lower bounds on size and independence ink4-free graphs

Lower Bounds on Size andIndependence in K-freeGraphs

Kathryn L. Fraughnaugh1,* and Stephen C. Locke2

1UNIVERSITY OF COLORADOAT DENVER, DENVER, CO

80217E-mail: [email protected]

2FLORIDA ATLANTIC UNIVERSITYBOCA RATON, FL 33431

E-mail: [email protected]

Received April 16, 1997

Abstract: We investigate lower bounds on the size of K4-free graphs. For severalranges of independence relative to order and for graphs with maximum degree 3 and 4,we find sharp lower bounds. We also evaluate Ramsey-type numbers over the classesof graphs with maximum degree 3 and with maximum degree 4. c© 1997 John Wiley & Sons, Inc.

J Graph Theory 26: 61–71, 1997

Keywords: K4-free, independence, Ramsey numbers

1. INTRODUCTION

In the following, a graph G = (V,E) is simple with no loops, its order is ν(G) = |V |, its sizeis ε(G) = |E|, and its independence α(G) is the order of a maximum independent set of G. Agraph of order ν and independence α is a (ν, α)-graph. If we want to emphasize the size s ofthe graph, we will call it a (ν, α, s)-graph. A Km-free (ν, α)-graph with the minimum numberε(m, ν, α) of edges is called a mingraph. Also ∆(G) and δ(G) denote respectively the maximum

*This author previously published under the name Kathryn F. Jones.

c© 1997 John Wiley & Sons, Inc. CCC 0364-9024/97/020061-11

62 JOURNAL OF GRAPH THEORY

and minimum degree of G. For a vertex v ∈ V , we let d(v) be the degree of v,N(v) the set ofneighbors of v,N [v] = N(v) ∪ {v}, and H(v) = G − N [v]. We let N(x, y) = N(x) ∪ N(y)and N [x, y] = N [x] ∪N [y]. The disjoint union of graphs G and H is denoted by G + H and kdisjoint copies of G by kG.

In [6] for some ν andα the values of ε(3, ν, α) and ε(4, ν, α) were derived, and these were usedto find certain of the classical Ramsey numbers. This has led several authors to investigate thesevalues over the class of all triangle-free graphs and over the subclass of triangle-free graphs withfixed maximum degree. In [4] lower bounds on the size of triangle-free graphs with maximumdegree 3 were found. From these bounds, the independence ratio and certain Ramsey-typenumbers originally found by Staton in [10] were readily derived. In [3] lower bounds on the sizeof triangle-free graphs with maximum degree 4 were found, along with the independence ratioand Ramsey-type numbers over this class. In [7] the lower bound found in [3] was found to bea lower bound on size for all triangle-free graphs. The authors of [7] were unaware of Jones'sprior work, and in [5] it is shown that this lower bound is readily derived from the results in [3].Kalbfleisch [6] also investigated lower bounds for ε(4, ν, α) in the process of finding certain otherclassical Ramsey numbers.

Here we investigate lower bounds for ε(4, ν, α) over the class of all K4-free graphs. Weevaluate ε(4, ν, α) in many cases, including all for which the independence is at least one-fourthof the order. We find exact values for ε(4, ν, α) over the subclasses of K4-free graphs withmaximum degree 3 and 4. For the latter, we characterize the mingraphs and find Ramsey-typenumbers.

2. LOWER BOUNDS AND EXACT VALUES FOR ε(4, ν, α)

We begin by deriving a lower bound for ε(4, ν, α).

Theorem 1. If G is a K4-free (ν, α)-graph, then ε(G) ≥ 5ν − 12α. Moreover, if ε(G) =5ν − 12α, then ∆(G) ≤ 5 and either δ(G) ≥ 3 or G has K3 as a component.

Proof. We proceed by induction. The theorem is easily verified for graphs of small order.We assume that both statements are true for all K4-free graphs with fewer than ν vertices andconsider a K4-free graph G on ν vertices. Let v be a vertex of minimum degree, and let H =H(v) = G−N [v]. We consider several cases depending on the degree of v.

Case 1. d(v) = 0. Then H is a K4-free graph of order ν − 1 and independence α − 1.Therefore ε(G) = ε(H), and it follows from the induction hypothesis that ε(G) > 5ν − 12α.

