Lower bounds for identifying codes in some infinite grids

# Lower bounds for identifying codes in some infinite grids

Ryan Martin  Brendon Stanton
Department of Mathematics
Iowa State University
Ames, IA 50010
Research supported in part by NSA grant H98230-08-1-0015 and NSF grant DMS 0901008
###### Abstract

An -identifying code on a graph is a set such that for every vertex in , the intersection of the radius- closed neighborhood with is nonempty and unique. On a finite graph, the density of a code is , which naturally extends to a definition of density in certain infinite graphs which are locally finite. We present new lower bounds for densities of codes for some small values of in both the square and hexagonal grids.

## 1 Introduction

Given a connected, undirected graph , we define –called the ball of radius centered at to be

 Br(v)={u∈V(G):d(u,v)≤r}.

A subset of is called an -identifying code (or simply a code, when is understood) if it has the properties:

1. , for all

The elements of a code are called codewords. When is understood, we define . We call the identifying set of .

Vertex identifying codes were introduced in [5] as a way to help with fault diagnosis in multiprocessor computer systems. Codes have been studied in many graphs, but of particular interest are codes in the infinite triangular, square, and hexagonal lattices as well as the square lattice with diagonals (king grid). For each of these graphs, there is a characterization so that the vertex set is . Let denote the set of vertices with and . We may then define the density of a code by

 D(C)=limsupm→∞|C∩Qm||Qm|.

Our first two theorems, Theorem 1 and Theorem 2, rely on a key lemma, Lemma 4, which gives a lower bound for the density of a code assuming that we are able to show that no codeword appears in “too many” identifying sets of size 2. Theorem 1 follows immediately from Lemma 4 and Lemma 5 while Theorem 2 follows immediately from Lemma 4 and Lemma 6.

###### Theorem 1

The minimum density of a 2-identifying code of the hex grid is at least 1/5.

###### Theorem 2

The minimum density of a 2-identifying code of the square grid is at least .

Theorem 2 can be improved via Lemma 7, which has a more detailed and technical proof than the prior lemmas. The idea the lemma is that even though it is possible for a codeword to be in 8 identifying sets of size 2, this forces other potentially undesirable things to happen in the code. We use the discharging method to show that on average a codeword can be involved in no more than 7 identifying sets of size 2. Lemma 7 leads to the improvement given in Theorem 2.

###### Theorem 3

The minimum density of a 2-identifying code of the square grid is at least .

The paper is organized as follows: Section 2 focuses on some key definitions that we use throughout the paper, provides the proof of Lemma 4 and provides some other basic facts. Section 3 states and proves Lemma 5 from which Theorem 1 immediately follows. It is possible to also use this technique to show that the density of a 3-identifying code is at least 3/25, but the proof is long and the improvement is minor so we will exclude it here. (The proof of this fact will appear in the second author’s dissertation [6]). Section 4 gives the proofs of Lemma 6 and 7. Finally, in Section 5, we give some concluding remarks and a summary of known results.

## 2 Definitions and General Lemmas

Let denote the square grid. Then has vertex set and

 E(GS)={{u,v}:u−v∈{(0,±1),(±1,0)}},

where subtraction is performed coordinatewise.

Let represent the hex grid. We will use the so-called “brick wall” representation, whence and

 E(GH)={{u=(i,j),v}:u−v∈{(0,(−1)i+j+1),(±1,0)}}.

Consider an -identifying code for a graph . Let be distinct. If for some we say that

1. forms a pair (with ) and

2. witnesses a pair (that contains ).

For , we define the set of witnesses of pairs that contain . Namely,

 P(c)={v:Ir(v)={c,c′}, for some c≠c′}.

We also define . In other words, is the set of all vertices that witness a pair containing and is the number of vertices that witness a pair containing . Furthermore, we call a -pair codeword if .

We start by noting two facts about pairs which are true for any code on any graph.

###### Fact 1

Let be a codeword and be a set of vertices such that for each , witnesses a pair containing . If and , then does not witness a pair containing .

Proof.  Suppose witnesses a pair containing . Hence, for some . Then and so for some . But then . But since , , contradicting the fact that witnesses a pair. Hence does not witness such a pair.

###### Fact 2

Let be a codeword and be any set with . If and

 B2(v)⊂⋃s∈Ss≠vB2(s)

then at most vertices in witness pairs containing .

