Markov Information Binary Sequences Based on Piecewise Linear Chaotic Maps

Fajar Purnama
10 min readJan 24, 2021

0. Note

This is the third assignment from my Masters Applied Digital Information Theory Course which has never been published anywhere and I, as the author and copyright holder, license this assignment customized CC-BY-SA where anyone can share, copy, republish, and sell on condition to state my name as the author and notify that the original and open version available here.

1. Introduction

On the first assignment we produced a chaotic sequence based skew tent map by showing that output sequence is uncontrollable as on the chaos theory. A large sequence produced by skew tent map is uniformly distributed. On the second assignment we produce a memoryless binary sequence based on the first assignment’s skew tent map based chaotic sequence. The probability of 0, 1, 00, 01, 10, 11, and the Markov chain is analyze. Finally the entropy is calculated based on the critical points of each data and find the correlation between the entropy and expected compression ratio. This time the same method on assignment 2 is used but change the assignment 1 of not using skew tent map but piecewise linear map.

2. Piecewise Linear Chaotic Map

Figure 1. Markov’s Chain

The design procedure of Markov Source we choose 4 values p1, and p2 based on Figure 1. Then can obtain the value c (steady state probability) with the formula c=P(0)=P2/(P1+P2), and then we can find the slope a=1/(1-(p1+p2)). The limitation here is we cannot choose the value that satisfy p1+p2 = 1. We are now ready to design the chaotic map. Basic knowledge on line equation is necessary here the one that refers to the equation y = ax + d where a is the slope that we calculated. There will be 3 lines and at least a positive and negative must be use. Each line we calculate its slope. For Figure 2 we calculate slope a = (y2 — y1)/(x2 — x1). For a=(c-0)/(c-c1)from point x,y of (c1,0) and (c,c). From the equation we can define c1=c-c/a. There’s also another slope for a=(1-c)/(c2-c)from point (c,c) and (c2,1) which then we can define c2=c+(1-c)/a. With the first line as negative slope the length proportion of c1:d1 = 1:1-c thus d1 = c1(1-c), and for the third line also negative slope 1-c2:1-d2 = 1:c which…

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Fajar Purnama

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