Fault-Tolerant Detection Systems on the King’s Grid

A detection system, modeled in a graph, uses “detectors” on a subset of vertices to uniquely identify an “intruder” at any vertex. We consider two types of detection systems: open-locating-dominating (OLD) sets and identifying codes (ICs). An OLD set gives each vertex a unique, non-empty open neighborhood of detectors, while an IC provides a unique, non-empty closed neighborhood of detectors. We explore their fault-tolerant variants: redundant OLD (RED:OLD) sets and redundant ICs (RED:ICs), which ensure that removing/disabling at most one detector guarantees the properties of OLD sets and ICs, respectively. This paper focuses on constructing optimal RED:OLD sets and RED:ICs on the inﬁnite king’s grid, and presents the proof for the bounds on their minimum densities; (cid:2) 310 , 13 (cid:3) for RED:OLD sets and (cid:2) 311 , 13 (cid:3) for RED:ICs.


Introduction
Let G = (V (G), E(G)) be an (undirected) graph, with vertices V (G) and edges E(G), modelling a system or facility with detectors to recognize a possible problem, traditionally referred to as an "intruder". For example, the vertices of the graph can represent sections of a shopping mall, the intruder could be a shoplifter, and the detectors can be video surveillance equipment or motion, magnetic, or RFID sensors. The goal here is to identify the exact location/vertex of the intruder by placing the minimum number of detectors in the facility/graph.
A distinguishing set ⊆ P(V (G)) is a collection of subsets of vertices which covers V (G) and provides the ability to separate any vertex from the rest of the vertices, and therefore it can be used to pinpoint the location of the intruder. A more practical method, however, is to consider a subset of vertices S ⊆ V (G) called a "detection system", where each vertex in S is installed with a specific type of detector or sensor for locating the intruder. Definition 1. For detector set S ⊆ V (G), the detection region of v ∈ S, denoted R det (v) ⊆ V (G), is the region in which v can detect an intruder. In a detection system, the detector installed at vertex v is capable of detecting an intruder if and only if it is within R det (v). Note that R det (v) has also been referred to as the "watching zone" of v in other papers, but with different notation [1,2]. Using an appropriate selection of detectors S ⊆ V (G) for a graph, we can generate {R det (v) : v ∈ S}, which forms a distinguishing set. Ideally, we would like to use the smallest number of detectors possible while managing to generate a distinguishing set for V (G) that can be used to pinpoint the location of the intruder. Many types of detection systems with various properties have been explored throughout the years. Over 470 papers have been published on various types of detector-based sets, distinguishing sets, and other related concepts [14]. One such system is the Locating-Dominating (LD) set, where each detector identifies an intruder in their closed neighborhood and also has the ability to distinguish the vertex itself from its neighbors [25,24]. Of particular interest in this paper are open-locating-dominating (OLD) sets [19], which are detection systems using R det (v) = N (v), and identifying codes (IC) [12], which are detection systems using R det (v) = N [v]. Note that both OLD sets and ICs generate distinguishing sets which cover the entire graph using detectors that detect intruders at distance at most 1. The only difference is that OLD detectors cannot detect an intruder at their own vertex while IC detectors must do so. For examples of minimum a OLD set and IC on the same graph, see Figure 1 (a) and (c), respectively.
An OLD set is a form of open-dominating set, which simply requires that ∪ v∈S N (v) = V (G) where S ⊆ V (G) is the open-dominating set [19]. However, open-domination alone is insufficient to generate a distinguishing set, which is why OLD sets add additional uniqueness requirements in order to distinguish vertices. Similarly, an IC is a form of dominating set, which simply requires that In the next section, we introduce the general concept of a detection system-which is similar to a "watching system" as used in other papers [1,2]-and fault-tolerant variants known as a k-redundant detection systems. In Section 4 we introduce the concept of a share argument, which will be used for later proofs. Finally, in Sections 5 and 6 we explore upper and lower bounds for optimal 1-redundant variants of OLD sets and ICs known as RED:OLD sets and RED:ICs, respectively.

Detection Systems
In a detector-based distinguishing set we use what is known as the "locating code" of a vertex to identify the intruder based on which sensors have detected the intruder in their detection region.
