Each of the three input values has two possible outputs, so their union has at most six elements. The preceding union formula also gives this bound. Choose noise values $0,3$, whose two translated input alphabets are disjoint. With columns ordered $0,1,2,3,4,5$ the transition matrix isThus $|\mathcal{Y}|=6$. Every output determines its input, explicitly $X=Y\mathrm{mod}3$, so $H(X\mid Y)=0$. Consequently $I(X;Y)=H(X)\le {\log }_{2}3$, with equality for a uniform input. ThereforeThe extra uncertainty of which noise value occurred does not reduce information about the input: the two possible outputs for each input remain distinguishable from those of every other input.

Translate the output alphabet by $-\min (z_1,z_2)$ and put $d=|z_2-z_1|\ge 1$. The reachable outputs becomeFor $d=1$ the two sets overlap in two points; for $d=2$ in one; for $d\ge 3$ they are disjoint. In particular $|\mathcal{Y}|\ge 4$, and choosing noise values $0,1$ attains four.

With rows indexed by inputs $0,1,2$ and columns by outputs $0,1,2,3$, independence and uniform two-point noise giveEvery conditional output distribution has information entropy one bit, so $I(X;Y)=H(Y)-1\le {\log }_{2}4-1=1$. The input probability mass function $(1/2,0,1/2)$ makes all four outputs equiprobable, attaining that bound. ThusUsing a zero-probability input in a channel-capacity maximization is legitimate; it is unrelated to the strictly positive source weights needed for the tree-shape claims in Question 2.

Assume a full-support finite distribution. The preceding structural arguments show that every internal vertex has three tree children, except possibly one tree parent at depth $L-1$ with two tree children. Let $N_3$ and $N_2$ count these two kinds of internal vertex. Part iv gives $N_2\in \{0,1\}$; no vertex has one tree child.

A finite rooted tree with $m$ tree leaves has $m+N_2+N_3-1$ edges, while counting outgoing edges gives $2N_2+3N_3$. Equating the counts yieldsFor an odd integer $m$, the right side forces $N_2$ to be an even integer. Since it is at most one, $N_2=0$. Thus no deepest pair has only two sibling codewords, and the optimal code tree is a full ternary tree. This leaf-count identity also directly proves that a full ternary tree has an odd number of tree leaves.

Again the unrestricted zero-weight version is false: $(0,10,11)$ is an optimal prefix code for $p=(1,0,0)$, with odd integer $m=3$, but its two deepest tree leaves form a two-sibling pair. There is no contradiction with the proof because its tree root has only two tree children, which the full-support finite distribution assumption would forbid above the deepest level.

Let $L$ again denote the largest codeword length, and assume a full-support finite distribution. The missing-branch exchange from part iii forces every internal vertex at depth at most $L-2$ to have three tree children. At depth $L-1$, all tree children are tree leaves. Part ii rules out one tree child there; the hypothesis that there is no pair of two deepest sibling codewords rules out two tree children. Thus each such tree parent also has three tree children. Every internal vertex has exactly three tree children, which is a full ternary tree, called semi-complete in this paper. The same last-level argument covers $L=1$.

Without the assumption of a full-support finite distribution the implication is false: for $p=(1,0,0)$, the optimal prefix code $(0,10,110)$ has no deepest pair of two sibling codewords, but its code tree contains vertices with a single tree child. Empty unused tree children are not vertices of the code tree.

With a full-support finite distribution, suppose two distinct tree parents at depth $L-1$ each have exactly two tree children, both tree leaves. Move one tree leaf from the second pair into the vacant third tree child slot of the first tree parent. Its length remains $L$, so this relocation does not change expected codeword length. The second tree parent now has a single tree child, which is a tree leaf. Replace that remaining tree child's codeword by the tree parent's word. This preserves the prefix code property and shortens a positive-probability codeword by one. Expected codeword length strictly decreases, a contradiction. Thus there is at most one deepest pair with exactly two sibling codewords.

If zero-probability symbols are retained without a reduction convention, the claim can fail. The probability mass function $(1,0,0,0,0)$ with prefix code $(0,100,101,110,111)$ has the optimal expected codeword length one, yet contains the two deepest pairs $(100,101)$ and $(110,111)$. The contraction would only shorten a zero-probability word.

Suppose a tree parent at depth $n-1$, with $n<L$, has an unused outgoing digit. The missing tree child gives an empty code tree branch at depth $n$. Move any deepest tree leaf into that position, retaining all other codewords and deleting the abandoned empty branch. The new word has no old word as a prefix, because it lies in an unused branch, and it is not a prefix of any retained word. Its length decreases from $L$ to $n$. With a full-support finite distribution this reduces expected codeword length by $p_i(L-n)>0$, contradicting optimality. Hence all three tree children exist, and each has a descendant codeword. In fact every internal vertex above depth $L-1$ has all three tree children. Applying this to the tree parent of the specified prefix gives the claimed three extensions.