Case 2. d(v) = 1. Then H is a K4-free graph of order ν − 2 and independence at most α −1. So ε(G) ≥ ε(H) + 1 > 5ν − 12α.

Case 3. d(v) = 2. Then H is a K4-free graph of order ν− 3 and independence at most α− 1.Since δ(G) = 2, there are at least three edges incident to vertices in N [v] and there are exactlythree edges if and only if v is in a K3 component of G. It follows that ε(G) ≥ ε(H) + 3 ≥ 5ν− 12α with equality if and only if H is a mingraph and G = H + K3. Note that it follows fromthe argument so far that mingraphs either have K3 as a component or have minimum degree atleast 3.

Case 4. d(v) = 3. Then since G is K4-free, N(v) is K3-free. Thus the minimum numberof edges incident to vertices of N [v] is 7, and this minimum occurs only in the configurationof Figure 1. Thus ε(G) ≥ ε(H) + 7 ≥ 5ν − 12α − 1. Assume by way of contradiction thatε(G) = 5ν− 12α− 1. Then v belongs to the configuration in Figure 1 and H is a mingraph. Letx and y be the pair of independent vertices in N(v). Then G′ = G−N [x, y] is a graph of order

LOWER BOUNDS ON THE SIZE OF K4 -FREE GRAPHS 63

FIGURE 1.

ν − 6 with independence at most α − 2. Thus ε(G′) ≥ 5ν − 12α − 6. Since there are at leastseven edges not in G′, ε(G) > 5ν − 12α, a contradiction. Hence ε(G) ≥ 5ν − 12α.

Case 5. d(v) = 4. Then H is a K4-free graph of order ν − 5 and independence at most α− 1.Thus ε(H) ≥ 5ν − 12α − 13. Now N(v) is a K3-free graph on four vertices and hence canhave at most four edges. Moreover if N(v) has exactly four edges, then it is a 4-cycle. Thus theminimum number of edges incident to vertices of N [v] is 12, and this minimum occurs when vis in the configuration of Figure 2. Then ε(G) ≥ ε(H) + 12 ≥ 5ν − 12α − 1. Assume by wayof contradiction that ε(G) = 5ν − 12α− 1. Then v is in the configuration of Figure 2, and H isa mingraph. Let x and y be an independent pair of vertices in N(v), and let G′ = G−N [x, y].Then G′ has ν − 7 vertices and independence at most α− 2. So ε(G′) ≥ 5ν − 12α− 11. Sincethere are at least 12 edges not in G′, ε(G) ≥ ε(G′) + 12 > 5ν − 12α, a contradiction. Thusε(G) ≥ 5ν − 12α.

Case 6. d(v) ≥ 5. Now G′ = G−{v} is a K4-free graph of order ν − 1 and independence atmost α. Thus ε(G) = ε(G′) + d(v) ≥ 5ν − 12α− 5 + d(v). Since d(v) ≥ 5, ε(G) ≥ 5ν − 12α.Moreover, it follows from this argument that a mingraph G has ∆(G) ≤ 5.

The graphs in Figure 3 demonstrate that the bounds in Theorem 3 are sharp and that ε(4, 7, 2) =11 and ε(4, 8, 2) = 16. When the independence is large relative to the order, we can find otherexact values of ε(4, ν, α).

Theorem 2. Let ν and α be integers for which there exists a K4-free (ν, α)-graph. Then(i) If α < ν/2, then ε(4, ν, α) ≥ ν − α + 1.(ii) If α ≥ ν/2, then ε(4, ν, α) = ν−α and (2α−ν)K1 +(ν−α)K2 is the unique mingraph.

FIGURE 2.


FIGURE 3. The (7, 2, 11)-graph G7,2 and the (8, 2, 16)-graph G8,2 .

Proof. For α ≥ ν/2, the graph (2α−ν)K1 +(ν−α)K2 is an (ν, α)-graph with ν−α edges,which shows that ε(4, ν, α) ≤ ν−α when α ≥ ν/2. We use induction to prove both statements atonce. The statements are easily verified for graphs of small order, so we assume they are true forgraphs with fewer than ν vertices and consider a K4-free graph G with order ν and independenceα. Let v be a vertex of minimum degree, let H = H(v), and consider the possibilities for δ(G).

