Partitioning Harary graphs into connected subgraphs containing prescribed vertices

A graph G is arbitrarily partitionable (AP for short) if for every partition ( n 1 , n 2 , . . . , n p ) of | V ( G ) | there exists a partition ( V 1 , V 2 , . . . , V p ) of V ( G ) such that each V i induces a connected subgraph of G with order n i . If, additionally, k of these subgraphs ( k ≤ p ) each contains an arbitrary vertex of G prescribed beforehand, then G is arbitrarily partitionable under k prescriptions (AP+ k for short). Every AP+ k graph on n vertices is ( k + 1) -connected, and thus has at least (cid:100) n ( k +1)2 (cid:101) edges. We show that there exist AP+ k graphs on n vertices and (cid:100) n ( k +1)2 (cid:101) edges for every k ≥ 1 and n ≥ k .


Introduction
We denote by V (G) and E(G) the sets of vertices and edges, respectively, of a graph G.The order (resp.size) of G is the cardinality of the set V (G) (resp.E(G)).If X is a subset of V (G), then G[X] denotes the subgraph of G induced by X.
In the late 1970s, the following well-known result was proved.
In this paper, we consider a more general partition problem resulting from the combination of the notion of arbitrarily partitionable graphs [1] with the constraint of prescribing a set of vertices from Theorem 1.Let G be a connected graph of order n.A sequence τ = (n 1 , n 2 , . . ., n p ) of positive integers is admissible for G if it performs a partition of n, that is if p i=1 n i = n.If, additionally, we can partition V (G) into p parts (V 1 , V 2 , . . ., V p ) such that each V i induces a connected subgraph of G with order n i , then τ is realizable in G, the partition (V 1 , V 2 , . . ., V p ) being a realization of τ in G.If every admissible sequence for G is also realizable in G, then G is arbitrarily partitionable (AP for short).The interested reader is referred to [1,2,6,8] for a review of some results on AP graphs.Now suppose that we still want to partition G into an arbitrary number, say p, of connected subgraphs G 1 , G 2 , . . ., G p of prescribed orders, but in such a way that for each i ∈ {1, 2, . . ., k} with fixed k ∈ {1, 2, . . ., p}, the subgraph G i contains a vertex v i of G arbitrarily chosen beforehand.To model this additional requirement, the definition of AP graphs can be strenghtened as follows [3].A k-prescription of G is a k-tuple P = (v 1 , v 2 , . . ., v k ) of k distinct vertices of G.We say that a sequence τ = (n 1 , n 2 , . . ., n p ) with p ≥ k elements is realizable in G under P if there exists a realization (V 1 , V 2 , . . ., V p ) of τ in G such that the vertex v i belongs to V i for every i ∈ {1, 2, . . ., k}.Notice that we have adopted the convention that the elements of τ associated with the prescribed vertices are the first elements of τ .We say that G is (p, k)-partitionable if every sequence admissible for G consisting of exactly p elements is realizable in G under every k-prescription.Finally, the graph G is arbitrarily partitionable under k prescriptions (AP+k for short) if G is (p, k)-partitionable for every p ∈ {k, k + 1, . . ., n}.
According to these definitions, an AP+0 graph is an AP graph.Stated differently, Theorem 1 asserts that every k-connected graph is (k, k)-partitionable.In the same flavour, note that every k-connected graph with k ≥ 2 is trivially (k, k − 1)-partitionable.Hence, when dealing with a k-connected graph, we only consider sequences with strictly more than k elements throughout this paper.It also has to be known that deciding whether a sequence is realizable in a graph under a prescription is NP-complete in general, even when the sequence or the prescription has a fixed number of elements [4].
Only a few classes of AP+k graphs are known.For every k ≥ 1, the set of complete graphs on at least k vertices is a trivial class of AP+k graphs, these graphs having the largest possible size.Regarding graphs with less edges, it was proved in [3] that k th powers of paths (resp.cycles) are AP+(k − 1) (resp.AP+(2k − 1)) for every k ≥ 1, these results being tight (i.e.we cannot always partition these graphs when more prescriptions are requested).
In this work, we investigate the least possible size of an AP+k graph.In this scope, we focus on optimal AP+k graphs, i.e. on AP+k graphs with the least possible number of edges regarding their order and connectivity.This is done by studying the family of well-known Harary graphs.After having introduced some notation and preliminary results in Section 2, we prove some more results regarding the partition of powers of paths or cycles in Section 3.These results are then used to show, in Section 4, that every (k + 1)-connected Harary graph is an optimal AP+k graph for every k = 2.We finally deal with 3connected Harary graphs in Section 5.In particular, we show that these graphs are not necessarily AP+2.We however provide another class of optimal AP+2 graphs instead.All these results imply that, for every k ≥ 1 and n ≥ k, every optimal AP+k graph with order n has size n(k+1) 2 .

