2015/7/16, Pondicherry India, Degree Pairs of a Graph, (International Conference on Graph Theory and its Application, 2015/7/15-16)

The degree pairs of a graph (Preliminary Report)

Chih-wen Weng

Department of Applied Mathematics, National Chiao Tung University

10:30-11:30, July 15, 2015

Let G be a simple connected graph with vertex set VG ={1, 2, . . . , n} and edge set EG. Let d_i and m_i be thedegree andaverage 2-degreeof the vertex i∈ VG respectively, define as follows.

d_i:=|N(i)|, m_i:=1

d_i

∑

ji∈EG

d_j,

where N(i) means the set{j ∈ VG | ji ∈ EG} of neighbors of i.

The sequence of degree pairs (d

, m

)

r r

r r

r r

(3,⁷₃)

(2,⁵₂) (2, 3)

(2, 2)

(2,⁵₂) (2,⁵₂)

r r

r r

r r

Figure: Two graphs whose sequences of degree pairs (di, mi) are different.

i i

r r

r r

r r r r (3,¹³₃)

(7, 3) r

r

r r

r r r r (3,¹³₃ )

Figure: Two graphs have the same sequence of degree pairs.

Motivation

A graph G isk-regular if d_i = k for all vertices i∈ VG, and is pseudo k-regularif m_i= k for all vertices i∈ VG.

In a two-side communication network, a node i of course knows the numberd_i of nodes which are adjacent to i.

A node i might not know exactly how may nodes adjacent to each of its neighbors, but has rough idea of the mean numberm_i of neighbors of its adjacent nodes.

Motivation

The pair (d_i, mi) appears often in the study of maximum eigenvalue ℓ₁(G) of the Laplacian matrixL = D− A associated with G.

(i) In 1998, Merris gave the following bound [1998M]:

ℓ₁(G)≤ max

i∈VG{di+ mi} .

(ii) Also in 1998, Li and Zhang gave the following bound [1998LZ]:

ℓ₁(G)≤ max

ij∈EG

{d_i(di+ mi) + dj(dj+ mj) d_i+ dj

} . (iii) In 2001, Li and Pan gave the following bound [2001LP]:

ℓ₁(G)≤ max

i∈VG

{√2di(di+ mi) }

. (iv) In 2004, Das gave the following bound [2004D]:

{ }

(v) Also in 2004, Zhang gave the following bounds [2004Z]:

(va)

ℓ₁(G)≤ max

ij∈EG

{ 2 +

√

di(di+ mi− 4) + dj(dj+ mj− 4) + 4 }

.

(vb)

ℓ₁(G)≤ max

i∈VG

{ di+√

dimi

} .

(vc)

ℓ₁(G)≤ max

ij∈EG

{√

di(d_i+ m_i) + d_j(d_j+ m_j) }

.

Motivation

For this moment, we rearrange the vertices of G by 1, 2,· · · , n such that m₁ ≥ m2≥ · · · ≥ mn. Let a₁(G) is the maximum eigenvalue ofadjacency matrixA associated with G. Then

(i) a₁(G)≤ m1. (A simple application of Perron-Frobenius Theorem) (ii) (2011, Chen, Pan and Zhang [2011CPZ]) Let

a := max{di/dj | 1 ≤ i, j ≤ n} . Then a₁(G)≤ m₂− a +√

(m₂+ a)²+ 4a(m₁− m2)

2 .

(iii) (2014, Huang and Weng [2014HW]) For any b≥ max {di/dj | ij ∈ EG} and 1 ≤ ℓ ≤ n,

m − b +√

(m + b)²+ 4b∑_l−1

(m − m )

This talk emphasizes more on combinatorics than linear algebra.

It is easy for a graph (resp. a pair of prime numbers) to generate its sequence of degree pairs (resp. its product), but much harder for the reverse.

Can we determine which graphs G to have the prescribed sequence of the pairs (d_i(G), m_i(G)) = (d_i, m_i).

( d_i m_i

)

=

( 3 2 2 2 1

5 3

5 2

5

2 2 3

) ,

r r

r r r

( 3 2 2 2 1

2 ⁵₂ ⁵₂ 2 2 )

r r

r r r

Figure: Two graphs uniquely determined by their sequences of degree pairs.

.Lemma 0.1 ..

...

∑

i∈VGd_im_i =∑

i∈VGd²_i. .Proof.

..

...

∑

i∈VG

d_im_i = ∑

i∈VG

d_i

∑

ji∈EGd_j

d_i = ∑

j∈VG

∑

ij∈EG

d_j= ∑

j∈VG

d²_j.

Another feasible condition

Like a property of degree sequence, we have the following.

.Lemma 0.2 ..

...There are even number of odd values dⁱm_i among i∈ VG.

