, MONICA M.Y. WANG, AND SHIQIANG ZHANG …IS 4 DAVID G.L. WANG†‡, MONICA M.Y. WANG, AND SHIQIANG ZHANG Abstract. It is known that the problem of computing the edge dimension of

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THE EDGE DIMENSION OF THE GENERALIZED PETERSEN GRAPH P (N, 3)

IS 4

DAVID G.L. WANG†‡, MONICA M.Y. WANG, AND SHIQIANG ZHANG

Abstract. It is known that the problem of computing the edge dimension of a graph is NP-hard,and that the edge dimension of any generalized Petersen graph P (n, k) is at least 3. We prove thatthe graph P (n, 3) has edge dimension 4 for n ≥ 11, by showing semi-combinatorially the nonexistenceof an edge resolving set of order 3 and by constructing explicitly an edge resolving set of order 4.

Contents

1. Introduction 1

2. Strategy of our proof 3

3. A formula for the vertex-edge distances in P (n, 3) 4

4. Proof of Proposition 2.2 10

5. Proof of Proposition 2.3 19

6. Concluding remarks 26

References 26

1. Introduction

Let n ≥ 3 and 1 ≤ k < n/2. The generalized Petersen graph, denoted P (n, k), is the graph withvertex set {uj, vj : j ∈ Zn} and edge set {ujuj+1, vjvj+k, ujvj | j ∈ Zn}, where Zn is the additivegroup of integers modulo n. The generalized Petersen graphs was introduced in 1950 by Coxeter [9]and was given its name in 1969 by Watkins [30] in a consideration of a conjecture of Tutte [29]. Forextensive surveys on the Petersen graph P (5, 2), see [7,16], which also involve all kinds of variantionsof the Petersen graph, including the family P (n, k). Many structural and algorithmic properties of thegeneralized Petersen graphs have been extensively investigated, among which [1–5,10,13,22,23,26,28]are oft-cited work that are used to introduce the family P (n, k). Recent advances in the field include[11,12,17,19,21,31,34], most of which are concerned with some specific graph theoretical concept forP (n, k).

For NP-hard and NP-complete problems, research on the graph P (n, 3) often attracts attentionand usually costs a considerable effort. For instance, Hlineny [15] showed that deteriming the crossingnumber of a cubic graph is NP-hard; Richter and Salazar [25] found a formula for the crossing number

2010 Mathematics Subject Classification. 05C30.Key words and phrases. generalized Petersen graph, metric dimension, resolving set, Floyd-Warshall algorithm.Corresponding author: David G.L. Wang.This paper was supported by National Natural Science Foundation of China (Grant No. 11671037).

1

http://arxiv.org/abs/2005.08038v3

2 D.G.L. WANG, M.M.Y. WANG, AND S.Q. ZHANG

for P (n, 3) according to the residuce of n modulo 3. As another example, Bresara and Sumenjakb [6]showed the NP-completeness of the 2-rainbow domination problem; Xu [32] enhanced the upper boundof the 2-rainbow domination number of P (n, 3) according to the residuce of n modulo 16.

In 2018, Kelenc, Tratnik, and Yero [20] introduced the concept of edge dimension for a graph, andshowed the NP-hardness of computing the edge dimension of a graph. They also pointed out thatthe edge dimension has applications in network security surveillance. In fact, an intruder accesses anetwork through edges can be identified by an edge resolving set. This paper is concerned with theedge dimension of the graph P (n, 3).

For any list w = (v1, v2, . . . , vk) of vertices and any vertex v in a connected graph G, the represen-tation of v with respect to w is the list

(

d(v, v1), d(v, v2), . . . , d(v, vk))

, where d(x, y) is the distancebetween the vertices x and y. The set {v1, v2, . . . , vk} is said to be a resolving set for G if everytwo vertices of G have distinct representations. It was Slater [27] who firstly considered the min-imum cardinality of a resolving set for G, called the metric dimension of G. In analog, the edgedimension for a graph is the minimum cardinality of a vertex set {v1, v2, . . . , vk} such that the lists(

d(e, v1), d(e, v2), . . . , d(e, vk))

for all edges e are distinct, where d(e, vi) is the distance between theedge e and the vertex vi. For instance, the graph P (8, 3) with an edge resolving set {u0, u1, u2, v3}is illustrated in Fig. 1.1. The exact values of the edge dimension for some classes of graphs were

u0

u7

u6

u5

u4

u3

u2

u1

v0v7

v6

v5v4

v3

v2

v1

Figure 1.1. The generalized Petersen graph P (8, 3) with an edge resolving set {u0, u1, u2, v3}.

known, while bounds are given to some other graph classes. The edge dimension of an n-vertex graphis at most n − 1; see [35, 36]. For a rich resource of different kinds of resolving sets of graphs withapplications, see Kelenc, Kuziak, Taranenko and Yero [18]. More progress on the edge dimension canbe found from [24, 33].

Fillipovic, Kartelj and Kratica [14] showed that the edge dimension of the generalized Petersengraph P (n, k) is at least 3, and confirmed that the edge dimension of P (n, k) equals 3 for k ∈ {1, 2}and n ≥ 10. Here is our main result.

Theorem 1.1. For n ≥ 11, the edge dimension of the generalized Petersen graph P (n, 3) is 4.

This paper is organized as follows. In § 3, we give a formula for the distance between any vertex andany edge in the graph P (n, 3), which serves as the basis for all arguments in the sequel. In § 4 and 5,we prove that the edge dimension of P (n, 3) has lower and upper bound 4, respectiely. Moreover, thevertex tetrad {u0, u1, x, y} with {x, y} presented in Table 1.1 is an edge resolving set of P (n, 3) forn ≥ 19.

THE EDGE DIMENSION OF P (n, 3) IS 4 3

Table 1.1. Vertex pairs {x, y} such that the tetrad {u0, u1, x, y} is an edge resolvingset of the graph P (n, 3), where n ≥ 19, and r′n is the residue of n modulo 6.

r′n 0, 1, 3 2 4 5

{x, y}{

v2, u⌊n/2⌋−1

} {

un/2−3, vn/2−2

} {

v2, un/2+3

} {

u⌊n/2⌋−1, v⌊n/2⌋}

2. Strategy of our proof

The edge dimension is all about the distance between a vertex and an edge. It is known that thetime complexity of finding the shortest paths between any two given vertices in a graph is O(n3); seethe Floyd-Warshall algorithm in [8, §25.2]. In order to establish Theorem 1.1, however, we furtherneed a formula for the distance between a given vertex and a given edge. Denote

eui = uiui+1, evi = vivi+3, and esi = uivi.

For n ∈ Z, define qn, rn ∈ Z by n = 3qn + rn, where rn ∈ {0, 1, 2}.

Theorem 2.1. Let n ≥ 13 and 0 ≤ i ≤ n− 1. In the graph P (n, 3), we have

d(u0, eui ) =

{

min(i, n− 1− i), if i ≤ 2 or i ≥ n− 3,

min(

⌈i/3⌉+ 2, ⌈(n− i− 1)/3⌉+ 2)

, otherwise;

d(u0, esi ) =

{

min(i, n− i), if i ≤ 2 or i ≥ n− 2,

min(

qi + ri + 1, qn − qi + r + 1)

, otherwise;

d(u0, evi ) =

{

min(i + 1, n− i+ 1), if i ≤ 1 or i = n− 1,

min(qi + ri + 1, qn − qi + r), otherwise;

d(v0, eui ) =

{

min(qi + ⌊ri/2⌋+ 1, qn − qi + ⌊rn/2⌋ − ⌊ri/2⌋+ 1), if ri = 0 or (rn, ri) = (0, 2),

min(qi + 2, qn − qi + 1), otherwise;

d(v0, esi ) =

min(qi, qn − qi), if ri = rn = 0,

min(qi + rn + 1, qn − qi), if ri = rn 6= 0,

min(qi, qn − qi + rn + 1), if ri = 0 6= rn,

min(qi + 2, qn − qi + 2), if ri = 1 6= rn,

min(qi + 3, qn − qi + rn), if ri = 2 6= rn;

d(v0, evi ) =

min(qi, qn − qi − 1), if ri = rn = 0,

min(qi + ri + 2, qn − qi + rn − ri − 1), if ri = rn 6= 0,

min(qi, qn − qi + rn + 1), if ri = 0 6= rn,

min(qi + 3, qn − qi + 2), if ri = 1 6= rn,

min(qi + 4, qn − qi + rn + 1), if ri = 2 6= rn.

Here r = |rn − ri|, except when (rn, ri) = (0, 2) and r = 0.

In § 3, we will establish Theorem 2.1 in the following 3 steps.

Step 1: Figure out the structure of a shortest path between 2 vertices and obtain Theorem 3.2.Step 2: Find a formula between any 2 vertices in P (n, 3); see Theorem 3.4.Step 3: Compare the distances d(x, y) and d(x, z) for any vertex x and any edge yz in P (n, 3), and

obtain the desired formula in Theorem 2.1.

Theorem 2.1 will serve as the basis of verifying and computing the distance between a vertex and anedge in P (n, 3).


Now we clarify the strategy of proving Theorem 1.1. We use the word triad (resp., tetrad) to denotea set of order 3 (resp., 4), and the word triple (resp., quadruple) to mean an ordered list of 3 (resp., 4)elements. Theorem 1.1 is an immediate consequence of the combination of Propositions 2.2 and 2.3.

Proposition 2.2. For n ≥ 11, the graph P (n, 3) has no edge resolving triads.

Proposition 2.3. For n ≥ 11, the graph P (n, 3) has an edge resolving tetrad.

First of all, we point out that it suffices to show Propositions 2.2 and 2.3 for n ≥ 100. In fact,for n < 100, one may quickly check that every vertex triad is not an edge resolving set with the aidof a computer verification. Precisely speaking, we can suppose that R = {α0, βy, γz} by symmetry,where α, β, γ ∈ {u, v} and y, z ∈ Zn. With Theorem 2.1 in hand, the time complexity of checking thatany vertex triad is not an edge resolving set is O(n4), since there are O(n2) choices for R and O(n2)choices for an edge pair. In practice, on a laptop with CPU clock rate 2.3GHz and 8 kernels, it costsless than 3 minutes to confirm Proposition 2.2 for all 11 ≤ n < 100. Similarly, the time complexityof checking that every tetrad is or is not an ecge resolving set is O(n5). In practice, with the aid ofTable 1.1, verifying Proposition 2.3 for n < 100 costs less than 3 hours.

From now on, we can suppose that n ≥ 100. This assumption saves us from considering an annoyingnumber of singular cases appeared for small n. Though it can be made even earlier before provingTheorem 2.1, our proof of Theorem 2.1 works well for n < 100.

