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Research article Special Issues

Examining the rare disease assumption used to justify HWE testing with control samples

  • Many statistical methods for analyzing genetic data, such as those used in genome-wide association studies, assume Hardy-Weinberg Equilibrium (HWE). Therefore, to use such methods, one must check whether the HWE assumption is valid. For a case-control study, researchers have recognized that Hardy Weinberg proportions will be distorted if the marker being tested happens to be associated with the disease. To alleviate this problem, many studies carry out HWE testing on controls only. A number of papers in the literature have justified this practice by making the rare disease assumption without providing rigorous theoretical basis for this justification. Even though many of the diseases studied today are common, whether it is justifiable to use controls to test for HWE when the disease is indeed rare remains an outstanding issue. In this study, we address the rare disease assumption as well as potential problems associated with testing for HWE using controls only, regardless of the prevalence of the disease. We carried out theoretical derivations and numerical studies; the latter were performed using simulated genotypes as well as data from the 1000 Genomes Project. The results from our study are striking: the type Ⅰ error can be severely inflated, regardless of whether the disease being investigated is rare or common. This study shows that, based on the common practice of using controls only to test for HWE, many genetic variants will be discarded erroneously, wasting valuable information and hindering the ability to detect disease-associated variants.

    Citation: Virginia L. Ma, Shili Lin. Examining the rare disease assumption used to justify HWE testing with control samples[J]. Mathematical Biosciences and Engineering, 2020, 17(1): 73-91. doi: 10.3934/mbe.2020004

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  • Many statistical methods for analyzing genetic data, such as those used in genome-wide association studies, assume Hardy-Weinberg Equilibrium (HWE). Therefore, to use such methods, one must check whether the HWE assumption is valid. For a case-control study, researchers have recognized that Hardy Weinberg proportions will be distorted if the marker being tested happens to be associated with the disease. To alleviate this problem, many studies carry out HWE testing on controls only. A number of papers in the literature have justified this practice by making the rare disease assumption without providing rigorous theoretical basis for this justification. Even though many of the diseases studied today are common, whether it is justifiable to use controls to test for HWE when the disease is indeed rare remains an outstanding issue. In this study, we address the rare disease assumption as well as potential problems associated with testing for HWE using controls only, regardless of the prevalence of the disease. We carried out theoretical derivations and numerical studies; the latter were performed using simulated genotypes as well as data from the 1000 Genomes Project. The results from our study are striking: the type Ⅰ error can be severely inflated, regardless of whether the disease being investigated is rare or common. This study shows that, based on the common practice of using controls only to test for HWE, many genetic variants will be discarded erroneously, wasting valuable information and hindering the ability to detect disease-associated variants.


    Let G be a simple, non-trivial connected graph with vertex set V(G). For any two distinct vertices u and v in G, u-v geodesic is a shortest walk between u and v without repetition of vertices. Two vertices are said to be adjacent if there is an edge between them, and they are also called neighbors of each other. The collection of all the neighbors of a vertex v in G is called the (open) neighborhood of v, denoted by N(v).

    A vertex v of G distinguishes a pair (x,y) of distinct vertices of G, if the number of edges in v-x geodesic is different from the number of edge in v-y geodesic. If (x,y) is a pair of neighbors in G, then v is said to be adjacently distinguish the pair (x,y). Equivalently, a vertex v adjacently distinguishes a pair (x,y) of two neighbors if the difference between the number of edges in v-x geodesic and the number of edges in v-y geodesic is one.

    A set DV(G) is a distinguishing set (metric generator) for G if the members of D distinguish every pair of distinct vertices in G. The cardinality of a smallest distinguishing set for G is called the metric dimension of G, denoted by dim(G) [7,23]. The concept of distinguishing set was introduced, very firstly, by Blumenthal [5] in the general context of metric spaces. It was later rediscovered and studied, in the context of graphs, by Slater with the name locating set/reference set [23]. Independently, Harary and Melter studied distinguishing set as resolving set (metric generator) [7,20]. Applications of this notion to the navigation of robots in networks are discussed in [13,21], and applications to pharmaceutical chemistry in [10,11]. For more details about the theory and applications of this notion, we refer the readers to the papers cited in [3,5,8,9,12,13,14,15,19,22] and the references therein.