.Proof.

..

...

Since∑

i∈VGd_i is even, there are even number of odd d_i, and so does d²_i. Hence∑

i∈VGd_im_i=∑

i∈VGd²_i is even.

.Corollary 0.3 ..

...

∑

i∈VG

m²_i ≥ ∑

i∈VG

d²_i with equality iff m_i= di= k for all i.

.Proof.

..

...

(∑

i∈VG

d²_i)(∑

i∈VG

m²_i)≥ (∑

i∈VG

d_im_i)² = (∑

i∈VG

d²_i)²

and equality iff m_i = cdi, where c = 1 by the above lemma. This is also equivalent to that all neighbors of a vertex of minimum degree k also have degree k.

Degrees give hints of graph properties, e.g. ∑

i∈VGd_i= 2|EG|.

Degree pairs give more of the graph structure.

.Proposition 0.4 ..

...If maxⁱ∈VGd_im_i ≥ n then the graph has girth at most 4.

.Proof.

..

...

If the graph has girth at least 5 then

n− 1 = |VG| − 1 ≥ |G1(i)| + |G2(i)| = dim_i. for any i∈ VG.

( d_i m_i

)

=

( 3 2 2 2 1

5 3

5 2

5

2 2 3

) ,

r r

r r r

( 3 2 2 2 1

2 ⁵₂ ⁵₂ 2 2 )

r r

r r r

Figure: Two graphs uniquely determined by their sequence of degree pairs.

max d_im_i≥ 5 = n ⇒ ∃K3 or C₄.

Let G² be the square of G, i.e.

VG²= VG and EG² ={xy | d(x, y) ≤ 2}.

The coloring of G² applies to solve data aggregation problem and collision�avoidance problem in a wireless sensor network G.

Using probability method, we have the following.

.Proposition 0.5 ..

...

α(G²)≥ ∑

i∈VG

1 1 + dim_i,

where α(G²) is the independence number of the square of G.

.Proof.

..

...

If a vertex is picked equally in random then the probability of a vertex i appears before those vertices in G₁(i)∩ G2(i) is (1 +|Gi(i)| + |G2(i)|)⁻¹. Hence the expected size of a set consisting of these i is

∑

i∈VG(1 +|Gi(i)| + |G2(i)|)⁻¹, which is at least ∑

i∈VG 1 1+dim_i.

We now turn to the study of pseudo k-regular graph, i.e. m_i= k for all i.

Pseudo k-regular graphs for k = 3, 4, 5

We try to find some theories for pseudo k-regular graphs.

From the definition of pseudo k-regular graphs, k∈ Q, but indeed we have the following.

.Proposition 0.6 ..

...If G is pseudo k-regular then k ∈ N.

.Proof.

..

...

Let A be the adjacency matrix of G, and note that (d1, d2, . . . , dn)A = k(d1, d2, . . . , dn).

Being a zero of the characteristic polynomial of A, k is an algebraic integer.

Since k is also a positive rational number, k is indeed a positive integer.

It is natural to ask when a pseudo k-regular graph attains the maximum number of edges when the order n of a graph is given.

.Theorem 0.7 ..

...

A pseudo k-regular graph has at most nk/2 edges, and the maximum is obtained iff the graph is regular.

.Proof.

..

...

From

2k|EG| = ∑

i∈VG

d_im_i = ∑

i∈VG

d²_i ≥ (∑

i∈VG

d_i)²/n = 4|EG|²/n,

we have|EG| ≤ nk/2 and equality is obtained iff di is a constant.

The next is to ask when a pseudo k-regular graph attains the minimal number of edges when the order n of a graph is given.

.Definition 0.8 ..

...

Let T_k be the tree of order k³− k²+ k + 1 whose root has degree k²− k + 1 and each neighbor of the root has k − 1 children as leafs.

r

r r r

r r r

Figure: The tree T2.

r r r r r r r r r r r r r r

r r r r r r r

r

Figure: The tree T3.

The first two cases of pseudo k-regular graphs are easy to settle.

.Lemma 0.9 ..

...If G is connected pseudo 1-regular then G is K2. .Lemma 0.10

..

...If G is connected pseudo 2-regular then G is a cycle or T2. .Proof.

..

...

Note that ∆(G) = 2 or 3, and the first implies that G is a cycle and the latter implies that G = T₂.

We shall study the connected pseudo k-regular graphs of order n which attain the minimum number of edges, i.e. pseudo k-regular trees if it exists.

We also want to find a connected pseudo k-regular graph of order n whose maximum degree is maximal among all connected pseudo k-regular graph of order n.

It turns out that both problems have the same graph as their solutions.

The following is a technical but useful proposition.

.Lemma 0.11 ..