The detection region function R det is only required to be defined over the detector set S ⊆ V (G). However, if it defined for every vertex-that is, a hypothetical detection region for a detector being placed at any given vertex-then it is said to be uniform.
Theorem 2.1 is not strictly necessary to compute locating codes, but when applicable it can be more convenient than Definition 4. Notably, OLD sets use R det (v) = N (v) and ICs use R det (v) = N [v] for every detector regardless of its position; thus they are uniform. And for undirected graphs-which this paper assumes-N (v) and N [v] are also symmetric. Thus, for OLD sets and ICs, Depending on the problem, we might need the detection system to be fully functional even when one of the detectors goes offline or is broken. For this purpose, we could instead use a fault-tolerant variant of a detection system known as a k-redundant detection system, which is a detection system for which removing at most k detectors is still a detection system. In this paper, we will explore redundant open-locating-dominating (RED:OLD) sets and redundant identifying codes (RED:ICs), which are 1-redundant detection systems [10,20].
Definition 8. A k-redundant detection system is a detection system S ⊆ V (G) such that for any D ⊆ S with |D| ≤ k, S − D is a detection system.
By Definition 8, a detection system is technically a 0-redundant detection system. Next, we present characterizations that will be useful in constructing detection systems.
In this paper, when quantifiers such as "at least" are used with k-dominated, they apply to the value of k. For instance, "at least 1-dominated" means k-dominated for some k ≥ 1.
The notation dom S (v) introduced by Definition 11 is technically equivalent to |L S (v)| but is used in this paper for semantic clarity when only the count is needed. Proof. Consider the properties given in Definition 7 and apply the following equivalences: Proof. Let S ⊆ V (G) be a k-redundant detection system and v ∈ V (G). Suppose |L S (v)| ≤ k. Consider the new set S = S − L S (v). Then L S (v) = ∅, implying S is not a detection system, a contradiction. Thus, |L S (v)| ≥ k +1 and we have that S at least (k +1)-dominates all vertices. Let u ∈ V (G) with u = v and let D = L S (u) L S (v). Suppose that |D| ≤ k. Consider the new set S = S − D. We see that L S (u) L S (v) = ∅, so S is not a detection system, a contradiction. Thus, |D| ≥ k + 1 and we have that S is (k + 1)-distinguishing.
For the converse, suppose S ⊆ V (G) is (k + 1)-distinguishing and it at least (k + 1)dominates all vertices. These properties satisfy the requirements of Theorem 2.2, so we know that S is a detection system. By hypothesis, S is (k + 1)-distinguishing, so ∀u, v ∈ V (G) with u = v, |L S (u) L S (v)| ≥ k + 1. Additionally, S is at least (k + 1)-dominating, so ∀v ∈ V (G), |L S (v)| ≥ k + 1. Let D be an arbitrary subset of S with |D| ≤ k. Consider the new set S = S − D. The removal of |D| elements from S will remove at most |D| elements from any locating code; thus, ∀v ∈ V (G), |L S (v)| ≥ 1, implying S at least 1-dominates all vertices. Additionally, the removal of |D| elements from S will remove at most |D| differences in any pair of locating codes, thus ∀u, v ∈ V (G) with u = v, |L S (u) L S (v)| ≥ 1, so S is 1-distinguishing. Therefore, by Theorem 2.2, S is a detection system, implying that S is a k-redundant detection system. Note that Theorem 2.3 has been introduced and proven for the specific case where k = 1 [23], but we have provided a generalized form here.
Theorem 2.4. Suppose S ⊆ V (G) at least p-dominates all vertices and is q-distinguishing. Then any superset S ⊇ S also at least p-dominates all vertices and is q-distinguishing.
Corollary 2.1. Any superset of a k-redundant detection system is also a k-redundant detection system.
Disregarding the specific choice of S used in Theorem 2.4, the theorem gives us an important fact: if there exists a non-trivial detection system or k-redundant detection system on some graph then there also exists a larger one. Therefore, what we are interested in is finding the smallest possible set on a given graph.