The strategy of proving Proposition 2.2 is as follows. First, we concentrate on the pairs of adjacentedges on the outer cycle and determine the sets

A0 = {i ∈ Zn : d(u0, eui−1) = d(u0, e

ui )} and D0 = {i ∈ Zn : d(v0, e

ui−1) = d(v0, e

ui )}.

As will be seen, it turns out quite fortunately that A0 ⊆ D0. Second, using the derived expressionsof A0 and D0 and by a symmetry argument, we can reduce Proposition 2.2 to that for each pair(a, b) ∈ Sn, where

Sn ={

(a, b) ∈ Z2 : 1 ≤ a ≤ ⌊n/3⌋, 2a ≤ b ≤ ⌊(n+ a)/2⌋

}

,

no triad in the collectionT (a, b) =

{

{α0, βa, γb} : α, β, γ ∈ {u, v}}

is an edge resolving set. Since |T (a, b)| = 8, it is plausible that the number of triads that we have toconsider is still considerably large. Born under a lucky star, in the third step, we manage to provethat A0 ∩ Aa ∩ Ab 6= ∅ for almost all pairs (a, b) ∈ Sn, where At = {x + t : x ∈ A0}; see § 4.1. Thisleads us to a collection of 22 sporadic cases of pairs (a, b), for which we handle in § 4.2.

We prove Proposition 2.2 according to the residuce r′n of n modulo 6. We adopt the same strategyto deal with each of the residues, which is as follows. First, we show that the edges whose distancesfrom u0 are less than 3 are distinguishable. To do this, we list all such edges e1 and compute thedistances d(v, e1) for each of the prescribed vertex v ∈ {u0, v0, x, y}; see Table 1.1. This step is easyto be done by using Theorem 2.1. Second, for any fixed integer d ≥ 3, we find out all edges e2 whosedistances from u0 are d. The number of such edges e2 for each residue case is about 20. Computingthe distances d(v, e2) for each prescibed vertices v ∈ {v0, x, y}, we obtain a table, from which one maysee that the edge metric representations with respect to the prescribed vertex tetrad are distinct.

3. A formula for the vertex-edge distances in P (n, 3)

Let G = (V,E) be a connected graph. The distance between a vertex x ∈ V and an edge e = uv ∈ Eis the integer d(x, e) = min{d(x, u), d(x, v)}, where d(x, y) is the distance between x and y. We saythat x resolves two edges e and e′ if d(x, e) 6= d(x, e′). A vertex set R = {x1, x2, . . . , xk} is said tobe an edge resolving set if every two distinct edges in G are resolved by a member of R. The edgedimension for G is the minimum cardinality of an edge resolving set of G. For any edge e ∈ E, thelist

(

d(e, x1), d(e, x2), . . . , d(e, xk))

is called the edge metric representation of e with respect to R.


In the generalized Petersen graph, the edges uivi are said to be spokes, the subgraph induced by thevertices u0, u1, . . . , un−1 is said to be the outer cycle, and that induced by the vertices v0, v1, . . . , vn−1

is said to be the inner cycle(s).

Lemma 3.1. Consider the graph P (n, k). Let i ∈ Z, p0 ∈ {u0, v0} and w ∈ {u, v}. Then

d(p0, wi) = d(p0, w−i) and d(u0, vi) = d(v0, ui).

Moreover, for any i, j ∈ Z, we have

d(ui, euj ) = d(u0, e

uj−i) = d(u0, e

ui−j−1),

d(ui, esj) = d(u0, e

sj−i) = d(u0, e

si−j),

d(ui, evj ) = d(u0, e

vj−i) = d(u0, e

vi−j−k).

Proof. All the results follow from the symmetry of the graph P (n, k). �

Let p = w0w1 · · ·wl be a path in P (n, k), where wi are vertices of P (n, k). Denote the length l by ℓp.We can rewrite p = p0p1 · · · ps, where each pj is a path contained entirely in either the outer cycle oran inner cycle. Accordingly we call pj an outer section or inner section. We say that an edge e isclockwise if e is of the form ujuj+1 or vjvj+k, and counterclockwise otherwise. A section pj is clockwise(resp., counterclockwise) if every edge in pj is so. The path p clockwise (resp., counterclockwise) ifevery section pj is so. We call p undeviating if it is either clockwise or counterclockwise. For any1 ≤ j ≤ s, we call the subpath pj−1pj a turn of p if one of the subpaths pj−1 and pj is clockwise andthe other is counterclockwise.

Theorem 3.2. For any vertices x and y in the grpah P (n, 3), there is a shortest path from x to ywhich is undeviating and contains at most 2 spokes.

Proof. Without loss of generality we can suppose that x ∈ {u0, v0} and y = wt, where w ∈ {u, v} and0 ≤ t ≤ n− 1. Let p be a path from x to y of length ℓp = d(x, y), with the minimum number of turns.If there is no undeviating path from x to y, then p has at least one turn. We will show that this isimpossible by contradiction. Let qq′ be the first turn in p, where q and q′ are sections of p.

Case 1. q is a clockwise outer section. Then we can suppose that q = uj−suj−s+1 · · ·uj for somes ≥ 1 and p = αqvjβ, where α is a clockwise subpath and β is a path starting from vj−3.

If s ≥ 2, then the path

p′ =

{

αuj−suj−s+1 · · ·uj−3β, if s ≥ 3

αuj−3β, if s = 2

is shorter than p, contradicting the choice of p. When s = 1, the path p reduces to p = αuj−1ujvjβ.

• If α contains a non-spoke arc, then α = α′vj−4vj−1 for some clockwise path α′, and the pathα′vj−4uj−4uj−3β is shorter than p, the same contradiction.

• If α has no non-spoke arcs, then the path αuj−1uj−2uj−3β has the same length as p and a lessnumber of turns than p.

This proves that Case 1 is impossible.

Case 2. q is a counterclockwise outer section. We define a reflection f on the vertex set of P (n, k) byf(wt) = w−t, where w ∈ {u, v}. It extends naturally to act on paths. The path f(p), which is fromf(x) to f(y), has length ℓf(p) = d(f(x), f(y)) and the minimum number of turns. Since f(q)f(q′) isthe first turn in f(p), and f(q) is a clockwise outer section, we know that Case 2 is impossible by theimpossibility of Case 1.

Case 3. q′ is a counterclockwise outer section. Then we can suppose that q′ = ujuj−1 · · ·uj−s andp = αvjq

′β, where s ≥ 1, α is a clockwise subpath ending at vj−3, and β is a subpath.


If s ≥ 2, then the path

p′ =

{

αuj−3uj−4 · · ·uj−sβ, if s ≥ 3

αuj−3uj−2β, if s = 2

is shorter than p, contradicting the choice of p. When s = 1, p reduces to p = αvjujuj−1β.

• If β contains no non-spoke arcs, then the path αuj−3uj−2uj−1β has the same length as p and aless number of turns than p.

• If β contains a non-spoke arc, then β = vj−1β′ for some path β′ starting from vj−4 or from

vj+2. In the former case, the path αuj−3uj−4β′ is shorter than p; in the latter case, the path

αvjujuj+1uj+2β′ has the same length as p and a less number of turns than p.

This proves that Case 3 is impossible.

Case 4. q′ is a clockwise outer section. Since f(q)f(q′) is the first turn in the path f(p), and f(q′) isa counterclockwise outer section, we know that Case 4 is impossible by the impossibility of Case 3.

This proves the existence of an undeviating path from x to y of length d(x, y). Let p be such apath. It remains to show that the number of spokes in p is at most 2. By symmetry, we can supposethat p is clockwise. Assume that p has at least 3 spokes.

If x = v0, then there exists h ≥ 0 and i, j ≥ 1 such that

p = v0v3 · · · v3hu3hu3h+1 · · ·u3h+iv3h+iv3h+i+3 · · · v3h+i+3jβ,

where β is a clockwise path starting from u3h+i+3j . Then

ℓp = h+ 1 + i+ 1 + j + 1 + ℓβ = h+ i+ j + ℓβ + 3.

Suppose that 3h+ i+ 3j = 3q + r, where q ≥ 1 and r ∈ {0, 1, 2}. The path

p′ = v0v3 · · · v3qu3qu3q+1 · · ·u3q+r−1β.

has length ℓp′ = q + r + 1 + ℓβ, and

ℓp − ℓp′ = 2 + (i− r) + (h+ j − q) = 2( i

3+ 1−

r

3

)

≥ 2(1

3+ 1−

2

3

)

=4

3,

contradicting the choice of p.

Otherwise x = u0. Then there exists h ≥ 0 and i, j ≥ 1 such that

p = u0u1 · · ·uhvhvh+3 · · · vh+3iuh+3iuh+3i+1 · · ·uh+3i+jβ,

where β is a clockwise path starting from vh+3i+j . Then

ℓp = h+ 1 + i+ 1 + j + 1 + ℓβ = h+ i+ j + ℓβ + 3.

Suppose that h+ 3i+ j = 3q + r, where q ≥ 1 and r ∈ {0, 1, 2}. The path

p′ = v0v3 · · · v3qu3qu3q+1 · · ·u3q+r−1β.

has length ℓp′ = q + r + 1 + ℓβ, and

ℓp − ℓp′ = 2 + (i− q) + (h+ j − r) = 2(h+ j

3+ 1−

r

3

)

≥ 2(1

3+ 1−

2

3

)

=4

3,

contradicting the choice of p. This completes the proof. �

Now we enter Step 2. Let x be a vertex in the graph P (n, 3) and let y be a vertex or an edge inP (n, 3). Denote by P(x, y) the set of shortest undeviating paths from x to y. Let P−(x, y) be theset of clockwise paths in P(x, y), and ℓ−(x, y) the length of any path in P−(x, y), called the clockwisedistance from x to y. Let P+(x, y) be the set of counterclockwise paths in P(x, y), and ℓ+(x, y) thelength of any path in P+(x, y), called the counterclockwise distance from x to y.

Lemma 3.3. Let 3 ≤ i ≤ n− 3. Then we have the following in the graph P (n, 3).


(1) ℓ−(u0, ui) = qi + ri + 2 and ℓ+(u0, ui) = qn − qi + r + 2.(2) ℓ−(u0, vi) = qi+ri+1 and ℓ+(u0, vi) = qn−qi+r+1. Furthermore, |P−(u0, vi)| = |P+(u0, vi)| = 1.(3) The clockwise and counterclockwise distances from v0 to vi are respectively

ℓ−(v0, vi) =

{

qi, if ri = 0

qi + ri + 2, if ri 6= 0and ℓ+(v0, vi) =

{

qn−i, if rn = ri

qn−i + rn−i + 2, if rn 6= ri.