Proof.  The result follows immediately from Fact 1. If each vertex in witnesses a pair, then cannot witness a pair. Hence, either does not witness a pair or some vertex in does not witness a pair.

Lemma 4 is a general statement about vertex-identifying codes and has a similar proof to Theorem 2 in [5].

###### Lemma 4

Let be an -identifying code for the square or hex grid. Let for any codeword. Let represent the density of , then if is the size of a ball of radius centered at any vertex ,

 D(C)≥62br+4+k.

Proof.  We first introduce an auxiliary graph . The vertices of are the vertices in and if and only if forms a pair with . Then we clearly have . Let denote the induced subgraph of on . It is clear that if then .

The total number of edges in by the handshaking lemma is

 12∑c∈Γ[C∩Gm]degΓ[C∩Gm]≤(k/2)|C∩Gm|.

But by our observation above, we note that the total number of pairs in is equal to the number of edges in . Denote this quantity by . Then

 Pm≤(k/2)|C∩Gm|.

Next we turn our attention to the grid in question. The arguments work for either the square or hex grid. Note that if is a code on the grid, may not be a valid code for . Hence, it is important to proceed carefully. Fix . By definition, is a subgraph of . Further, for each vertex , . Hence must be able to distinguish between each vertex in .

Let and . Let be the vertices of and let be our codewords. We consider the binary matrix where if and otherwise. We count the number of non-zero elements in two ways.

On the one hand, each column can contain at most ones since each codeword occurs in for at most vertices. Thus, the total number of ones is at most .

Counting ones in the other direction, we will only count the number of ones in rows corresponding to vertices in . There can be at most of these rows that contain a single one and at most of these rows which contain 2 ones. Then there are left corresponding to vertices in and so there must be at least 3 ones in each of these rows. Thus the total number of ones counted this way is at least . Thus

 brK≥−2K+3|Gm−r|−Pm. (1)

But since , this gives

 brK≥−2K+3|Gm−r|−(k/2)K.

Rearranging the inequality and replacing with gives

 |C∩Gm||Gm−r|≥62br+4+k.

Then

 D(C) = limsupm→∞|C∩Gm||Gm| = limsupm→∞|C∩Gm||Gm−r|⋅limsupm→∞|Gm−r||Gm| ≥ 62br+4+k⋅limsupm→∞32(m−r)2+32(m−r)+132m2+32m+1 = 62br+4+k.

## 3 Lower Bound for the Hexagonal Grid

Lemma 5 establishes an upper bound of 6 for the degree of the graph formed by a code in the hex grid, which allows us to prove Theorem 1.

###### Lemma 5

Let be a 2-identifying code for the hex grid. For each , .

Proof.  Let be a code and is an arbitrary member. Let and be the neighbors of and let .

Case 1:

There exists some with . Without loss of generality, assume that . Since at most one of witnesses a pair containing .

Now, unless each of witnesses a pair.

If and each witness a pair, then we have for ; otherwise and so and are not distinguishable by our code. Thus, there must be some . This forces and so either or . Hence, one of these cannot witness a pair and still be distinguishable from . This ends case 1.

Case 2:

First note that itself does not witness a pair.

If witnesses a pair, then there is some and so either or and so one of these cannot witness a pair and still be distinguishable from . Hence at most two of can witness a pair.

Likewise at most at most two of and can witness a pair. Thus . This ends both case 2 and the proof of the lemma.

Proof of Theorem 1.  Using Lemmas 4 and 5, if is a -identifying code in the hexagonal grid, then

 D(C)≥62b2+4+6=630=15.

## 4 Lower Bounds for the Square Grid

Lemma 6 establishes an upper bound of 8 for the degree of the graph formed by a code in the square grid, which allows us to prove Theorem 2. Then we prove Lemma 7, which bounds the average degree of by 7, allowing for the improvement in Theorem 3.

###### Lemma 6

Let be a 2-identifying code for the square grid. For each , .

Proof.  Let , a -identifying code in the square grid. Without loss of generality, we will assume that .

Case 1: witnesses a pair.

This case implies immediately that . The other codeword in , namely , is in one of the following 4 sets, the union of which is . See Figure 1.