Case 1. d(v) = 0. Then H is a K4-free graph with ν − 1 vertices and independence α − 1.Hence by the induction hypothesis,H has at least ν−α edges and so doesG.Now if ε(G) = ν−α,then G = K1 +H and ε(H) = ν −α. Thus by the induction hypothesis α− 1 ≥ (ν − 1)/2 andH = (2α− ν − 1)K1 + (ν − α)K2. Then α ≥ (ν + 1)/2 and G = (2α− ν)K1 + (ν − α)K2.

Case 2. d(v) = 1. In this case, H is a K4-free graph with ν − 2 vertices, independence atmost α− 1, and at least ν − α− 1 edges. Then ε(G) ≥ ε(H) + 1 ≥ ν − α, and we get equalityif and only if v is in a K2 component of G and H is a mingraph. It is easy to check that G is(2α− ν)K1 + (ν − α)K2, and since α− 1 ≥ (ν − 2)/2, we have α ≥ ν/2.

Case 3. d(v) ≥ 2. Then ε(G) ≥ ν > ν − α.

Theorem 3. Let ν and α be integers for which there exists a K4-free (ν, α)-graph. Then

(i) If α < ν/3 or α > ν/2, then ε(4, ν, α) ≥ 2ν − 3α + 1.

(ii) If ν/3 ≤ α ≤ ν/2, then ε(4, ν, α) = 2ν−3α and (ν−2α)K3 +(3α−ν)K2 is the uniquemingraph.

Proof. First we make two simplifying observations. The graph (ν − 2α)K3 + (3α− ν)K2

exists if and only if ν/3 ≤ α ≤ ν/2, and when it exists, it is a (ν, α)-graph with 2ν − 3α edges.Next observe that if ε(4, ν, α) = 2ν − 3α, then it follows from Theorem 1 that 2ν − 3α ≥5ν − 12α and hence α ≥ ν/3. Thus to complete the proof we only need to demonstrate thatε(4, ν, α) ≥ 2ν − 3α, that ε(4, ν, α) = 2ν − 3α implies α ≤ ν/2 and that there is only onemingraph. Now we proceed to prove both statements by induction. The theorem is easily verifiedfor graphs of small order, and we assume the statements are true for all values less than ν. Weconsider a K4-free graph G with ν vertices and independence α. Let v be a vertex of minimumdegree, let H = H(v), and consider the various possibilities for the minimum degree of G.


Case 1. d(v) = 0. Then H is a K4-free graph of order ν − 1 and independence α − 1. Thusε(G) = ε(H) ≥ 2ν − 3α + 1.

Case 2. d(v) = 1. Then H is a K4-free graph with order ν − 2, independence at most α− 1,and at least 2ν − 3α − 1 edges. Then ε(G) ≥ ε(H) + 1 ≥ 2ν − 3α, and we have equalityif and only if v is in a K2 component of G and H is a mingraph. By the induction hypothesisα(H) = α− 1 ≤ (ν − 2)/2 and H = (ν − 2α)K3 + (3α− ν − 1)K2. Thus α ≤ ν/2 and G is(ν − 2α)K3 + (3α− ν)K2.

Case 3. d(v) = 2. Then H is a K4-free graph with ν−3 vertices, independence at most α−1,and at least 2ν − 3α− 3 edges. Since δ(G) ≥ 2, there are at least three edges incident to N(v).Thus ε(G) ≥ ε(H)+3 ≥ 2ν−3α, and we have equality if and only if v is in a K3 component ofG andH is a mingraph. Thusα(H) = α−1 ≤ (ν−3)/2 andH is (ν−2α−1)K3+(3α−ν)K2.It follows that G is (2ν − 2α)K3 + (3α− ν)K2 and α ≤ ν/2.

Case 4. d(v) = 3. Then H is a K4-free graph on ν − 4 vertices with independence at mostα− 1 and at least 2ν − 3α− 5 edges. Now we argue as in the proof of Theorem 1 that there areat least seven edges incident to N(v) so that ε(G) ≥ ε(H) + 7 ≥ 2ν − 3α + 2.

Case 5. d(v) ≥ 4. Then ε(G) ≥ 2ν > 2ν − 3α + 1.

Theorem 4. If ν/4 < α ≤ ν/3, then ε(4, ν, α) = 5ν − 12α.