Definitions, notation, and preliminary results
We also say that G is spanned by H. Given an integer k ≥ 1, the k th power of G, denoted by G k , is the graph with the same vertex set as G, two vertices of G k being adjacent if they are at distance at most k in G.We denote by P n (resp.C n ) the path (resp.cycle) on n vertices.The vertices of P n or C n are consecutively denoted by v 0 , v 1 , . . ., v n−1 .Regarding P n , the vertices v 0 and v n−1 are its first and last vertices, respectively, sometimes called its endvertices.We use the same terminology to deal with the vertices of P k n (resp.C k n ) according to its natural spanning P n (resp.C n ).Let k ≥ 1 and n ≥ k be two integers.The k-connected Harary graph on n vertices, denoted by H k,n , has a vertex set {v 0 , v 1 , . . ., v n−1 } and the following edges: • if k = 2r is even, then two vertices v i and v j are linked if and only if i − r ≤ j ≤ i + r; where the subscripts are taken modulo n.Three examples of Harary graphs are given in Figure 1.When k is odd, the neighbours of a vertex v i of H k,n which are at distance strictly more than r from v i in the underlying C n (there are at most two of them) are called the antipodal neighbours of v i .In particular, the vertex v i has two antipodal neighbours if and only if i = 0, and k and n are both odd.A diagonal edge of H k,n is an edge linking two vertices each of which is an antipodal neighbour of the other one.
If G is a graph with a natural ordering of its vertices (like powers of paths and cycles, or Harary graphs), then, for every vertex v of G, we denote by v + (resp.v − ) the neighbour of v succeeding (resp.preceding) v in this ordering.Every power of path P k n with underlying path P n = v 0 v 1 . . .v n−1 is considered to be depicted in a "usual" way, i.e. from its leftmost vertex v 0 to its rightmost vertex v n−1 .By uGv we refer to the graph G[{u, u + , (u + ) + , . . ., v − , v}] for every two vertices u and v of G. Assuming P is a prescription of G, a prescribed block One important property of AP graphs is the following.
Recall that a graph is traceable if it admits a Hamiltonian path.Since every path is AP, Observation 2 implies the following result.

Corollary 3. Every traceable graph is AP.
We now point out the following property of AP+k graphs, from which we deduce a bound on the size of an optimal AP+k graph.Observation 4. Let k ≥ 1.Every AP+k graph is (k + 1)-connected.Therefore, an optimal AP+k graph on n vertices has at least n(k+1) 2 edges.
Proof: Assume G is a graph with order n.If there exist k vertices v 1 , v 2 , . . ., v k such that G−{v 1 , v 2 , . . ., v k } is not connected, then the sequence (1, 1, . . ., 1, n − k) with the value 1 appearing k times cannot be realized in G under (v 1 , v 2 , . . ., v k ).Therefore, a necessary condition for G to be AP+k is to be (k + 1)-connected.The lower bound then follows.
As mentioned by Corollary 3, paths are AP+0, while it is easy to check that cycles are AP+1.Baudon et al. generalized these observations to powers of paths and cycles [3].

