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==== Decoding with unreadable characters ====
Suppose the same scenario, but the received word has two unreadable characters [ 1 {{color|red|0}} 0 ? 1 1 ? 0 0 {{color|red|1}} 1 0 1 0 0 ]. We replace the unreadable characters by zeros while creating the polynom reflecting their positions <math>\Gamma(x) = \left(\alpha^8x - 1\right)\left(\alpha^{11}x - 1\right).</math> We compute the syndromes <math>s_1=\alpha^{-7}, s_2=\alpha^{1}, s_3=\alpha^{4}, s_4=\alpha^{2}, s_5=\alpha^{5},</math> and <math>s_6=\alpha^{-7}.</math> (Using log notation which is independent on GF(2<sup>4</sup>) isomorphisms. For computation checking we can use the same representation for addition as was used in previous example. Hexadecimal description of the powers of <math>\alpha</math> are consecutively 1,2,4,8,3,6,C,B,5,A,7,E,F,D,9 with the addition based on bitwise xor.)
Let us make syndrome polynomial
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Therefore,
:<math>
Let <math>\Lambda(x)=\alpha^{3}+\alpha^{-5}x+\alpha^{6}x^2.</math> Don't worry that <math>\lambda_0\neq 1.</math> Find by brute force a root of <math>\Lambda.</math> The roots are <math>\alpha^2,</math> and <math>\alpha^{10}</math> (after finding for example <math>\alpha^2</math> we can divide <math>\Lambda</math> by corresponding monom <math>\left(x - \alpha^2\right)</math> and the root of resulting monom could be found easily).
Let
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Let us look for error values using formula
:<math>e_j = -\frac{\Omega \left
where <math>\alpha^{-i_j}</math> are roots of <math>\Xi(x).</math> <math>\Xi'(x)=\alpha^{2}x^2.</math> We get
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