Proof. It is elementary to compute that

(qn−i, rn−i) =

(qn − qi, rn − ri), if rn ≥ ri

(qn − qi − 1, 1), if (rn, ri) = (0, 2)

(qn − qi − 1, 2), otherwise

(3.1)

(1) Consider p ∈ P−(u0, ui). By Theorem 3.2, the path p has either 0 or 2 spokes. In the formercase, ℓp = i; in the latter case, ℓp = qi + ri + 2 for p has qi steps on an inner cycle between the twospokes and ri steps on the outer cycle. Since i ≥ 3, we can deduce that

ℓ−(u0, ui) = min(i, qi + ri + 2) = qi + ri + 2.

By symmetry and by Eq. (3.1), ℓ+(u0, ui) = qn−i + rn−i + 2 = qn − qi + r + 2.

(2) Consider p ∈ P−(u0, vi). By Theorem 3.2, the path p has exactly one spoke. Then

p = u0u1 · · ·urivrivri+3 · · · vi

is unique, with length ℓp = qi + ri + 1. Indeed, the unique spoke in p must be immediately after thefirst ri steps on the outer cycle. By symmetry, |P+(u0, vi)| = 1. By Eq. (3.1),

ℓ+(u0, vi) = qn−i + rn−i + 1 = qn − qi + r + 1.

(3) Consider p ∈ P−(v0, vi). By Theorem 3.2, the path p has either 0 or 2 spokes. If ri = 0, thenp = v0v3 · · · vi is unique and ℓp = qi. Otherwise rn ∈ {1, 2}. If p has no spokes, then ℓp = i; if p hasexactly two spokes, then ℓp = qi + ri + 2 for p has qi steps on an inner cycle between the two spokesand ri steps on the outer cycle. This proves the first desired formula. It is direct to obtain the otherformula by symmetry. �

Note that ℓ±(u0, ui) = ℓ±(u0, vi) + 1, and the path obtained by adding the arc viui to the uniquepath in P±(u0, vi) belongs to P±(u0, ui). For n ∈ Z, define h ∈ Z and r′n ∈ {0, 1, . . . , 5} by n = 6h+r′n.Define Mn = max{j < ⌊n/2⌋ : rj = rn}.

Theorem 3.4. Let 0 ≤ i ≤ ⌊n/2⌋. Then we have the following in the graph P (n, 3).

d(u0, ui) =

qi + ri, if i ≤ 2

qi + ri + 1, if (r′n, i) = (5, ⌊n/2⌋)

qi + ri + 2, otherwise,

d(u0, vi) =

{

qi + ri, if (r′n, i) = (5, ⌊n/2⌋)

qi + ri + 1, otherwise,

d(v0, vi) =

qi + ri − 1, if (r′n, i) = (5, ⌊n/2⌋)

qi + ri, if r′n ∈ {2, 4} and i = Mn, or ri = 0

qi + ri + 1, if r′n ∈ {1, 5} and i = Mn

qi + ri + 2, otherwise.

Proof. Let 0 ≤ i ≤ ⌊n/2⌋. We show them individually.


Consider d(u0, ui). If i ≤ 2, it is easy to check that d = qi + ri. Let 3 ≤ i ≤ ⌊n/2⌋. Suppose thatℓ+(u0, ui) < ℓ−(u0, ui). By Lemma 3.3, qn − qi + r + 2 < qi + ri + 2. It is elementary to show that(r′n, i) = (5, ⌊n/2⌋). In this case, (qn, rn, qi, ri, r) = (2h+ 1, 2, h, 2, 0) and

d(u0, ui) = qn − qi + r + 2 = 2h+ 1− h+ 0 + 2 = qi + ri + 1.

Since rn−i = 0, the set P consists of a unique path, which has no edge in the outer cycle.

Consider d(u0, vi). Suppose that ℓ+(u0, vi) < ℓ−(u0, vi). By Lemma 3.3 and the previous case, wefind (r′n, i) = (5, ⌊n/2⌋) and d(u0, ui) = qi + ri.

Consider d = d(v0, vi). Suppose that ℓ+(v0, vi) < ℓ−(v0, vi). By Lemma 3.3, it is elementary toshow that rn = ri 6= 0, d = qn−i, and

(3.2) qn − qi ≤ qi + ri + 1.

We proceed according to the value of r′n.

• If n = 6h + 1, then i ≤ 3h − 2 since ri = 1. By Ineq. (3.2), we find i = 3h − 2 = Mn andd = h+ 1 = qi + ri + 1.

• If n = 6h+ 2, then i ≤ 3h− 1 since ri = 2. By Ineq. (3.2), we find i = Mn and d = qi + ri.• If n = 6h+ 4, then i ≤ 3h+ 1 since ri = 1. By Ineq. (3.2), we find i = Mn and d = qi + ri.• If n = 6h+5, then i ≤ 3h+2 since ri = 2. By Ineq. (3.2), we find that either i = 3h+2 = ⌊n/2⌋ or

i = 3h−1 = Mn. In the former case, d = h+1 = qi+ri−1; in the latter case, d = h+2 = qi+ri+1.

This completes the proof. �

Now we start Step 3.

Proposition 3.5. Let i, j ∈ Z and j > i. We have the following equivalence:

qj + rj < qi + ri ⇐⇒ j = i+ 1 and ri = 2.

In this case, qi+1 + ri+1 + 1 = qi + ri.

Proof. It is elementary and we omit the proof. �

Lemma 3.6. Let 0 ≤ i ≤ ⌈n/2⌉ − 1. Then the following equivalences hold:

d(u0, ui+1) < d(u0, ui) ⇐⇒ i ∈ {5, 8, 11, . . . , 3⌊n/6⌋ − 1},

d(v0, ui+1) < d(v0, ui) ⇐⇒ i ∈ {2, 5, 8, . . . , 3⌊n/6⌋ − 1}.

Proof. We show the two equivalences individually.

Let d1 = d(u0, ui) and d2 = d(u0, ui+1). Suppose that d2 < d1. We proceed by contradiction.

Case 1. i ≤ 2. Then (d1, d2) = (i, i+ 1), contradicting the premise d2 < d1.

Case 2. r′n = 5 and i = ⌊n/2⌋. Then n is odd and d2 = d(u0, un−i−1) = d1.

Case 3. (r′n, i) 6= (5, ⌊n/2⌋ − 1). Then (n, i) = (6h+ 5, 3h+ 1). By Theorem 3.4, d1 = h+ 3 = d2.

Case 4. None of the above. By Theorem 3.4, the premise d2 < d1 reduces to the inequaltiy inProposition 3.5, which is equivalent to ri = 2; in this case, d2 = d1 − 1 by Proposition 3.5.

Rearranging the above results, we obtain the first desired equivalence.

Now, let d1 = d(v0, ui) and d2 = d(v0, ui+1). Suppose that d2 < d1. We treat 3 cases.

Case 1. i = ⌈n/2⌉ − 1. If n is odd, then d1 = d2 by symmetry, a contradiction. Suppose that n iseven. By Theorem 3.4, the assumption d2 < d1 reduces to the inquality in Proposition 3.5, which isequivalent to ri = 2. Since i = n/2− 1, we find r′n = 0.


Case 2. (r′n, i) = (5, ⌈n/2⌉ − 2). Then i = 3h+ 1 and d1 = h+ 2 = d2, a contradiction.

Case 3. None of the above. Since i+1 ≤ ⌊n/2⌋, by Theorem 3.4 and Proposition 3.5, the assumptiond2 < d1 reduces to ri = 2.

Rearranging the above results, we obtain the second desired equivalence. �

Lemma 3.7. Let 0 ≤ i ≤ ⌊n/2⌋. Then the following equavilences hold:

d(u0, ui) < d(u0, vi) ⇐⇒ i ∈ {0, 1, 2},

d(v0, vi) < d(v0, ui) ⇐⇒ ri = 0 or (r′n, i) ∈ {(5, ⌊n/2⌋), (2,Mn), (4,Mn)}.

Proof. Direct from Theorem 3.4. �

Lemma 3.8. Let 0 ≤ i ≤ ⌈n/2⌉ − 2, d1 = d(u0, vi) and d2 = d(u0, vi+3).

(1) If (r′n, i) = (2, n/2− 2), then d1 > d2 = qi + ri.(2) Otherwise, d2 ≥ d1 = qi + ri + 1.

Proof. Let 0 ≤ i ≤ ⌈n/2⌉ − 2 and j = n − i − 3. Then d2 = d(u0, vj) by symmetry. Suppose thatd2 < d1. We treat 3 cases.

Case 1. i = ⌈n/2⌉ − 2. Then j = ⌊n/2⌋ − 1. If n is odd, then j = i and d1 = d2, a contradiction.Otherwise n is even, then j = i+1. By Theorem 3.4, the assumption d2 < d1 reduces to the inequaltiyin Proposition 3.5, which is equivalent to ri = 2. It follows that r′n = 2 and d2 = h+ 1 = qi + ri.

Case 2. i = ⌈n/2⌉ − 3 and n is odd. Then j = ⌊n/2⌋ = i + 2. If r′n = 5, then (ri, rj) = (0, 2).On the other hand, by Theorem 3.4, the assumption d2 < d1 reduces to qj + rj < qi + ri + 1, i.e.,qj + 1 < qi, which is impossible since j = i+ 2. Otherwise r′n ∈ {1, 3}. Then (ri, rj) = {(1, 0), (2, 1)}and rj = ri − 1. By Theorem 3.4, the assumption d2 < d1 reduces to qj + rj + 1 < qi + ri + 1, i.e.,qj ≤ qi. Since j = i+ 2, we find qj = qi. It follows that rj ≥ ri, a contradiction.

Case 3. None of the above. Then i+ 3 ≤ ⌊n/2⌋. By Theorem 3.4,

(3.3) d2 − d1 = qi+3 + ri+3 − qi − ri = 1,

contradicting the assumption d2 < d1.

Therefore, by Theorem 3.4, as if (r′n, i) 6= (2, n/2− 2), d2 ≥ d1 = qi + ri + 1. �

Lemma 3.9. Let 0 ≤ i ≤ ⌈n/2⌉− 2, d1 = d(v0, vi) and d2 = d(v0, vi+3). Suppose that d2 < d1. Thenone of the following is true.

(1) r′n = 2 and (i, d2) = (Mn, qr + ri − 1).(2) r′n ∈ {1, 5} and (i, d2) = (Mn, qr + ri).(3) r′n ∈ {2, 4} and (i, d2) = (Mn − 3, qr + ri + 1).