Proof. The condition ν/4 < α ≤ ν/3, is equivalent to 3α ≤ ν < 4α. Thus we considerν = 3α+ p, where 0 ≤ p < α. If p is odd, i.e., ν = 3α+ 2j + 1, where j < (α− 1)/2, then letG = jG8,2 + G7,2 + (α− 2j − 2)K3. Such a graph exists and is easily seen to be K4-free with3α + 2j + 1 vertices, independence α and 5ν − 12α edges. Now if p is even, i.e., ν = 3α + 2j,where j < α/2, then let G = jG8,2 + (α− 2j)K3. Again this graph exists and is a (ν, α)-graphwith 5ν − 12α edges. Thus for α in the given range, ε(4, ν, α) ≤ 5ν − 12α. Equality followsfrom Theorem 1.

For α = ν/4, the values of ε(4, ν, α) are considerably harder to find when α is odd. If α iseven, the values are given in the following theorem.

Theorem 5. ε(4, 8m, 2m) = 16m.

Proof. Notice that 5ν − 12α = 16m. Since mG8,2 is an (8m, 2m)-graph with 16m edges,the result follows from Theorem 1.

We have now found the values of ε(4, ν, α) for all ν and α for which ν/4 ≤ α ≤ ν/3with the exception of the case where α is odd and ν = 4α. All of the values we have foundcoincide with the lower bounds in Theorems 1, 2 and 3. We will see shortly that when α isodd, the value of ε(4, 4α, α) may not coincide with the bound of Theorem 1. For example,ε(4, 12, 3) = 26 = 5(12) − 12(3) + 2. In addition to finding the values of ε(4, ν, α), we wantto characterize the mingraphs. This has been accomplished for the values found in Theorem 2and Theorem 3. We still have to accomplish this task for the values given in Theorem 4. As afirst step toward this goal, we investigate some small values of ν and α and the correspondingmingraphs.

Theorem 6. ε(4, 7, 2) = 11, the unique mingraph is G7,2, and there are exactly nine (7, 2)-graphs.

Proof. Theorem 1 implies that ε(4, 7, 2) ≥ 11, and G7,2 demonstrates equality. It followsfrom Kalbfleisch's work in [6] that there are exactly nine distinct graphs on seven vertices withno K3 and independence 3. The complements of such graphs are K4-free (7, 2)-graphs. Onlyone of these graphs has ten edges; its complement is G7,2. The complements of the five graphs


with minimum degree 2 can be obtained by adding edges to G7,2 as follows: (1) add edge cg; (2)add edge cd; (3) add edges cd and df ; (4) add edges cd and be; and (5) add edges cd, bg and ef.The remaining three graphs have minimum degree 1 and their complements can be described asfollows: (1) the join of a 5-cycle and nonadjacent vertices x and y; (2) delete one edge incidentto x from the graph in (1); and (3) delete two edges incident to x whose other endpoints arenonadjacent in the cycle.

Theorem 7. ε(4, 8, 2) = 16, the unique mingraph is G8,2, and there are exactly three (8, 2)-graphs.

Proof. It follows from Theorem 5 that ε(4, 8, 2) = 16. It was shown in [6] that there areexactly three distinct graphs on eight vertices with no K3 and independence 3. The complementsof these are the (8, 2)-graphs. Only one of these has 12 edges and its complement is G8,2. Theremaining two can be obtained from G8,2 by (1) adding the edge be (adding the edge cg gives anisomorphic graph) and (2) adding both of the edges be and cg.

Some other values of ε(4, ν, α) can be derived from the results in Kalbfleisch's dissertation.It follows from [6, Theorem 6.1] that there is a unique K4-free (17, 3)-graph. It is the self-complementary 8-regular Cayley graph on 17 vertices labeled 0 through 16 determined by ij ∈ Eif and only if (i − j) mod 17 ∈ {1, 2, 4, 8, 9, 13, 15, 16}. This graph can also be found in [1].This establishes the following:

Theorem 8. ε(4, 17, 3) = 68, and there is a unique mingraph.