...

d_i≤ mi(mj− 1) + 1

for any j with ji∈ EG and dj ≤ mi. Moreover the above equality holds iff d_j = mi and all neighbors of j have degree 1 except the neighbor i of j.

.Proof.

..

...

Pick j such that ji∈ EG and dj ≤ mi. Then d_jm_j≥ di+ (d_j− 1) · 1. Hence m_i(mj− 1) + 1 ≥ dj(mj− 1) + 1 ≥ di.

.Theorem 0.12 ..

...

Let G be a connected graph with m_i≤ k (for example G is a pseudo k-regular graph) for all i∈ VG, where k ∈ N. Then

∆(G)≤ k²− k + 1.

Moreover the following (i)-(iv) are equivalent.

(i) ∆(G) = k²− k + 1.

(ii) G is the tree T_k.

(iii) G is a pseudo k-regular tree.

(iv) G has a vertex j such that d_j= mj= k and all neighbors of j have degree 1 with one exception.

Proof of the Theorem 0.12

Choose i such that d_i= ∆(G). Then by Lemma 0.11,

∆(G) = d_i≤ mi(m_j− 1) + 1 = k²− k + 1 for any j withji∈ EG and d_j ≤ mi. Moreover ∆(G) = k²− k + 1 iff d_j = mj = mi= k and d_z= 1 for all neighbors z̸= i of j. Hence (i) and (ii) are equivalent.

The implications of (ii)⇒(iii) and (iii)⇒(iv) are clear.

Assume that (iv) holds, and let i be the unique neighbor of j with degree d_i ̸= 1. Then k² = djm_j = (k− 1) + di to conclude that d_i= k²− k + 1. By the first statement of the theorem, ∆(G) = k²− k + 1. This proves (i).

Just ten days ago when we presented the last theorem in the conference

“2015 International Conference on Graph Theory and Combinatorics &

Eighth Cross-strait Conference on Graph Theory and Combinatorics”, Professor Hou, Yaoping (Hunan Normal University) told us that the theorem had been proved in the following article:

S. Grünewald, Harmonic Trees, Applied Mathematics Letters, 15 (2002), 1001-1004.

Pseudo regular graphs are called harmonic graphsin that paper.

Let G be a pseudo k-regular graph.

The unique neighbor of a vertex of degree 1 of course has degree k in G.

We have seen in the previous proof that any neighbor of a vertex of degree k²− k + 1 also has degree k in G.

We are interested in what other vertices have their neighbors of the same degree k.

Lemma 0.13 ..

...

Let G be a pseudo k-regular graph. Let ij be an edge with 2≤ dj< k.

Then

2≤ di≤ k²− 3k + 4,

with the second equality iff all neighbors of j except i have degree d_j = 2.

.Proof.

..(i) is clear.

Note that d_i ̸= 1, otherwise dj = k, a contradiction. Indeed dz̸= 1 for any neighbors z of j. Hence

d_i+ 2(dj− 1) ≤ djm_j = djk.

Hence

d_i≤ dj(k− 2) + 2 ≤ k²− 3k + 4.

.Corollary 0.14 ..

...

Let G be a pseudo k-regular graph of order n with a vertex of degree d_i ≥ k²− 3k + 5. Then

(i) Every neighbor j of i has degree d_j= k;

(ii) The order of G is at least f(k) :=⌈

(5k⁴− 31k³+ 94k²− 140k + 100)/k²⌉ .

Note that for k = 3, k²− 3k + 5 = 5 and f(3) = 11.

.Proof

..(i) From Lemma 0.13(i) d_j ̸= 1, and from Lemma 0.13(ii) dj≥ k. This is true for all neighbors j of i. Hence d = k.

.Proof ..

...

(ii) From∑

w∈VGd²_w=∑

w∈Gd_wm_w, d²_i + dik²+ ∑

w̸∈{i}∪G1(i)

d²_w= kdi+ k²d_i+ ∑

w̸∈{i}∪G1(i)

kd_w.

Hence

k⁴− 7k³+ 22k²− 35k + 25 ≤ ∑

w̸∈{i}∪G¹(i)

d_w(k− dw)

≤ (k

2 )2

(n− 1 − (k²− 3k + 5)).

The family E

of pseudo k-regular graphs

LetEk be a family of graphs constructed as the following. Firstly pick a bipartite (k− 1)-regular graph of order 2(2k − 1) with bipartition X ∪ Y, where|X| = |Y| = 2k − 1. Then add a new vertex connecting to all vertices of X. One can check thatgraphs inEk are pseudo k-regular of order 4k− 1 with maximum degree 2k − 1.

ri

X, 2k− 1

Y, 2K− 1 Y, 2K− 1 Y, 2K− 1 (k− 1)-regular

r r r r r

r r r r r

r

From Corollary 0.14(ii), we know a pseudo 3-regular graph with maximum degree at least 5 has at least f(3) = 11 vertices. All the graphs in E3 are extremal for this property.