    A set AV(G) is a neighbor-distinguishing set (local metric generator) for G if the members of A adjacently distinguish every pair of neighboring (adjacent) vertices in G. The cardinality of a smallest neighbor-distinguishing set for G is called the adjacency (local) metric dimension of G, and we denote it by dima(G).

    The problem of distinguishing every two neighbors with the aid of distance (the number of edges in a geodesic) in a connected graph was introduced and studied by Okamoto et al. in 2010 [16]. Then, up till now, this notion endlessly received remarkable interest of many researchers working with distance in graphs. In 2015 and 2018, every two neighbors in the corona product of graphs are distinguished [6,18], while this problem for strong product and lexicographic product of graphs was solved in 2016 [4] and in 2018 [2,6], respectively. Using the neighbor-distinguishing problem of primary subgraphs, this problem was solved for the super graphs of these subgraphs in 2015 [17]. In 2018, Salman et al. proposed linear programming formulation for this problem and distinguished neighbors in two families of convex polytopes [19]. Recently, in 2019, split graphs of complete and complete bipartite graphs have been considered in the context of this problem [1]. Due to this noteworthy attention of researchers to this problem, we extend this study towards a very renowned family of generalized Petersen graphs in this article. Next, we state two results, proved by Okamoto et al. [16], and Salman et al. [19], respectively, which will be used in the sequel.

    Theorem 1. [16] Let G be a non-trivial connected graph of order n. Then dima(G) =n1 if and only if G is a complete graph, and dima(G)=1 if and only if G is a bipartite graph.

    Proposition 2. [19] A subset A of vertices in a connected graph G is a neighbor-distinguishing set for G if and only if for every uV(G) and for each vN(u), the pair (u,v) adjacently distinguished by some element of A.

    Watkins, in 1969 [24], generalized the eminent Petersen graph, and proposed the notation P(n,m) to this generalized family, where n3 and 1mn12. P(n,m) is a cubic graph having the set

    V(P(n,m))={u1,u2,,un,v1,v2,,vn}

    as the vertex set, and the set

    E(P(n,m))=ni=1{uix,viy;xN(ui),yN(vi)}

    as the edge set, where N(ui)={ui+1,ui1,vi} and N(vi)={ui,vi+m,vim} for each 1in, and the indices greater than n or less than 1 will be taken modulo n. Vertices ui and vi (1in) are called the outer vertices and inner vertices, respectively, in P(n,m). Figure 1 depicts graphs to two different families of generalized Petersen graphs.

    Figure 1.  Generalized Petersen graphs P(15,4) and P(16,7).

    The rest of the paper is divided into two sections: one is on the family of generalized Petersen graphs P(n,4); and the second is on the family of generalized Petersen graphs P(2n,n1). These families have been considered in the context of metric dimension problem by Naz et al. [15] and Ahmad et al. [3], respectively. Here, we solve the neighbor-distinguishing problem for these families.

    In the next result, we show that only two vertices of P(n,4) perform the neighbor-distinguishing.

    Theorem 3. For n9, let G be a generalized Petersen graph P(n,4), then a neighbor-distinguishing set for G is a 2-element subset of V(G).

    Proof. For n=9, it is an easy exercise to see that the set A={v1,v2} is a neighbor-distinguishing set for G. For n10, let A be a 2-element subset of V(G). Then, according to Proposition 2, we would perform neighbor-distinguishing for each pair (x,y), where xV(G) and yN(x). Note that, if xA, then (x,y) is adjacently distinguished, because the number of edges in yx geodesic is 1, while the number of edges in xx geodesic is 0. Now, we discuss the following eight cases:

    Case 1: (n=8k with k2)

    Let A={v1=a1,v3=a2}, then

    ● the number of edges in u1a2 geodesic is 3,

    ● the number of edges in u2a2 geodesic is 2,

    ● the number of edges in v1a2 geodesic is 4,

    ● the number of edges in v2a2 geodesic is 3.