2
, an optimal AP+k graph on n vertices may have less edges than P k+1 n .On the contrary, every graph C k n is 2k-regular and hence is an edge-minimal 2k-connected graph.According to Observation 4, it follows that the set of k th powers of cycles is a set of optimal AP+(2k − 1) graphs for every k ≥ 1.

Partitioning powers of paths and cycles under prescriptions
As pointed out in Theorem 5, recall that k th powers of paths and cycles are AP+(k − 1) and AP+(2k − 1), respectively.This result is tight according to Observation 4, in the sense that we cannot always prescribe more vertices while partitioning these graphs.In this section, we exhibit situations under which these graphs can be partitioned under more prescriptions than indicated by their connectivity.
The following first result asserts that k th powers of paths can be partitioned under k-prescriptions when either the first or the last vertex is prescribed.
n with p ≥ k elements is realizable in P k n under P .In the next result, we prove that k th powers of paths are also partitionable under k-prescriptions when the prescribed vertices do not form a prescribed block with size k.
If the prescribed vertices do not form a prescribed block with size k, then every sequence τ = (n 1 , n 2 , . . ., n p ) admissible for P k n with p ≥ k elements is realizable in P k n under P .
On the other hand, if n 1 ≤ i 1 , then let V 1 be a subset of {v 0 , v 1 , . . ., v i1 } obtained as follows.First, we set V 1 = {v i1 } and we then repeatedly add to V 1 the vertex located at distance 2 on the left of the last vertex added to V 1 as long as |V 1 | < n 1 and v 0 is not reached.If there is no vertex at distance 2 on the left of the last vertex added to If such a value of r does not exist, then let r be such that and v a ∈ P be the nearest neighbour of v i1−1 located on the right of v i1 .Such a vertex necessarily exists since the opposite assumption would imply that our k prescribed vertices are located consecutively along G.Moreover, either v a or v i2 is the first vertex of G − {v 0 , v 1 , . . ., v i1 }.We then obtain a realization These two realizations exist according to Lemma 6.
We now strengthen Lemma 6 by showing that k th powers of paths are partitionable under (k + 1)prescriptions when their endvertices are prescribed.
If i 1 = 0 and i k+1 = n−1, then every sequence τ = (n 1 , n 2 , . . ., n p ) admissible for P k n with p ≥ k + 1 elements is realizable in P k n under P .
Proof: We prove this claim by induction on k.For k = 1, the result is obvious.We thus now suppose that k ≥ 2 and that the claim holds for every k < k.
).This realization necessarily exists according to Lemma 6 since Suppose now that n 1 > i 2 .Observe that {0, 1, . . ., k − 1} − k j=2 {i j mod k} is not empty, so let us denote by r one of its elements.The subset V 1 of the realization is constructed as follows.It first contains all the vertices between v 0 and Finally, as long as |V 1 | < n 1 , we repeatedly add to V 1 the vertex at distance k on the right from the last vertex added to V 1 , unless it is equal to v n−1 , i.e. v a+k , then v a+2k , and so on.According to our choice of r, these vertices are not prescribed ones and, at any moment of the procedure, the subgraph G − V 1 is spanned by the (k − 1) th power of a path, and the subgraph G[V 1 ] is connected.
On the one hand, if |V 1 | = n 1 holds after the procedure, then (V 1 , V 2 , . . ., V p ) is a realization of τ under P , where (V 2 , V 3 , . . ., V p ) is a realization of (n 2 , n 3 , . . ., n p ) in G − V 1 under the prescription (v i2 , v i3 , . . ., v i k+1 ) which necessarily exists by the induction hypothesis since v i2 and v i k+1 are the endvertices of G − V 1 .
