Isomorphism of Regular Trees and Words

Abstract

The computational complexity of the isomorphism problem for regular trees, regular linear orders, and regular words is analyzed. A tree is regular if it is isomorphic to the prefix order on a regular language. In case regular languages are represented by NFAs (DFAs), the isomorphism problem for regular trees turns out to be EXPTIME-complete (resp. P-complete). In case the input automata are acyclic NFAs (acyclic DFAs), the corresponding trees are (succinctly represented) finite trees, and the isomorphism problem turns out to be PSPACEcomplete (resp. P-complete). A linear order is regular if it is isomorphic to the lexicographic order on a regular language. A polynomial time algorithm for the isomorphism problem for regular linear orders (and even regular words, which generalize the latter) given by DFAs is presented. This solves an open problem byÉsik and Bloom.

Isomorphisms of scattered automatic linear orders

“…Courcelle [7] initiated the study of regular words, i.e., labeled linear orders derived from frontiers of regular trees. Thomas proved that the isomorphism problem for these words is decidable [29], the complexity of this problem was determined by Lohrey and Mathissen [24]. Based on techniques and results from [2], we will show that, given a regular language L, it is decidable whether (L; ≤ lex ) is rigid.…”

Compressed Tree Canonization

“…Membership in PTIME follows immediately from Lemma 4, Lemma 5, and Theorem 6. Moreover, PTIME-hardness already holds for dags, i.e., SLT grammars where all nonterminals have rank 0, as shown in [18]. ⊓ ⊔…”

“…The upper bound follows from Lemma 9, Lemma 10, and Corollary 7. Hardness for PTIME follows from the PTIME-hardness for dags [18] and the fact that isomorphism of rooted unordered trees can be reduced to isomorphism of unrooted unordered trees by labelling the roots with a fresh symbol. ⊓ ⊔…”

Tree Compression Using String Grammars

“…Proof. For the upper bound we use the following lemma from [35]: If a function f : Σ * → Γ * is PSPACE-computable and L ⊆ Γ * belongs to NSPACE(log k (n)) for some constant k, then f −1 (L) belongs to PSPACE. Given an SLP A for the tree t = val(A), one can compute the tree t by a PSPACE-transducer by computing the symbol t[i] for every position i ∈ {1, .…”