    Further, Tables 1 and 2 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 1.  List of the number of edges in geodesics from the outer vertices to aA in P(n,4).
    GeodesicThe number of edges in the geodesic
    uiai0 (mod 4)i1 (mod 4)i2 (mod 4)i3 (mod 4)
    n=8k with k2, and A={v1=a1,v3=a2}
    uia1, 1i4k+1i+84i+34i+64i+94
    uia1, 4k+2inni+84ni+54ni+104ni+114
    uia2, 3i4k+3i+44i+74i+64i+14
    uia2, 4k+4inni+124ni+134ni+104ni+74
    n=8k+1 with k2, and A={v1=a1,v4=a2}
    uia1, 1i4k+1i+84i+34i+64i+94
    uia1, 4k+2inni+114ni+84ni+54ni+104
    uia2, 4i5k+1i4i+34i+64i+54
    uia2, 5k+2inni+114ni+84ni+134ni+144
    n=8k+2 with k1, and A={v1=a1,v2=a2}
    uia1, 1i4k+2i+84i+34i+64i+94
    uia1, 4k+3inni+104ni+114ni+84ni+54
    uia2, 2i4k+3i+84i+74i+24i+54
    uia2, 4k+4inni+64ni+114ni+124ni+94
    n=8k+3 with k1, and A={v1=a1,v3=a2}
    uia1, 1i4k+2i+84i+34i+64i+94
    uia1, 4k+3inni+54ni+104ni+114ni+84
    uia2, 3i5ki+44i+74i+64i+14
    uia2, 5k+1inni+134ni+104ni+74ni+124
    n=8k+4 with k1, and A={v1=a1,v4=a2}
    uia1, 1i4k+3i+84i+34i+64i+94
    uia1, 4k+4inni+84ni+54ni+104ni+114
    uia2, 3i5k+2i4i+34i+64i+54
    uia2, 5k+3inni+84ni+134ni+144ni+114
    n=8k+5 with k1, and A={v1=a1,v2=a2}
    uia1, 1i5k+1i+84i+34i+64i+94
    uia1, 5k+2inni+114ni+84ni+54ni+104
    uia2, 2i5k+2i+84i+74i+24i+54
    uia2, 5k+3inni+114ni+124ni+94ni+64
    n=8k+6 with k1, and A={v1=a1,v2=a2}
    uia1, 1i4k+2i+84i+34i+64i+94
    uia1, 4k+6inni+104ni+114ni+84ni+54
    uia2, 1i4k+3i+84i+74i+24i+54
    uia2, 4k+7inni+64ni+114ni+124ni+94
    n=8k+7 with k1, and A={v1=a1,v3=a2}
    uia1, 1i4k+3i+84i+34i+64i+94
    uia1, 4k+6inni+54ni+104ni+114ni+84
    uia2, 2i5k+1i+44i+74i+64i+14
    uia2, 5k+4inni+134ni+104ni+74ni+124

