nothing fascinates me as the amazing structures generated by a chaotic processes that describe (hyperbolic) fractals.
It's fairly easy to draw a fractal image:
• The input
• A small set of very easy function (linear transformations)
• A probability to choose one of the aforementioned functions.
• The process
• Choose randomly a starting point.
• Choose randomly one of the above functions according to the above defined probability.
• Map the starting point using the above selected function.
• iterate as long as you want :)
• Admire the meaning of the life!
• just plot the points
Wherever you are, whatever decision you make, despite the chaos, the overall picture will be always the same. More you will move, more clear the picture will be!

When I approached for the first time fractals,  after the reading of the great Barnley's book (Fractals Everywhere) I opened the laptop and I wrote the routine to draw my first IFS!
...By mistake I also plotted the lines that join the points:

...It's quite a messy, isn't it?
Sometimes it's better to give up to connect the points, just relax and wait until everything will be clear. If we focus on every single detail we risk to lose what really counts!

What is a fractal (Iterated Fractal System)?
There are many formal definitions to describe an IFS, in my opinion the most effective describe a fractal as the fixed point of the Markov process (mentioned above).
The process converges to the fractal :)

The powerful of the fractal
In literature there are plenty articles in which fractals are used to solve complex problems (financial analysis, stock exchange preditictions, description of biological processes,...).
Also in Computer Science fractals fond a good collocation, think about the image compression: the ivy leaf image takes around 25MB of space, all the information to obtain it thru the IFS takes less 1KB!!
The problem
Given the ivy leaf image, what are the parameters of the Markov process to generate it?
A bit more formally what we have to estimate is:
• A set of functions: $\{ \tau_1, \cdots,\tau_N \}$, where $\tau_i(p_0)=A_1 \cdot p_0 +Q_1$, where:
$A= \begin{bmatrix}\alpha & \beta\\ \gamma & \delta \end{bmatrix} Q= \begin{bmatrix} \epsilon\\ \zeta \end{bmatrix}$
• We need also the estimate the probability to choose each $\tau_j$
The combination of the collage theorem and the box counting approach is the most common technique to solve the problem.
When I was student I approached the problem from a different angle. I have to say that the results obtained were partials but I still think that something more can be done :)
Before to start we need one more notion: the contractive maps (have a look at Banach Theorem)
Under certains conditions the iteration  $\tau_j$ led to a fixed point:
 Example of contractive map applied to an ellipsoid. It converges to a fixed pointed.

First conjecture:

• An IFS is characterised by a fixed point that lies on its convex hull.
• From a fixed point that lies on the border of the IFS, the iterations of the contractive maps that generate the fractal return the convex hull of the fractal.

 The ivy leaf IFS is generated by 4 contractive maps. Each color describes the map used to generate the point.
The above animated gif shows the meaning of the conjecture.
An experimental evidence about the fact that at least one fixed point lies on the convex hull of the fractal can be obtained changing the params of the maps:
 The light blue points depict the fixed point of the maps used to plot the fractals.
Despite the changes in the maps, the fixed point on top of the leaf still lays on the convex hull of the fractal.
In the next post I'll show you a nice recursive algorithm I found to obtain different levels of convex hull for an IFS.
Stay tuned
Cristian.

## Friday, May 2, 2014

### Waiting Time Polynomials: tech explanation - last part

This is the last step to complete the explanation of the waiting time polynomials formula.

Unfortunately it's a bit technical, but I'm sure that it can be understood without deep math knowledge.
At the very end if you can't explain it simply you don't understand it well enough! (A. Einstein)

The trick
Last time we left with the tally of overall waiting time $w(x_i) = \phi(x_i)-|x_i|$ where $\phi(x_i)$ returns just the position of the last $|x_i|$ in the vector $V$.
Let's have a look at the following example that will be used during the remaining steps.
There are two questions that might be answered:

• given $|x_1|= i, |x_2|= j, |x_3|= k$ what are the tuples $\{w(x_1),w(x_2),w(x_3)\}$?
• given a $\{w(x_1)=I,w(x_2)=J,w(x_3)\}=K$, how many vectors $V$ can be built?
To answer to the first question, I noticed that the three coloured cells are the only ones that really count.
The idea is the following:

1. consider the three cells as placeholders
2. analyse the admitted values for each of them
3. replace the placeholders with all the possible permutations of the alphabet $\{x_1,x_2,x_3\}$.