Proof. Suppose that rn = 0. When i+3 ≤ n/2, we can deduce Eq. (3.3) by Theorem 3.4, contradictingthe premise d2 < d1. Consider the other case n/2 − 3 < i ≤ ⌈n/2⌉ − 2. If n = 6h, then i = 3h − 2and d1 = h + 2 < h + 3 = d2, the same contradiction. Otherwise n = 6h + 3. Then i = 3h− 1 andd1 = d2 = h+ 3, the same contradiction.

Below we can suppose that rn 6= 0. When i+ 3 < Mn, we obtain the same contradiction Eq. (3.3)to the premise d2 < d1. It remains to compute the pair (d1, d2) for Mn − 3 ≤ i ≤ ⌈n/2⌉ − 2. Weobserve that the upper bound can be further improved to ⌊n/2⌋ − 2 when n is odd since d1 = d2 bysymmetry when i = (n+ 1)/2− 2. By Theorem 3.4, we find Table 3.1, from which we see that

• if r′n ∈ {1, 5}, then (i, d2) = (Mn, qi + ri);


• if r′n = 2, then (i, d2) = (Mn, qi + ri − 1) or (i, d2) = (Mn − 3, qi + ri + 1); and• if r′n = 4, then (i, d2) = (Mn − 3, qi + ri + 1).

Table 3.1. The distances d1 and d2 when Mn − 3 ≤ i ≤ ⌊n/2⌋ − 2.

i

n 3h 3h− 1 3h− 2 3h− 3 3h− 4 3h− 5

6h+ 1 d1 h+ 1 h− 1 h+ 2 h+ 1d2 h h h+ 3 h+ 1

6h+ 2 d1 h+ 1 h+ 2 h− 1 h+ 2d2 h h+ 3 h h+ 1

6h+ 4 d1 h+ 3 h+ 3 h+ 2d2 h+ 3 h+ 4 h+ 1

6h+ 5 d1 h h+ 2 h+ 2 h− 1 h+ 2d2 h+ 1 h+ 1 h+ 3 h h+ 2


Now, combining Lemmas 3.6 to 3.9, it is elementary to show Theorem 2.1 for when i < n/2, andeasy to derive the formulas for i ≥ n/2 by symmetry. We leave the proof details to the readers.

4. Proof of Proposition 2.2

This section is devoted to proving that the edge dimension of the graph P (n, 3) is at least 4.Looking for a pair of edges that have the same distance to a given vertex, we focus on pairs of edgesof the form (eui−1, e

ui ).

Lemma 4.1. Let n ≥ 100 and A0 = {i ∈ Zn : d(u0, eui−1) = d(u0, e

ui )}. Then

A0 =

{

{±i ∈ Z : 5 ≤ i < n/2 and ri 6= 1} ∪ {0, n/2}, if n is even;

{±i ∈ Z : 5 ≤ i < n/2 and ri 6= 1} ∪ {0}, if n is odd.

Proof. By symmetry, n/2 ∈ A0 when n is even. By Theorem 2.1, it is direct to verify that 0 ∈ A0

and {±1, ±2, ±3, ±4} ∩ A0 = ∅. Consider 5 ≤ i < n/2. By Theorem 2.1,

d(u0, eui−1) = 2 +min

(

⌈(i− 1)/3⌉, ⌈(n− i)/3⌉)

= 2 + ⌈(i − 1)/3⌉,

d(u0, eui ) = 2 +min

(

⌈i/3⌉, ⌈(n− i− 1)/3⌉)

= 2 + ⌈i/3⌉.

Hence we derive the equivalences i ∈ A0 ⇐⇒ ⌈(i − 1)/3⌉ = ⌈i/3⌉ ⇐⇒ ri 6= 1. By symmetry, weknow that −i ∈ A0 ⇐⇒ ri 6= 1. This completes the proof. �

Lemma 4.2. Let n ≥ 100 and B0 = {i ∈ Zn : d(v0, eui−1) = d(v0, e

ui )}. Then

B0 =

{

{±i ∈ Z : 0 ≤ i < n/2 and ri 6= 1} ∪ {n/2}, if n is even;

{±i ∈ Z : 0 ≤ i < n/2 and ri 6= 1}, if n is odd.

As a consequence, A0 ⊆ B0.

Proof. It is clear that 0 ∈ B0. By symmetry, n/2 ∈ B0 when n is even. Consider 1 ≤ i < n/2. Letd1 = d(v0, e

ui−1) and d2 = d(v0, e

ui ). By Theorem 2.1, it is direct to check that for 0 ≤ i ≤ ⌈n/2⌉− 1,

d(v0, ui+1) < d(v0, ui) ⇐⇒ i ∈ {2, 5, 8, . . . , 3⌊n/6⌋ − 1}.

Therefore, if ri = 0, then d1 = d(v0, ui) = d2 and i ∈ B0.


Suppose that ri = 1. Then (qi−1, ri−1) = (qi, 0). Since 3qn + rn = n > 2i = 6qi + 2, we findqn > 2qi, which implies that

(4.1) qn − qi ≥ qi + 1.

By Theorem 2.1,

d1 = min(qi−1 + ⌊ri−1/2⌋+ 1, qn − qi−1 + ⌊rn/2⌋ − ⌊ri−1/2⌋+ 1)

= min(qi + 1, qn − qi + ⌊rn/2⌋+ 1) = qi + 1, and

d2 = min(qi + 2, qn − qi + 1) = qi + 2.

Therefore, d2 6= d1. This proves that i 6∈ B0.

Suppose that ri = 2. Then (qi−1, ri−1) = (qi, 1). Along the same line we can derive Ineq. (4.1).When rn = 0, one may enhence it to qn − qi ≥ qi + 2 since 3qn = n > 2i = 6qi + 4. By Theorem 2.1,

d1 = min(qi−1 + 2, qn − qi−1 + 1) = min(qi + 2, qn − qi + 1) = qi + 2,

d2 =

{

min(qi + ⌊ri/2⌋+ 1, qn − qi + ⌊rn/2⌋ − ⌊ri/2⌋+ 1), if rn = 0

min(qi + 2, qn − qi + 1), if rn 6= 0

=

{

min(qi + 2, qn − qi), if rn = 0

min(qi + 2, qn − qi + 1), if rn 6= 0

= qi + 2.

By symmetry, we obtain the desired formula for B0. As a consequence, A0 ⊆ B0 by Lemma 4.1. �

For t ∈ Zn, define At = A0 + t and Bt = B0 + t. From the definitions, we see that

d(ut, eui−1) = d(ut, e

ui ) for i ∈ At, and

d(vt, euj−1) = d(vt, e

uj ) for j ∈ Bt.

By Lemma 4.2, we obtain

(4.2) At ⊆ Bt for any t ∈ Zn.

Lemma 4.3. Let n ≥ 100. If the graph P (n, 3) has an edge resolving triad R, then the vertices in Rhave distinct subscripts.

Proof. Suppose that P (n, 3) has an edge resolving set R = {αx, βy, γz}, where α, β, γ ∈ {u, v} andx, y, z ∈ Zn. Let T = {x, y, z}. Then |T | ∈ {2, 3}.

Assume that |T | = 2. By symmetry, we can suppose that R = {u0, v0, γt}, where 1 ≤ t ≤ ⌊n/2⌋.By Lemmas 4.1 and 4.2, it suffices to show that

• A0 ∩B0 ∩ At 6= ∅ if γ = u, and that• A0 ∩B0 ∩Bt 6= ∅ if γ = v.

In view of (4.2), it suffices to show A0 ∩ At 6= ∅. In fact, by Lemma 4.1, it is easy to read out anelement in the intersection; see Table 4.1. �

Table 4.1. An element in the intersection A0 ∩ At, when n ≥ 100.

t rt = 0 rt = 1 rt = 2

≤ ⌊n/2⌋ t+ 6 t+ 5 t+ 6

> ⌊n/2⌋ t− 6 t− 5 t− 5


Lemma 4.3 leads us to consider vertex triads with distinct subscripts. We say that two triplesT = (x, y, z) ∈ Z

3n and T ′ ∈ Z

3n are equivalent if T ′ = (x + a, y + a, z + a) for some a ∈ Zn, or

T ′ = (2t − x, 2t − y, 2t − z) for some t ∈ Zn, i.e., if T ′ is a translation or reflection of T . It isclear that a vertex triad {αx, βy, γz}, where α, β, γ ∈ {u, v} and where x, y, z ∈ Zn are distinct, is anedge resolving set if and only if so is every vertex set {αx′ , βy′ , γz′}, where (x′, y′, z′) is equivalent to(x, y, z). Considering the representatives of the equivalence classes, we produce Proposition 4.4.

Proposition 4.4. For any 3 distinct elements x, y, z ∈ Zn, there exist (a, b) ∈ Sn such that the triples(x, y, z) and (0, a, b) are equivalent.

Proof. For any x, y ∈ Zn, define d(x, y) = min(

|x′ − y′|, n− |x′ − y′|)

, where x′ and y′ are the leastnonnegative residue of x and y modulo n, respectively. By symmetry, we can suppose that

0 = x < y < z < n and d(x, y) ≤ min(

d(y, z), d(z, x))

.

It follows that d(0, y) = y and 2y ≤ z ≤ n− y. Thus y ≤ ⌊n/3⌋. If z ≤ ⌊(n+ y)/2⌋, then we can take(a, b) = (y, z). It remains to consider when

(4.3) z ≥ ⌊(n+ y)/2⌋+ 1.

In this case, we translate the triple (0, y, z) to T ′ = (−y, 0, z − y) by substituting y from eachcoordinate, and reflect T ′ to T ′′ = (y, 0, y − z) about the central axis going through 0. Let a = yand b = y − z + n. Then T ′′ = (0, a, b) forms an edge resolving set. It is routine to verify that1 ≤ a ≤ ⌊n/3⌋ and 2a ≤ b. If b ≤ ⌊(n+ a)/2⌋, then we are done. Otherwise b ≥ ⌊(n+ a)/2⌋+ 1, i.e.,

(4.4) y − z + n ≥ ⌊(n+ y)/2⌋+ 1.

Adding Ineqs. (4.3) and (4.4) yields t ≥ 2⌊t/2⌋+ 2 for t = y + n, which is absurd. �

By Proposition 4.4, it suffices to show that for each (a, b) ∈ Sn, no triad in the set T (a, b) is anedge resolving set. We produce Lemma 4.5, which works for the |T (a, b)| = 8 triads altogether.

Lemma 4.5. Let n ≥ 100 and (a, b) ∈ Sn. If A0 ∩ Aa ∩ Ab 6= ∅, the no triad in the set T (a, b) is anedge resolving set.