For finite graphs, we use the notations OLD(G), RED:OLD(G), IC(G), and RED:IC(G) to denote the cardinality of the smallest possible such sets on graph G, respectively. However, for infinite graphs, we measure via the density of the subset, which is defined as the ratio of the size of the subset to the size of the whole set [11,19,23]. We use the notations OLD%(G), RED:OLD%(G), IC%(G), and RED:IC%(G) to denote the lowest density of any possible such set on G [11,19,23]. Note that density is also defined for finite graphs.  One common method for constructing vertex subsets for an infinite graph G is via tessellation, in which we take a non-empty region R ⊆ V (G) of the graph and tile it to cover V (G) with no overlapping of tiles. Let T be the set of all tiles required to cover V (G). T must have the following properties:

Detection Systems on Infinite Graphs
Therefore, T is a partition of V (G). As |A| = |R| for any tile A ∈ T , there exists a bijective function f A : A → R mapping from any vertex in tile A back into the original tile R. In tessellation, we consider every tile to be a copy of R, so f A maps any vertex in tile A back to its corresponding geometric position in the original tile R. Additionally, as A ∩ B = ∅ for any two distinct tiles A, B ∈ T and ∪ A∈T A = V (G) there exists a surjective function g : V (G) → T mapping any vertex in the graph to its (only) enclosing local tile. That is, ∀v ∈ V (G), v ∈ g(v) ∈ T . Combining these functions, we can create a function F : (v) mapping from any vertex in the graph back to its corresponding geometric position in the original tile R. Once this tessellation has been constructed, we can use it to create the desired structured subsets of the infinite graph. Let S ⊆ R be the desired subset of the original tile R. Then the tessellated subset Figure 2 shows two examples of tessellated constructions for an optimal OLD-set [13] and a RED:OLD set [23], respectively, on the infinite triangular grid (TRI). Let R ⊆ V (G) be the original tile, and let S ⊆ R. One of the benefits of tessellation is that, by their very construction, the density of any generated subset Ψ S in V (G) is equal to the density of S in R. Thus, as Figure 2 shows optimal solutions we see that OLD%(TRI) = 4 13 and RED:OLD%(TRI) = 6 16 = 3 8 . In this paper, we consider RED:OLD sets and RED:ICs on the infinite king's grid (K)also known as the infinite king's graph-an 8-regular graph inspired by chess, with vertices representing positions on an infinite chessboard and edges representing all the normal movements of a king. Several papers have been published exploring various graphical parameters on the king's grid. The first such paper-by Cohen, Honkala, and Lobstein [5]-covered the creation of ICs; importantly, they proved that 2 9 ≤ IC%(K) 4 17 . This was soon improved by Charon, Hudry, and Lobstein [4], who demonstrated that IC%(K) = 2 9 . Honkala and Laihonen [8] were the first to study locating-dominating sets on K, proving that the minimum density is 1 5 . Many papers have since been published covering various properties of ICs [3,6,7,9,17] and LD sets [15,18,16] on finite and infinite king's grids. We will explore upper and lower bounds for RED:OLD%(K) and RED:IC%(K). Depending on the graph, directly proving a lower bound for the minimum density of a detector-based distinguishing set can be difficult. Instead, it's often easier to use what's known as a share argument, established by Slater [25] and used extensively by numerous papers on related concepts of distinguishing sets [11,23,22,19,21]. The inverse of average share is equal to the density. Thus, if we find an upper bound for the average share, its inverse is a lower bound for density.

Lower Bounds
Let G be a graph, . We will also introduce the following compact notation: As an example of calculating shares and partial shares, consider the open-dominating set S = {v 2 , v 4 , v 6 } on the G 8 graph in Figure 3. Consider v 2 ∈ S, which is adjacent to And finally, we have v 6 ∈ S with sh S (v 6 ) = 1 2 + 1 + 1 = 5 2 . The average share is 1 3 3 + 5 2 + 5 2 = 8 3 and indeed we see that the density of this solution is 3 8 . The graphs shown in subsequent figures use circles to represent "unknowns" (vertices which may or may not be detectors), shaded squares to represent detectors, and non-shaded squares to represent non-detectors. Additionally, any vertex labeled simply with number k will be referred to as v k in the text. For instance, in Figure 4, {x, v 1 } are detectors, {v 4 , v 6 } are non-detectors, and everything else is currently unknown.
