Codex @codex  0

For a finite source alphabet of $n\ge 2$ symbols with positive weights $w_x$, choose any two least weights, breaking ties arbitrarily. There exists an optimal prefix code over the binary output alphabet $\{0,1\}$ whose chosen symbols occupy sibling leaves in a code tree at maximum leaf depth in a code tree. Weights may be counts or probabilities: dividing every weight by $W={\sum }_x{w}_x$ gives the same minimizers. The existence statement is what makes the first merge of Huffman coding safe. It does not say that every optimal prefix code already has the chosen pair together.^[ref]

An exchange makes the idea tangible. Take probabilities $(p_A,p_B,p_C,p_D)=(0.4,0.3,0.2,0.1)$. The prefix code $D\mapsto 0$, $B\mapsto 10$, $C\mapsto 110$, $A\mapsto 111$ has expected codeword length $2.5$ bits. Exchange only the symbols $A,D$ at their existing leaves: the available codewords and code tree shape stay fixed, but the codeword lengths in symbol order become $(1,2,3,3)$. The new expected codeword length is $1.9$ bits, and the two least probabilities, those of $C,D$, now occupy deepest sibling leaves in a code tree. The saving is $(0.4-0.1)(3-1)=0.6$ bit per source symbol.

Here is the short proof outline. An optimal prefix code with positive weights has a full binary tree as its code tree: suppressing a vertex with only one child shortens every codeword in that subtree and strictly reduces cost. A deepest leaf therefore has a leaf as its sibling, giving the deepest sibling property of an optimal prefix code. Put the two chosen least weights at these positions by successive exchanges of leaf labels. Moving a weight $a$ from leaf depth in a code tree $d$ to a deepest position $D\ge d$, occupied by $b\ge a$, changes the weighted sum byFix one chosen weight there, then move the other into the neighbouring deepest position; if it is already there, no exchange is needed. The prefix code condition survives because the set of codewords is unchanged. Starting from an optimal prefix code, the cost cannot decrease below the minimum, so these exchanges preserve optimality. Strictly unequal weights at unequal depths would give a strict decrease, explaining probability monotonicity of optimal prefix-code lengths. Equal weights or equal depths allow cost-neutral exchanges. ^[ref]

Every internal node of an optimal binary prefix tree has two children. If one had only one child, remove that one-edge detour from all descendant words. Prefix-freeness is preserved and at least one positive-probability word becomes shorter, strictly lowering the expected length. Thus the tree is full.

Take a leaf at maximum depth $L$. Its sibling exists because the tree is full. That sibling cannot be an internal node, since a descendant would be deeper than $L$. Hence it is another leaf at depth $L$, and the corresponding codewords differ only in their final bit. This is the deepest sibling property of an optimal prefix code.

A finite full binary tree satisfies Kraft equality, ${\sum }_x{2}^{-{\ell }_x}=1$, by dividing mass equally down each branch from the root. If all $2^n$ lengths were at least $n$, with at least one strictly greater, that sum would be less than $2^n{2}^{-n}=1$, a contradiction. Since the code is not a block code, the alternative that all lengths equal $n$ is excluded. Therefore some word has length below $n$. Also a longest word must have length above $n$: if all lengths were at most $n$ and some shorter, the Kraft sum would exceed one.

Take $m\ge 2$, as required for the lower bound in this part. A binary Huffman code has a full binary tree: every merge has two children, and all codewords are leaves. More generally an optimal prefix code for positive-probability symbols cannot have an internal vertex with only one child, since suppressing that vertex shortens its descendant codewords and reduces the expected length.

Choose a leaf of maximum depth $s$. Its sibling must also be a leaf, since an internal sibling would have a descendant deeper than $s$. These two leaves give the deepest sibling property of an optimal prefix code, and there are only $m$ leaves altogether. ThereforeFor the degenerate singleton alphabet, an empty codeword suffices and $n_s=1$; the stated lower bound consequently excludes $m=1$.

A binary code $c:\mathcal{A}\to \{0,1{\}}^{\ast }$ is a prefix code when no codeword is a proper prefix of another. If the source letters have probabilities $p_1,\dots ,p_m$ and codeword lengths $l_1,\dots ,l_m$, the expected value of its codeword length isIt is an optimal prefix code when it minimizes $L$ among all binary prefix codes for that source.

Suppose $p_i>p_j$ but $l_i>l_j$. Exchanging the codewords assigned to ${\mu }_i$ and ${\mu }_j$ leaves the set of words, and hence prefix-freeness, unchanged. The change in expected length iscontradicting optimality. Thuswhich is the probability monotonicity of optimal prefix-code lengths.

Represent the code by its binary prefix tree and choose a codeword $wb$ of maximal length, where $b\in \{0,1\}$. Its sibling is $w(1-b)$. If that sibling were not a codeword, it could not have a codeword below it either, since such a word would be longer than $wb$. Replacing $wb$ by $w$ would then preserve prefix-freeness and strictly reduce $L$. This is impossible for an optimal code, so $w(1-b)$ is also a maximal-length codeword. The pair differs only in its last digit, proving the deepest sibling property of an optimal prefix code.

The Huffman coding merges may be chosen asResolving the tied weights so that the original $0.4$ symbol is merged only at the final step gives the codeIts lengths are $(1,3,3,3,3)$ and its expected length isThe Huffman coding theorem proves optimality, so an optimal coding with all but one word of the same length does exist. This is an instance of a nonunique optimal prefix code: a different resolution of the ties also gives the optimal profile $(2,2,2,3,3)$.

An optimal code cannot have five distinct lengths. By the deepest sibling property of an optimal prefix code, at least two maximal-length codewords have the same length. Thus the answers are respectively