Hamilton Paths in Certain Arithmetic Graphs

For each integer m ≥ 1, consider the graph Gm whose vertex set is the set N = {0, 1, 2, . . . } of natural numbers and whose edges are the pairs xy with y = x + m or y = x − m or y = mx or y = x/m. Our aim in this note is to show that, for each m, the graph Gm contains a Hamilton path. This answers a question of Lichiardopol. For each integer m ≥ 1, consider the graph Gm whose vertex set is the set N = {0, 1, 2, . . . } of natural numbers and whose edges are the pairs xy with y = x+m or y = x−m or y = mx or y = x/m. We show that, for each m, the graph Gm contains a Hamilton path. Here, by ‘Hamilton path’ we mean a ‘oneway infinite Hamilton path’, i.e. a sequence x0, x1, x2, . . . of vertices of Gm such that each vertex appears precisely once and, for all i, the vertices xi and xi+1 are adjacent. We shall use this to answer a question of Lichiardopol [1] about two-way infinite Hamilton paths in graphs defined similarly but with vertex set the set Z of integers. The case m = 1 is trivial so we begin at m = 2. The construction of the Hamilton path in the graph G2 is similar in spirit to those used later, but this case is much easier. Proposition 1. The graph G2 contains a Hamilton path. Proof. Our approach is to define inductively a strictly increasing sequence x0, x1, x2, . . . of natural numbers with x0 = 0, and show that, for each i = 0, 1, 2, . . . , there is a Hamilton path in G2[xi, xi+1] from xi to xi+1; putting these paths end-to-end gives the required Hamilton path in G2. Now, take • x0 = 0; • x1 = 3; • xi = 2xi−1 + 5 (i ≥ 2). ∗Department of Pure Mathematics and Mathematical Statistics, Centre for Mathematical Sciences, Wilberforce Road, Cambridge CB3 0WB, England. †[email protected] Our path in G2[x0, x1] is simply 0,2,1,3. To show that there is such a path in G2[xi, xi+1] for i ≥ 1, it suffices to exhibit a Hamilton path in G2[x, 2x + 5] for odd x > 0; such a path is given by x, 2x, 2x− 2, 2x− 4, . . . , x + 1, 2x + 2, 2x + 4, x + 2, x + 4, x + 6, . . . , 2x + 5. We next consider the case of even m > 2. The approach is similar to that used for the graph G2, but instead of splitting N up into intervals we need to use slightly more complicated sets. Proposition 2. For all even m > 2, the graph Gm contains a Hamilton path. Proof. Define inductively a strictly increasing sequence x0, x1, x2, . . . of natural numbers by • x0 = 0; • xi = m(xi−1 + 2) (i ≥ 1). Note that each xi is divisible by m. For i = 0, 1, 2, . . . , let G m be the graph G m = Gm [([xi, xi+1]−mN) ∪ ([mxi,mxi+1] ∩mN)] . Note that, for all i, the sets V (G m ) and V (G (i+1) m ) intersect only at mxi+1; and for all i and j with |i − j| > 1, the sets V (G m ) and V (G m ) are disjoint. Moreover, the union of the sets V (G m ) (i = 0, 1, 2, . . .) is the whole of N. Hence it is enough to construct, for each i, a Hamilton path in G m from mxi to mxi+1; putting these paths end-to-end again gives a Hamilton path in Gm as required. So, fix i. Observe that, for each j = 1, 2, . . . , m − 1, there is a path Pj in G (i) m from m(xi + j) to m(xi+1 − m + j) whose internal vertices are precisely those vertices of G m which are congruent to j (mod m), namely the path m(xi + j), xi + j, xi + m + j, xi + 2m + j, . . . , xi+1 −m + j, m(xi+1 −m + j). Note that the V (Pj) (1 ≤ j ≤ m − 1) partition V (G m ) except for the vertices mxi, m(xi +m), m(xi +m+1), m(xi +m+2), . . . , m(xi+1−m), mxi+1 which are missed. Moreover, the first (last) vertex of Pj is adjacent to the first (last) vertex of Pj+1 (1 ≤ j ≤ m−2). Hence it is possible to join these paths together to make the required Hamilton path in G m , namely mxi, P1,m(xi+1 −m),m(xi+1 −m− 1), . . . , m(xi + m), Pm−1, P −1 m−2, Pm−3, . . . , P −1 2 ,mxi+1 (where P−1 denotes the path obtained by traversing the path P in reverse). This only leaves us to deal with odd m. The construction used in Proposition 2 will not work here as, since m is odd, we would have to finish by traversing the path P2 forwards, and so we would be unable to reach the point