The Cost of Distinguishing Graphs

Boutin, Debra L.; Imrich, Wilfried; Ceccherini‐Silberstein, Tullio; Salvatori, Maura; Sava-Huss, Ecaterina

doi:10.1017/9781316576571.005

Cited by 8 publications

(15 citation statements)

References 16 publications

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“…If G is a locally finite graph with linear growth and v is a vertex in G then there is a constant k such that

false| S_{v} (n) false| = k

for infinitely many values of n . (This is observed by Boutin and Imrich in their paper [, Fact 2 in the proof of Proposition 13].) From this we deduce that the vertex‐degree of an end of G is at most equal to k , since each ray in G must pass through all but finitely many of the spheres

S_{v} false(n false)

.…”

Section: A Dichotomy Results For Automorphism Groupssupporting

confidence: 69%

“…In particular a connected graph with linear growth and a countably infinite autormorphism group cannot have one end. Thus one can strengthen [, Theorem 22] and get: Theorem Every locally finite connected graph with linear growth and countably infinite automorphism group has 2 ends.…”

Section: A Dichotomy Results For Automorphism Groupsmentioning

confidence: 98%

“…As a corollary we can answer a question posed by Boutin and Imrich in . In order to state this question, we first need some notation.…”

Section: A Dichotomy Results For Automorphism Groupsmentioning

confidence: 99%

“…Furthermore one can in [, Theorem 18] remove the assumption that the graph is 2‐ended, since it is implied by the other assumptions.…”

Section: A Dichotomy Results For Automorphism Groupsmentioning

confidence: 99%

See 3 more Smart Citations

On tree‐decompositions of one‐ended graphs

Carmesin

Lehner

Möller

2018

Mathematische Nachrichten

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A graph is one-ended if it contains a ray (a one way infinite path) and whenever we remove a finite number of vertices from the graph then what remains has only one component which contains rays. A vertex dominates a ray in the end if there are infinitely many paths connecting to the ray such that any two of these paths have only the vertex in common. We prove that if a one-ended graph contains no ray which is dominated by a vertex and no infinite family of pairwise disjoint rays, then it has a tree-decomposition such that the decomposition tree is one-ended and the treedecomposition is invariant under the group of automorphisms. This can be applied to prove a conjecture of Halin from 2000 that the automorphism group of such a graph cannot be countably infinite and solves a recent problem of Boutin and Imrich. Furthermore, it implies that every transitive one-ended graph contains an infinite family of pairwise disjoint rays. K E Y W O R D S graph automorphism, infinite graph, tree decomposition M S C ( 2 0 1 0 ) 20B27, 05C05, 05C40, 05C63

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“…If G is a locally finite graph with linear growth and v is a vertex in G then there is a constant k such that

false| S_{v} (n) false| = k

S_{v} false(n false)

.…”

Section: A Dichotomy Results For Automorphism Groupssupporting

confidence: 69%

Section: A Dichotomy Results For Automorphism Groupsmentioning

confidence: 98%

“…As a corollary we can answer a question posed by Boutin and Imrich in . In order to state this question, we first need some notation.…”

Section: A Dichotomy Results For Automorphism Groupsmentioning

confidence: 99%

“…Furthermore one can in [, Theorem 18] remove the assumption that the graph is 2‐ended, since it is implied by the other assumptions.…”

Section: A Dichotomy Results For Automorphism Groupsmentioning

confidence: 99%

See 2 more Smart Citations

On tree‐decompositions of one‐ended graphs

Carmesin

Lehner

Möller

2018

Mathematische Nachrichten

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“…Note that though Det(G) is a lower bound for Dist(G), it is not always a good lower bound. As seen in [6], the cost of 2-distinguishing can be an arbitrarily large multiple of the determining number.…”

Section: Introductionmentioning

confidence: 99%

The Cost of 2-Distinguishing Cartesian Powers

Boutin

2013

Electron. J. Combin.

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A graph $G$ is said to be $2$-distinguishable if there is a labeling of the vertices with two labels so that only the trivial automorphism preserves the label classes. The minimum size of a label class in any such labeling of $G$ is called the cost of $2$-distinguishing $G$ and is denoted by $\rho(G)$. The determining number of a graph $G$, denoted $\det(G)$, is the minimum size of a set of vertices whose pointwise stabilizer is trivial. The main result of this paper is that if $G^k$ is a $2$-distinguishable Cartesian power of a prime, connected graph $G$ on at least three vertices with $\det(G)\leq k$ and $\max\{2, \det(G)\} < \det(G^k)$, then $\rho(G^k) \in \{\det(G^k), \det(G^k)+1\}$. In particular, for $n\geq 3$, $\rho(K_3^n)\in \{ \left\lceil {\log_3 (2n+1)} \right\rceil$ $+1, \left\lceil {\log_3 (2n+1)} \right\rceil$ $+2\}$.

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Symmetry Parameters for Mycielskian Graphs

Boutin

Cockburn

Keough

et al. 2021

Association for Women in Mathematics Series

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Praeger-Xu graphs are connected, symmetric, 4-regular graphs that are unusual both in that their automorphism groups are large, and in that vertex stabilizer subgroups are also large. Determining number and distinguishing number are parameters that measure the symmetry of a graph by investigating additional conditions that can be imposed on a graph to eliminate its nontrivial automorphisms. In this paper, we compute the values of these parameters for Praeger-Xu graphs. Most Praeger-Xu graphs are 2-distinguishable; for these graphs we also proved the cost of 2-distinguishing.

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The Cost of Distinguishing Graphs

Cited by 8 publications

References 16 publications

On tree‐decompositions of one‐ended graphs

On tree‐decompositions of one‐ended graphs

The Cost of 2-Distinguishing Cartesian Powers

Symmetry Parameters for Mycielskian Graphs

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