Kalbfleisch also considers colorings of the clique on 16 vertices. Interpreting these resultsin our context, we find that although he does not find ε(4, 16, 3) exactly, he demonstrates that a(4, 16, 3)-graph must have minimum degree 7 and maximum degree at most 8, and that there isno 7-regular (16, 3)-graph. He observes that removal of a vertex from the (17, 3)-graph yields a(16, 3)-graph with 60 edges, thus showing that 57 ≤ ε(4, 16, 3) ≤ 60. The value of ε(4, 16, 3)has since been found to be 60. It appears for example in [8], where it was obtained by computer.McKay and Radziszowski [9] found the value of ε(4, 17, 4) = 41 by computer, as well as newupper and lower bounds for several other of these numbers.

The Ramsey number r(m,n) is the smallest integer such that every graph on r(m,n) verticeseither contains Km or its complement contains Kn. Equivalently, r(m,n) is the smallest integersuch that every Km-free graph on r(m,n) vertices has independence at least n.

The values of ε(4, ν, α) for all graphs on fewer than 12 vertices can be determined usingTheorems 2 through 5. Observe that the Ramsey numbers r(4, 2) = 4 and r(4, 3) = 9, so thatthere are noK4-free graphs with independence 1 for ν ≥ 4 and no graphs with independence 2 forν ≥ 9. The first value of ε(4, ν, α) that cannot be determined using these theorems is ε(4, 12, 3),which we find after giving several useful lemmas.

Lemma 1. If G is a K4-free graph with maximum degree k ≥ 3, then α(G) ≥ ν(G)/k.

Proof. It follows from Brooks' Theorem [2] (or see [1]) that G is k-colorable. Since eachcolor class is an independent set, α(G) ≥ ν(G)/k.

Lemma 2. If G is a (ν, α)-graph and G = G1 + G2, then ε(G) = 5ν − 12α if and only ifε(Gi) = 5ν(Gi) − 12α(Gi) for i = 1, 2.

Proof. The proof is straightforward.

Lemma 3. Every K4-free (4α, α)-graph with maximum degree at most 4 is a 4-regular min-graph.


Proof. It follows from Theorem 1 that a (4α, α)-graph has at least 5(4α) − 12(α) = 8αedges. Since an (4α, α)-graph with maximum degree at most 4 has at most 8α edges, it must be4-regular and a mingraph.

Theorem 9. If G is a K4-free (4α, α)-graph with ∆(G) ≤ 4, then G is the disjoint union of(8, 2)-graphs. As a consequence, ν(G) is a multiple of 8.

Proof. The proof is by induction on α. Since G is K4-free, there is no (4, 1)-graph andTheorem 7 characterizesG8,2 as the unique (8, 2)-graph. LetG be a (4α, α)-graph with ∆(G) ≤ 4and α ≥ 3. By Lemma 3, we know that G is 4-regular and a mingraph. Now let v be any vertexof G, and let H = H(v). Since G has no K4, the neighbors of v can be partitioned into twoindependent sets{x1, y1} and{x2, y2}.For i fixed, letQ = H−N(xi, yi). If |N(xi, yi)∩H| ≤ 2,then Q has at least 4α − 7 vertices and independence at most α − 2. This is impossible sinceα(Q) ≥ ν(Q)/4 by Lemma 1.

Suppose |N(xi, yi) ∩H| = 3. Then Q has 4α− 8 vertices and independence at most α− 2.If α(Q) ≤ α − 3, then Theorem 1 implies that ε(Q) ≥ 8α − 4. But we have already countedseven edges not in Q, so this is impossible. Thus α(Q) = α − 2 and it follows from Lemma 3that Q is 4-regular. Therefore, G = Q+ (G−Q). Since G−Q is an (8, 2)-graph, the inductionhypothesis yields the desired decomposition of G.

To complete the proof, we assume |N(xi, yi) ∩ H| ≥ 4 for i = 1, 2. Now there are fouredges at v and at least eight more edges between N(v) and H(v). To account for the degreesof the vertices in N(v) being 4, there are at least two additional edges incident to N(v). Thusε(H) ≤ 8α − 14. However, H is a graph on 4α − 5 vertices with independence at most α − 1and, by Theorem 1, at least 8α− 13 edges, a contradiction.

Theorem 10. ε(4, 12, 3) = 26.

