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Among the good asymptotic bounds in coding theory in the MRRW bound. It is obtained by using the linear programming problem of Delsarte's and providing a solution. The LP problem is

Suppose $C \subset \mathbb{F}_2^n $ is a code such $d(C)\ge d$. Let $\beta(x) = 1+ \sum_{k=1}^{n} y_k K_k (x)$ be a polynomial such that $y_k \ge 0$ but $\beta(j) \le 0$ for $j=d, d+1,\dots ,n$. Then, we have that $|C| \le \beta(0)$.

Here $K_k(x)$ are the Kravchuk polynomials. In the proof of the MRRW bound, upto scaling, they basically come up with the following polynomial $\beta$ for a general $n$.

$$\beta(x) =\frac{1}{a-x} \left[ K_t(a) K_{t+1}(x) - K_{t+1}(a)K_{t}(x) \right]^{2}$$

After using the Christoffel-Darboux formula the values of $t$ and $a$ are adjusted to make it optimal.

There is no justification for why such a polynomial was chosen other than that it works. Is there anything more that can be said over why this polynomial was chosen?

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    $\begingroup$ You might find a partial answer to your question in: A. Samorodnitsky, "On the optimum of Delsarte's linear program," J. Comb. Th. Ser. A, 96(2), 261-287, 2001. $\endgroup$ – Moshe Schwartz Jan 21 '19 at 19:44
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For linear programming type bounds it is sometimes only possible to give effective bounds (that is bounds that work and are manageable) and what is surprising is that the primitive method often gives in fact optimal bounds.

The LP problem from McEliece, Rodemich, Rumsey and Welch's paper that is cited in the question requires the auxiliary function $\beta(x)$ satisfy $\beta(j) = 1+ \sum_{k=1}^{n} y_k K_k (j)\leq 0$ for all $j=d,d+1,\dots,n$. The supplied $\beta(x)$ is designed to meet this requirement by making it change sign at value $a$, so that $\beta(x)\leq 0$ for $x\geq a$. This is the first point.

The Krawtchouk polynomials already appear in the LP problem, so that there are no questions of why the might appear in the auxiliary function $\beta$, but just to emphasize the importance of the Krawtchouk polynomials, they are used in discrete linear programming problems due to the positive definiteness criterion associated to them, namely that for a polynomial $$f(z)=\sum\limits_{i=0}^{n} a_{i}z^{i}$$ the matrix $$f(d(x,y)),\ x,y\in\mathbb{F}^{n}$$ is non-negative definite if and only if all coefficients $\lambda_{i}$ of the expansion $f(z)=\sum\limits_{i=0}^{n} \lambda_{i} K_{i}(z)$ over Krawtouck polynomials are nonnegative.

Now, to keep all the coefficients $y_k$ positive as required in the LP program, it makes sense to introduce the square, but in order to preserve the sign change in $\beta(x)$ at $x=a$, one divides by $(a-x)$.

Finally, the choice of the exact expression inside the square works, because as was noted, the Christoffel-Darboux formula allows for rewriting

$$K_t(a) K_{t+1}(x) - K_{t+1}(a)K_{t}(x)=\frac{2(a-x)}{t+1}\binom{n}{t}\sum\limits_{k=0}^t \frac{K_{k}(x)K_{k}(a)}{\binom{n}{k}}$$

so that one may check quickly that $\beta(x)$ has expansion coefficients $y_{k}$ in the Krawtouck polynomials non-negative. Optimizing in $a$ and $t$ as noted give the MRRW upper bound $M_{LP}(n,d)\leq \binom{n}{t}\frac{(n+1)^2}{2(t+1)}$.

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  • $\begingroup$ So the overall reason is that they saw the Chrisotoffel-Darboux expressions and had a clever inspiration? Maybe it's really just that. Let's see if someone has a better answer. $\endgroup$ – Breakfastisready Dec 27 '18 at 5:04
  • $\begingroup$ @Breakfastisready Yes, essentially this is what is being suggested. This tool and the fact that the function $\beta(x)$ has the desired property of switching signs at $x=a$ seem to be ample motivation for choosing the above function. $\endgroup$ – Josiah Park Dec 27 '18 at 5:11

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