Pseudo 3-regular graphs

Now we restrict our attention to pseudo 3-regular graph G.

Note that the maximum degree 3≤ ∆(G) ≤ k²− k + 1 = 7 and the case

∆ = 7 is solved in by Theorem 0.12.

.Lemma 0.15 ..

...

Let G be a pseudo 3-regular graph with a vertex i of degree d_i= 6. Then all neighbors j of i have degree d_j= 3, and the neighbors of j have degree sequence (6, 1, 2).

.Lemma 0.16 ..

...

Let G be a pseudo 3-regular graph with a vertex i of degree d_i= 5. Then all neighbors j of i have degree d_j= 3, and the neighbors of j have degree sequence (5, 2, 2) or (5, 3, 1).

.Lemma 0.17

..Let G be a pseudo 3-regular graph. Then the neighbor degree sequence of a vertex of degree 4 is (3, 3, 3, 3), (4, 3, 3, 2), or (4, 4, 2, 2).

Pseudo 3-regular graph of order at most 10

We have shown that a pseudo 3-regular graph with maximum degree at least 5 must have at least 11 vertices.

We will list all the pseudo 3-regular graph of order at most 10.

.Lemma 0.18 ..

...

Let G be a connected pseudo 3-regular graph with ∆(G) = 4 and a_j :=|{i ∈ VG | di= j}|

for j = 1, 2, 3, 4. Then (i) a₁+ a2= 2a4, (ii) |VG| = a3+ 3a₄, (iii) a₁≤ a3,

(iv) a₁, a2, a3 have same parity.

.Proof.

..

...

(i) and (ii) follow from solving

0 = ∑

i∈VG

(mi− di)di = ∑

i∈VG

(3− di)di= a₁· 2 + a2· 2 + a4(−4).

(iii) follows since there exists an injection from the set of degree one vertices into set of degree 3 vertices. Since there are even number of vertices of odd degrees, a₁+ a3 is even. The remaining follows from (i) and (ii). This proves (iv).

|VG| = 7:

r r

r r

r

r r

r r r

r r r

r

Figure: Graphs with sequence (n, a4, a₄, a₃, a₂) = (7, 1, 4, 2, 0).

|VG| = 8:

r r r r

r r r r

Figure: The graph with sequence (n, a4, a4, a3, a2) = (8, 2, 2, 2, 2).

|VG| = 9(i):

r r

r r

r

r r

r r r

r r

r r r

r

r r

Figure: Graphs with sequence (n, a4, a4, a3, a2) = (9, 1, 6, 2, 0).

|VG| = 9(ii):

r r

r r r r

r

r r

r r

r r r r

r

r r

r r r r

r

r r

|VG| = 9(iii):

r r

r r r

r r r r

Figure: The graph with sequence (n, a4, a4, a3, a2) = (9, 2, 3, 3, 1).

|VG| = 10(i):

r r r r

r r r r

r r

Figure: The graph with sequence (n, a4, a4, a3, a2) = (10, 2, 4, 0, 4).

|VG| = 10(ii):

r r r r

r r r r

r r

r r r r

r r r r

r r

r r r r

r r r r

r r

r r r r

r r r r

r r

Figure: Graphs with sequence (n, a , a , a , a ) = (10, 2, 4, 4, 0).

References

[2011CPZ] Y. Chen and R. Pan, and X. Zhang, Two sharp upper bounds for the signless Laplacian spectral radius of graphs, Discrete Mathematics, Algorithms and Applications, 3(2011), 185-191.

[2004D] K.C. Das, A characterization on graphs which achieve the upper bound for the largest Laplacian eigenvalue of graphs, Linear Algebra and its

Applications, 376(2004), 173-186.

[2014HW] Y. P. Huang and C. W. Weng, Spectral radius and average 2-degree sequence of a graph, Discrete Mathematics, Algorithms and Applications, 6(2014).

[2001LP] J.S. Li and Y.L. Pan, De Caen’s inequality and bounds on the largest Laplacian eigenvalue of a graph, Linear Algebra and its Applications, 328(2001), 153-160.

[1998LZ] J.S. Li and X.D. Zhang, On Laplacian eigenvalues of a graph, Linear Algebra and its Applications, 285(1998), 305-307.

[1998M] R. Merris, A note on Laplacian graph eigenvalues, Linear Algebra and its Applications, 285(1998), 33-35.

[2004Z] X.D. Zhang, Two sharp upper bounds for the Laplacian eigenvalues, Linear

Thank you for your attention.