# Chaotic and Memoryless Binary Sequences Based on Skew Tent Maps

# 0. Note

This is the second 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 previous assignment we generate a chaotic sequence based on skew tent maps, and this time we generate a binary sequence (either ‘0’ or ‘1’). On this occasion we generate the chaotic binary sequence based on the threshold function.

bin = 0 if (x < c)

bin = 1 if (x ≥ c)

When a sequence value ‘x’ is less than critical point ‘c’ the binary value will be ‘0’ and when a sequence value ‘x’ is equal or greater to critical point ‘c’ the binary value will be ‘1’. For example the critical value is set as ‘c’ = 0.4 and chaotic sequence ‘x’ = [0.750000 0.416667 0.972222 0.046296 0.115741 0.289352 0.723380], the binary sequence will be ‘bin’ = [1 1 1 0 0 0 1] as Figure 1.

Figure 1. Chaotic binary sequence

# 2. Probability of binary sequence

Using the chaotic sequence based on skew tent map to generate binary sequence we compute the probability the sequence value will be ‘0’ or ‘1’ as ‘P(0) = c’ and ‘P(1) = 1-c’. Using the skew tent map we can compute ‘P(00) = c x c’, ‘P(01) = c(1-c)’, ‘P(10) = (1-c)c’, and ‘P(11) = (1-c)(1-c)’. On Markov’s chain the probability that ‘0’ will remain ‘0’ is ‘P(0|0) = c’, ‘0’ will become ‘1’ is ‘P(1|0) = 1-c’, and so-on ‘P(0|1) = c’, ‘P(1|1) = 1-c’.

We continue the assignment using the previous assignment to generate the binary sequence and calculate the probability of ‘0’ and ‘1’ theoretically and the actual number of ‘0’ and ‘1’ on the binary sequence generated. We used the initial value ‘x(1) = 0.3’ and trials of ‘c = 0.1, 0.2, 0.301, 0.4, 0.6, 0.7, 0.8, 0.9’. The total sequence generated is 2000000 values (2 Millions). The result is on Table 1.