On the other hand, if |V 1 | < n 1 holds once the procedure is achieved, then each vertex from Once again, such a realization necessarily exists according to the induction hypothesis.
We now prove an analogous result concerning cycles to the power of at least 2. Let G = C k n for some k ≥ 2 and n ≥ 2k, the sequence τ = (n 0 , n 1 , . . ., n p−1 ) be admissible for G, and P = (v i0 , v i1 , . . ., v i 2k−1 ) be a 2k-prescription of G, with p ≥ 2k and 0 For every j ∈ {0, 1, . . ., 2k − 1}, we denote by D j the set {v In particular, we have If the prescribed vertices are not organized into two maximal prescribed blocks with size k, then every sequence τ = (n 0 , n 1 , . . ., n p−1 ) admissible for C k n with p ≥ 2k elements is realizable in C k n under P .
Proof: Let k ≥ 2 be fixed, and G = C k n for some value of n ≥ 2k.We prove that every partition τ = (n 0 , n 1 , . . ., n p−1 ) of n with p ≥ 2k elements is realizable in G under every 2k-prescription P = (v i0 , v i1 , . . ., v i 2k−1 ) with 0 ≤ i 0 < i 1 < . . .< i 2k−1 ≤ n − 1 when the prescribed vertices do not form two maximal prescribed blocks with size k.For every j ∈ {0, 1, . . ., 2k − 1}, let q j = j+k−1 =j d and n , where the indices are counted modulo 2k.In other words, the value q j is the order of the graph v + ij−1 Gv i j+k−1 = G[{i + j−1 , (i + j−1 ) + , . . ., i j+k−1 }] including the k prescribed vertices v ij , v ij+1 , . . ., v i j+k−1 , and s j is the amount of vertices needed by the subgraphs containing these prescribed vertices in a realization of τ in G under P .Note that there necessarily exists a j such that s j ≤ q j since having To prove the claim, we distinguish several cases depending on the relationship between s j 's and q j 's.Case 1. s j = q j for some j ∈ {0, 1, . . ., 2k − 1}.
In this situation, a realization of τ in G under P is deduced as follows.Assume j = 0 without loss of generality, and set Therefore, there exists a realization (V k , V k+1 , . . ., V p−1 ) of (n k , n k+1 , . . ., n p−1 ) under (v i k , v i k+1 , . . ., v i 2k−1 ) in this graph by Lemma 6.The partition (V 0 , V 1 , . . ., V p−1 ) is then a realization of τ in G under P .
In particular, there exists a value of j for which s j > q j and s j+1 < q j+1 .Suppose j = 0 without loss of generality.
Case 2.1.There exists a set A realization of τ in G under P can be obtained as follows.Firstly, let In such a situation, we have Besides, since n 's and d 's are strictly greater than 0, we get A realization of τ in G under P is then obtained as follows.On the one hand, let On the other hand, let n 0 = (n 0 − n 0 ) + 1 (note that n 0 ≥ 1), and let (V 0 , V k+1 , V k+2 , . . ., V p−1 ) be a realization of (n 0 , n k+1 , n k+2 , . . ., is the k th power of a path with last vertex v i0 , and k prescribed vertices are specified.The partition (V 0 ∪ V 0 , V 1 , V 2 , . . ., V p−1 ) is then a realization of τ in G under P since G[V 0 ] and G[V 0 ] are connected and both contain the vertex v i0 (which is actually the only vertex appearing in both these subgraphs).Case 3. s j < q j for every j ∈ {0, 1, . . ., 2k − 1}.
We distinguish two subcases.
Case 3.1.There are two consecutive prescribed vertices.Assume v i0 = v + i 2k−1 without loss of generality, with i 0 = 0 and i 2k−1 = n − 1. Case 3.1.1.There exists r ∈ {2k, 2k + 1, . . ., p − 1} such that s 0 + r =2k n = q 0 .In this situation, we can deduce a realization of τ in G under P as follows.Firstly, let this graph being the k th power of some path.Secondly, let be a realization of which exists for the same reason as previously since is then a realization of τ in G under P .Case 3.1.2.Such r does not exist.

