 Huffman code

A Huffman code is the prefix code returned by Huffman coding; the algorithm chooses a code tree, while the code specifies the actual symbol-to-codeword assignment. For the probability distribution $(0.4,0.3,0.2,0.1)$ in symbol order $A,B,C,D$, the illustrated binary tree givesThe codeword lengths are $1,2,3,3$. This is a prefix code because no complete codeword is an initial segment of another complete codeword. Thus a decoder can start at the root of a tree, follow one edge per bit, emit a symbol on reaching its leaf, then restart at the root of a tree. In particular, 011010111 splits as 0 | 110 | 10 | 111 and decodes to $A,C,B,D$. No separators are transmitted: the leaf tells the decoder where a codeword ends.

The expected codeword length isbits per source symbol. Four symbols would require $2$ bits each if all codewords had the same length, so this Huffman code saves $0.1$ bit per symbol on average, or $5\%$ of that fixed-length payload. A particular message can grow: $A,C,B,D$ uses nine bits here instead of eight. These lengths count the encoded symbols only; transmitting a code description or message framing can add further cost.

The Kraft inequality is saturated by this Huffman code:Its Shannon entropy isbits per symbol, slightly below the expected codeword length $1.9$. The entropy lower bound for prefix codes applies to every binary prefix code for this probability distribution. It need not be attainable with integer codeword lengths for individual symbols; Huffman coding reaches the best mean within that class. ^[ref]

A Huffman code need not be unique. In the first example, the tied $B$ and $CD$ weights can exchange their $0/1$ edge labels, changing codewords but preserving their codeword lengths. Ties can also change the lengths themselves. For a second probability distribution $(0.4,0.2,0.2,0.2)$, first merge $C,D$ to weight $0.4$. The next merge can join $B$ with either $A$ or the $CD$ subtree, since both have weight $0.4$. Joining $B$ with $A$ permits $A\mapsto 00,B\mapsto 01,C\mapsto 10,D\mapsto 11$, with all lengths $2$. Joining $B$ with the $CD$ subtree permits $A\mapsto 0,B\mapsto 10,C\mapsto 110,D\mapsto 111$, with lengths $1,2,3,3$. Both are optimal prefix codes with expected codeword length $2$: for the latter, $0.4\cdot 1+0.2\cdot 2+0.2\cdot 3+0.2\cdot 3=2$. Tie choices can change the code tree without changing the minimum mean.

// Target: algebra.bigb