# Length of shortest codeword in Huffman encoding

Under Huffman Encoding, if one character occurs more than 1/3rd of the time, is it guaranteed that there will be at least one character whose codeword is of length 1?

I thought of 2 cases where this might be true, but they're not exactly proofs -

Case 1: There is only one character which occurs more than 1/3rd of the time. In this case, a good huffman encoding scheme will ensure that this character corresponds to a single-character encoding, because there are no competing characters. Even though there may be characters that occur with high probability (but still less than 0.33), they must have length greater than 1 to ensure the prefix-free nature of the encoding.

Case 2: There are multiple characters that occur more than 33% of the time. We can also make the stronger assertion that in this case, there can only be at most 2 such characters. This can be further subdivided into cases where 1) there are a total of 2 characters in the encoding scheme, and 2) where there are more than 2 characters in the scheme. In case 1), both the characters can correspond to a single-character encoding scheme (0 and 1 for each respectively). In case 2), at most 1 of the 2 largely prevalent/frequent characters must have an encoding of length of 1 (since we're considering a good huffman encoding scheme) while the other n-1 characters have a length of at least 2.

I'm afraid that this is not the right approach. Could someone please help me out?

Consider the distribution $$\Pr[a]=0.34,\Pr[b]=\Pr[c]=\Pr[d]=0.22$$.
Huffman's algorithm will first merge (say) $$c$$ and $$d$$ to create the symbol $$cd$$ with probability $$0.44$$. It will then merge $$a$$ and $$b$$ to create the symbol $$ab$$ with probability $$0.56$$. Finally, it will merge the remaining symbols.
You can check that the same example works up to $$\Pr[a] = 0.4$$ (we need $$\Pr[a] \leq 2\Pr[b] = (2/3)(1-\Pr[a])$$). What if $$\Pr[a] > 0.4$$? Consider the time at which Huffman's algorithm merges $$a$$ with some symbol $$X$$ (which may be composed of several original symbols). If $$a$$ and $$X$$ are the only symbols, then $$a$$ will be encoded using a symbol of length 1. Otherwise, there is some other symbol $$Y$$ of probability at least $$\max(\Pr[a],\Pr[X]) > 0.4$$. We see that $$a,X,Y$$ must be the only remaining symbols, and furthermore $$\Pr[a] > \Pr[X]$$ (otherwise there would 3 symbols whose probability exceeds 0.4).
Suppose that $$Y$$ was obtained by merging two other symbols $$Z,W$$. At that point, since $$X$$ itself or one of its constituents wasn't merged, we must have $$\Pr[X] \geq \max(\Pr[Z],\Pr[W]) \geq \Pr[Y]/2$$. But $$\Pr[Y] > 0.4$$ and so $$\Pr[X] > 0.2$$, and so the probabilities sum up to more than 1. This contradiction shows that $$Y$$ is an original symbol, and so in this case there is a codeword of length 1.