The positive-probability qualification attached only to the specified prefix does not by itself suffice. In the zero-weight example $(p,c)=((1,0,0),(0,10,110))$, choose the positive-probability word $0$, so $n=1<L=3$. There is no word starting with $2$. The deepest tree leaf available for relocation has probability zero, so the strict-improvement argument fails. This is why the full-support finite distribution assumption is used for the structural conclusions.

First assume a full-support finite distribution, as stated above. Let $L$ be the largest codeword length and take a tree leaf at depth $L$. Every existing tree child of its tree parent must also be a tree leaf: an internal tree child would have a descendant deeper than $L$. If the chosen tree leaf were its tree parent's only tree child, moving its symbol to that tree parent would shorten its codeword by one, preserve the prefix code property, and reduce expected codeword length by its positive probability. That contradicts optimality. The tree parent therefore has at least two tree children, both tree leaves. A second codeword has the same first $L-1$ digits and a different last digit, proving the required deepest sibling codeword property for a ternary prefix code.

For the literal wording allowing zero probabilities, take $p=(1,0,0)$ and the prefix code $(0,10,110)$. Its expected codeword length is one, which is minimal: with more than one message an empty codeword is impossible, so the positive-probability message must have length at least one. Yet the deepest tree leaf $110$ has no sibling codeword. Thus positivity or an additional convention removing redundant zero-weight branches is needed.

Exchange the two complete codewords while leaving the code tree unchanged. The prefix code property is preserved because the set of codewords is unchanged. If $p_i>p_j$ but $l_i>l_j$, the change in expected codeword length iscontradicting optimality. Hence $p_i>p_j$ implies $l_i\le l_j$. This probability monotonicity of optimal prefix-code lengths is valid even if some probabilities are zero; equal probabilities impose no order.^[ref]

Reference-distribution ambiguity: $\rho$ is not defined in the paper. If it is an arbitrary probability mass function, the printed inequality is false. For instance, choose $p=(1,0)$ and $\rho =(1/2,1/2)$. Every empirical distribution equals $p$, making the printed left side one bit and its right side zero. This is a counterexample for every $n\ge 1$.

Under the consistent interpretation $\rho =p$, the preceding leave-one-out identity for empirical distributions and convexity of relative entropy in its first argument give, for every sample,Deleting any one observation from $n+1$ independent and identically distributed random variables leaves $n$ independent observations with probability mass function $p$. Consequently each ${\hat{q}}_{n,i}$ has the same distribution as ${\hat{p}}_n$. They need not be independent of one another. Taking expected values and using linearity of expectation therefore provesZero coordinates of $p$ are never sampled, almost surely, so the calculation can be restricted to its probability support. In particular the expected values are finite on this finite alphabet.

The discrepancy cannot be removed by silently choosing any fixed reference. Whenever $\rho$ is positive on the probability support of $p$, direct subtraction and $\mathbb{E}{\hat{p}}_n=p$ give the expected relative entropy decompositionThus the valid arbitrary-reference version has $\rho$ on both sides; the printed mixed-reference version has an additional term whose sign prevents the proposed conclusion.

Index correction: the printed definition only permits $i\le n$, although the required average includes $i=n+1$. Extend the same deletion formula to $1\le i\le n+1$. Without that extension the final summand is undefined; there is no new sample or different estimator to supply.

For a fixed outcome of the sample and a fixed symbol $x$, put $A_j=1_{\{X_j=x\}}$. The leave-one-out empirical distribution then satisfiesDividing by $n+1$ gives ${\hat{p}}_{n+1}=(n+1{)}^{-1}{\sum }_{i=1}^{n+1}{\hat{q}}_{n,i}$. Each of the $n+1$ indicator functions was counted exactly $n$ times before division by $n$. This leave-one-out identity for empirical distributions is deterministic; independence plays no part in this part.

For probability mass functions on the finite alphabet, take logarithms to base two and define the relative entropy byA term with $q_i=0$ is zero, including $p_i=0$; a term with $q_i>0=p_i$ is $+\infty$. Fix $p_i>0$ and write $f_i(t)=t{\log }_2(t/p_i)$. For $t>0$,Thus $f_i$ is a convex function on $[0,\infty )$ by continuous extension to zero. For $0\le \lambda \le 1$, applying this scalar convexity to each coordinate and summing provesIf $p$ has zero coordinates, the finite-valued domain consists of probability mass functions supported on $\{i:p_i>0\}$; it is convex and the same proof applies there. Outside that domain the right side is infinite whenever an offending probability mass function has a positive mixture weight, so the extended-valued convexity assertion still holds. At mixture weights zero or one the inequality is an identity, without needing the expression $0\cdot \infty$.