 S1:={(1,0),(1,1),(1,−1),(2,0)}S2:={(0,1),(1,1),(−1,1),(0,2)}S3:={(−1,0),(−1,1),(−1,−1),(−2,0)}S4:={(0,−1),(1,−1),(−1,−1),(0,−2)}

If, however, , then no can witness a pair because and could not be distinguished from . Without loss of generality, assume that . Thus, all vertices witnessing pairs in are in the set

 R:={(x,y):(x,y)∈B2(c),x≥0}.

But because

 B2((1,0))⊆⋃s∈S1∪{c}B2(s),

Fact 1 gives that not all members of can witness a pair. See Figure 2.

Therefore, and, without loss of generality, and at least one of does not witness a pair. This ends Case 1.

Case 2: does not witness a pair.

This case implies immediately that either or .

First suppose . There must be two distinct codewords . If are in the same set for some , then for any and so no vertex in witnesses a pair. Thus, the only vertices which can witness a pair are in . There are only 7 of these, so . (See the gray vertices in Figure 2).

If and for some , then only one vertex in each of and can witness a pair. There are at most 5 other vertices not in and so .

Thus, if , then .

Second, suppose . We will define a right angle of witnesses to be subsets of 3 vertices of that all witness pairs and are one of the following 8 sets: , , , and . If a right angle is present then, without loss of generality, let it be . See Figure 3. In order for these all to be witnesses, then must have one codeword not in , which can only be . Since , none of those three vertices can witness a pair.

In addition, must contain a codeword not in , which can only be . See Figure 4. Since , the vertex cannot witness a pair.

Finally, it is not possible for all of to be witnesses because the only member of that is not in the union of the second neighborhoods of the others is the vertex , which cannot be a codeword in this case. Hence, at most 7 members of can witness a pair if has a right angle of witnesses.

Consequently, if does not witness a pair and , then and fails to have a right angle of witnesses. We can enumerate the remaining possibilities according to how many of the vertices are witnesses. If 1, 2 or 3 of them are witnesses and there is no right angle of witnesses, it is easy to see that there are at most 7 witnesses in and so .

The first remaining case is if 0 of them are witnesses, implying each of the eight vertices , , and are witnesses. The second remaining case is if 4 of them are witnesses. This implies that at most one of are witnesses and similarly for , and .

This ends both Case 2 and the proof of the lemma. So, with equality only if one of two cases in the previous paragraph holds.

Proof of Theorem 2.  Using Lemmas 4 and 6, if is a -identifying code in the square grid, then

 D(C)≥62b2+4+8=638=319.

###### Lemma 7

Let denote the induced subgraph of the square grid on the vertex set . Let be a code for the square grid. Then .

Proof.  Let be a code on and let

 R(c)={c′:I2(v)={c,c′} for some c∈C}.

Suppose that for some . We claim that one of the two following properties holds.

1. There exist distinct such that and for .

2. There exist distinct such that for all .

We will prove this by characterizing all possible -pair vertices, but first we wish to define 3 different types of codewords. The definition of each type extends by taking translations and rotations. So, we may assume in defining the types that .

We say that is a type 1 codeword if . See Figure 5.

We say that is a type 2 codeword if . See Figure 6.

We say that is a type 3 codeword if . See Figure 7.

Claim 1 shows that adjacent codewords do not need to be considered because they are in few pairs.

###### Claim 1

If is adjacent to another codeword, then .

Proof.  Without loss of generality, assume that and that is a codeword. Then

 (−1,0),(0,0),(0,1),(0,2),(1,0),(1,1),(−1,0),(−1,1)

are all at most distance 2 from both codewords and so at most 1 of them can witness a pair. Thus, the other 7 do not witness pairs containing . Since , . This proves Claim 1.

Claims 23 and 4 show that types 1, 2 and 3 codewords, respectively, are not in many pairs.

###### Claim 2

If is a type 1 codeword, then .

Proof.  Without loss of generality, let . We consider all vertices which are distance 2 from and either or . There are 11 such vertices and at most 2 of them can witness pairs, so . See Figure 5. This proves Claim 2.

###### Claim 3

If is a type 2 codeword, then .

Proof.  Without loss of generality, let . We consider all vertices which are distance at most 2 from and distance at most 2 from either or . There are 8 such vertices and at most 2 of them can witness pairs. The remaining 5 vertices are and the vertices in the set . But then and, by Fact 1 at most 4 of those remaining 5 vertices can witness pairs. Thus, . See Figure 6. This proves Claim 3.

###### Claim 4

If is a type 3 codeword, then .