Let's start with the case depicted in the above image, where we assumed that $\phi(x_1) < \phi(x_3) < \phi(x_2)$, then we have the following facts:
• $\phi(x_1)$ can take values between: $0 \leq \phi(x_1) \leq |x_1|+|x_2|+ |x_3|-2$
• $\phi(x_2)$ can take just one value: $|V|=|x_1|+|x_2|+ |x_3|$
• The upper bound of $\phi(x_3)$ is $|V|-1$ because it can slide till the second last element of $V$, that is $\phi(x_2)-1$
• what about the lower bound of $\phi (x_3)$? We have two cases depicted in the below image:

To sum up, so far we explained the following part of the formula (I hope you don't mind that I changed a bit the indexes notation):

We have now to consider that for each configuration of $\{w(x_1)=I,w(x_2)=J,w(x_3)\}=K$ we can have more than one vector $V$.
Do you like combinatorics? The remaining part it's just matter of tally, and the only formula we need is the formula for permutation with repetitions. I let you rub up the concept on your favourite website for trivial combinatorics.

The formula can be split in two chunks, because we have to blocks of cells to be filled

1. In how many ways can we fill the cells between the positions $[1,\phi(x_1)]$?
2. In how many ways can we fill the cells between the positions $[phi(x_1),phi(x_2)]$?

Let's answer the first question we have to find the values for the denominator of the following:
$\frac{(\phi(x_1)-1)!}{\#(x_1)!\#(x_2)!\#(x_3)!}$
• we have $|x_1|-1$ cells that can be filled.
• it contains all the instances of $x_1$ (except for the last the occupied $\phi(x_1)$)
• the number of $x_3$ instances depends by $\phi(x_1)$ and $\phi(x_3)$:

• the computation of the number of instances of $x_2$ in the first slot is straightforward, and it can easily derived by difference:
• $\frac{(\phi(x_1)-1)!}{(|x_1|-1)!(|x_3|-Min(|x_3|,\phi(x_3)))!\#(x_3)!}$;
• $(|x_1|-1)+(|x_3|-Min(|x_3|,\phi(x_3)))+\#(x_2)=\phi(x_1)-1$
• so $\#(x_2)= \phi(x_1)-|x_1|-(|x_3|-Min(|x_3|,\phi(x_3)))$
This explains the following boxed part of the formula:

The final step is to count in how many ways we can fill the slot 2 depicted by the interval $[phi(x_1),phi(x_2)]$ and to make the formula more readable let's rename $(|x_3|-Min(|x_3|,\phi(x_3))= \epsilon$.
As we did for the first slot we have to identify the values of the denominator of the below formula:
$\frac{(\phi(x_3)-\phi(x_1)-1)!}{\#(x_2)!\#(x_3)!}$

• Out of $|x_3|$ instances, $\epsilon$ have been placed in the slot 1, so the slot 2 contains exactly $|x_3|-1- \epsilon$.
• again by difference we can get the instances of $x_2$:
• the occurrences of $x_2$ before $\phi(x_3)$ are exactly $\phi(x_3)- (|x_1|+|x_3|)$
• the occurrences of $x_2$ in the slot 1 (listed above) are: $\#(x_2)= \phi(x_1)-|x_1|-\epsilon$
• that is : $\#(x_2)=\phi(x_3)- (|x_1|+|x_3|)- \phi(x_1)+|x_1|+ \epsilon$
• finally we have: $\#(x_2)=\phi(x_3)-\phi(x_1)-|x_3|+ \epsilon$

That's complete the proof of the formula.
It's quite easy now extend the formula to more variables. The only expedient to make it easier is to remove from the formula the operator $Min$ splitting the formulas in two branches.
I'll show it in paper.
What's the most painful point of this formula?
... The introduction of placeholders requires to apply the formula for each permutation of the variables involved. It means that having $k$ variables we need to calculate the formula $k!$
Anyway I don't expect to use the formula with large set of variables, after all the principle of selecting the right and small set of features is always recommended!

As usual Stay Tuned.
Cristian.

## Saturday, April 19, 2014

### Waiting Time Polynomials: how to derive the analytical formula: Part IV

Introduction before you start
I got many clarification requests about the Waiting Time Polynomials I published on the blog in the last three posts.
The paper is almost ready to be submitted for review, but I think that some technical explanation might be interesting also for not academic audience.
I consider myself a curious and hungry seasoned student, and I know how can be tedious read formulas and mathematical passages especially when it comes from a blog!!
So why technical explanations?
The answer is in the following quote of one of my favourite scientists, Gregory Chaitin. In "The quest for Omega" he wrote:

The books I loved were books where the author’s personality shows through, books with lots of words, explanations and ideas, not just formulas and equations! I still think that the best way to learn a new idea is to see its history, to see why someone was forced to go through the painful and wonderful process of giving birth to a new idea! To the person who discovered it, a new idea seems inevitable, unavoidable. The first paper may be clumsy, the first proof may not be polished, but that is raw creation for you, just as messy as making love, just as messy as giving birth! But you will be able to see where the new idea comes from. If a proof is “elegant”, if it’s the result of two-hundred years of finicky polishing, it will be as inscrutable as a direct divine revelation, and it’s impossible to guess how anyone could have discovered or invented it. It will give you no insight, no, probably none at all.

