Let $\Sigma$ be an alphabet of cardinal $n$. Let $T$ be the set of ordered binary tree whose nodes are labeled by words over $\Sigma$, such that each leaf is labeled by a letter $a\in \Sigma$ and the parent of two nodes is labeled by the concatenation of their labels. Given such a tree $t\in T$, we write $W_t$ for the set of words that appear as labels in $t$, and $|W_t|$ for its cardinal. Given $u\in\Sigma^*$, let $T_u$ be the subset of $T$ consisting of trees whose root is labeled by $u$. I am interested in bounds on $$f(n,l)=\max_{u\in\Sigma^l}\min_{t\in T_u}|W_t|$$ in terms of $l$ when $n$ is constant, or in terms of both $l$ and $n$.

This quantity is useful because if we replace the tree by the corresponding directed acyclic graph, $|W_t|$ corresponds to the number of nodes in the DAG, and the quantity therefore correspond to the maximal number of nodes one may need to represent a word of length $l$. I use this DAG structure in this answer.

For now I only know that:

  • $\boxed{l\le n \Rightarrow f(n,l)\ge l}$ If $l\le n$ then by taking a word $u$ that uses $l$ distinct letters, we get $|W_t|\ge l$ for any $t\in T_u$ by counting the leaves.

  • $\boxed{l=2^k\Rightarrow f(1,l)=\log_2(l)+1}$ If $n=1$ and $l=2^k$, then we can build a tree $t\in T_u$ that only uses the labels $a^{2^i}$ for $1\le i\le k$, and we have $|W_t|=k+1=\log_2(l)+1$ words.

Ideally, I would like a bound that either proves or disproves $f(2, l)=o(l)$.

  • $\begingroup$ I'm missing something. Why doesn't $T_u$ always have size $1$? $\endgroup$ Commented Jun 21, 2021 at 14:16
  • $\begingroup$ @mathworker21 consider the tree: (()())() and ()(()()) in other words two nodes from the root, down one with one leaf and the other with two leafs. Label from left to right the nodes as a,b,c. How can you tell the two trees apart? $\endgroup$
    – Phicar
    Commented Jun 21, 2021 at 14:23
  • $\begingroup$ @Phicar Thanks. I was thinking of "complete binary tree" $\endgroup$ Commented Jun 21, 2021 at 14:27

1 Answer 1


Unless I made a mistake, I think it's easy to show $f(2,l) = o(l)$.

It suffices to prove this for $l$ a power of $2$, say $l = 2^k$.

Take $u \in \{a,b\}^l$ and just do a complete binary tree of depth $l$ with leaves being the bits of $u$.

At depth $j$, there are $2^j$ strings each of length $2^{k-j}$. Since there are at most $2^{2^{k-j}}$ strings of length $2^{k-j}$, there are at most $\max(2^j,2^{2^{k-j}})$ distinct strings at depth $j$. So there are a total of at most $\sum_{j=0}^l \max(2^j,2^{2^{k-j}})$ strings that appear as nodes. We upper bound this sum by $\sum_{j=0}^{k-\frac{1}{2}\log k} 2^j+\sum_{j=k-\frac{1}{2}\log k}^k 2^{2^{k-j}}$. The first term is basically $2^{k-\frac{1}{2}\log k} = \frac{1}{\sqrt{k}}2^k$, and the second term is basically $2^{2^{1/2\log k}} = 2^{\sqrt{k}}$. So we get $f(2,l) = O(\frac{1}{\sqrt{k}}2^k)$.

A more interesting question is whether $\log f(2,l) = o(\log l)$, i.e., whether $f(2,2^k) = 2^{o(k)}$, but I think this can be disproved by looking at random strings, since all substrings of length $> \log n$ will be distinct.


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