Let S ⊆ V (K) be a RED:OLD set on K. Consider the scenario in Figure 4. Vertices v 2 , v 3 , v 7 , and v 8 are dominated by the same two vertices. If one of the four vertices is 2-dominated then the other three must be at least 4-dominated in order to be distinguished; 3 . This finding is useful, but it can easily be generalized to any graph and many possible configurations and types of sets: Proof. By hypothesis, D ⊆ S and D ⊆ ∩ u∈A R det (u). Because R det is uniform and symmetric, by Theorem 2.1, D ⊆ ∩ u∈A L S (u). Suppose ∃u ∈ A such that dom S (u) = |D|. Then L S (u) = D and we observe that By hypothesis, S is k-distinguishing, so this quantity must be at least k. Therefore, dom S (w) − |D| ≥ k, which of course means dom S (w) ≥ |D| + k.
In either case sh S [A](v) ≤ max |A| |D|+1 , 1 |D| + |A|−1 |D|+k . We will now characterize when this is |A| |D|+1 . For convenience, let a = |A| and d = |D|: We have found two non-isomorphic RED:OLD sets on the infinite king's grid K, as shown in Figure 5. These solutions can be formed via tessellation of rectangles (regions outlined in the figure). For Figure 5 (a), we see that 1 of the 3 vertices in the tile is a detector (shaded vertices), giving a density of 1 3 . Similarly, in Figure 5 (b) we see that 2 of the 6 vertices in the tile are used, also giving a density of 2 6 = 1 3 . It is also possible to form these solutions via tessellation of diagonal ribbons of infinite length. For the first solution (a), the ribbon consists of 1 diagonal of detectors and 2 diagonals of non-detectors, giving a density of 1 3 .
For the second solution (b), the ribbon consists of 2 diagonals of detectors and f diagonals of non-detectors, giving a density of 2 6 = 1 3 . As we have a RED:OLD set on K with density 1 3 , we have an upper bound for the optimal density: RED:OLD%(K) ≤ 1 3 .

Lower Bound
Let S ⊆ V (K) be a RED:OLD set on K, and let the notations sh, dom, and k-dominated implicitly use S. These will persist for the rest of Section 5.2.
S is a 1-redundant detection system, so Theorem 2.3 gives us that every vertex must be at least 2-dominated, so for any detector x ∈ S and any neighbor v ∈ N (x), sh[v](x) ≤ 1 2 . Additionally, K is an 8-regular graph, so |N (x)| = 8. Therefore, sh(x) ≤ 8 2 = 4, giving us a simple lower bound of RED:OLD%(K) ≥ 1 4 . We will now prove another, higher bound. Consider a detector x ∈ S and its as-of-yet unknown local region of K, as shown in Figure 6. Suppose that ∀w ∈ N (x), dom(w) = 2. All neighbors of x are dominated by x itself. Because the neighbors of x are all 2-dominated by hypothesis, they must all be dominated exactly one more time. However, if two neighbors of x are dominated by the same two detectors, then they will not be distinguished. If v 9 ∈ S ∨ v 10 ∈ S then {v 1 , v 2 } are not distinguished. Likewise, if v 1 ∈ S then {v 2 , v 3 } are not distinguished. And if v 2 ∈ S then {v 1 , v 3 } are not distinguished. By symmetry, the only vertices in the figure which can possibly be detectors are {v 11 , v 15 , v 19 , v 23 }, but this means that ∀w ∈ {v 1 , v 3 , v 5 , v 7 }, dom(w) = 1, contradicting that S is a RED:OLD set on K. Therefore, ∃w ∈ N (x) such that dom(w) ≥ 3, so sh[w](x) ≤ 1 3 . By Theorem 2.3, any RED:OLD set at least 2-dominates every vertex, thus, where u ∈ N (x), dom(u) ≥ 2, thus sh[u](x) ≤ 1 2 . Thus, sh S (x) ≤ 7 2 + 1 3 = 23 6 , meaning we have a higher lower bound: RED:OLD%(K) ≥ 6 23 . At this point, we have two lower bounds for RED:OLD%(K): 1 4 