This realization exists according to
In this situation, either v i k−1 or v i k belongs to a prescribed block with size at least k.
Then one can relabel the prescribed vertices so that v i0 and v i 2k−1 correspond to two consecutive prescribed vertices from this prescribed block, and use the procedures from Case 3.1.Since s j < q j for every j ∈ {0, 1, . . ., 2k − 1}, note that this time the two vertices v a and v b (if these vertices are needed) have to be adjacent since otherwise it would mean that the prescribed vertices form another prescribed block with size at least k, implying that there are two prescribed blocks with size k, contradicting the assumption of the lemma.
(n k+1 , n k+2 , . . ., n 2k−1 , n r2 , n r2+1 , n r2+2 , . . ., n p−1 ).Note that the elements of τ 3 sum up to |{v + i k , (v + i k ) + , . . ., v i 2k−1 }|.Remark that every element of τ has been associated with one of τ 1 , τ 2 and τ 3 , with at most two non-prescribed elements being split so that the τ i 's sum up exactly to the orders of some subgraphs of G.In the case where τ contains a "big" non-prescribed element, it is even possible that this element was split into three integers among τ 1 , τ 2 and τ 3 .To obtain the realization of τ in G under P , we realize τ 1 , τ 2 and τ 3 in vertex-disjoint subgraphs of G, and this in such a way that if an original element of τ was dispatched in several of the τ i 's, then the resulting connected subgraphs perform a whole connected subgraph when unified.The three realizations R 1 , R 2 and R 3 are obtained as follows.
), which exists according to Lemma 8 since v i1 and v i k are the endvertices of G[{v i1 , v + i1 , . . ., v i k }] and there are k + 1 prescribed vertices.
which is traceable by our choice of V 0 .Additionally request the realization to satisfy the prescription (v − i1 , v + i 2k−1 ) when τ 2 has at least two elements.Such a requirement is allowed according to Lemma 8.
. The existence of such a realization follows from Lemma 6 since G[{v The realization of τ in G under P is obtained by considering V 0 and the parts from R 1 , R 2 and R 3 , and unifying those parts whose sizes result from the split of a single element of τ , if necessary.By our choice of the prescribed vertices, these parts have neighbouring vertices (this follows from the facts that k ≥ 2, and that the prescribed vertices of P are not consecutive), and thus induce connected subgraphs.This completes the proof.

Partitioning Harary graphs under prescriptions
Harary graphs are trivially AP according to Corollary 3. We here show that we can always prescribe the largest possible number of vertices (with respect to their connectivity) while partitioning these graphs, except for 3-connected Harary graphs.We consider the three kinds of Harary graphs for this purpose.

Construction 1: k is even
The Harary graph H k,n with k even is isomorphic to C k/2 n which is AP+(k − 1) according to Theorem 5 for every k ≥ 2 and n ≥ 2k.We thus derive the following result.

