2001
DOI: 10.37236/1583
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General Bounds for Identifying Codes in Some Infinite Regular Graphs

Abstract: Consider a connected undirected graph $G=(V,E)$ and a subset of vertices $C$. If for all vertices $v \in V$, the sets $B_r(v) \cap C$ are all nonempty and pairwise distinct, where $B_r(v)$ denotes the set of all points within distance $r$ from $v$, then we call $C$ an $r$-identifying code. We give general lower and upper bounds on the best possible density of $r$-identifying codes in three infinite regular graphs.

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Cited by 39 publications
(43 citation statements)
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(8 reference statements)
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“…Assume that share is shifted from (4, 1) according to the rule 5 (see Figure 3(e)). Now we have C ∩ S 1 = {(2, 2), (4, 1)} and (4, 3) ∈ C. Therefore, since at most 1/12 units of share is shifted from (4, 1) according to the rule 2.1, we obtain that no more than 1/6 + 1/12 = 1/4 units of share can be shifted from S 1 to c. Similarly, if share is shifted from (5,1) according to the rule 6, then it can be shown that c receives at most 1/4 units of share from S 1 . In conclusion, at most 1/4 units of share is shifted from the vertices of S 1 to c. Analogously, this statement also holds for the vertices of S 2 .…”
Section: Proof Of Lemma 33mentioning
confidence: 95%
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“…Assume that share is shifted from (4, 1) according to the rule 5 (see Figure 3(e)). Now we have C ∩ S 1 = {(2, 2), (4, 1)} and (4, 3) ∈ C. Therefore, since at most 1/12 units of share is shifted from (4, 1) according to the rule 2.1, we obtain that no more than 1/6 + 1/12 = 1/4 units of share can be shifted from S 1 to c. Similarly, if share is shifted from (5,1) according to the rule 6, then it can be shown that c receives at most 1/4 units of share from S 1 . In conclusion, at most 1/4 units of share is shifted from the vertices of S 1 to c. Analogously, this statement also holds for the vertices of S 2 .…”
Section: Proof Of Lemma 33mentioning
confidence: 95%
“…Observe then that if u ∈ S 1 shifts share to c according to the rules 1-6 or their modifications, then we have I 2 (u) = {u}. Therefore, if two vertices of S 1 shift share, then one of these vertices is (4, 1) or (5,1). Assume that share is shifted from (4, 1) according to the rule 5 (see Figure 3(e)).…”
Section: Proof Of Lemma 33mentioning
confidence: 99%
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“…Given a vertex v ∈ V the set of codewords which r-cover v are denoted as: Although the aim of this paper is to construct identifying codes over G(M ) graphs, the problem over the infinite mesh will be considered before. Previous papers dealing with the problem of identification over the infinite mesh are, for example, [7], [10] and [6]. Therefore, let us define the infinite mesh as the Cayley graph Cay(Z 2 , E).…”
Section: Previous Definitions On Identificationmentioning
confidence: 99%