Proof. Suppose that A0 ∩Aa ∩Ab 6= ∅. By Lemma 4.1, the vertex triad {u0, ua, ub} is not a resolvingset. By (4.2), any other triad in T (a, b) is not a resolving set. �

Inspired by Lemma 4.5, we are going to determine, as more as possible, the pairs (a, b) ∈ Sn suchthat A0 ∩ Aa ∩Ab = ∅. To do this, we define a set Wn ⊆ Sn by Table 4.2.

Table 4.2. Definition of the set Wn for n ≥ 100, where r′n is the residue of n modulo 6.

r′n Wn |Wn|

0 ∅ 0

1 {(1, ⌊n/2⌋ − 2), (2, ⌊n/2⌋), (5, ⌊n/2⌋+ 3)} 3

2 ∅ 0

3 {(1, 2), (2, 4)} 2

4{

(1, 2), (1, n2 − 4), (1, n2 − 1), (2, n2 − 2), (2, n2 + 1), (4, n

2 − 1), (5, n2 + 1), (8, n2 + 4)}

8

5 {(1, 2), (1, 5), (1, 8), (2, 4), (2, 7), (2, 10), (4, 8), (4, 11), (7, 14)} 9

Now, to prove Proposition 2.2, it suffices to establish Proposition 4.6 and Lemma 4.7.

Proposition 4.6. Let n ≥ 100 and (a, b) ∈ Sn. If A0 ∩ Aa ∩Ab = ∅, then (a, b) ∈ Wn.


Lemma 4.7. Let n ≥ 100 and (a, b) ∈ Wn. For any R ∈ T (a, b), there exists a pair of edges with thesame edge metric representation with respect to R.

We will show Proposition 4.6 and Lemma 4.7 in § 4.1 and 4.2, respectively.

4.1. Proof of Proposition 4.6. Throughout this section, we suppose that n ≥ 100 and (a, b) ∈ Sn.Let J = A0∩Aa∩Ab. We will establish Proposition 4.6 for r′n ∈ {2, 5, 0} by combinatorial arguments,and for r′n ∈ {3, 1, 4} by explicitly constructing an element in J as if (a, b) 6∈ Wn. For any integers mand M , we denote [m,M ] = {m,m+ 1, . . . ,M}. When rm = rM , we define

g(m,M) = {m, m+ 3, m+ 6, . . . , M}.

Lemma 4.8. Let m < M be integers such that rm = rM . Suppose that g(m,M) ⊆ Ax ∩ Ay andz ∈ [m− 6, M + 9]. If Ax ∩ Ay ∩ Az = ∅, then

z ∈ {m− 4, m− 1, m+ 2, M − 2, M + 1, M + 4}.

Proof. Let J = Ax ∩ Ay ∩Az . Then we have the following.

• If rz−m = 0, then t ∈ J , where

t =

z, if z ∈ g(m,M);

z + 6, if z ∈ {m− 3, m− 6};

z − 6, if z ∈ {M + 3, M + 6}.

• If rz−m = 1 and m− 5 ≤ z ≤ M − 5, then z + 5 ∈ J .• If rz−m = 2 and m+ 5 ≤ b ≤ M + 5, then z − 5 ∈ J .

Since J = ∅ by premise, we obtain the desired result. �

To use Lemma 4.8 smoothly, we point out that for any (a, b) ∈ Sn,

(4.5) b ≤n+ a

2≤

2n

3< n− 8.

4.1.1. Proposition 4.6 is true if r′n = 2. Suppose that r′n = 2 and J = ∅. By Lemma 4.1,

(4.6) A0 ∩ Aa ⊇

g(a+ 5, n− 6) ∪ g(a+ 6, n− 5), if ra = 0;

g(a+ 5, n− 5), if ra = 1;

g(a+ 6, n− 6), if ra = 2.

By Lemma 4.8 and Ineq. (4.5), we deduce that b ≤ a+ 8.

For any t, j ∈ Zn, define

Ftj = t+ n/2 + {0, ±1, ±2, . . . , ±j}.

By Lemma 4.1, we obtain Ft2 ⊆ At. If x, y 6∈ Fz4, then |Ax ∩ Fz2| ≥ 3 and |Ay ∩ Fz2| ≥ 3. Since|Fz2| = 5, we deduce that Ax ∩ Ay ∩ Fz2 6= ∅, contradicting the premise J = ∅. Henceforth, we cansuppose that Ft4 ∩ T 6= ∅ for any t ∈ T . Since a ≤ n/3 < n/2 − 4, we deduce that a 6∈ F04. ThusF04 = {b}. From the definition, we obtain

(4.7) n/2 + a− 4 ≤ b ≤ ⌊(n+ a)/2⌋.

Hence n/2+ a− 4 ≤ a+8, contradicting the premise n ≥ 100. This proves Proposition 4.6 for r′n = 2.


4.1.2. Proposition 4.6 is true if r′n = 5. Suppose that r′n = 5 and J = ∅. By Lemma 4.1, we obtainthe relation (4.6). Now we use Lemma 4.8 with the aids of (4.6) and Ineq. (4.5). If ra = 0, then2a ≤ b ≤ a− 2, which is impossible. If a ∈ {1, 2, 4}, it is direct to deduce that (a, b) ∈ Wn. If a ≥ 5and ra = 2, then b ∈ {a+ 5, a+ 8}. It follows that a ∈ J , contradicting the premise J = ∅. If a ≥ 7and ra = 1, then b = a+ 7 and (a, b) = (7, 14) ∈ Wn. This completes the proof of Proposition 4.6 forr′n = 5.

4.1.3. Proposition 4.6 is true if r′n = 0. Suppose that r′n = 0 and J = ∅. Consider the general caseT = (x, y, z) ⊂ Zn and J = Ax ∩ Ay ∩ Az .

First, from Lemma 4.1, we see that Ft1 ⊆ At. If x, y 6∈ Fz4, then

|Ax ∩ Fz1| ≥ 2 and |Ay ∩ Fz1| ≥ 2.

Since |Fz1| = 3, we deduce that Ax ∩ Ay ∩ Fz1 6= ∅, contradicting the premise J = ∅. Henceforth, wecan suppose that Ft4 ∩ T 6= ∅ for any t ∈ T . As § 4.1.1, we can deriv Ineq. (4.7).

Second, we claim that the elements in T form a complete residue class modulo 3. Let

Yj = {i ∈ [ 0, n− 1] : ri = j} and Act = Zn\At.

Suppose that n = 6h for some h ∈ Z. By Lemma 4.1, we find

∣

∣Act ∩ Yj

∣

∣ =

{

2, if j = rt;

h+ 1, otherwise.

If rx = ry, then for j = rx,

|Acx ∩ Yj |+ |Ac

y ∩ Yj |+ |Acz ∩ Yj | ≤ 2 + 2 + (h+ 1) < 2h = |Yj |.

Therefore, J ∩ Yj 6= ∅, contradicting the premise J = ∅. This proves the claim.

Now we are in a position to show that i∗ ∈ J , where i∗ = a + 4 + ra. In fact, by Ineq. (4.7) andthe premise n ≥ 100, one may compute the distances

d(i∗, 0) = a+ 4 + ra, d(i∗, a) = 4 + ra, and d(i∗, b) = b− a− 4− ra.

By Ineq. (4.7) and the claim, we can deduce that

5 ≤ a+ 4+ ra 6≡ 1 (mod 3) =⇒ i∗ ∈ A0,

5 ≤ 4 + ra 6≡ 1 (mod 3) =⇒ i∗ ∈ Aa, and

5 ≤ (n/2− 4)− 4− 2 ≤ b− a− 4− ra ≡ 2 (mod 3) =⇒ i∗ ∈ Ab.

This completes the prooof of Proposition 4.6 for r′n = 0.

4.1.4. Proposition 4.6 is true if r′n ∈ {3, 1, 4}. For the remaining case that r′n ∈ {3, 1, 4}, we explicitlyconstruct an element in J as if (a, b) ∈ Sn\Wn; see Tables 4.3 to 4.5. It is routine to check them byusing Lemma 4.1.

4.2. Proof of Lemma 4.7. This section is to devoted to establish Lemma 4.7. In view of Table 4.2,we need to deal with 3+2+8+9 = 22 cases. We will handle them independently in Propositions 4.10to 4.13 and Tables 4.6 to 4.8.

Lemma 4.9. Let x, y, z be distinct elements in Zn. Suppose that the edges eui−1 and eui have the sameedge metric representation with respect to the vertex triad {ux, uy, vz}. Then these edges have thesame edge metric representation with respect to any one of the following triads:

{ux, vy, vz}, {vx, uy, vz}, {vx, vy, vz}.


Table 4.3. An element in A0 ∩ Aa ∩Ab, where (a, b) ∈ Sn\Wn and n = 6h+ 3 ≥ 100.

ra rb = 0 rb = 1 rb = 2

0 −6 b+ 5

{

3h− 4, if b ∈ {3h− 4, 3h+ 2}

3h− 1, otherwise

1

{

−5, if (a, b) = (1, 3)

b, otherwise−5

{

a+ 5, if b− a > 7

3h+ b− a, otherwise

2 −6

{

a, if a > 2

3h+ 6, otherwise

a, if a > 2

11, if a = 2, b ∈ {5, 11}

8, otherwise


ra rb = 0 rb = 1 rb = 2

0 −5

{

−5, if b ≤ 3h− 5

3h− 7, if b ≥ 3h− 2b

1 −5

−5, if b ≤ 3h− 5

3h+ 4, if b = 3h− 2

b, otherwise

{

3h+ 1, if b ≤ 3h− 4

−5, otherwise

2 b+ 5

{

3h+ 4, if b ≤ 3h− 2

b, otherwise−6


ra rb = 0 rb = 1 rb = 2

0 −5

{

−5, if b ≤ 3h− 5

3h− 7, otherwise−6

1 −5

−5, if b ≤ 3h− 5

b, if b ≥ 3h+ 4

b+ 6, otherwise

{

b+ 3h, if b ≤ 3h− 4

−5, otherwise

2 b + 5

{

3h+ 4, if b ≤ 3h− 2

3h− 4, otherwise−6

Proof. Recall from (4.2) that At ⊆ Bt. Since eui−1 and eui have the same distance from ux, we obtaini ∈ Ax. Thus i ∈ Bx and the edges have the same distance from vx. For the same reason, the edgeshave the same distance from vy. Hence they have the same edge metric representation with respectto each of the desired vertex sets. This completes the proof. �

Throughout this section, we suppose that (a, b) ∈ Wn.


4.2.1. Lemma 4.7 is true if r′n = 1. Suppose that r′n = 1. Then

Wn = {(1, 3h− 2), (2, 3h), (5, 3h+ 3)}

by definition. In Table 4.6, we exhibit a pair of edges with the same edge metric representation withrespect to a triad R in T (a, b), except {u0, va, vb}, {v0, ua, vb}, and {v0, va, vb}. By Lemma 4.9, this isenough to establish Lemma 4.7 for r′n = 1. It is routine to check the truth of Table 4.6 by Theorem 2.1.

