最大化霍夫曼编码平均长度：概率向量选择与极值相关技术问询

Great question! Let's unpack this thoroughly:

1. Yes, equal probabilities maximize the average code length #

You're exactly right—when every symbol in the probability vector has an equal chance ($p_i = 1/n$ for all $i$), the average code length of Huffman coding reaches its maximum possible value.

Take your $n=4$ example: with $p = [0.25, 0.25, 0.25, 0.25]$, the optimal Huffman code assigns each symbol a 2-bit code (e.g., 00, 01, 10, 11), giving an average code length of $2$. Compare this to an uneven distribution like $[0.5, 0.25, 0.125, 0.125]$: here, the high-probability symbol gets a 1-bit code, others get 2 or 3 bits, resulting in an average length of $10.5 + 20.25 + 30.125 + 30.125 = 1.75$, which is clearly smaller.

2. Why equal probabilities work #

Huffman coding's core design is to minimize average code length by assigning shorter codes to more frequent symbols. To reverse this and maximize the average length, we need to eliminate any "priority" for shorter codes. When all probabilities are equal, there's no justification for giving any symbol a shorter code than another—so all codes end up as long as possible (or as evenly matched as the Huffman algorithm allows). Any deviation from equal probabilities creates at least one symbol that can be assigned a shorter code, pulling down the overall average.

3. Yes, a maximum average code length exists #

The maximum average code length is achieved precisely when all symbols have equal probability. There's no way to get a longer average length than this—any other distribution will have at least one symbol with a shorter code, reducing the average.

4. Calculating the maximum average code length #

The formula depends on whether $n$ (the number of symbols) is a power of 2:

• If $n$ is a power of 2 ($n = 2^k$ for some integer $k ≥1$):
  The average code length equals the entropy of the distribution:
  $$L_{max} = \log_2(n)$$
  Every symbol gets exactly $k$-bit codes, so the average is perfectly uniform.
• If $n$ is not a power of 2:
  First find the largest integer $m$ such that $2^m < n < 2^{m+1}$. The maximum average code length is:
  $$L_{max} = (m+2) - \frac{2^{m+1}}{n}$$
  Alternatively, this can be written using the count of symbols with different code lengths: $(2^{m+1} - n)$ symbols will have $m$-bit codes, and the remaining $(2n - 2^{m+1})$ symbols will have $(m+1)$-bit codes. The average is:
  $$L_{max} = \frac{(2^{m+1}-n)m + (2n - 2^{m+1})(m+1)}{n}$$
  For example, $n=3$: $m=1$ (since $2^1=2 <3 <4=2^2$), so $L_{max} = (1+2) - 4/3 = 5/3 ≈1.666$, which matches the average of the Huffman codes (1 bit for one symbol, 2 bits for the other two).

This maximum value always sits between $\log_2(n)$ (the entropy) and $\log_2(n)+1$.

内容的提问来源于stack exchange，提问作者John Lexus