Construction 2: k is odd and n is even
Let k ≥ 2 and n ≥ 2k + 1 be two integers such that n is even.By construction, the Harary graph H 2k+1,n is spanned by H 2k,n and is thus AP+(2k − 1) according to Corollary 10.However, regarding the connectivity of H 2k+1,n , one could wonder whether H 2k+1,n is AP+2k.
Before proving that H 2k+1,n is indeed AP+2k, we first introduce the following lemma which deals with the traceability of a graph composed by two linked squares of paths.
are both spanned by the square of a path, and there exists an edge joining one vertex of V 1 and one of V 2 , then G is traceable.
Consider the following subpaths of G: It is then easy to check that P Qv a u b RS is a Hamiltonian path of G.
We are now ready to prove our main result.Theorem 12.For every k ≥ 2 and even n ≥ 2k + 1, the Harary graph H 2k+1,n is AP+2k.
Case 1.If the prescribed vertices are not organized into two maximal prescribed blocks with size k, then, because k ≥ 2, we can deduce a realization of τ in the spanning C k n of G under P , thanks to Lemma 9.Such a realization is naturally a realization of τ in G under P .Case 2. Suppose now that the prescribed vertices form two maximal prescribed blocks B 1 and B 2 with size exactly k in G.In this situation, note that G − P only remains connected thanks to some diagonal edges.Indeed, assume without loss of generality.Then the antipodal neighbours of v − i0 and v + i k−1 cannot both belong to P : since n ≥ 2k + 2, if this were the case then these two antipodal neighbours would belong to B 2 , and similarly for all antipodal neighbours of v i0 , v i1 , . . ., v i k−1 (according to our assumptions on the maximal prescribed blocks).We would then get that B 2 has size at least k + 2, a contradiction.P x,y (G) = w 2 x w 1 x if x = y, w 2 x w 1 x P x+1,y (G) otherwise.
P x,y (G) = w 1 x w 2 x if x = y, w 1 x w 2 x P x+1,y (G) otherwise.
The paths P x,y (G) and P x,y (G) of G are defined analogously from right to left when x ≥ y.For every α ∈ {1, 2}, we additionally define P α,→ x,y (G) (resp.P α,← x,y (G)) for x < y (resp.x > y) to be the path w α x w α x+1 . . .w α y (resp.w α x w α x−1 . . .w α y ) of G .For convenience, let us assume that P x,y (G) = P x,y (G) = P α,→ x,y (G) = ∅ (resp.P x,y (G) = P x,y (G) = P α,← x,y (G) = ∅) whenever x or y does not belong to the interval above or when x > y (resp.x < y).According to our terminology, note e.g. that uP 1,→ 1,4 (P r 10 )vP 2,← 4,1 (P r 10 ) and uP 1,4 (P r 10 )v are Hamiltonian paths of P r 10 .We are now ready to prove that every P r n graph is Hamiltonian-connected, and thus AP+2 according to Lemma 16.
Theorem 17.For every n ≥ 4, the graph P r n is Hamiltonian-connected.
Proof: Let G = P r n , and q = n−2 2 if n is even, or q = n−3 2 otherwise.Table 1 (resp.Table 2) exhibits, given two distinct vertices s and t of G, a Hamiltonian path P of G whose endvertices are s and t when n is even (resp.n is odd).In Table 1 (resp.Table 2), it is assumed that 1 ≤ i ≤ q when j is not defined (resp. 1 < i < q), and 0 ≤ i < j ≤ q otherwise (resp. 1 < i < j < q).Every Hamiltonian path which does not appear in these two tables can be deduced from another Hamiltonian path using the symmetries of G.
Corollary 18.For every n ≥ 4, the graph P r n is AP+2.

Conclusion
We summarize Corollaries 10 and 18 and Theorems 12 and 13 in this concluding theorem.
Theorem 19.For every k ≥ 1 and n ≥ k, there exists an optimal AP+k graph on n vertices and n(k+1) 2 edges.This result does not tell much about the number of optimal AP+k graphs on n vertices for some fixed values of k and n.However, this number is upper bounded by the number of edge-minimal (k + 1)connected graphs with order n according to Observation 4.

r− 1
=2k n ) and n r = n r − n r .Denote by v a the last non-prescribed vertex of G[ k−1 =0 D ], and by v b the first non-prescribed vertex of G − k−1 =0 D .Case 3.1.2.1.The vertices v a and v b are adjacent in G.
is odd and n is even, then H k,n is obtained by joining v i and v i+ n 2 in H 2r,n for every i ∈ {0, 1, . . ., n 2 − 1}; • if k = 2r + 1 and n are odd, then H k,n is obtained from H 2r,n by first linking v 0 to both v n