Proof.  Without loss of generality, let . We partition into 4 sets:

 T0:={(0,1),(0,2)}T1:={(−2,0),(−1,0),(−1,1)}T2:={(2,0),(1,0),(1,1)}T3:={(−1,−1),(0,−1),(1,−1),(0,−2)}

At most 1 vertex in witnesses a pair since .

At most 1 vertex in can witness a pair since every vertex in is at most distance 2 from . Likewise, at most 1 vertex in can witness a pair.

If all vertices in witness pairs, then since is the only vertex in which is not in for any other . But then is adjacent to another codeword, and by Claim 1, . So we may assume that at most 3 vertices in form pairs with .

Now, if does not itself witness a pair, these partitions give . If does witness a pair, then there must be another codeword for some . But then we see that no other vertex in can witness a pair, since every vertex in is at most distance two from . Thus, . See Figure 7. This proves Claim 4.

We are now ready to characterize the 8-pair codewords.

###### Claim 5

If witnesses a pair and , then satisfies property (P1).

Proof.  Without loss of generality, let . Recall Case 1 of the proof of Lemma 6. That is, and, without loss of generality, equality implies that there is a and at least one of does not witness a pair.

If , the proof is finished, so let us assume that and hence exactly one of the vertices in does not witness a pair. We will show that it is . So, suppose that does not witness a pair. Recall that .

If , then

 B2((1,0))⊆⋃s∈R−{(1,y),(1,0)}B2(s)

and, by Fact 1, neither nor witnesses a pair and .

If , then

 B2((1,1))⊆⋃s∈R−{(1,0),(1,1)}B2(s)

and, by Fact 1, neither nor witnesses a pair and . It follows that each vertex in must witness a pair containing .

Each vertex which is distance 2 or less from 2 vertices in cannot be a codeword. Thus, is the only vertex in other than which has not been marked as a non-codeword and so . Since , the vertex is the only possibility for a second codeword for and is the only possibility for a second codeword for . See Figure 8.

Then is a type 1 codeword and so it is in at most 4 pairs. Codewords and are both adjacent to another codeword, so they are in at most 6 pairs. Hence, satisfies Property (P1). This proves Claim 5.

###### Claim 6

If does not witness a pair and , then satisfies either property (P1) or property (P2).

Proof.  Without loss of generality, let . Recall Case 2 of the proof of Lemma 6. That is, and, without loss of generality, equality implies . Furthermore, one of the following two cases occurs:
(1) The eight witnesses are the vertices , , and .
(2) The witnesses include as well as exactly one of each of the following pairs: , , and
.

If case (1) occurs, then the eight witnesses are the vertices , , and . In this case, simply observe that is a subset of the other seven witnesses. This contradicts Fact 2 and so this case cannot occur.

So, we may assume that case (2) occurs. The vertex cannot be a codeword because and so at most one of these three vertices witness pairs, a contradiction to case (2). By symmetry, none of the following vertices witness pairs:

 (2,1),(1,2),(−1,2),(−2,1),(−2,−1),(−1,−2),(1,−2),(2,−1).

In order to distinguish from , the only vertex available to be a codeword is and symmetrically, , and are codewords. This implies that each of , , and witness pairs.

Then, for the other 4 pairs, there are exactly 3 choices for codewords which are not in the ball of radius 2 for any of our other pairs. See Figure 9.

 Vertex\omit\span\omit\span\omitOther % Codeword(1,1)c1∈{(3,1),(2,2),(1,3)}(−1,1)c2∈{(−1,3),(−2,2),(−3,1)}(−1,−1)c3∈{(−3,−1),(−2,−2),(−1,−3)}(1,−1)c4∈{(1,−3),(2,−2),(3,−1)}

For each , either is adjacent to another codeword or is a type 2 codeword. Claims 1 and 3 imply that, in either case, . It remains to show that one of the following holds: (1) There exist such that and , hence satisfies (P1). (2) There exists an such that , hence satisfies (P2).

First, suppose that there are , , such that , . If , then is a type 1 codeword and so . If , then both and are adjacent to another codeword and so and . Either (P1) or (P2) is satisfied, respectively.

If there is at most one such that for some , then we have three codewords of the form . Without loss of generality, assume that and are codewords. In this case, and are type 3 codewords and hence and . So again, (P2) is satisfied.

This proves Claim 6.