     | Show Table
    DownLoad: CSV
    Table 2.  List of the number of edges in geodesics from the inner vertices to aA in P(n,4).
    GeodesicThe number of edges in the geodesic.
    viai0 (mod 4)i1 (mod 4)i2 (mod 4)i3 (mod 4)
    n=8k with k2, and A={v1=a1,v3=a2}
    via1, 1i4k+1i+124i14i+104i+134
    via1, 4k+2inni+124ni+14ni+144ni+154
    via2, 3i4k+3i+84i+114i+104i34
    via2, 4k+4inni+164ni+174ni+144ni+34
    n=8k+1 with k2, and A={v1=a1,v4=a2}
    via1, 1i3k+1i+124i14i+104i+134
    via1, 3k+10inni+154ni+124ni+14ni+144
    via2, 4i4ki44i+74i+104i+94
    via2, 4k+6inni+154ni+44ni+174ni+184
    n=8k+2 with k1, and A={v1=a1,v2=a2}
    via1, 1i3k+2i+124i14i+104i+134
    via1, 3k+10inni+144ni+154ni+124ni+14
    via2, 2i3k+3i+124i+114i24i+94
    via2, 3k+11inni+24ni+154ni+164ni+134
    n=8k+3 with k1, and A={v1=a1,v3=a2}
    via1, 1i3k+3i+124i14i+104i+134
    via1, 3k+10inni+14ni+144ni+154ni+124
    via2, 3i4k+1i+84i+114i+104i34
    via2, 4k+8inni+174ni+144ni+34ni+164
    n=8k+4 with k1, and A={v1=a1,v4=a2}
    via1, 1i4k+3i+124i14i+104i+134
    via1, 4k+4inni+124ni+14ni+144ni+154
    via2, 3i5k+2i44i+74i+104i+94
    via2, 5k+3inni+44ni+174ni+184ni+154
    n=8k+5 with k1, and A={v1=a1,v2=a2}
    via1, 1i4k+1i+124i14i+104i+134
    via1, 4k+6inni+154ni+124ni+14ni+144
    via2, 2i4k+2i+124i+114i24i+94
    via2, 4k+7inni+154ni+164ni+134ni+24
    n=8k+6 with k1, and A={v1=a1,v2=a2}
    via1, 1i3k+2i+124i14i+104i+134
    via1, 3k+14inni+144ni+154ni+124ni+14
    via2, 1i3k+3i+124i+114i24i+94
    via2, 3k+15inni+24ni+154ni+164ni+134
    n=8k+7 with k1, and A={v1=a1,v3=a2}
    via1, 1i3k+3i+124i14i+104i+134
    via1, 3k+14inni+14ni+144ni+154ni+124
    via2, 2i4k+1i+84i+114i+104i34
    via2, 4k+12inni+174ni+144ni+34ni+164

     | Show Table
    DownLoad: CSV

    Case 2: (n=8k+1 with k2)

    Let A={v1=a1,v4=a2}, then

    ● the number of edges in u1a2 geodesic is 3,

    ● the number of edges in u2a2 geodesic is 3,

    ● the number of edges in u3a2 geodesic is 2.

    Further, Tables 13 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 3.  λi is the number of edges in a via1 geodesic, and λj is the number of edges in a vja2 geodesic.
    i3k+22(mod 4)3k+33(mod 4)3k+40(mod 4)3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)
    λi5k43k+1643k+1643k+445k445k+845k+843k+84
    j1234k+11(mod 4)4k+22(mod 4)4k+33(mod 4)4k+40(mod 4)4k+51(mod 4)
    λj443k+1k+3k+3kk

     | Show Table
    DownLoad: CSV

    Case 3: (n=8k+2 with k1)

    Let A={v1=a1,v2=a2}, then

    ● the number of edges in u2a2 geodesic is 2.

    Further, Tables 1, 2 and 4 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 4.  λi is the number of edges in a via1 geodesic, and λj is the number of edges in a vja2 geodesic.
    i3k+33(mod 4)3k+40(mod 4)3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)
    λi5k43k+1643k+443k+1645k445k+843k+84
    j123k+40(mod 4)3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)3k+102(mod 4)
    λj305k43k+1643k+445k+845k445k+843k+84

     | Show Table
    DownLoad: CSV

    Case 4: (n=8k+3 with k1)

    Let A={v1=a1,v3=a2}, then

    ● the number of edges in u1a2 geodesic is 3,

    ● the number of edges in u2a2 geodesic is 2.

    Further, Tables 1, 2 and 5 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 5.  λi is the number of edges in a via1 geodesic, and λj is the number of edges in a vja2 geodesic.
    i3k+40(mod 4)3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)
    λi5k43k+443k+1645k+845k443k+84
    j124k+22(mod 4)4k+33(mod 4)4k+40(mod 4)4k+51(mod 4)4k+62(mod 4)4k+73(mod 4)
    λj43k+1kk+3k+3kk+1

     | Show Table
    DownLoad: CSV

    Case 5: (n=8k+4 with k1)

    Let A={v1=a1,v4=a2}, then

    ● the number of edges in v1a2 geodesic is 4,

    ● the number of edges in v2a2 geodesic is 4,

    ● the number of edges in u1a2 geodesic is 3,

    ● the number of edges in u2a2 geodesic is 3.

