Juxtaposition

Give an (s1, N, d1)-code C1 and an (s2, N, d2)-code C2, an (s1 + s2, N, d1 + d2)-code over the same field can be constructed. Therefore a linear orthogonal array OA(bs1+s2−n, s1 + s2, Sb, k1 + k2 + 1) can be constructed from a linear OA(bs1−n, s1, Sb, k1) and a linear OA(bs2−n, s2, Sb, k2).

Construction for Linear Codes

Let C1 = {x1,…,xN} and C2 = {y1,…yN}, then the new code C is given by

C = {(xi,yi)  :  i = 1,…, N}.

If the Ci are linear [si, n, di]-codes with generator matrices Gi, then C is an [s1 + s2, n, d1 + d2]-code with generator matrix

($\displaystyle \begin{array}{cc} \vec{G}_{1} & \vec{G}_{2})\end{array}$.

If Gi = (InAi) and Hi = (BiImi) with Bi = – AiT and mi = si−n is a parity check matrix of Ci, then the generator matrix of C is given by

($\displaystyle \begin{array}{cccc} \vec{I}_{n} & \vec{A}_{1} & \vec{I}_{n} & \vec{A}_{2}\end{array}$)

and its parity check matrix by

$\displaystyle \left(\vphantom{\begin{array}{cccc} \vec{B}_{1} & \vec{I}_{m_{1}}… …{0}_{n\times m_{1}} & -\vec{I}_{n} & \vec{0}_{n\times m_{2}}\end{array}}\right.$$\displaystyle \begin{array}{cccc} \vec{B}_{1} & \vec{I}_{m_{1}} & \vec{0}_{m_{1… …} & \vec{0}_{n\times m_{1}} & -\vec{I}_{n} & \vec{0}_{n\times m_{2}}\end{array}$$\displaystyle \left.\vphantom{\begin{array}{cccc} \vec{B}_{1} & \vec{I}_{m_{1}}… …{0}_{n\times m_{1}} & -\vec{I}_{n} & \vec{0}_{n\times m_{2}}\end{array}}\right)$

see ([1, Problem 10.1]).

See Also

References

[1]A. S. Hedayat, Neil J. A. Sloane, and John Stufken.
Orthogonal Arrays.
Springer Series in Statistics. Springer-Verlag, 1999.
[2]F. Jessie MacWilliams and Neil J. A. Sloane.
The Theory of Error-Correcting Codes.
North-Holland, Amsterdam, 1977.

Copyright

Copyright © 2004, 2005, 2006, 2007, 2008, 2009, 2010 by Rudolf Schürer and Wolfgang Ch. Schmid.
Cite this as: Rudolf Schürer and Wolfgang Ch. Schmid. “Juxtaposition.” From MinT—the database of optimal net, code, OA, and OOA parameters. Version: 2015-09-03. http://mint.sbg.ac.at/desc_CJuxtaposition.html

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