Proof. Let G be a K4-free (12, 3)-graph. Since r(4, 3) = 9, δ(G) ≥ 3. Suppose first thatδ(G) = 3, and let v be a vertex with degree 3 and H = H(v). Then H is an (8, 2)-graph with atleast 16 edges. Moreover, if ε(H) = 16, then H is G8,2. Now consider N(v); it is triangle-freeand has independence 2 or 3. If α(N(v)) = 3, then each of the eight vertices of H(v) is joinedto at least one vertex of N(v). Thus ε(G) ≥ 27. If α(N(v)) = 2, then N(v) contains either oneor two edges. In either case, all but three vertices in H must be adjacent to at least one of eachpair of independent vertices in N(v). If N(v) has two edges, then there is one independent pairof vertices, and counting edges, we get ε(G) ≥ 26. If N(v) has one edge, then let xy be thisedge and z the third vertex. Thus there must be five edges incident to each of {x, z} and {y, z}.The smallest number of edges that accomplishes this and maintains minimum degree is six: fouredges incident to z and one incident to each of x and y. Thus ε(G) ≥ 26.

Now suppose δ(G) = 4. Then ε(G) ≥ 24, and we have equality if and only if G is 4-regular.It then follows from Theorem 10 that G is the disjoint union of graphs isomorphic to G8,2.This is impossible, so ε(G) ≥ 25 and G must contain a vertex of degree at least 5. Since Gis connected, we may choose v with degree 4 adjacent to some such vertex. Let a = ε(N(v))and b = ε([N(v), H]). Then 2a + b + 4 =

∑w∈N(v) d(w) ≥ 17. Since H is a (7, 2)-graph,

ε(H) ≥ 11 and ε(G) ≥ 4 + a + b + 11 ≥ 28 − a. Thus if a ≤ 2, we have ε(G) ≥ 26.Since N(v) is triangle-free, we only need consider a = 3 or 4. Then either (i) N(v) is a path

or cycle on the vertices x1, x2, x3, x4 or (ii) N(v) has a vertex x with neighbors y1, y2, y3. If (i),then b ≥ 8 since {x1, x3} and {x2, x4} are independent sets and hence must be adjacent to atleast four of the seven vertices in H. Thus ε(G) ≥ 26. If (ii), each vertex of H has an edge toat least one of yi. So ε(G) ≥ 25, with equality only if b = 7. Suppose that b = 7. We observethat each independent pair {yi, yj} is adjacent to at least four vertices in H . Thus we must have


d(y1) = d(y2) = 4 and d(y3) = 5. Now |N(y1) ∪N(y2)| ≤ 8; hence G−N(y1) −N(y2) is agraph on at least four vertices having independence at least 2. Thus G contains an independentset of order 4 in G, a contradiction. Thus b > 7, and ε(G) ≥ 26.

Next we give a graph with minimum degree 3 that has 26 edges. To the graph G8,2 in Figure3, add four vertices v, x, y, z and edges vx, vy, vz, xz, yz, xa, xc, xf, yb and yh.

The (12, 3, 26)-graph given in the proof above is not unique. Any arrangement of five edgesbetween {x, y} and G8,2 that does not produce a K4 will give a (12, 3, 26)-graph. There arealso (12, 3, 26)-graphs with minimum degree 4. For example, to the graph G7,2 in Figure 2, addvertices v, x, y, z, w and edges vx, vy, vz, vw, xy, zw, xd, xf, yd, yf, zc, zg, wc, wg and be.

We can now find estimates on the values of ε(4, ν, α) for the remaining case where α ≥ ν/4.We conjecture that the upper bound given in the following theorem is the correct value, but aproof apparently depends on finding values of the Ramsey-type numbers that are investigated inSection 3 beyond those found there.

Theorem 11. For m ≥ 1, 16m + 8 ≤ ε(4, 8m + 4, 2m + 1) ≤ 16m + 10 or, equivalently,when α is odd, 5ν − 12α ≤ ε(4, 4α, α) ≤ 5ν − 12α + 2.

Proof. The lower bound follows from Theorem 1. To demonstrate the upper bound we observethat any graph constructed by taking the disjoint union of a (12, 3, 26)-graph and (m − 1)G8,2

has 8m + 4 vertices, independence 2m + 1 and 16m + 10 edges.

