🤖 AI Summary
This paper investigates the avoidance of “tangrams”—nonempty factors in infinite words where every letter occurs an even number of times—in words with cutting number ≤ k, and determines the minimum alphabet size t(k) required. Using combinatorics on words, pattern-avoiding constructions, backtracking search, and inductive reasoning, the authors establish for the first time that t(3) = t(4) = 4, resolving an open problem posed by Dębski et al. They further prove that a four-letter alphabet suffices to construct an infinite word avoiding tangrams of cutting number ≤ 4, and show this bound is tight. Together with prior results t(1) = t(2) = 3, this completes the exact characterization of t(k) for k = 1,…,4. The work provides a foundational benchmark for avoidance problems driven by cutting-number–based complexity measures.
📝 Abstract
A tangram is a word in which every letter occurs an even number of times. Thus it can be cut into parts that can be arranged into two identical words. The emph{cut number} of a tangram is the minimum number of required cuts in this process. Tangrams with cut number one corresponds to squares. For $kge1$, let $t(k)$ denote the minimum size of an alphabet over which an infinite word avoids tangrams with cut number at most~$k$. The existence of infinite ternary square-free words shows that $t(1)=t(2)=3$. We show that $t(3)=t(4)=4$, answering a question from Dk{e}bski, Grytczuk, Pawlik, Przybyl{}o, and 'Sleszy'nska-Nowak.