A new class of Fibonacci sequence based error correcting codes

Abstract

A new class of matrices is introduced for use in error control coding. This extends previous results on the class of Fibonacci error correcting codes. For a given integer p, a (p+1)×(p+1) binary matrix M _p is given whose nonzero entries are located either on the superdiagonal or the last row of the matrix. The matrices ${M^{n}_{p}}$ and $M^{-n}_{p}$, the nth power of M _p and its inverse, are employed as the encoding and decoding matrices, respectively. It is shown that for sufficiently large n, independent of the message matrix M, relations exist among the elements of the encoded matrix $E=M\times {M_{p}^{n}}$. These relations play a key role in the error detection and correction.

Appendix

This appendix provides a proof of the existence of a unique real solution for the system of equations given by (28). By setting u = α ₁ α ₂⋯α _p , the system of equations given by (28) is changed to

$$\left\{ \begin{array}{ccl} \frac{1}{\alpha_{1}} &=& 1+u \, ,\\ &&\\ \frac{1}{ \alpha_{1} \alpha_{2}} &=& 1+u+u^{2} \, ,\\ &&\\ \frac{1}{\alpha_{1} \alpha_{2} \alpha_{3}}&=& 1+u+u^{2}+u^{3}\, , \\ &&\\ {\vdots} & & {\vdots} \\ &&\\ \frac{1}{\alpha_{1} \alpha_{2} {\cdots} \alpha_{p}}&=& 1+u+u^{2}+{\cdots} +u^{p}\, , \end{array} \right. $$

so that in general

$$\frac{1}{\alpha_{1} \alpha_{2} {\cdots} \alpha_{k}}=1+u+u^{2}+{\cdots} +u^{k},~~~1\leq k\leq p. $$

It follows from $\frac {1}{\alpha _{1} \alpha _{2} {\cdots } \alpha _{p}}= 1+u+u^{2}+{\cdots } +u^{p}$ that $\frac {1}{u}= 1+u+u^{2}+{\cdots } +u^{p}$ and hence u ≠ 1. From

$$\alpha_{k}=\frac{\frac{1}{\alpha_{1} \alpha_{2} {\cdots} \alpha_{k-1}}}{\frac{1}{\alpha_{1} \alpha_{2} \cdots \alpha_{k-1} \alpha_{k}}} =\frac{1+u+u^{2}+{\cdots} +u^{k-1}}{1+u+u^{2}+{\cdots} +u^{k}}, $$

we have that

$$\alpha_{k}=\frac{(u-1)(1+u+{\cdots} +u^{k-1})}{(u-1)(1+u+{\cdots} +u^{k})} =\frac{u^{k}-1}{u^{k+1}-1}. $$

This results in

$$u=\alpha_{1} \alpha_{2} {\cdots} \alpha_{p}=\frac{u-1}{u^{2}-1} \times \frac{u^{2}-1}{u^{3}-1}{\cdots} \frac{u^{p}-1}{u^{p+1}-1}= \frac{u-1}{u^{p+1}-1}, $$

and hence

$$u^{p+2}=2u-1. $$

A real solution for this equation is u = 1 which is not acceptable as mentioned above. For an even number p, the function y ₁(u) = u ^p+2 is concave, and the two functions y ₁(u) = u ^p+2 and y ₂(u) = 2u−1 are equal at u = 1 and a number in (0, 1), so the latter is the desired solution. If p is odd, y ₁(u) = u ^p+2 is negative for u < 0, and y ₁(u) = u ^p+2 and y ₂(u) = 2u−1 intersect at u = 1, a number in (0,1) and a number in $(-\infty , 0)$. As the α _i are positive, a negative solution for (28) is not acceptable, so (28) has a unique real solution. This argument provides the following method for solving (28). First find a real solution for u ^p+2−2u+1=0 in the interval (0,1), and then compute α _k by making use of $\alpha _{k}=\frac {u^{k}-1}{u^{k+1}-1}$.