Content deleted Content added
m fmt |
m fmt and finish |
||
Line 8:
Now if we have α a root of ''x''<sup>4</sup>+''x''+1, ''m''<sub>1</sub>(''x'')=''x''<sup>4</sup>+''x''+1. Now ''m''<sub>1</sub> is minimal for α since
:''m''<sub>1</sub>(''x'')=(''x''-α)(''x''-α<sup>2</sup>)(''x''-α<sup>4</sup>)(''x''-α<sup>8</sup>)=''x''<sup>4</
If we wish to construct a code to correct 1 error we use ''m''<sub>1</sub>(''x''). Our codewords will be
Line 55:
So then
:''S''<sub>1</sub>=''R''(α)=''C''(α)+α<sup>''i''</sup>=α<sup>i</sup>
:''S''<sub>3</sub>=''R''(α<sup>3</sup>)=
:: =(α<sup>i</sup>)<sup>3</sup>=''S''<sub>1</sub><sup>3</sup>
so we can recognize one error. A change in the bit position shown by α's power will aid us correct that error.
If there are two errors
:'''r'''='''c'''+'''e'''<sub>i</sub> +'''e'''<sub>j</sub>
then
:''S''<sub>1</sub>=''R''(α)=''C''(α)+α<sup>''i''</sup>+α<sup>j</sup>
:''S''<sub>3</sub>=''R''(α<sup>3</sup>)=C(α<sup>3</sup>)+(α<sup>3</sup>)<sup>i</sup>+(α<sup>3</sup>)<sup>j</sup>
:: = (α<sup>3</sup>)<sup>i</sup>+(α<sup>3</sup>)<sup>j</sup>
which is '''not''' the same as ''S''<sub>1</sub><sup>3</sup> so we can recognize ''two'' errors. Further algebra <!-- add when have time --> can aid us in correcting these two errors.
| |||