and 6 23 . These are both still relatively far from our upper bound of 1 3 . To help bridge the gap between upper and lower bounds, we would now like to prove a significantly more ambitious lower bound of 3 10 , which will require demonstrating an upper bound for average share of 10 3 . Theorem 5.1. In any RED:OLD set S on K, the average share over all vertices in S is no larger than 10 3 . Proof. Let x ∈ S be an arbitrary vertex in the RED:OLD set S. We will explore all non-isomorphic configurations and show that the average share it gives rise to is at most 10 3 . We begin by considering when x is dominated by two neighbors in a "bent" shape, as in Figure 7. We will demonstrate that in any possible sub-configuration sh(x) ≤ 13 4 < 10 3 . From the figure we clearly see that Next, we consider when x is dominated by vertex v 1 directly above it, as in Figure 8. We will demonstrate that in any possible subconfiguration sh(x) ≤ 13 4 < 10 3 . From Case 1 we know that if {v 4 , v 6 } ∩ S = ∅ then sh(x) ≤ 13 4 and we are done. Therefore, we assume that neither of these is a detector. Suppose that ∀w ∈ {v 4 , v 5 , v 6 }, dom(w) = 2. If v 17 ∈ S then {v 4 , v 5 } are impossible to distinguish due to both being 2-dominated by hypothesis; thus, v 17 / ∈ S. By the same logic, v 16 / ∈ S and v 3 / ∈ S. Likewise, if v 18 ∈ S then {v 5 , v 6 } are impossible to distinguish. By the same logic, v 7 / ∈ S. We therefore find that dom(v 5 ) = 1, contradicting that S is a RED:OLD set. Therefore, ∃w ∈ {v 4 , v 5 , v 6 } such that dom(w) ≥ 3. Additionally, by Lemma 4.1, We now consider when x is dominated by four neighbors in an "x" shape, as shown in Figure 9. From Cases 1-2 we see that if

Case 4
We now consider the case where x is dominated by three neighbors in a "lambda" shape, as in Figure 10.  Now, we assume that dom(v 3 ) = 4 ≤ 5 = dom(v 5 ). We initially observe that if ∃w ∈ {v 1 , v 2 , v 4 , v 6 , v 7 , v 8 } such that dom(w) ≥ 3 then sh(x) ≤ 1 5 + 1 4 + 1 3 + 5 2 = 197 60 < 10 3 . Thus, we need only consider when ∀w ∈ {v 1 We also see that dom(v 4 ) = 2, which means {v 13 , v 14 , v 15 } ∩ S = ∅. This means v 12 ∈ S because dom(v 3 ) = 4, and v 11 / ∈ S because dom(v 2 ) = 2. This leads to the sub-configuration shown in Figure 11. We see that sh(x) = 1 5 + 1 4 + 6 2 = 69 20 > 10 3 . However, we observe that x is adjacent to {v 2 , v 4 , v 6 } ⊆ S, all of which have "bent" shape. As demonstrated in Case 3, ∀w ∈ {v 2 , v 4 , v 6 }, sh(w) ≤ 13 4 < 10 3 . Additionally, the only other detectors these 3 vertices are adjacent to are {v 12 , v 16 , v 18 }, all of which are adjacent via vertical or horizontal edges. Thus, the findings of Cases 1-4 tell us that ∀w ∈ {v 12 , v 16 , v 18 }, sh(w) ≤ 13 4 < 10 3 . Therefore, we are safe to use {x, v 2 , v 4 , v 6 } in an averaging argument; consider the average share:  18 , v 19 } must be a detector. We will now consider two cases: v 19 ∈ S or v 19 / ∈ S. If v 19 ∈ S then v 18 / ∈ S. By hypothesis, dom(v 5 ) = 5, so v 16 ∈ S, and by hypothesis dom(v 4 ) = 2, requiring v 15 / ∈ S. This leads to the sub-configuration shown in Figure 12 (a). sh(x) = 2 4 + 6 2 = 7 2 > 10 3 . However, as demonstrated previously in this case we can safely average with the adjacent "bent" shape vertices {v 2 , v 4 }, which have share no more than = 10 3 . Thus in this terminal sub-configuration, we still have that the average share is no more than 10 3 . If v 19 / ∈ S then v 18 ∈ S. Then, by similar logic to the previous paragraph, v 