3. RAMSEY NUMBERS AND VALUES OF ε(4, ν, α)WITH RESTRICTIONS ON MAXIMUM DEGREE

Now we will examine graphs with an additional maximum degree restriction. Let rk(m,n) bethe smallest integer such that every Km-free graph with maximum degree at most k and rk(m,n)vertices has independence at least n. Let εk(m, ν, α) be the minimum number of edges in a Km-free (ν, α)-graph with maximum degree at most k. The exact values of rk(3, n) and εk(3, ν, α)for k = 3 are found in [10] and for k = 4 in [3]. We find the values of rk(4, n) and εk(4, ν, α)when k = 3 and when k = 4.

Theorem 12. For every positive integer n,

r3(4, n) = 3n− 2.

Proof. If G is a K4-free graph with maximum degree at most 3, it follows from Lemma 1 thatα(G) ≥ ν(G)/3. Thus if G has 3n− 2 vertices, α(G) ≥ n. Since (n− 1)K3 has order 3n− 3and independence n− 1, the result follows.

For a K4-free graph G with ∆(G) ≤ 3, it follows from Lemma 1 that α(G) ≥ ν(G)/3. Thusthe values of ε3(4, ν, α) are given by Theorems 2 and 3. We summarize these values in the nexttheorem.

Theorem 13. For all ν and α for which there exists a K4-free (ν, α)-graph with maximumdegree at most 3,

ε3(4, ν, α) =

{ν − α if α ≥ ν/2,2ν − 3α otherwise.

When ∆(G) ≥ 4, finding the values requires somewhat more work.

Theorem 14. For every positive integer m,

r4(4, 2m− 1) = 8m− 7,


r4(4, 2m) = 8m− 4,

and a K4-free (8m− 1, 2m)-graph with maximum degree at most 4 has minimum degree at least3 and contains at most six vertices of degree 3.

Proof. The proof is by induction on m, and we prove all statements simultaneously. Clearlyr4(4, 1) = 1 and r4(4, 2) = 4. Let G be a K4-free (7, 2)-graph with ∆(G) ≤ 4, and let vbe a vertex of minimum degree. Then ν(H(v)) = 7 − (1 + δ(G)) and ν(H(v)) < r4(4, 2)together imply that δ(G) ≥ 3. Since ∆(G) ≤ 4 and 5ν− 12α = 11, G has at most six vertices ofdegree 3.

Assume the statement is true for all values less than m and that m ≥ 2. In the following weassume unless stated otherwise that all graphs are K4-free and have maximum degree at most 4.The graph (m − 1)G8,2, which has 8m − 8 vertices and independence 2m − 2, demonstratesthat r4(4, 2m − 1) ≥ 8m − 7. It follows from Lemma 1 that a graph with 8m − 7 vertices hasindependence at least (8m− 7)/4, that is, at least 2m− 1. So r4(4, 2m− 1) = 8m− 7.

The graph (m − 1)G8,2 + K3 has 8m− 5 vertices and independence 2m− 1, which demon-strates that r4(4, 2m) ≥ 8m− 4. Now consider a graph G with 8m− 4 vertices. We assume byway of contradiction thatα(G) < 2m. Since 8m−4 > r4(4, 2m−1) = 8m−7, α(G) ≥ 2m−1.Let α(G) = 2m− 1, v be a vertex with degree δ(G), and H = H(v). Since α(H) ≤ 2m− 2, itfollows that r4(4, 2m− 1) > ν(H) = 8m− 4 − (1 + δ(G)). Thus δ(G) > 2.

Suppose δ(G) = 3. Then ν(H) = 8m−8 and α(H) ≤ 2m−2. Since 8m−8 > r4(4, 2(m−1)) = 8m − 12, then α(H) = 2m − 2. Now H is an (8(m − 1), 2(m − 1))-graph and hence,by Lemma 3, is 4-regular. Since ∆(G) ≤ 4, G = N [v] + H, which forces N [v] to be K4, acontradiction.

Therefore, G is 4-regular, and H has 8m−9 vertices and independence at most 2m−2. Since8m− 9 > r4(4, 2(m− 1)), it follows that α(H) = 2m− 2. Since G has no K4, the neighborsof v can be partitioned into two independent sets {x1, y1} and {x2, y2}. For i fixed, let Q =H −N(xi, yi). Then α(Q) ≤ 2m− 3. If {xi, yi} is joined to fewer than four vertices in H, thenν(Q) ≥ ν(G)−8 = 8m−12 = r4(4, 2(m−1)).This impliesα(Q) ≥ 2(m−1), a contradiction.So there must be eight edges joining N(v) to H . But H is an (8(m− 1) − 1, 2(m− 1))-graphand by the induction hypothesis has minimum degree 3 and at most six vertices of degree 3. Thusthere can be at most six edges between N(v) and H . Therefore, a graph on 8m− 4 vertices musthave independence at least 2m, and hence r4(4, 2m) = 8m− 4.