Here the asserted bound is actually an identity, so equality always holds whenever the displayed quantities are defined. Expand the same mutual information in two orders:Equating these expressions and using symmetry of mutual information and conditional mutual information givesThere is no further independence condition. Treating this part as a strict inequality would obscure its cancellation.

The chain rule for information entropy converts the two differences into conditional entropies. Their gap, in the order right minus left, isThis is additional conditioning reduces entropy. The relative entropy representation of conditional mutual information identifies exactly when that gap vanishes: equality holds if and only if $Y$ and $Z$ have conditional independence given $X$. Knowledge of $Y$ then changes no conditional probability of $Z$ after $X$ is known.

Expand the two observations in order using the chain rule for mutual information:The remaining conditional mutual information isRelative entropy nonnegativity proves the inequality. Equality requires every positive-weight relative entropy to vanish, which is equivalent to $p(y,z\mid x)=p(y\mid x)p(z\mid x)$ for every $x$ of positive probability. Thus equality holds exactly when $Y$ and $Z$ have conditional independence given $X$. No unconditional independence is required.

The chain rule for conditional entropy givesIndeed, for every conditioning value $(x,z)$ of positive probability, each summand $-p(y\mid x,z){\log }_{2}p(y\mid x,z)$ is nonnegative. Its sum is zero exactly when the conditional probability mass function is concentrated at one value. Averaging over $(X,Z)$ therefore proves both the inequality and its equality criterion: equality holds exactly when $Y=f(X,Z)$ almost surely for some function $f$. Conditioning values of probability zero impose no restriction. This is a statement about discrete random variables; a differential entropy need not be nonnegative.

For the fundamental representation and the two-index symmetric representation, the Dynkin indices are $1$ and $5$. Therefore the specified discrete gauge anomaly condition for ${\mathbb{Z}}_4$ isAll ordered solutions areOnly $(1,3)$ and $(3,1)$ give a faithful ${\mathbb{Z}}_4$ action on this pair of Weyl fermions; faithfulness was not imposed. As in solution, cancellation here concerns the stipulated discrete gauge anomaly, not cancellation of the pure cubic SU(3) gauge anomaly. The integer normalization matters: one must not replace $c$ by a half-integer trace index inside an unchanged modular congruence.

Use the integer normalization $c(3)=1$ from solution. For two fundamental representations, the specified discrete gauge anomaly coefficient is $\mathcal{A}=q_1+q_2$. Consequently all solutions in the finite cyclic group of charges areThere are $n$ ordered charge pairs. The trivial pair is included: the problem does not require a faithful action. If a faithful action on these Weyl fermions were additionally required, one would retain precisely the $q$ coprime to $n$.

This cancels the discrete gauge anomaly defined in the paper. Two left-handed fundamental representations by themselves have a nonzero pure cubic SU(3) gauge anomaly; a complete gauge theory would need additional fermions to cancel it. That separate condition does not alter the requested discrete congruence.

The two-index symmetric representation of SU(3) is $6$, with Dynkin index $5$. Thus $n$ copies have $c(r_S)=5n$, and the supplied beta function isThe strict sign condition for asymptotic freedom is $n<132/5=26.4$, givingIndeed $132-5(26)=2>0$, while $132-5(27)=-3<0$.

The Dynkin index is additive over a direct sum of group representations, so $n$ copies of the fundamental representation have $c(r_S)=n$. With $c(\mathrm{ad})=6$, the supplied beta function becomesFor positive gauge coupling $g$, asymptotic freedom at this order requires $\beta (g)<0$, hence $n<132$. For nonnegative integer $n$,At $n=132$ the one-loop coefficient vanishes, so this formula alone does not establish asymptotic freedom. This count follows the scalar multiplicity convention of the printed beta function; changing how a real or complex scalar field is counted requires changing its coefficient too.

Now the doublet invariant tensor would satisfy $D_1(Q{)}^{T}L{D}_2(Q)=L$. Its four entries acquire factors ${\alpha }^3$, ${\alpha }^{-1}$, $\alpha$, and ${\alpha }^{-3}$, respectively. None equals $1$, so $L=0$. Equivalently, $R_2\otimes R_3=R_2\oplus R_3$ has no trivial representation. The mixed doublet and third-family terms are again forbidden because their tensor product of group representations contain no trivial representation. Only the product of the two $R_1$ fields survives: