Justesen code: Difference between revisions

Revision as of 00:56, 9 May 2011

Revision as of 00:58, 9 May 2011

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Suppose <math>C_{out}(m_1) = (c_1^1,c_2^1,..,c_N^1)</math> and <math>C_{out}(m_2) = (c_1^2,c_2^2,..,c_N^2)</math>.

Recall that <math>\{ C_{in}^i \}_{1 \le i \le N}</math> is a [http://en.wikipedia.org/wiki/EnsembleWonzencraft_Ensemble Wonzencraft ensemble]. Due to "Wonzencraft ensemble theorem", there are at least <math>(1-\varepsilon)N</math> linear codes <math>C_{in}^i</math> that have distance <math>H_q^{-1}(\frac{1}{2}-\varepsilon) \cdot 2k</math>.

So if for some <math>1 \le i \le N</math>, <math>c_i^1 \ne c_i^2</math> and <math>C_{in}^i</math> code has distance <math>\ge H_q^{-1}(\frac{1}{2}-\varepsilon) \cdot 2k</math>, then <math>\Delta(C_{in}^i(c_i^1),C_{in}^i(c_i^2)) \ge H_q^{-1}(\frac{1}{2}-\varepsilon) \cdot 2k</math>.

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Now we want to estimate <math>\left| S \right|</math>. Obviously <math>\left| S \right| = \Delta(C_{out}(m_1),C_{out}(m_2)) \ge (1-R)N</math>.

Due to the [http://en.wikipedia.org/wiki/EnsembleWonzencraft_Ensemble Wonzencraft Ensemble Theorem], there are at most <math>\varepsilon N</math> linear codes having distance less than <math>H_q^{-1}(\frac{1}{2}-\varepsilon) \cdot 2k</math>, so <math>T \ge \left| S \right| - \varepsilon N \ge (1-R)N - \varepsilon N = (1-R-\varepsilon)N</math>.

Finally,we have