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The Power Word Problem in Graph Products

AbstractThe power word problem for a group $$\varvec{G}$$ G asks whether an expression $$\varvec{u_1^{x_1} \cdots u_n^{x_n}}$$ u 1 x 1 ⋯ u n x n , where the $$\varvec{u_i}$$ u i are words over a finite set of generators of $$\varvec{G}$$ G and the $$\varvec{x_i}$$ x i binary encoded integers, is equal to the identity of $$\varvec{G}$$ G . It is a restriction of the compressed word problem, where the input word is represented by a straight-line program (i.e., an algebraic circuit over $$\varvec{G}$$ G ). We start by showing some easy results concerning the power word problem. In particular, the power word problem for a group $$\varvec{G}$$ G is $$\varvec{\textsf{uNC}^{1}}$$ uNC 1 -many-one reducible to the power word problem for a finite-index subgroup of $$\varvec{G}$$ G . For our main result, we consider graph products of groups that do not have elements of order two. We show that the power word problem in a fixed such graph product is $$\varvec{\textsf{AC} ^0}$$ AC 0 -Turing-reducible to the word problem for the free group $$\varvec{F_2}$$ F 2 and the power word problems of the base groups. Furthermore, we look into the uniform power word problem in a graph product, where the dependence graph and the base groups are part of the input. Given a class of finitely generated groups $$\varvec{\mathcal {C}}$$ C without order two elements, the uniform power word problem in a graph product can be solved in $$\varvec{\textsf{AC} ^0[\textsf{C}_=\textsf{L} ^{{{\,\textrm{UPowWP}\,}}(\mathcal {C})}]}$$ AC 0 [ C = L UPowWP ( C ) ] , where $$\varvec{{{\,\textrm{UPowWP}\,}}(\mathcal {C})}$$ UPowWP ( C ) denotes the uniform power word problem for groups from the class $$\varvec{\mathcal {C}}$$ C . As a consequence of our results, the uniform knapsack problem in right-angled Artin groups is $$\varvec{\textsf{NP}}$$ NP -complete. The present paper is a combination of the two conference papers (Lohrey and Weiß 2019b, Stober and Weiß 2022a). In Stober and Weiß (2022a) our results on graph products were wrongly stated without the additional assumption that the base groups do not have elements of order two. In the present work we correct this mistake. While we strongly conjecture that the result as stated in Stober and Weiß (2022a) is true, our proof relies on this additional assumption.

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Pumping Lemmas Can be “Harmful”

AbstractA pumping lemma for a class of languages $$\varvec{\mathcal {C}}$$ C is often used to show particular languages are not in $$\varvec{\mathcal {C}}$$ C . In contrast, we show that a pumping lemma for a class of languages $$\varvec{\mathcal {C}}$$ C can be used to study the computational complexity of the predicate “$$\in \varvec{\mathcal {C}}$$ ∈ C ” via highly efficient many-one reductions. In this paper, we use extended regular expressions (EXREGs, introduced in Câmpeanu et al. (Int. J. Foundations Comput. Sci. 14(6), 1007–1018, 2003)) as an example to illustrate the proof technique and establish the complexity of the predicate “is an EXREG language” for several classes of languages. Due to the efficiency of the reductions, both productiveness (a stronger form of non-recursive enumerability) and complexity results can be obtained simultaneously. For example, we show that the predicate “is an EXREG language” is productive (hence, not recursively enumerable) for context-free grammars, and is Co-NEXPTIME-hard for context-free grammars generating bounded languages. The proof technique is easy to use and requires only a few conditions. This suggests that for any class of languages $$\varvec{\mathcal {C}}$$ C having a pumping lemma, the language class comparison problems (e.g., does a given context-free grammar generate a language in $$\varvec{\mathcal {C}}$$ C ?) are almost guaranteed to be hard. So, pumping lemmas sometimes could be “harmful” when studying computational complexity results.

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