Table 4.6. A pair of edges together with the same edge metric representation withrespect to the vertex triad {α0, βa, γb}, where n = 6h+ 1 ≥ 100.

(a, b)

(α, β, γ) (1, 3h− 2) (2, 3h) (5, 3h+ 3)

(u, u, u){

eu3h+3, es3h+3

} {

eu2 , es2

} {

eu5 , es5

}

(h+ 1, h+ 2, 4) (2, 0, h+ 1) (4, 0, h+ 1)

(u, u, v){

eu3h, eu3h+1

} {

eu3h+1, eu3h+2

} {

eu3h+4, eu3h+5

}

(h+ 2, h+ 2, 2) (h+ 2, h+ 2, 2) (h+ 1, h+ 2, 2)

(u, v, u){

eu3h−1, es3h−1

} {

eu3h+4, es3h+5

} {

eu3h+7, es3h+8

}

(h+ 2, h+ 1, 1) (h+ 1, h+ 1, 4) (h, h+ 1, 4)

(v, u, u){

eu3h+2, es3h+3

} {

eu4 , es4

} {

eu7 , es7

}

(h+ 1, h+ 2, 4) (3, 2, h+ 1) (4, 2, h+ 1)

(v, v, u){

eu−3, eu−2

} {

eu−1, eu0

} {

eu2 , eu3

}

(2, 2, h+ 2) (1, 2, h+ 2) (2, 2, h+ 2)

4.2.2. Lemma 4.7 is true for r′n = 3. Suppose that r′n = 3. Then Wn = {(1, 2), (2, 4)} by definition.

Proposition 4.10. Let n ≥ 100. For any triad R ∈ T (1, 2), the edges eu−5 and eu6 have the same edgemetric representation with respect to R.

Proof. By Theorem 2.1, for any t ∈ {0, 1, 2} and any e ∈ {eu−5, eu6}, one may compute that d(ut, e) = 4

and d(vt, e) = 3. �

Proposition 4.11. Let n = 6h+3 ≥ 100 for some h ∈ Z. For any triad R ∈ T (2, 4), the edges eu3h+2

and eu3h+4 have the same edge metric representation with respect to R.

Proof. By Theorem 2.1, for any t ∈ {0, 2, 4} and any e ∈ {eu3h+2, eu3h+4}, one may compute that

d(ut, e) = h+ 2 and d(vt, e) = h+ 1. �

By Propositions 4.10 and 4.11, Lemma 4.7 is true for r′n = 3.

4.2.3. Lemma 4.7 is true for r′n = 4. Suppose that r′n = 4. By definition, we have

Wn = {(1, 2), (1, 3h− 2), (1, 3h+ 1), (2, 3h), (2, 3h+ 3), (4, 3h+ 1), (5, 3h+ 3), (8, 3h+ 6)}.

Proposition 4.12. Let n = 6h+ 4 ≥ 100 for some h ∈ Z. Then we have the following.

(1) For any triad R ∈ T (1, 3h + 1), the edges eu−6 and eu5 have the same edge metric representationwith respect to R.

(2) For any triad R ∈ T (2, 3h + 3), the edges eu−5 and eu6 have the same edge metric representationwith respect to R.


Proof. By Theorem 2.1, we have the following.

(1) For any edge e ∈ {eu−6, eu5},

d(u0, e) = d(u1, e) = 4, d(u3h+1, e) = h+ 1, d(v0, e) = d(v1, e) = 3, and d(v3h+1, e) = h.

(2) For any edge e ∈ {eu−5, eu6},

d(u0, e) = d(u2, e) = 4, d(u3h+3, e) = h+ 1, d(v0, e) = d(v2, e) = 3, and d(v3h+3, e) = h.


In view of Propositions 4.10 and 4.12, we need to consider another 5 pairs (a, b) ∈ Wn. ByTheorem 2.1, it is routine to check Table 4.7. By Lemma 4.9, we completes the proof of Lemma 4.7

Table 4.7. A pair of edges together with the same edge metric representation withrespect to the triad {α0, βa, γb}, when n = 6h+ 4 ≥ 100.

(a, b)

(α, β, γ) (1, 3h− 2) (2, 3h) (4, 3h+ 1) (5, 3h+ 3) (8, 3h+ 6)

(u, u, u){

eu3h+3, es3h+3

} {

eu3h+5, es3h+5

} {

eu5 , es5

} {

eu3h+8, es3h+8

} {

eu3h+11, es3h+11

}

(h+ 2, h+ 3, 4) (h+ 2, h+ 2, 4) (4, 1, h+ 1) (h+ 1, h+ 2, 4) (h, h+ 2, 4)

(u, u, v){

eu3h−1, eu3h

} {

eu3h+1, eu3h+2

} {

eu3h+3, eu3h+4

} {

eu3h+4, eu3h+5

} {

eu3h+7, eu3h+8

}

(h+ 2, h+ 2, 2) (h+ 3, h+ 2, 2) (h+ 2, h+ 2, 2) (h+ 2, h+ 2, 2) (h+ 1, h+ 2, 2)

(u, v, u){

eu3h−1, es3h−1

} {

eu3h+5, es3h+5

} {

eu5 , es5

} {

eu3h+8, es3h+8

} {

eu3h+11, es3h+11

}

(h+ 2, h+ 1, 1) (h+ 2, h+ 1, 4) (4, 2, h+ 1) (h+ 1, h+ 1, 4) (h, h+ 1, 4)

(v, u, u){

eu3h+3, es3h+3

} {

eu4 , es4

} {

eu3h+3, es3h+3

} {

eu7 , es7

} {

eu10, es10

}

(h+ 1, h+ 3, 4) (3, 2, h+ 1) (h+ 1, h+ 2, 2) (4, 2, h+ 1) (5, 2, h+ 1)

(v, v, u){

eu−3, eu−2

} {

eu−1, eu0

} {

eu1 , eu2

} {

eu2 , eu3

} {

eu5 , eu6

}

(2, 2, h+ 2) (1, 2, h+ 2) (2, 2, h+ 2) (2, 2, h+ 2) (3, 2, h+ 2)

for r′n = 4.

4.2.4. Lemma 4.7 is true for r′n = 5. Suppose that r′n = 5. By definition, we have

Wn = {(1, 2), (1, 5), (1, 8), (2, 4), (2, 7), (2, 10), (4, 8), (4, 11), (7, 14)}.

Proposition 4.13. Let n = 6h+ 5 ≥ 100 for some h ∈ Z. Then we have the following.

(1) For any set R ∈ T (4, 8) ∪ T (4, 11), the edges ev3h+4 and ev3h+6 have the same edge metric repre-sentation with respect to R.

(2) For any set R ∈ T (7, 14), the edges ev3h+7 and ev3h+9 have the same edge metric representationwith respect to R.

Proof. By Theorem 2.1, we have the following.

(1) For any edge e ∈ {ev3h+4, ev3h+6},

d(u0, e) = d(u4, e) = d(u8, e) = d(v11, e) = h+ 1,

d(u11, e) = d(v4, e) = h and

d(v0, e) = d(v8, e) = h+ 2.


(2) For any edge e ∈ {ev3h+7, ev3h+9},

d(u0, e) = d(u14, e) = d(v7, e) = h and d(u7, e) = d(v0, e) = d(v14, e) = h+ 1.


In view of Propositions 4.10 and 4.13, we need to consider another 5 pairs (a, b) ∈ Wn. ByTheorem 2.1, it is routine to check Table 4.8.

Table 4.8. A pair of edges together with the same edge metric representation withrespect to the triad {α0, βa, γb}, when n = 6h+ 5 ≥ 100.

(α, β, γ) (a, b) = (1, 5) (a, b) = (1, 8) (a, b) = (2, 4)

(u, u, u){

eu1 , es1

} {

eu1 , es1

} {

ev3h+2, ev3h+4

}

(1, 0, 3) (1, 0, 4) (h+ 1, h+ 1, h+ 1)

(u, u, v){

eu3h+5, es3h+3

} {

eu3h+5, es3h+3

} {

eu3h+7, es3h+5

}

(h+ 2, h+ 2, h+ 1) (h+ 2, h+ 2, h) (h+ 1, h+ 2, h+ 2)

(u, v, u){

eu2 , es2

} {

eu5 , es5

} {

ev3h+2, ev3h+4

}

(2, 2, 2) (4, 3, 2) (h+ 1, h, h+ 1)

(u, v, v){

eu3h+5, es3h+3

} {

eu3h+5, es3h+3

} {

eu3h+7, es3h+5

}

(h+ 2, h+ 1, h+ 1) (h+ 2, h+ 1, h) (h+ 1, h+ 1, h+ 2)

(v, u, u){

eu1 , es1

} {

eu1 , es1

} {

eu4 , es4

}

(2, 0, 3) (2, 0, 4) (3, 2, 0)

(v, u, v){

eu3h+5, es3h+3

} {

eu3h+5, es3h+3

} {

eu3h+7, es3h+5

}

(h+ 1, h+ 2, h+ 1) (h+ 1, h+ 2, h) (h, h+ 2, h+ 2)

(v, v, u){

eu3h+4, es3h+7

} {

eu3h+7, es3h+10

} {

eu3h+4, es3h+3

}

(h+ 1, h+ 2, h+ 2) (h, h+ 1, h+ 2) (h+ 1, h+ 2, h+ 2)

(v, v, v){

eu3h+5, es3h+3

} {

eu3h+5, es3h+3

} {

eu3h+7, es3h+5

}

(h+ 1, h+ 1, h+ 1) (h+ 1, h+ 1, h) (h, h+ 1, h+ 2)

(α, β, γ) (a, b) = (2, 7) (a, b) = (2, 10)

(u, u, u){

ev3h+2, ev3h+4

} {

ev3h+2, ev3h+4

}

(h+ 1, h+ 1, h) (h+ 1, h+ 1, h− 1)

(u, u, v){

eu3h+7, es3h+5

} {

eu3h+7, es3h+5

}

(h+ 1, h+ 2, h+ 1) (h+ 1, h+ 2, h)

(u, v, u){

ev3h+2, ev3h+4

} {

ev3h+2, ev3h+4

}

(h+ 1, h, h) (h+ 1, h, h− 1)

(u, v, v){

eu3h+7, es3h+5

} {

eu3h+7, es3h+5

}

(h+ 1, h+ 1, h+ 1) (h+ 1, h+ 1, h)

(v, u, u){

eu1 , eu2

} {

eu1 , eu2

}

(2, 0, 4) (2, 0, 5)

By Lemma 4.9, we completes the proof of Lemma 4.7 for r′n = 5.