    Moreover, Tables 1 and 2 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Case 6: (n=8k+5 with k1)

    Let A={v1=a1,v2=a2}, then

    ● the number of edges in u1a2 geodesic is 2,

    ● the number of edges in v1a2 geodesic is 3.

    Further, Tables 1, 2 and 6 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 6.  λi is the number of edges in a via1 geodesic, and λj is the number of edges in a vja2 geodesic.
    i4k+22(mod 4)4k+33(mod 4)4k+40(mod 4)4k+51(mod 4)
    λik+1k+4k+4k+1
    j4k+33(mod 4)4k+40(mod 4)4k+51(mod 4)4k+62(mod 4)
    λjk+1k+4k+4k+1

     | Show Table
    DownLoad: CSV

    Case 7: (n=8k+6 with k1)

    Let A={v1=a1,v2=a2}, then

    ● the number of edges in v3k+3a1 geodesic is 5k+443(mod 4),

    ● the number of edges in v3k+4a2 geodesic is 5k+440(mod 4),

    ● the number of edges in u4k+3a1 geodesic is k+2,

    ● the number of edges in u4k+4a1 geodesic is k+3,

    ● the number of edges in u4k+5a1 geodesic is k+2,

    ● the number of edges in u4k+4a2 geodesic is k+2,

    ● the number of edges in u4k+5a2 geodesic is k+3,

    ● the number of edges in u4k+6a2 geodesic is k+2.

    Further, Tables 1, 2 and 7 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 7.  λi is the number of edges in a via1 geodesic, and λj is the number of edges in a vja2 geodesic.
    i3k+40(mod 4)3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)3k+102(mod 4)3k+113(mod 4)3k+120(mod 4)3k+131(mod 4)
    λi3k+1643k+443k+1645k43k+2043k+845k+645k445k+843k+124
    j3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)3k+102(mod 4)3k+113(mod 4)3k+120(mod 4)3k+131(mod 4)3k+142(mod 4)
    λj3k+1643k+443k+1645k43k+2043k+845k+845k445k+843k+124

     | Show Table
    DownLoad: CSV

    Case 8: (n=8k+7 with k1)

    Let A={v1=a1,v3=a2}, then

    ● the number of edges in v1a2 geodesic is 4,

    ● the number of edges in u4k+4a1 geodesic is k+2,

    ● the number of edges in u4k+5a1 geodesic is k+2,

    ● the number of edges in u1a2 geodesic is 3,

    ● the number of edges in u5k+2a2 geodesic is 3k+1242(mod 4),

    ● the number of edges in u5k+3a2 geodesic is 5k+443(mod 4).

    Moreover, Tables 1, 2 and 8 provide the lists of number of edges in xa1 and xa2 geodesics for all xV(G)A.

    Table 8.  λi is the number of edges in a via1 geodesic, and λj is the number of edges in a vja2 geodesic.
    i3k+40(mod 4)3k+51(mod 4)3k+62(mod 4)3k+73(mod 4)3k+80(mod 4)3k+91(mod 4)3k+102(mod 4)3k+113(mod 4)3k+120(mod 4)3k+131(mod 4)
    λi5k+443k+443k+1643k+2045k43k+843k+845k+1245k+845k44
    j4k+22(mod 4)4k+33(mod 4)4k+40(mod 4)4k+51(mod 4)4k+62(mod 4)4k+73(mod 4)4k+80(mod 4)4k+91(mod 4)4k+102(mod 4)4k+113(mod 4)
    λjk+2kk+3k+4k+1k+1k+4k+3kk+2

     | Show Table
    DownLoad: CSV

    In all of these eight cases, for yN(x), if we denote

    ● the number of edges in xa1 geodesic by α1,

    ● the number of edges in ya1 geodesic by β1,

    ● the number of edges in xa2 geodesic by α2,

    ● the number of edges in ya2 geodesic by β2,

    then it can be seen that

    either|α1β1|=1wheneverα2=β2,or|α2β2|=1wheneverα1=β1,

    which implies that either a1A or a2A adjacently distinguishes the pair (x,y). Hence, A is a neighbor-distinguishing set for G.

    Theorem 4. For n9, if G is a generalized Petersen graph P(n,4), then dima(G)=2.

