# Problem 77: Anchored squares

$$\def\sa#1{\tt{#1}} \def\dd{\dot{}\dot{}}$$

When searching, in a divide-and-conquer way, a word for square factor it is natural to look for squares in the product of two square-free words. The problem deals with the latter question and extends to a square-freeness test running in $O(n\log n)$ time on a word of length $n$.

The method based on prefix tables (see Problem 74) achieves the same goal but requires tables of size $O(n)$ while the present solution needs only a few variables in addition to input.

Let $y$ and $z$ be two square-free words. Algorithm Right tests if $yz$ contains a square centred in $z$ only. Other squares in the product can be found symmetrically.

The algorithm examines all possible periods of a square. Given a period $p$ (see picture), it computes the longest common suffix $u'=z[j\dd p-1]$ between $y$ and $z[0\dd p-1]$. Then it checks if $z[0\dd j-1]$ occurs at position $p$ on $z$, scanning $z$ from the right position $k-1$ of the potential square. If it is successful, a square is found.

Right$(y,z \textrm{ non-empty square-free strings})$
\begin{algorithmic}
\STATE $(p,\mathit{end})\leftarrow (|z|,|z|)$
\WHILE{$p \gt 0$}
\STATE $j\leftarrow \min\{q \mid z[q..p-1] \mbox{ suffix of } y\}$
\IF{$j = 0$}
\RETURN{True}
\ENDIF
\STATE $k\leftarrow p+j$
\IF{$k \lt \mathit{end}$}
\STATE $\mathit{end}\leftarrow \min\{q \mid z[q..k-1] \mbox{ suffix of } z[0..j-1]\}$
\IF{$\mathit{end} = p$}
\RETURN{True}
\ENDIF
\ENDIF
\STATE $p\leftarrow \max\{j-1,\lfloor p/2 \rfloor\}$
\ENDWHILE
\RETURN{False}
\end{algorithmic}

Show both that Algorithm Right returns true if and only if the word $yz$ contains a square centred in $z$ and that its running time is $O(|z|)$ with only constant extra space.

In Algorithm Right the role of variable $\mathit{end}$ and instructions at lines 7 and 11 are crucial to get the running time announced in the question. Note that it does not depend on the length of $y$.

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