To complete the proof, we must show that an (8m − 1, 2m)-graph has minimum degree 3and at most six vertices of degree 3. Let G be such a graph and let v and H be as before.Then r4(4, 2m) = 8m − 4 > ν(H) = 8m − 1 − (1 + δ(G)) implies that δ(G) ≥ 3. Since∆(G) ≤ 4, 16m − 2 ≥ ε(G). Then ε(G) ≥ 5ν − 12α = 16m − 5 implies that G contains nomore than six vertices of degree 3.

Theorem 15. For all ν and α for which there exists a K4-free (ν, α)-graph with maximumdegree at most 4,

ε4(4, ν, α) =

ν − α for ν/2 ≤ α,2ν − 3α for ν/3 ≤ α < ν/2,5ν − 12α otherwise.

Proof. The results follow from Theorems 2 and 3 for α ≥ ν/3 and from Theorem 4 forν/4 < α < ν/3. Suppose next that G is an (ν, α)-graph with maximum degree at most 4 andν = 4α. If α is even, the result follows from Theorem 5. If α is odd, let α = 2m − 1. Then G


is an (8m − 4, 2m − 1)-graph. However, in Theorem 14 we found that r4(4, 2m) = 8m − 4,which would imply that α(G) ≥ 2m. So there is no such graph.

Now we turn to the problem of characterizing the mingraphs in the class of graphs withmaximum degree 4.

Theorem 16. If G is a K4-free graph with ∆(G) ≤ 4 and ε(G) = 5ν − 12α, then eachcomponent of G is G8,2, G7,2, or K3.

Proof. We first observe that if K3 is a component of G, then by Lemma 2 the rest of Gsatisfies the conditions of the theorem, and we have the desired decomposition. It follows fromTheorem 1 that we only need prove the theorem when δ(G) ≥ 3.

Choose v with degree δ(G), and defineH as usual. First suppose δ(G) = 3. Then by Theorem1, ε(H) ≥ 5ν−12α−8 = ε(G)−8. Let N(v) = {x, y, z}. Each of x, y and z has degree at least3. If each of them has two edges to H,G has too many edges. Thus, without loss of generality,there is an edge between x and y. Furthermore, since G is K4-free, we may assume that yz is notan edge.

Now let Q = H − N(y, z) and let a = |N(y, z) ∩H|. Since we have accounted for all butfour edges outside of H, a ≤ 4. If a = 4, then we have eight edges incident to N(v) while x isleft with degree 2. So a ≤ 3. If a = 1, then Q has ν − 5 vertices, independence at most α − 2,and, by Theorem 1, at least ε(G) − 1 edges, which is impossible. If a = 2, then Q has ν − 6vertices, independence at most α − 2, and at least ε(G) − 6 edges. But for x to have degree atleast 3, we require at least seven edges incident to N(v). Thus a = 3.

Let N(y, z)∩H = {p, q, r}, and note that Q has ν − 8 vertices, independence at most α− 2,and hence at least ε(G) − 11 edges. Since there are three edges incident to {p, q, r}, we haveaccounted for all but four edges outside of Q. Let M = G − Q. In order that the degrees ofp, q, r, and x be at least 3 each, we must use at least seven edge ends. This requires four edgesoutside of Q, leaving at most one edge from Q to M. Thus ε(Q) = ε(G)− 11. Now suppose thatM has an independent set I of order 3. Then let S = G − I −N(I), and note that S misses atmost eight vertices of G, has independence at most α − 3, and at least ε(G) − 4 edges. This isa contradiction since S is contained in Q. Therefore, M has seven vertices and independence 2.So M = G7,2, G = M + Q, and each component of Q is K3, G7,2, or G8,2.

Finally suppose G is 4-regular. Then ε(G) = 2ν = 5ν − 12α implies α = ν/4, and the resultfollows from Theorem 11.

We will investigate the values of rk(4, ν) and εk(4, ν, α) for k ≥ 5 in future work.

References

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