Now, by Proposition 4.6 and Lemma 4.7, we obtain Proposition 2.2.


5. Proof of Proposition 2.3

This section is devoted to proving that the edge dimension of the graph P (n, 3) is at most 4. Weproceed according to the residue r′n.

5.1. Proposition 2.3 is true if rn = 0. By definition, r = 1 if ri = 1, and r = 0 otherwise. First ofall, the edges whose distances from u0 are less than 3 are distinguishable; see Table 5.1 for their edgemetric representations with respect to the tetrad {u0, u1, v2, u⌊n/2⌋−1}, where an asterisk ∗ entrymeans that it is unnecessary to be computed for the purpose of distinguishing the edge indicated bythe row the entry lies in.

Table 5.1. The edge metric representations of edges eli whose distances from u0 areless than 3, when n = 3qn.

l i d(

u0, eli

)

d(

u1, eli

)

d(

v2, eli

)

d(

u⌊n/2⌋−1, eli

)

u 0 0 0 ∗ ∗1 1 0 1 ∗2 2 1 1 ∗−3 2 3 3 ⌊qn/2⌋+ 2−2 1 2 2 ∗−1 0 1 2 ∗

s 0 0 1 3 ∗1 1 0 2 ∗2 2 1 0 ∗3 2 2 2 ∗−3 2 3 4 ⌊qn/2⌋+ 2−2 2 2 3 ∗−1 1 2 1 ∗

v 0 1 2 3 ∗1 2 1 3 ⌈qn/2⌉3 2 3 3 ≤ ⌊qn/2⌋−6 2 3 4 ⌊qn/2⌋−4 2 3 1 ∗−3 1 2 4 ∗−2 2 1 3 ⌈qn/2⌉+ 1−1 2 2 0 ∗

While most entries in Table 5.1 can be obtained directly from Theorem 2.1, we explain the u⌊n/2⌋−1-coordinate of the row ev3, which is derived in the following way. Consider

i = ⌈n/2⌉+ 4 =

{

3h+ 4, if n = 6h;

3h+ 6, if n = 6h+ 3.

It follows that

(qi, ri, r) =

{

(h+ 1, 1, 1), if n = 6h;

(h+ 2, 0, 0), if n = 6h+ 3.

Therefore,

d(

u⌊n/2⌋−1, ev3

)

= d(

u0, ev⌈n/2⌉+4

)

= min(qi + ri + 1, qn − qi + r) =

{

h, if n = 6h

h− 1, if n = 6h+ 3

is at most ⌊qn/2⌋.


Table 5.2. The edges eli whose distances from u0 are d, with the corresponding lowerbounds of qn and d, when n = 3qn.

l i lower bound of qn lower bound of d

u 3d− 6 2d− 4 33d− 7 2d− 4 43d− 8 2d− 5 45− 3d 2d− 4 36− 3d 2d− 4 47− 3d 2d− 5 4

s ±(3d− 3) 2d− 2 3±(3d− 5) 2d− 4 3±(3d− 7) 2d− 4 4

v 3d− 3 2d− 1 33d− 5 2d− 3 33d− 7 2d− 3 3−3d 2d− 1 32− 3d 2d− 3 34− 3d 2d− 3 3

Let d ≥ 3 be an integer. By using Theorem 2.1, one may solve out the set of edges whose distancefrom u0 is d, under some lower bound conditions on qn and d; see Table 5.2. The lower bound of dworks for all integers n ≥ 18, since it is derived by requiring that distance from the edge eli to u0

is computed by using the second expression in each of the first three formulas in Theorem 2.1. Itremains to show that the edge metric representations are distinct.

From Table 5.2, we see that the minimum lower bound of qn is 2d − 5, which is attained by theedges eu3d−8 and eu7−3d. In fact, these two edges coincide with each other, because

(3d− 8)− (7− 3d) = 6d− 15 = 3(2d− 5) = 3qn = n.

Using the same idea, we see 4 edges for which the lower bound of qn is 2d− 4:

eu3d−6 = eu6−3d, eu3d−7 = eu5−3d, es3d−5 = es7−3d, and es3d−7 = es5−3d.

They have distances d− 1, d− 2, d, and d− 3 from v2 respectively, and thus distinguishable.

Below we can suppose that qn ≥ 2d−3. By Theorem 2.1, we compute the metric representations ofedges in Table 5.2 with respect to the remaining vertex triple (u1, v2, u⌊n/2⌋−1); see Table 5.3, whereIeven(qn) equals 1 if qn is odd and 0 if qn is even. Note that

3d− 5 ≡ 4− 3d (mod n) when qn = 2d− 3,

3d− 7 ≡ 2− 3d (mod n) when qn = 2d− 3,

3d− 3 ≡ 3− 3d (mod n) when qn = 2d− 2,

3d− 3 ≡ −3d (mod n) when qn = 2d− 1.

From Table 5.3, we see that every edge is recognizable, as desired.

5.2. Proposition 2.3 is true if r′n = 1. By definition, r = |1− ri|. Let d ≥ 0 be an integer. Ford ≤ 2, all entries except the non-asterisks in the last column in Table 5.1 keep invariant. The non-asterisks are listed in Table 5.4, From which we see that the edges whose distance d from u0 is at most2 are distinguishable by the quadruple (u0, u1, v2, u⌊n/2⌋−1).


Table 5.3. The metric representations of the edges whose distances from u0 are thesame number d, when n = 3qn ≥ 3(2d− 3).

l i qn d(u1, eli) d(v2, e

li) d(u⌊n/2⌋−1, e

li)

u 3d− 6 ≥ 2d− 3 2 (d = 3) d− 1 ≤ ⌊qn/2⌋ − d+ 4d (d ≥ 4)

3d− 7 2d− 3 d d− 2 02d− 2 2≥ 2d− 1 ∗

3d− 8 ≥ 2d− 3 d− 1 d− 2 ∗5− 3d 2d− 3 d+ 1 d− 1 ∗

≥ 2d− 2 d ⌈qn/2⌉ − d+ 46− 3d ≥ 2d− 3 d d ∗7− 3d ≥ 2d− 3 d d− 1 ⌈qn/2⌉ − d+ 5

s 3d− 3 ≥ 2d− 2 d+ 1 d ≤ ⌊qn/2⌋ − d+ 33d− 5 ≥ 2d− 3 d− 1 d ≤ ⌈qn/2⌉ − d+ 33d− 7 2d− 3 d− 1 d− 3 1

≥ 2d− 2 ⌈qn/2⌉ − d+ 33− 3d 2d− 2 d+ 1 d ∗

2d− 1 d+ 1 2≥ 2d d+ 2 ⌊qn/2⌋ − d+ 3

5− 3d 2d− 3 d d− 2 2≥ 2d− 2 d+ 1 d− 1 ⌊qn/2⌋ − d+ 3 + 2Ieven(qn)

7− 3d ≥ 2d− 3 d− 1 d ⌊qn/2⌋ − d+ 5

v 3d− 3 2d− 1 d+ 1 d+ 1 1≥ 2d ∗

3d− 5 2d− 3 d− 1 d+ 1 ∗2d− 2 2≥ 2d− 1 ⌈qn/2⌉ − d+ 2

3d− 7 2d− 3 d− 1 d− 3 2≥ 2d− 2 ⌈qn/2⌉ − d+ 2

−3d 2d− 1 d+ 1 d+ 1 ∗2d d+ 2 2≥ 2d+ 1 d+ 3 ∗

2− 3d 2d− 3 d− 1 d− 3 ∗2d− 2 d d− 2 1≥ 2d− 1 d+ 1 d− 1 ⌊qn/2⌋ − d+ 2 + 2Ieven(qn)

4− 3d ≥ 2d− 3 d− 1 d+ 1 ⌊qn/2⌋ − d+ 4

Table 5.4. The edge metric representations for d ≤ 2 that contains the vertexu⌊n/2⌋−1 in an edge resolving set, when n = 6h+ 1 ≥ 19.

l i d(

u0, eli

)

d(

u1, eli

)

d(

v2, eli

)

d(

u⌊n/2⌋−1, eli

)

u −3 2 3 3 h+ 2s −3 2 3 4 h+ 2v 1 2 1 3 h

3 2 3 3 h−6 2 3 4 h+ 1−2 2 1 3 h+ 1


Table 5.5. The metric representations of the edges eli whose distances from u0 ared, when n = 6h+ 1.

l i d d(

u1, eli

)

d(

v2, eli

)

d(

u⌊n/2⌋−1, eli

)

u 3d− 6 h+ 2 d (d ≥ 4) d− 1 h− d+ 3h+ 1 h− d+ 2[3, h] 2 (d = 3) h− d+ 4

d (d ≥ 4)3d− 7 [4, h+ 2] d d− 2 ∗3d− 8 h+ 2 d− 1 d− 2 h− d+ 2

[4, h+ 1] ∗5− 3d h+ 2 d d− 1 ∗

[3, h+ 1] d+ 1 d h− d+ 56− 3d h+ 2 d d− 1 h− d+ 4

[4, h+ 1] d ∗7− 3d [4, h+ 2] d d− 1 h− d+ 5

s 3d− 3 [3, h+ 1] d+ 1 d ≤ h− d+ 33d− 5 [3, h+ 1] d− 1 d ≤ h− d+ 33d− 7 h+ 2 d− 1 d− 3 h− d+ 2

[4, h+ 1] h− d+ 33− 3d h+ 1 d d+ 1 ∗

[3, h] d+ 1 d+ 2 ∗5− 3d [3, h+ 1] d+ 1 d− 1 h− d+ 47− 3d h+ 2 d− 1 d− 2 h− d+ 4

[4, h+ 1] d h− d+ 4

v 3d− 3 h+ 1 d+ 1 d− 1 ∗[3, h] d+ 1 ∗

3d− 5 h+ 1 d− 1 d+ 1 h− d+ 3[3, h] h− d+ 2

3d− 7 h+ 2 d− 1 d− 3 h− d+ 3[3, h+ 1] h− d+ 2

−3d h+ 1 d− 1 d+ 1 ∗[3, h] d+ 1 d+ 3 ∗

2− 3d [3, h+ 1] d+ 1 d− 1 h− d+ 34− 3d h+ 2 d− 1 d− 3 ∗

h+ 1 d− 1 ∗[3, h] d+ 1 h− d+ 3

Let d ≥ 3. Note that

3d− 3 ≡ 2− 3d and 3d− 5 ≡ −3d (mod n) when d = h+ 1.

