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Special task from NASA to best MSE minds (joking:).

Consider two polynoms over $F_2$

$$ f(x)=1+x+x^3+x^4+x^6,\qquad g(x)=1+x^3+x^4+x^5+x^6 $$

Prove or disprove that for any polynom $p(x) \in F_2[x]\setminus\{0\}$ $$ |pf|+|pg|\ge 10,$$ where $|\cdot|$ is number of non-zero monoms.

Remark 1: Clearly bound is sharp - take $p(x)=1$ we get $|f|+|g|=10$

Remark 2: See this discussion for motivation, history, etc.

Remark 3: This question is more for fun, I think it is quite nice since it is "everyone can understand" but demonstrates real problems in the theory of convolutional codes, so might be used as exercise in teaching, that it is why I would like to share it. I am afraid that there is no nice solution exists - one should consider many cases... What would be really great to find some "idea-based and short" solution to this puzzle, but may be there is no such... (Some solution (probably not nice) should be somewhere in the literature on convolutional codes).

P.S.

All good in question is from Jyrki Lahtonen, all bad from me.

P.S.P.S.

The sense of the question is about: prove that certain error-correcting code is "good".

Think of coefficients of $p(x)$ as information bits you want to transmit. Encoder sends $p(x)$ to pair $(pf, pg)$ so you add redundancy bits before transmission. If equality is true means that code is able to correct $4$ errors ($4+4<10$).

Gossips say that this code has been used in Voyager mission to transmit these pretty pictures. Thanks for "draks" for link to pictures.

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    How is Voyager related? And which Voyager: Voyager 1&2, [Voyager 6](http://en.wikipedia.org/wiki/Star_Trek:_The_Motion_Picture#Plot), Star Trek:Voyager ...2012-08-01
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    @draks Quote from http://mathoverflow.net/questions/103497/what-are-best-polynoms-fx-gx-of-degree-n-i-e-ideal-generated-by-them-is/103509#103509 "IIRC this code was used during the Voyager mission to transmit data (e.g. the pretty images) back to Earth"2012-08-01
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    Ah ok, I didn't see...but I remember the [pretty pictures](http://en.wikipedia.org/wiki/Pale_Blue_Dot)...2012-08-01
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    A teaser is that $g(x)$ is a factor of $x^{15}+1$ and $f(x)$ is a factor $x^{63}+1$, so a suitable choice of $p$ will make $|pf|$ or $|pg|$ equal to two.2012-08-01
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    I hazard a guess that [these pictures](http://voyager.jpl.nasa.gov/image/saturn.html) were also transmitted using this code.2012-08-01
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    Look for algorithms for finding the _free distance_ of a convolutional code. In most cases, substantial computation is required, and there are no easy answers or proofs, and no _useful_ bounds that are easy to compute. I definitely recommend against using this as an _exercise_ in teaching, unless the intent is to get the students to develop their own algorithms for the task.2012-08-01
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    @DilipSarwate Thank you for yours comment. May be you right, exercise a bad word, may be complicated puzzle is better... What is not quite clear for me about "no easy answers or proofs, and no useful bounds " is that - is it known that any bound will be complicated or just this is open problem to find good bound ? E.g. may I ask you to look at http://mathoverflow.net/questions/103497/what-are-best-polynoms-fx-gx-of-degree-n-i-e-ideal-generated-by-them-is/103509#103509 any comments are welcome !2012-08-01
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    @Alexander Chervov See my answer on mathoverflow. For linear binary _block_ codes, finding the minimum distance (even approximately, i.e. to within a constant factor) is a NP-hard problem. The same should presumably be true of the free distance problem.2012-08-01
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    @DilipSarwate Thank you for yours comments ! Probably a part of my question is complicated and may be NP, but there is the following YES/NO sub-question: Is it true that Max_{f,g, deg = n} MinDist(f,g) > n ? I think yes, what is sharper bound ? may be n +log(n) or n+sqrt(n) ? What is known about it ?2012-08-01

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