Planar Graphs have Independence Ratio at least 3/13
aa r X i v : . [ m a t h . C O ] S e p Planar graphs have independence ratio at least 3 / Daniel W. Cranston ∗ Landon Rabern † Submitted: June 6, 2015; Accepted: August 15, 2016; Published: XXMathematics Subject Classifications: 05C69, 05C10
Abstract
The 4 Color Theorem (4CT) implies that every n -vertex planar graph has an independentset of size at least n ; this is best possible, as shown by the disjoint union of many copies of K . In 1968, Erd˝os asked whether this bound on independence number could be proved moreeasily than the full 4CT. In 1976 Albertson showed (independently of the 4CT) that every n -vertex planar graph has an independent set of size at least n . Until now, this remainedthe best bound independent of the 4CT. Our main result improves this bound to n . An independent set is a subset of vertices that induce no edges. The independence number α ( G ) of a graph G is the size of a largest independent set in G . Determining the independencenumber of an arbitrary graph G is widely-studied and well-known to be NP-complete. In fact,this problem remains NP-complete, even when restricted to planar graphs of maximum degree 3(see, for example, [5, Lemma 1]). Thus, much work in this area focuses on proving lower boundsfor the independence number of some special class of graphs, often in terms of | V ( G ) | . The independence ratio of a graph G is the quantity α ( G ) | V ( G ) | .An immediate consequence of the 4 Color Theorem [2, 3] is that every planar graph hasindependence ratio at least ; simply take the largest color class. In fact, this bound is bestpossible, as shown by the disjoint union of many copies of K . In 1968, Erd˝os [4] suggestedthat perhaps this corollary could be proved more easily than the full 4 Color Theorem. And in1976, Albertson [1] showed (independently of the 4 Color Theorem) that every planar graph hasindependence ratio at least . Our main theorem improves this bound to . Theorem 1.
Every planar graph has independent ratio at least . The proof of Theorem 1 is heavily influenced by Albertson’s proof. One apparent differenceis that our proof uses the discharging method, while his does not. However, this distinction islargely cosmetic. To demonstrate this point, we begin with a short discharging version of thefinal step in Albertson’s proof, which he verified using edge-counting. Although the argumentsare essentially equivalent, the discharging method is somewhat more flexible. In part it was thisadded flexibility that allowed us to push his ideas further. ∗ Department of Mathematics and Applied Mathematics, Virginia Commonwealth University, Richmond, VA,23284. email: [email protected] † LBD Data Solutions, Lancaster, PA, 17601. email: [email protected] . the electronic journal of combinatorics (2016), he proof of our main result has the following outline. The bulk of the work consists in show-ing that certain configurations are reducible , i.e., they cannot appear in a minimal counterex-ample to the theorem. The remainder of the proof is a counting argument (called discharging ),where we show that every planar graph contains one of the forbidden configurations; hence, itis not a minimal counterexample.In the discharging section, we give each vertex v initial charge d ( v ) −
6, where d ( v ) is thedegree of v . By Euler’s formula the sum of the initial charges is −
12. Our goal is to redistributecharge, without changing the sum, (assuming that G contains no reducible configuration) so thatevery vertex finishes with nonnegative charge. This contradiction proves that, in fact, G mustcontain a reducible configuration. To this end, we want to show that G contains a reducibleconfiguration whenever it has many vertices of degree at most 6 near each other, since verticesof degree 5 will need to receive charge and vertices of degree 6 will have no spare charge to giveaway. (We will see in Lemma 6 that G must have minimum degree 5.) Most of the work in thereducibility section goes into proving various formalizations of this intuition.Typically, proofs like ours present the reducibility portion before the discharging portion.However, because many of our reducibility arguments are quite technical, we make the unusualchoice to give the discharging first, with the goal of providing context for the reducible config-urations. (Usually the process of finding a proof switches back and forth between dischargingand reducibility. By necessity, though, the proof must present one of these first.)We start with definitions. A k -vertex is a vertex of degree k ; similarly, a k − -vertex (resp. k + -vertex) has degree at most (resp. at least) k . A k -neighbor of a vertex v is a k -vertex that isa neighbor of v ; and k − -neighbors and k + -neigbors are defined analogously. A k -cycle is acycle of length k . A vertex set V in a connected graph G is separating if G \ V has at leasttwo components. A cycle C is separating if V ( C ) is separating. An independent k -set is anindependent set of size k . When vertices u and v are adjacent, we write u ↔ v ; otherwise u v .For a vertex v , let H v denote the subgraph induced by the 5-neighbors and 6-neighbors of v .Throughout the proof we consider a (hypothetical) minimal counterexample G , which will be atriangulation. In Lemma 2, we show that G has no separating 3-cycle. These properties togetherimply that, for every vertex v , the subgraph induced by the neighbors of v is a cycle. If some w ∈ V ( H v ) has d H v ( w ) = 0, then w is an isolated neighbor of v ; otherwise w is a non-isolated neighbor. A non-isolated 5-neighbor of a vertex v is crowded (with respect to v ) if it has two6-neighbors in H v . We use crowded 5-neighbors in the discharging proof to help ensure that7-vertices finish with sufficient charge, specifically to handle the configuration in Figure 1. v + + Figure 1: A 7-vertex v gives no charge to any crowded 5-neighbor. the electronic journal of combinatorics (2016), (2016),