We list the edges whose distance from u0 is d, with their metric representations with respect tothe remaining vertex triple (u1, v2, u⌊n/2⌋−1) as in Table 5.5, From which we see that all edges aredistinguishable by the tetrad {u0, u1, v2, u⌊n/2⌋−1}, as desired.

5.3. Proposition 2.3 is true if r′n = 2. From definition, r = |1− ri|. For d ≤ 2, we compute theedge metric representations as in Table 5.6, in which the columns for d(u0, e

li) and d(u1, e

li) are same

to those in Table 5.1. Let d ≥ 3. Note that

3d− 5 ≡ 5− 3d and 3d− 7 ≡ 3− 3d (mod n) when d = h+ 2.


Table 5.6. The metric representations of edges eli whose distances from u0 are atmost 2, when n = 6h+ 2.

l i d(

u0, eli

)

d(

u1, eli

)

d(

un/2−3, eli

)

d(

vn/2−2, eli

)

u 0 0 0 ∗ ∗1 1 0 h+ 1 ∗2 2 1 h+ 1 h−3 2 3 h+ 2 h+ 1−2 1 2 h+ 2 h+ 1−1 0 1 h+ 2 ∗

s 0 0 1 h+ 1 ∗1 1 0 h ∗2 2 1 h+ 1 h− 13 2 2 h ∗−3 2 3 h+ 2 h−2 2 2 h+ 1 h+ 2−1 1 2 h+ 2 h

v 0 1 2 h ∗1 2 1 h− 1 ∗3 2 3 h− 1 ∗−6 2 3 h+ 1 h− 1−4 2 3 h+ 1 h−3 1 2 h+ 1 ∗−2 2 1 h ∗−1 2 2 h+ 1 h− 1

We list the edges whose distance from u0 is d, with their metric representations with respect to theremaining vertex triple in Table 5.7, from which we see that all edges are distinguishable, as desired.

5.4. Proposition 2.3 is true if r′n = 4. By definition, r = |1− ri|. For d ≤ 2, all entries except thenon-asterisks in the last column in Table 5.1 keep invariant. The non-asterisks are listed in Table 5.8,from which we see that the edges whose disctances from u0 are at most 2 are distinguishable by(u0, u1, v2, un/2+3).

Let d ≥ 3. Note that

3d− 7 ≡ 7− 3d and 3d− 8 ≡ 6− 3d (mod n) when d = h+ 3,

3d− 3 ≡ 5− 3d and 3d− 5 ≡ 3− 3d (mod n) when d = h+ 2.

We list the edges whose distance from u0 is d, with their metric representations with respect to theremaining vertex triple as in Table 5.9, from which we see that all edges are distinguishable, as desired.

5.5. Proposition 2.3 is true if r′n = 5. Note that in this case the parameter r = |1− ri|. For d ≤ 2,we compute the edge metric representations as in Table 5.10. All entries for the first four columns inTable 5.1 keep invariant.

Let d ≥ 3. We note that d = h+ 3 happens only when l = u. Since n = 6h+ 5 = 6d− 13, amongthe 18 edges there are only three distinct edges have the same distance d from u0, that is,

eu3d−6 = eu7−3d, eu3d−7 = eu6−3d, and eu3d−8 = eu5−3d.

Their distances from u(n−3)/2 are respectively 2, 1, and 0. Thus these three edges are distinguishable.For d ≤ h+ 2, we compute out Table 5.11, from which we see that all edges are distinguishable.



l i d d(

u1, eli

)

d(

un/2−3, eli

)

d(

vn/2−2, eli

)

u 3d− 6 h+ 2 d (d ≥ 4) h− d+ 4 ∗h+ 1 h− d+ 1 ∗[3, h] 2 (d = 3) h− d+ 3 ∗

d (d ≥ 4) ∗3d− 7 h+ 2 d h− d+ 3 ∗

h+ 1 h− d+ 2 ∗[4, h] h− d+ 4 ∗

3d− 8 h+ 2 d− 1 h− d+ 2 ∗h+ 1 h− d+ 3 h− d+ 3[4, h] h− d+ 4

5− 3d [3, h+ 2] ≥ d h− d+ 5 ∗6− 3d [4, h+ 2] d h− d+ 6 h− d+ 47− 3d [4, h+ 2] d h− d+ 6 h− d+ 5

s 3d− 3 h+ 2 d− 1 h− d+ 6 h− d+ 3h+ 1 d+ 1 h− d+ 3 ∗h h− d+ 1 ∗[3, h− 1] h− d+ 2 ∗

3d− 5 h+ 2 d− 1 h− d+ 4 ∗h+ 1 h− d+ 1 ∗[3, h] h− d+ 2 ∗

3d− 7 h+ 2 d− 1 h− d+ 3 ∗h+ 1 h− d+ 3 h− d+ 2[4, h] h− d+ 4 h− d+ 2

3− 3d h+ 2 d− 1 h− d+ 3 ∗[3, h+ 1] d+ 1 h− d+ 4 h− d+ 2

5− 3d h+ 2 d− 1 h− d+ 4 ∗[3, h+ 1] d+ 1 h− d+ 4 (d = 3)

h− d+ 5 (d ≥ 4)7− 3d [4, h+ 2] d− 1 h− d+ 6 h− d+ 5

v 3d− 3 h+ 1 d h− d+ 4 ∗h d+ 1 h− d+ 2 ∗[3, h− 1] h− d+ 1 ∗

3d− 5 h+ 1 d− 1 h− d+ 2 ∗[3, h] h− d+ 1 ∗

3d− 7 [3, h+ 1] d− 1 h− d+ 3 h− d+ 1−3d h+ 1 d h− d+ 3 h− d+ 1

[3, h] d+ 12− 3d h+ 1 d h− d+ 3 ≥ h− d+ 4

[3, h] d+ 14− 3d [3, h+ 1] d− 1 h− d+ 4 (d = 3) ∗

h− d+ 5 (d ≥ 4) ∗

Now, we obtain a complete proof of Proposition 2.3. Recall that Theorem 1.1 is true for n < 100.By Propositions 2.2 and 2.3, we complete the proof of Theorem 1.1.


Table 5.8. The metric representations for edges with d ≤ 2 that need the vertexun/2+3 as a resolving set, when n = 6h+ 4.

l i d(u0, eli) d(u1, e

li) d(v2, e

li) d(un/2+3, e

li)

u −3 2 3 3 h+ 1s −3 2 3 4 h+ 1v 1 2 1 3 h+ 1

3 2 3 3 h+ 2−6 2 3 4 h−2 2 1 3 h


l i d d(u1, eli) d(v2, e

li) d(un/2+3, e

li)

u 3d− 6 [3, h+ 2] 2 (d = 3) d− 1 h− d+ 6d (d ≥ 4)

3d− 7 [4, h+ 3] d d− 2 ∗3d− 8 [4, h+ 3] d− 1 d− 2 h− d+ 65− 3d h+ 2 d+ 1 d h− d+ 3

h+ 1 h− d+ 2[3, h] h− d+ 4

6− 3d h+ 3 d− 1 d− 2 ∗[4, h+ 2] d d ∗

7− 3d h+ 3 d d− 2 ∗h+ 2 d− 1 h− d+ 2[4, h+ 1] h− d+ 4

s 3d− 3 h+ 2 d+ 1 d− 1 ∗[3, h+ 1] d h− d+ 5

3d− 5 [3, h+ 2] d− 1 d h− d+ 53d− 7 [4, h+ 3] d− 1 d− 3 h− d+ 53− 3d h+ 2 d− 1 d ∗

[3, h+ 1] d+ 1 d+ 2 ∗5− 3d h+ 2 d+ 1 d− 1 ∗

h+ 1 h− d+ 2[3, h] h− d+ 3

7− 3d h+ 3 d− 1 d− 3 ∗h+ 2 d− 1 ∗[4, h+ 1] d h− d+ 3

v 3d− 3 h+ 1 d+ 1 d h− d+ 4[3, h] d+ 1 ∗

3d− 5 [3, h+ 1] d− 1 d+ 1 h− d+ 43d− 7 [3, h+ 2] d− 1 d− 3 h− d+ 4−3d h+ 1 d d+ 2 ∗

[3, h] d+ 1 d+ 3 ∗2− 3d h+ 1 d+ 1 d− 1 h− d+ 3

[3, h] h− d+ 24− 3d h+ 2 d− 1 d− 2 h− d+ 3

h+ 1 d h− d+ 2[4, h] d+ 1 h− d+ 2


Table 5.10. The metric representations of edges eli whose distances from u0 are atmost 2, when n = 6h+ 5.

l i d(u0, eli) d(u1, e

li) d(u(n−3)/2, e

li) d(v(n−1)/2, e

li)

u 0 0 0 ∗ ∗1 1 0 h+ 2 ∗2 2 1 h+ 2 h+ 1−3 2 3 h+ 3 ∗−2 1 2 h+ 3 ∗−1 0 1 h+ 3 ∗

s 0 0 1 h+ 2 ∗1 1 0 h+ 1 ∗2 2 1 h+ 2 h3 2 2 h+ 1 ∗−3 2 3 h+ 2 ∗−2 2 2 h+ 2 h+ 2−1 1 2 h+ 2 h+ 1

v 0 1 2 h+ 1 ∗1 2 1 h ∗3 2 3 h ∗−6 2 3 h+ 1 h− 1−4 2 3 h+ 1 h+ 1−3 1 2 h+ 2 h−2 2 1 h+ 1 ∗−1 2 2 h+ 2 h

6. Concluding remarks

The graphs P (7, 3) and P (7, 2) are isomorphic to each other. By a result of Filipovic et al. [14], weknow that P (7, 2) has edge dimension 4. It is routine to check that the Mobius-Kantor graph P (8, 3)has edge dimensions 4, and both the graphs P (9, 3) and P (10, 3) have edge dimension 3. We think ita challenge to find out the edge dimension of P (n, k) for k ≥ 4.

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Table 5.11. The metric representations of the edges eli whose distance from u0 ared, when n = 6h+ 5 and d ≤ h+ 2.

l i d d(u1, eli) d(u(n−3)/2, e

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[4, h+ 1] h− d+ 55− 3d [3, h+ 2] d+ 1 h− d+ 5 ∗6− 3d [4, h+ 2] d h− d+ 6 h− d+ 47− 3d [4, h+ 2] d h− d+ 6 h− d+ 5

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†School of Mathematics and Statistics, Beijing Institute of Technology, 102488 Beijing, P. R. China,‡Beijing Key Laboratory on MCAACI, Beijing Institute of Technology, 102488 Beijing, P. R. China

E-mail address: [email protected]

School of Mathematics and Statistics, Beijing Institute of Technology, 102488 Beijing, P. R. China


School of Mathematics and Statistics, Beijing Institute of Technology, 102488 Beijing, P. R. China


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