Finally, we can finish the proof of Lemma 7 by way of the discharging method. (For a more extensive application of the discharging method on vertex identifying codes, see Cranston and Yu [3].) Let denote an auxiliary graph with vertex set for some . There is an edge between two vertices and if and only if for some . For each vertex in our auxiliary graph , we assign it an initial charge of . Note that . We apply the following discharging rules if .

1. If is adjacent to one vertex of degree at most 4 and two of degree at most 6 (condition (P1)), then discharge 2/3 to a vertex of degree at most 4 and 1/6 to two vertices of degree at most 6.

2. If is adjacent to 6 vertices of degree at most 6 (condition (P2)), then discharge 1/6 to 6 neighbors of degree at most 6.

We have proven that one of the above cases is possible. Let be the charge of each vertex after discharging takes place. We show that for each vertex in .

If , then our initial charge was . In either of the two cases, we are discharging a total of 1 unit to its neighbors. Since no degree 8 vertex receives a charge from any other vertex, we have .

If then its initial charge is 0 and it neither gives nor receives a charge and so .

If , then its initial charge was at most . Since this vertex has at most 6 neighbors and can receive a charge of at most from each of them, this gives .

If , then its initial charge was at most . Since this vertex has at most 2 neighbors and can receive a charge of at most from each of them, this gives .

Since no vertex can have degree more than 8, this covers all of the cases. Then we have

 ∑c∈C∩Gm(p(c)−7)=∑v∈Γ(degΓ(v)−7)=∑v∈Γe(v)≤0.

Therefore, it follows that .

Proof of Theorem  3.  Consider and let be a code for and . Recall inequality (1) from Theorem 4. In this case, and Lemma 7 shows that

 Pm≤12∑c∈C∩Gmp(c)≤72|C∩Gm|.

Substituting the above inequality into inequality (1) and rearranging gives

 |C∩Gm||Gm−r|≥637.

Taking the limit as gives the desired , completing the proof.

## 5 Conclusions

Below is a table noting our improvements.

 \omit\span\omit\span\omit\span\omitHex % Gridrprevious lower boundsnew lower boundsupper % bounds22/11≈0.1818~{}\@@cite[cite]{[\@@bibref{}{Karpovsky1998% }{}{}]}1/5=0.24/19≈0.2105~{}\@@cite[cite]{[\@@bibref{}{% Charon2002}{}{}]}32/17≈0.1176~{}\@@cite[cite]{[\@@bibref{}{Charon2001}{}% {}]}3/25=0.12~{}\@@cite[cite]{[\@@bibref{}{StantonPending}{}{}]}1/6≈0.1667~{}\@@cite[cite]{[\@@bibref{}{Charon2002}{}{}]}\omit\span\omit\span\omit\span\omitSquare Grid23/20=0.15~{}\@@cite[cite]{[\@@bibref{}{Charon2001}{}{}]}6/37≈0.16225/29≈0.1724~{}\@@cite[cite]{[\@@bibref{}{% Honkala2002}{}{}]}

This technique works quite well for small values of , but we note that grows quadratically in , so the denominator in Lemma 4 would grow quadratically. But the known the lower bounds for -identifying codes is proportional to in all of the well-studied grids (square, hexagonal, triangular and king). Therefore, our technique is less effective as grows.

## References

• [1] Irène Charon, Iiro Honkala, Olivier Hudry, and Antoine Lobstein. General bounds for identifying codes in some infinite regular graphs. Electron. J. Combin., 8(1):Research Paper 39, 21 pp. (electronic), 2001.
• [2] Irène Charon, Olivier Hudry, and Antoine Lobstein. Identifying codes with small radius in some infinite regular graphs. Electron. J. Combin., 9(1):Research Paper 11, 25 pp. (electronic), 2002.
• [3] Daniel W. Cranston and Gexin Yu. A new lower bound on the density of vertex identifying codes for the infinite hexagonal grid. Electron. J. Combin., 16(1):Research Paper 113, 16, 2009.
• [4] Iiro Honkala and Antoine Lobstein. On the density of identifying codes in the square lattice. J. Combin. Theory Ser. B, 85(2):297–306, 2002.
• [5] Mark G. Karpovsky, Krishnendu Chakrabarty, and Lev B. Levitin. On a new class of codes for identifying vertices in graphs. IEEE Trans. Inform. Theory, 44(2):599–611, 1998.
• [6] Brendon Stanton. PhD thesis, Iowa State University, in progress.
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