    Proof. Since dima(G)=1 if and only if G is a bipartite graph, by Theorem 1. So dima(G)2, because G is not a bipartite graph. Hence, we get the required result, by Theorem 3.

    The results of this section provide the solution of the problem of neighbor-distinguishing in the generalized Petersen graphs P(2n,n1).

    Theorem 5. For all n3, if G is a generalized Petersen graph P(2n,n1), then the set A={u1,vn1} is a neighbor-distinguishing set for G.

    Proof. According to Proposition 2, we have to perform neighbor-distinguishing for each pair (x,y), where xV(G) and yN(x). When xA, then the pair (x,y) adjacently distinguished by x, because the number of edges in yx geodesic is 1 while the number of edges in xx geodesic is 0. Further, Table 9 provides the list of number of edges in xu1 and xvn1 geodesics for all xV(G)A.

    Table 9.  List of the number of edges in xa geodesics for xV(G)A and aA.
    GeodesicThe number of edges in the geodesics
    When n=2k+1, k1
    For i/geodesicsuiu1uivn1viu1 i is oddviu1 i is evenvivn1 i is oddvivn1 i is even
    1ik1i1i+2iii+3i+1
    kik+1i12ki+1ii2ki+22ki
    i=k+2n+12k1n+12n+122ki+22ki
    k+3i2k2ki+42ki+12ki+32ki+32ki+22ki
    i=2k+1i2k+2i2k+1i2k+1i2k+1i2k+2i2k+2
    i=2k+24in+233i2ki2k
    2k+3i3ki2ki2k+1i2k1i2k1i2k+2i2k
    i=3k+1i2kin+1inink+2k
    i=3k+2n+12kinink+1k1
    3k+3i4k2ni+14ki+22ni+22ni+24ki+34ki+1
    i=2n1233344
    i=2n122211
    When n=2k, k2
    1ik2i1i+2iii+3i+1
    i=k1i1niiikk
    kik+1i1niii2ki12ki+1
    i=k+2i1nin2n2ni1ni+1
    k+3i2k12ki+32ki2ki+22ki+22ki12ki+1
    i=2k322233
    i=2k+1433322
    2k+2i3k2i2k+1i2k+2i2ki2ki2k+1i2k+3
    i=3k1i2k+1i2k+2i2ki2kkk
    3ki3k+12ni+14kii2ki2k4ki+14ki1
    i=3k+2k1k2n2n24ki+14ki1
    3k+3i4k22ni+14ki2ni+22ni+24ki+14ki1
    i=2n1233344
    i=2n122211

     | Show Table
    DownLoad: CSV

    Now, for any yN(x), if we denote

    ● the number of edges in xu1 geodesic by α1,

    ● the number of edges in yu1 geodesic by β1,

    ● the number of edges in xvn1 geodesic by α2,

    ● the number of edges in yvn1 geodesic by β2,

    then it can be seen that

    either|α1β1|=1wheneverα2=β2,or|α2β2|=1wheneverα1=β1,

    which implies that either u1A or vn1A adjacently distinguishes the pair (x,y). Hence, A is a neighbor-distinguishing set for G.

    Theorem 6. For n3, if G is a generalized Petersen graph P(2n,n1), then dima(G)=2.

    Proof. Since dima(G)=1 if and only if G is a bipartite graph, by Theorem 1. So dima(G)2, because G is not a bipartite graph. Hence, we get the required result, by Theorem 5.

    Distinguishing every two vertices in a graph is an eminent problem in graph theory. Many graph theorists have been shown remarkable interest to solve this problem with the aid of distance (the number of edges in a geodesic) from last four decades. Using the technique of finding geodesics between vertices, we solved the problem of distinguishing every two neighbors in generalized Petersen graphs P(n,4) and P(2n,n1). We investigated that, in both the families of generalized Petersen graphs, only two vertices are adequate to distinguish every two neighbors.

    The authors are grateful to the editor and anonymous referees for their comments and suggestions to improve the quality of this article. This research is supported by Balochistan University of Engineering and Technology Khuzdar, Khuzdar 89100, Pakistan.

    The authors declare that they have no conflict of interest.



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