16 / ∈ S ∧ v 15 ∈ S. We now have the sub-configuration shown in Figure 12 (b). This again has sh(x) = 2 4 + 6 2 = 7 2 > 10 3 . However, we also still have 2 "bent" shape neighbors of x: {v 2 , v 6 }, which are safe to use for averaging by the same logic, as demonstrated previously in this case. Thus, we once again have the average share 1   Figure 14 represents the final possible sub-configuration for Case 5, and we find that sh(x) = 7 2 > 10 3 . We observe that x has two neighbors, {v 2 , v 6 }, of "bent" shape, which in Case 1 was found to have share no more than 13 4 . Additionally, these two vertices are only adjacent to two other detector vertices {v 12 , v 20 } ⊆ S, and are connected via a horizontal edge, which Cases 1-2 have proven to have share no more than 13 4 . Thus {v 2 , v 6 } are not adjacent to any other vertices with share exceeding 10 3 and so are safe to use with averaging. Consider the average share of {x, v 2 , v 6 }: avg ≤ 1 In the final case, we consider when x is dominated by two neighbors in a "wedge" shape like Figure 15. Applying the cumulative results of Cases 1-5 we see that if {v 1 , v 3 , v 5 , v 6 , v 7 , v 8 } ∩ S = ∅ then the average share is no more than 10 3 . We will proceed by expanding every possible state for the neighborhood of v 7 . Vertex v 7 must be at least 2-dominated, so we consider all sub-cases for 2 ≤ dom(v 7 ) ≤ 4.
First, we consider when dom(v 7 ) = 4, which requires {v 20 From Theorem 5.1, we have that the average share for any RED:OLD set on K is no more than 10 3 . Thus, 3 10 is a lower bound for RED:OLD%(K). 6 Bounds for RED:IC%(K) 6

.1 Upper Bound
We have found three RED:ICs on K, with density 1 3 , as shown in Figure 16. Solutions (a) and (b) are identical to the RED:OLD solutions in Figure 5-that is, these two solutions are both RED:OLD and RED:IC. However, Figure 16

Lower Bound
Let S ⊆ V (G) be a RED:IC on K, and let the notations sh, dom, and k-dominated implicitly use S. These will persist for the rest of Section 6.2.
S is a 1-redundant detection system, so Theorem 2.3 gives us that every vertex is at least 2-dominated. Additionally, K is 8-regular, so |N [v]| = 9. Therefore, ∀x ∈ S, sh(x) ≤ 9 2 . Thus, we have a simple lower bound: RED:IC%(K) ≥ 2 9 . Theorem 6.1. The share of any given detector in S is at most 11 3 . Proof. Just as in the previous section for proving a lower bound for RED:OLD%(K), we consider a detector x ∈ S and its local region of K. Vertex x must be at least 2-dominated and dominates itself, so we consider the following two cases for the other detector being horizontal/vertical to x or being diagonal to x, and show sh(x) ≤ 11 3 .  16 , v 17 , v 18 } ∩ S = ∅, which results in dom(v 5 ) = 1, a contradiction. Therefore, ∃w ∈ {v 4 , v 5 , v 6 } such that dom(w) ≥ 3, making sh(x) ≤ 4 3 + 1 3 + 4 2 = 11 3 and we are done.
From Theorem 6.1, we have that the average share for any RED:IC on K is no more than 11 3 . Thus, 3 11 is a lower bound for RED:IC%(K). Corollary 6.1. RED:IC%(K) ≥ 3 11

Conclusion
We have now demonstrated that 3 10 ≤ RED:OLD%(K) ≤ 1 3 and 3 11 ≤ RED:IC%(K) ≤ 1 3 . Interestingly, two of the solutions achieving the upper bound of 1 3 for RED:OLD and RED:IC were the same subset: (a) and (b) from Figures 5 and 16. For many graphs, RED:IC%(G) < RED:OLD%(G), but in this case, the upper bounds we obtained are equal. Although it is possible that RED:IC%(K) = RED:OLD%(K), it would not be surprising if the provided RED:ICs were found to be sub-optimal.