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    $\begingroup$ What happens if, during the running of this algorithm, the "none is a substring of another" condition is violated by $s_{i1},...,s_{ik}, b_{n-k+1}$, where the $b$ string is the newly formed big string? Or is that the point? Gerhard "Sometimes Not Quick On Uptake" Paseman, 2015.10.06 $\endgroup$ Commented Oct 6, 2015 at 17:08
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    $\begingroup$ That can't really happen during the execution (except at the beginning). Suppose for contradiction that this happens first time when we merge $s_i$ and $s_j$ to obtain $b$, and now $s_k$ is contained in $b$. Now either $s_k$ was already contained in $s_i$ or $s_j$ (contradicting that it happened the first time), or the overlap of $s_k$ with $s_i$ is larger than the overlap between $s_i$ and $s_j$, contradicting that we executed the correct step. (my previous comment - now deleted - where I wrote that this can happen was wrong) $\endgroup$ Commented Oct 7, 2015 at 9:50