On the Minimal Uncompletable Word Problem

If S is not complete, Σ * \ Fact(S * ) is not empty and a word in this set of minimal length is called a minimal uncompletable word (with respect to the non-complete set S).

The problem of finding minimal uncompletable words and their length was introduced by Restivo [4], who conjectured that there is a quadratic upper bound for the length of a minimal uncompletable word for S in terms of the maximal length of words in S.

A more general related question of deciding whether a given regular language L satisfies one of the properties Σ * = Fact(L), Σ * = Pref(L), Σ * = Suff(L) has been recently considered by Rampersad et al. in [3], where the computational complexity of the aforesaid problems in case L is represented by a DFA or NFA is studied. In the particular case L = S * for S being a finite set of words -which is the case that is of interest for us -the authors mention that the complexity of deciding whether or not Σ * = Fact(S * ) is still an open problem.

In this note, we show by mean of an example that the length of a minimal uncompletable word for a set S whose longest word is of length k seems to grow as 3k 2 asymptotically and at least gets larger than 2k 2 for effectively computed values, thus improving on a previous example given by Antonio Restivo [4]. The computations of a minimal uncompletable word for the successive values of k in the parametrized example were made on the Vaucanson platform for computing automata [5]. This result is briefly mentioned in [1].

The previous attempts to studying non-complete sets of words lead us to the following formulation.

Let S ⊆ Σ * and denote by

and by

In fact we shall be interested by the case of binary alphabet, and we write

). The problem is to find upper and lower bounds for U W L(k).

2 Bounds on the length of minimal uncompletable words Proposition 2.1. [4] Let k be an integer and let S be a finite set of words whose maximal length is k and such that there exists a word u of length k with the property that no element of S is a factor of u. Then S is non-complete and the word

with a ∈ Σ is an uncompletable word for S.

A direct consequence of this statement is then Corollary 2.2. [4] For any integer k ≥ 2 and any word u in Σ k , the set S = Σ k \ {u} is non-complete.

Actually, if S = Σ k \ {u} and u is an unbordered word, it can be proved that the uncompletable word from Proposition 2.1 is also the shortest such word: Proposition 2.3. [2] For any integer k ≥ 2 and any unbordered word u ∈ Σ k , a shortest uncompletable word of S = Σ * \ {u} has length

Of course, if S contained in Σ * is non-complete and if S ∪ T is also contained in Σ * and noncomplete, any uncompletable word for S ∪ T is uncompletable for S and uwl(S ∪ T ) ≥ uwl(S).

The “game” is thus to start from a set S of the form Σ * \ {u} and to find a subset T of words of length shorter than k such that S ∪ T remains non-complete and the length of minimal uncompletable words increases as much as possible. This is the way that the bound k 2 + k -1 was already improved in [4]: Note that in this example the shortest uncompletable word maintains the structure uv 1 uv 2 • • • uv k-1 u of the uncompletable word from Proposition 2.1, but the intermediate words v i this time have length k -1. This example led Restivo to conjecture that U W L(k) ≤ 2k 2 . More precisely:

Conjecture. [4] If S is a non-complete set and k is the maximal length of words in S, there exists an uncompletable word of length at most 2k 2 . Moreover this word is of the form uv 1 uv 2 u • • • uv k-1 u, where u is the suitable word of length k and v 1 , v 2 , . . . , v k-1 are words of length less than or equal to k.

is a minimal uncompletable word for S k . Thus U W L(k) ≥ 2k 2 -2k + 1 for 5 ≤ k ≤ 12. Using a similar technique as in [2], it can be proved that this word is uncompletable for each k ≥ 5, but we are not aware whether this word is minimal uncompletable for k > 12.

Unfortunately, it is not true in general that U W L(k) ≤ 2k 2 . Indeed, we have Example 3. Let k > 6 and let

k is obtained from the set S k considered in Example 2 by adding just the word b 4 .

Let u be an unbordered word of length k, and S = Σ k \ {u}. Any uncompletable word for S must contain the word u as a factor, and any word that contains an unbordered factor u can be uniquely written under the form

Actually, we can say a little bit more on the structure of minimal uncompletable words.

Proposition 3.1. Let u be an unbordered word of length k and S = Σ k \ {u}. Then u is both a prefix and a suffix of any minimal uncompletable word for S, that is, any minimal uncompletable word for S is of the form

Proof. Let w be any minimal uncompletable word for S. Arguing by contradicti