A Lemma on Almost Regular Graphs and an Alternative Proof for Bounds on t k mP P

Gravier et al. established bounds on the size of a minimal totally dominant subset for graphs . This paper offers an alternative calculation, based on the following lemma: Let k m P P  , k r so and . Let 3 k  2 r  H be an regular finite graph, and put . 1) If a perfect totally dominant subset exists for , then it is minimal; 2) If and a perfect totally dominant subset exists for , then every minimal totally dominant subset of must be perfect. Perfect dominant subsets exist for when and satisfy specific modular conditions. Bounds for , for all follow easily from this lemma. Note: The analogue to this result, in which we replace “totally dominant” by simply “dominant”, is also true. r


Introduction
Let be a graph.In this paper, each edge of a graph must have two different endpoints; also, two vertices may be linked by at most one edge.A subset Z of vertices is said to totally dominate G if every vertex of G has a neighbor in Z.We say Z perfectly totally dominates if every vertex has exactly one neighbor in Z. Next, suppose that G is finite.In this case, we say a totally dominant subset Z is minimal if Z is the smallest size possible among all dominant subsets.This minimal size is denoted by

For
, we say that a graph G is r-regular if every vertex is the endpoint of exactly r edges.Suppose G is regular.A subset Z which perfectly totally dominates is clearly minimal.If a perfect dominant set does not exist, we can search for minimality among dominant subsets Z by counting "overlaps".That is, for each Z -overlaps is strictly less than the sum of 2 Zoverlaps.
These elementary links between minimality, perfection and overlaps may fail if G is not regular.For arbitrary theorists, a challenge is to specific assertions that apply to a broad family of graphs.
The following conventions graphs, all sorts of behavior is possible.For graph will be used here.(1a) For k   , 2 k  , let k P , the k -path be the graph whose ce the numbers . Two vertices   The co ally dominant subset, determines Z is a sum over  n alternate construction of a lower bound.The bound is met for these perfect subsets.Next, using these subsets, one can establish a general upper bound for k m P P  for all m .

A Tie with Perfection
Gravier [1] proves that the Z consisting of the middle row of 3 n P P  , for any n , s a minimal totally dominant subset.Obviously, this choice of minimal i For each integer 1 j k   , put subset pro blocks, duces many overlaps.By rotating 3 3  we can produce other minimal dominant se th fewer overlaps, as in Figure 3. Furthermore, if n is a multiple of 4, there is a variation which is a perfect total domination of 3 n P C  , as in Figure 4.The flexibility in the number of s which are dominated by more than one member of ts wi vertice Z reflects the presence of vertices of two degrees, namely 3 and 4.
In this example, the size of a minimal, imperfect to

Weights
ts of theorems based on series.tally dominant subset "ties" the size of a perfect totally dominant set.Can a minimal subset be smaller than a perfect one?We prove that a tie is rare, and that beating is impossible.

We have two se
Definition 1 Let r be a real number.Let   r   i a be the set of infinite sequences of real numbers is a real vector space, and the fu real, let i be the unique member of or G a graph and Z a dominant (but possibly not totally) subset, and and v e such that and , For each 1 j k   , 0 (4a) j   , and is odd and is fer to

Suppose is an regular graph, and put
A trivial consequence of this theorem and the prece ume the hypothesis of Theorem 4.
ding lemma is:

Modeled with Matrices
linear algebra model.Our results are based on a simple For convenience, (5) For k   , let   .We identify the real vector space k  with leng olumn vectors.We use trans-th k c pose notation to write these horizontally: We denote the zero vector by d let H be a finite, In particular, lly dominates, then each of these expr   , M r k to be the Note that the case is covered in both parts of this conditional definition.atrix

Relevant Sequences convexity for functions of a i j 
As we shall see, the m

  , M r k
There is a discrete analogy to single real variable.We recall some basics.
be a sequence of real numbers, starting at index We say that the sequence is convex if We say a the sequence is strictly convex if

Proof. ge
We may interchange u and v without loss of nerality.Hence, assume u v  .For each i   , put For eac fo h index i in the last sum, the term has the rmat p q b b  where > p q. Therefore 0 0.
) must be 0. Wh Then every term uppose in the final sum of (8 en 1 i  , we get e first remark is that the sign c eparated from the magnitude.
Lemma 1 an be s Lemma 11 Each member of   the two identities to induct on b  r , and Proof.In what follows, a sum from any integer to k m v 1  is defined to be 0.For this proof, we abbre iate m The sequences i t and , and agree on the first two indices.H e sequence.This gives the equality of (10).
□ Lemma 15 Let ence, they are e sam r be a real number, and let , , 1 , 1 .

The Inverse Matrices
We can now prove Lemma 16 Let k   and .The matric pr duct the lemma by comparing the v en and  times the u, v entry of the ident here a t of cases.ity matrix.T lo Case re a Then and .

Case
. For any , this is Now suppose .Then , and

For any
There are three subcases here.First, suppos .


The recursive definition states that  Hence, th ex- Next, suppose .Then Again, the recursive definition implies that this expression is 0.
There remains only the subcase .
By Lemma 15, this equals , , In the previo first sum is determined by all the parameter is us line, the Lemma 14. Rec r  , n r ot Proof.We start with Part (D), as that is our motivation.Given , , and , , , , , and ssume let

When r = 2
Th s to add some secon-e numerical calculations allow u ry comments on th p Sections a 1 and 1.2.x 2 r  , and put Consequently, a perfect totally dominant subset ist if n is odd.However, since 0 j   for j even domina bsets who , th s th ere may be totally nt su e size "ties" e estimate for a perfect subset.In the case 3 k  , the set consisting of the middle row has row-count   0, , 0 n .

Its image under
The ent re integral i if 1 k  di .
Unlike the case hen k is odd, a bijection.fine the extend nctigraph on this data to be G in which , and where each j  is a w j of v .Remark.A variation on total d inati n no neighbors in Z, or Z. rfect mple Perfect Behavior ts: first, exhibit a subite.

Figure
Figure 1.On ce, k odd.e color dominan

Figure
Figure 3. Two ways to totally dominate 3 n P P  .

0
Let r be a real number.Then Trivial.□ Th Proof. let [2] grid graphs.The program begins with the work herself, Molland and Payan[2]on the tiling question.The solution generates perfectly dominant subsets on n with the "Ma hatten metric" notion of the edge: two les t  r of G have gree  .The vertices of the degree 1 r  form two conn subgraphs.A crude bound for minimal totally dominant subset of G is ected a   2 k H r  .However, this bound is too low by a positiv es e number tim H .We find a subtler minimal bo usin und he t subset is minimal, and bset cannot have fewer members is odd, if a perfect totally e nclusions follow from a formula which, for Z a r  and n minant subset exists, th n every minimal subset is perfect.
Lemma 15, this is the stated formula.□ At last, we introduce weights.Define