2 consecutive messages have errors. We have 9 messages from $ GF(2^6, x^6+x+1) $. Messages were encoded with $ (x+1)(x+a)(x+a^2)\sum_{i=1}^6X_ix^{i-1}=\sum_{l=1}^9Y_lx^{l-1} $ , where $ a=(0,0,0,0,1,0) $. I have 9 messages $ \tilde{Y} $ from that $ GF(2^6 $, 2 (consecutive) of which have errors. So I wonder how to solve such problem - get all $ Y $ s corrected?
I wonder if it is possible to turn my 9 items $ \tilde{Y} $ into 3 items $ \tilde{K} $ from some $ GF( 2^{18} ) $ and solve Reed-Solomon code $ (3, 18) $ following solution described here, than turn item from that $ GF( 2^{18} ) $ into 3 items from my $ GF(2^6) $ and have all my 2 errors corrected? So how to turn from $ GF(2^6, x^6+x+1) $ into $ GF(2^{18}, something?) $ and back from that $ GF(2^{18}) $ into my "home" $ GF(2^{6}) $?
Let me introduce an image that explains what I meant by converting 9 messages from $ GF(2^6) $ by turning tham into 3 messages in $ GF(2^{18}) $. My Idea is: if it is any how possible to to turn from $ GF(2^6, x^6+x+1) $ into $ GF(2^{18}, something?) $ and back than it is probably possible to turn from $ GF(2^6, x^6+x+1) $ into $ GF(2^{24}, something?) $ and back. If we could turn our 9 messages into 3 from $ GF(2^{24}) $ and 3 from $ GF(2^{18}) $ we would always have to fix only one message with error in that field. With simple algorithm, and with out iterations all over blocks.
If such cnversion (from one GF into another, group of messages into message) is possible, than what would be $ a $ in new field, what would be irreducible polynomial for that $ GF(2^{24}) $ or $ GF(2^{18}) $ ?
ofcourse we can not bother with $ GF(2^{24}) $ and simply use something like:
So real problem: we have ( as Dilip Sarwate calls them) symbols $ \tilde{Y} $. (in $ GF(2^6, {\alpha}^{6}+\alpha+1) $)
$ [1, 0, 1, 1, 0, 0], $$[1, 1, 1, 1, 1, 1], $$[0, 0, 1, 1, 1, 0], $$[1, 0, 0, 1, 1, 1], $$[0, 1, 0, 0, 1, 1], $$[1, 1, 0, 1, 0, 1], $$[0, 0, 0, 1, 1, 0], $$[1, 0, 0, 0, 1, 0], $$[0, 0, 0, 0, 1, 0] $
There are $0$ or $1$ or $2$ errors in the 9 symbols. If there are $2$ then they occur in consecutive positions.
I calculated $ S(\alpha^2) $ , $ S(\alpha) $ and $ S(1) $
$S(\alpha^2) = \alpha^4+\alpha$ $S(\alpha) = \alpha^5+\alpha^4+\alpha^2 $ $S(1) = \alpha^4+\alpha^3+1 $
So $ \dfrac{S(\alpha)}{S(1)} = \alpha^5 $ and $ \dfrac{S(\alpha^2)}{S(\alpha)} = \alpha^5+\alpha^4+\alpha $ and so $ \dfrac{S(\alpha)}{S(1)} $ is $ \neq $ to $ \dfrac{S(\alpha^2)}{S(\alpha)} $ and $ \dfrac{S(\alpha)}{S(1)} - \dfrac{S(\alpha^2)}{S(\alpha)} = \alpha^4+\alpha$
I tried to do a brute-force search for solutions to the quadratic equation $ S(\alpha^2) + [(1+\alpha)S(\alpha)]x + [\alpha S(1)]x^2 = 0 $ by substituting $ x = 1, \alpha, \alpha^2, \ldots, \alpha^8 $ (as Dilip Sarwate sad) and got next results: $ {\alpha}^{4}+{\alpha}^{3}+{\alpha}^{2}+\alpha+1 $ $ {\alpha}^{5}+\alpha $ $ {\alpha}^{4}+{\alpha}^{3}+{\alpha}^{2}+\alpha+1 $ $ {\alpha}^{5}+{\alpha}^{4}+{\alpha}^{2}+\alpha $ $ {\alpha}^{5}+{\alpha}^{2} $ $ {\alpha}^{3}+{\alpha}^{2} $ $ {\alpha}^{5}+{\alpha}^{3}+{\alpha}^{2}+\alpha+1 $ $ {\alpha}^{5}+{\alpha}^{3}+{\alpha}^{2}+\alpha+1 $ $ {\alpha}^{5}+{\alpha}^{4}+{\alpha}^{3}+\alpha+1 $ (Here is maple file I created for playing aroung GF - there is simplified API, tests for it and this task at the end of document...) The book talls me errors shall be located at $6$th and $7$th symbols but I can not find them=(