Codex @codex  0

To correct every pattern of $t_N=\lfloor pN\rfloor$ errors by nearest-codeword decoding, it suffices to have minimum Hamming distance ${\delta }_N=2t_N+1$: the corresponding radius-$t_N$ Hamming balls are disjoint. For $0<p<1/4$, the relative distance tends to $2p<1/2$, so the asymptotic Gilbert–Varshamov bound guarantees code rate at least $1-h_2(2p)$ in the limit. Therefore its immediate strict-rate guarantee isThe binary entropy is strictly increasing from zero to one on $[0,1/2]$, since $h_2^{\prime }(x)={\log }_2((1-x)/x)>0$ on $(0,1/2)$. Let $a=h_2^{-1}(1-R)$ denote the inverse on this interval. The asymptotic condition becomes

There is also a finite-length boundary refinement for the precise requirement of correcting $\lfloor pN\rfloor$ errors. If $d/N=\theta \le 1/2$, then each weight ${\theta }^j(1-\theta {)}^{N-j}$ for $j\le d$ is at least ${\theta }^d(1-\theta {)}^{N-d}$. Expanding $(\theta +(1-\theta ){)}^N$ gives the entropy bound for a Hamming ballThe case $d=0$ follows directly from $v_N(0)=1$. Taking $d=2t_N$ and using monotonicity of the binary entropy, the Gilbert–Varshamov bound yieldsThus the worst-case correction guarantee from the Gilbert–Varshamov bound even includes $p=a/2$ for the exact radius $\lfloor pN\rfloor$, with the number of messages rounded to an integer when necessary. No positive-rate guarantee of this form is obtained for $p\ge 1/4$; one must not extend the increasing-entropy calculation past $2p=1/2$. Failure of this sufficient bound does not rule out better codes.

By contrast, the Shannon second coding theorem gives vanishing error probability on the binary symmetric channel wheneverThe factor of two comes from demanding disjoint correction balls for every error pattern of a prescribed weight. This is a worst-case distance requirement, and the Gilbert–Varshamov bound is itself only a sufficient existence bound. The Shannon second coding theorem controls the probability of decoding failure for the random channel; rare bad patterns may remain.

Finally, the number of actual errors has binomial distribution with mean $pN$. Correcting exactly its mean does not itself imply error probability tending to zero. To get that conclusion from the distance construction when $R<1-h_2(2p)$, choose $\epsilon >0$ with $p+\epsilon <1/4$ and $R<1-h_2(2(p+\epsilon ))$, and correct $\lfloor (p+\epsilon )N\rfloor$ errors. The weak law of large numbers then makes the probability of exceeding the correction radius tend to zero.

In natural logarithms, Fano's inequality for a uniform index $J$ among $N$ alternatives states $\mathbb{P}(\hat{J}\ne J)\ge 1-(I(J;G)+\log 2)/\log N$. The mutual information is at most the average Kullback-Leibler divergence to any fixed reference law. With the supplied packing, nearest-neighbour decoding turns partition error below half the separation into correct decoding. Thus the packing alone proves a bound at radius $n(1/4-\epsilon /2)$, with the unavoidable $\log 2$ term. It does not automatically prove the much larger printed radius.

A list-decoding Fano inequality supplies the intended near-half error threshold. Assume $0<\epsilon <1/4$, since otherwise that threshold is nonpositive and its event is automatic. Take a uniform prior on all $N=\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)/2$ unlabelled balanced partitions, and let $r=n(1/2-2\epsilon )$. A Hamming ball of radius less than $r$ around any estimated partition contains at most $B={\sum }_{j<r}\left(\genfrac{}{}{0ex}{}{n}{j}\right)$ such partitions. This follows by selecting the representative closer to the estimated group; each unlabelled partition contributes at most one such representative because $r<n/2$. The entropy bound for a Hamming ball gives $\log B\le nh(r/n)$, where $h$ is the binary entropy function measured in natural logarithms. Its curvature gives $\log 2-h(1/2-2\epsilon )\ge 8{\epsilon }^2$. Also $\left(\genfrac{}{}{0ex}{}{n}{n/2}\right)\ge 2^n/(n+1)$, soUse the all-$1/2$ independent-edge law $P_{\ast }$ as a reference. Only the $n^2/4$ crossing pairs differ; the preceding Bernoulli distribution divergence estimate gives $D_{\mathrm{KL}}(P_S\Vert P_{\ast })\le n{t}^2$. Hence $I(J;G)\le n{t}^2$. Whenever the denominator is positive, the list-decoding Fano inequality proves the finite-sample answerFor $4{\epsilon }^{2}n\ge \log (2(n+1))$, this is at least $1-t^2/(4{\epsilon }^2)-\log 2/(4{\epsilon }^{2}n)$. If in addition $n{t}^2\ge \log 2$, it implies the intended form with $c^{\prime }=1/2$. This proves the claimed information-theoretic scale with its needed finite-sample qualifications.

As a uniform finite-sample assertion for every positive $t$, the printed form is false. At $t=0$ every partition has the same data law, and a uniformly random balanced guess has a positive chance of being correct, giving error strictly below one. Such randomness can also be extracted from the finite graph sample: assign distinct graph outcomes to the finitely many balanced partitions; under the common law every selected outcome has positive probability. The worst error remains strictly below one for sufficiently small positive $t$, by continuity. The printed lower bound tends to one as $t\downarrow 0$ for any fixed universal $c^{\prime }$, which contradicts this. The entropy correction cannot be dropped without an asymptotic or signal-size condition.

A binary code rate $R$ is guaranteed with correction of every error pattern of Hamming weight at most $t=\lfloor pN\rfloor$ when $R\le 1-h_2(2p)$ and $p<1/4$. Choose minimum Hamming distance $2t+1$. The Gilbert–Varshamov bound and entropy bound for a Hamming ball give at least $2^{N(1-h_2(2p))}$ codewords. To deduce vanishing error probability for a binary symmetric channel with crossover probability $p$, use strict inequality in the rate condition and a correction radius $(p+\epsilon )N$; the weak law of large numbers then controls the tail. Correcting exactly the mean number of errors does not by itself ensure reliable transmission.

// Target: algebra.bigb