That's the spirit that leads the following explanation!

Definition of the problem
Given an alphabet of 3 elements $\{X_1,X_2,X_3\}$, the function $w(X_i)$ counts  the number of failed trials before the last event $X_i$.
Consider now the following configuration: $\{\left\vert{X_1}\right\vert =i , \left\vert{X_2}\right\vert =j,\left\vert{X_3}\right\vert =k\}: i+j+k= Z \wedge i,j,k>0$

• What are the admitted sequences  $\{w(X_1),w(X_2),w(X_3)\}$ ?

Step I: Find all the possible configurations of events
How can we list the sequences of length $Z$ that can be built with $\{\left\vert{X_1}\right\vert =i , \left\vert{X_2}\right\vert =j,\left\vert{X_3}\right\vert =k\}: i+j+k= Z \wedge i,j,k>0$ ?

 Example of overall waiting time $w(x_i)$  in a succession of events.
• once we set the values of the first two variables, the third it's determined by $Z-i-j$.
• we imposed that all the variables occur at least once, so we $X_1$ can assume all the values between $[1,Z-2]$.
• for each value of $X_1$ the variable $X_2$ can assume values between $[1,Z-i]$.
•  $p_i$ is the probability that $X_i$ occur in a Bernullian trial.
Now we have all the ingredients to make the cake:

$\sum_{i=1}^{Z}\sum_{j=1}^{Z-i}\sum_{k=1}^{Z-i-j}{p_1^ip_2^jp_3^k}$

In the first two summations,  $i$ assumes values between $[1,Z]$ just to keep the formula cleaned.
...I let you proof why the result doesn't change :).
last point about this step:the limit of the above summation $Z \rightarrow \infty = \frac{p_1 p_2 p_3}{\left(p_1-1\right) \left(p_2-1\right) \left(p_3-1\right)}$
Such limit will be used to build the probabilistic density function.
• The number of sequences that can be built with vectors of length $[3,Z]$ are $\binom{Z}{3}$
• The number of sequences that can be built with vectors of length $Z$ are $\binom{Z}{2}$
Step II: Waiting for an event!
What's the easiest way to describe the overall waiting time for an event in a finite succession?
There are many ways to get the $w(x_i)$, the easiest I found is given by the position of the last occurrence of $x_i$ minus the number of occurrences of $x_i$.
For instance, let's consider $w(x_1)$:
• The position of the last occurrence of $x_1= 8$;
•  $\left \vert{X_1} \right \vert = 4$
• $w(X_1)=4$
Where we are:
The first two steps explain the circled pieces of the formula:

What the "overall waiting time" for?
For each event $X_i$ we are counting the holes among all the occurrences, so smaller is the overall waiting time, closer each other are the events $X_i$: it's a measure of proximity for the occurrences of $X_i$.
What I did, is to extend such measure (it would be interesting to prove that it's really a measure!) to different kind of events (aleatory variables) ${X_1, X_2,...,X_n}$ over the discrete line of the time.
Applications
There are several area for which such kind of analysis might be helpful, I showed last time an its application as powerful document classifier, where each variable $X_i$ is a word of a document.
If we consider a document as a succession of $Z$ words, the proximity measure inducted by the waiting time polynomials is a sort of finger print for the document, since for similar documents we expect that the same words are characterised by similar overall waiting time.
Moreover, the dependency among the words are considered, since we are taking in account simultaneously an arbitrary number of words (the alphabet ${X_1, X_2,...,X_n}$).

In the next step I'll explain the logic to get the remaining pieces of the puzzle, that will make easier the generalisation of the approach to an arbitrary alphabet.
Stay Tuned!
cristian

## Tuesday, April 8, 2014

### Coefficients of Waiting Time Polynomials listed in OEIS Database!

I'm happy to announce that the sequence generated by coefficients of waiting time polynomial has been listed in the OEIS database (The Online Encyclopedia of Integer Sequences).
The sequence is: A239700.
In the next posts I'm going to explain how I derived the analytical formula of the polynomials.
As usual: Stay Tuned!
cristian

## Friday, March 28, 2014

### Coefficients of Waiting Time Polynomials: a nice representation

I was doing some simulation for the paper I'm writing about the waiting time polynomials, and I got something unexpected.
If we sort in lexicographic order all the polynomials generated for a given length of the events vector, and we consider the coefficients, what  you get is an apparently irregular series of integers.
To capture some regularity, I decided to plot such matrix:

 The matrix plot, obtained considering an alphabet of three words and vectors having length=35.

Isn't nice?
Sometimes, an appropriate graphical representation helps in capture interesting aspects of your data

Stay Tuned
Cristian