A well known result of Fraenkel and Simpson
states that the number of distinct squares in a word of length n is bounded by 2n since at each position there are at most two distinct squares whose last occurrence starts.
In this paper, we investigate squares in partial words with one hole,
or sequences over a finite alphabet that have a “do not know” symbol or “hole”.
A square in a partial word over a given alphabet has the form uv where u is compatible with v, and consequently, such square is compatible with a number of words over the alphabet that are squares.
Recently, it was shown that for partial words with one hole, there may be more than two squares that have their last occurrence starting at the same position. Here, we prove that if such is the case, then the length of the shortest square is at most half the length of the third shortest square.
As a result, we show that the number of distinct squares compatible with factors of a partial word with one hole of length n is bounded by