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explain how we can get a MDS code of length q+2 and dimension q-1 from a hyperoval

in a projective plane PG2(q) with q a power of 2?

HINT:a hyperoval Q is a set of q+2 points such that no three points in Q are collinear.

you are expected to get a [q+2,q-1,4] binary code from this.take points,one dimensional

subspaces and blocks as the lines 2-dimensional subspaces

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2 Answers 2

As an addition to the answer of Jyrki Lahtonen:

The standard way to get projective coordinates of the points of a hyperoval over $\mathbb F_q$ is to take the vectors $[1 : t : t^2]$ with $t\in\mathbb F_q$ together with $[0 : 1 : 0]$ and $[0 : 0 : 1]$.

Placing these vectors into the columns of a matrix, in the example $\mathbb F_4 = \{0,1,a,a^2\}$ (with $a^2 + a + 1 = 0$) a possible check matrix of a $[6,3,4]$ MDS code is $$ \begin{pmatrix} 1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & 1 & a & a^2 \\ 0 & 0 & 1 & 1 & a^2 & a \end{pmatrix} $$ This specific code is also called Hexacode.

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Hint: Form a check matrix with $q+2$ columns and three rows. Place the homogeneous coordinates of the points on the hyperoval into the columns. Prove that the code with this check matrix has no words of weight $\le3$. The points of the projective plane are distinct => no words of weight two. No three collinear => no words of weight three.

More precisely. If there is a word of weight two or three (or any number, really), this means that those two or three (or whatever) columns of the check matrix must be linearly dependent. Here the presence of two linearly dependent columns means that the two points on the hyperoval are equal. The presence of three linearly dependent columns means that the homgeneous coordinates of the corresponding three points satisfy a linear equation, i.e. they are collinear.

Comment: You cannot expect to get a binary code (the Griesmer bound forbids the existence of a binary code with these parameters when $q>2$). You do get a $q$-ary code, which is presumably want the question really asks about.

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