12. Our goal is to redistributecharge, without changing the sum, (assuming that G contains no reducible configuration) so thatevery vertex finishes with nonnegative charge. This contradiction proves that, in fact, G mustcontain a reducible configuration. To this end, we want to show that G contains a reducibleconfiguration whenever it has many vertices of degree at most 6 near each other, since verticesof degree 5 will need to receive charge and vertices of degree 6 will have no spare charge to giveaway. (We will see in Lemma 6 that G must have minimum degree 5.) Most of the work in thereducibility section goes into proving various formalizations of this intuition.Typically, proofs like ours present the reducibility portion before the discharging portion.However, because many of our reducibility arguments are quite technical, we make the unusualchoice to give the discharging first, with the goal of providing context for the reducible config-urations. (Usually the process of finding a proof switches back and forth between dischargingand reducibility. By necessity, though, the proof must present one of these first.)We start with definitions. A k -vertex is a vertex of degree k ; similarly, a k − -vertex (resp. k + -vertex) has degree at most (resp. at least) k . A k -neighbor of a vertex v is a k -vertex that isa neighbor of v ; and k − -neighbors and k + -neigbors are defined analogously. A k -cycle is acycle of length k . A vertex set V in a connected graph G is separating if G \ V has at leasttwo components. A cycle C is separating if V ( C ) is separating. An independent k -set is anindependent set of size k . When vertices u and v are adjacent, we write u ↔ v ; otherwise u v .For a vertex v , let H v denote the subgraph induced by the 5-neighbors and 6-neighbors of v .Throughout the proof we consider a (hypothetical) minimal counterexample G , which will be atriangulation. In Lemma 2, we show that G has no separating 3-cycle. These properties togetherimply that, for every vertex v , the subgraph induced by the neighbors of v is a cycle. If some w ∈ V ( H v ) has d H v ( w ) = 0, then w is an isolated neighbor of v ; otherwise w is a non-isolated neighbor. A non-isolated 5-neighbor of a vertex v is crowded (with respect to v ) if it has two6-neighbors in H v . We use crowded 5-neighbors in the discharging proof to help ensure that7-vertices finish with sufficient charge, specifically to handle the configuration in Figure 1. v + + Figure 1: A 7-vertex v gives no charge to any crowded 5-neighbor. the electronic journal of combinatorics (2016), (2016), uw x (a) Adjacent vertices u and v , with nonad-jacent common 5-neighbors w and x . uv xw (b) Adjacent vertices u and v , with non-adjacent common neighbors w and x , ofdegree 5 and 6. Figure 2: The two instances of configuration H . As a warmup to our main proof, in this section we give a short discharging proof that everyplanar triangulation with minimum degree 5 and no separating 3-cycle must contain a certainconfiguration, which Albertson showed could not appear in a minimal planar graph with inde-pendence ratio less than . (In fact, finding this proof helped encourage us to begin work onthe present paper.) Lemma A.
Let u and v be adjacent vertices, such that uvw and uvx are 3-faces and d ( w ) = 5 and d ( x ) ; call this configuration H . (See Figure 2.) If G is a plane triangulation withminimum degree 5 and no separating 3-cycle, then G contains a copy of H .Proof. Assume that G has minimum degree 5 and no separating 3-cycle, but also has no copyof H . This assumption leads to a contradiction, which implies the result. An immediate conse-quence of this assumption (by Pigeonhole) is that the number of 5-neighbors of each vertex v is at most j d ( v )2 k . Below, when we verify that each vertex finishes with nonnegative charge, weconsider both the degree of v and its number of 5-neighbors. We write ( a, b ) -vertex to denote avertex of degree a that has b v a charge ch( v ), where ch( v ) = d ( v ) −
6. Note that P v ∈ V ch( v ) =2 | E ( G ) | − | V ( G ) | . Since G is a plane triangulation, Euler’s formula implies that 2 | E ( G ) | − | V ( G ) | = −
12. Now we redistribute the charge, without changing the sum, so that each vertexfinishes with nonnegative charge. This redistribution is called discharging , and we write ch ∗ ( v )to denote the charge at each vertex v after discharging. Since each vertex finishes with non-negative charge, we get the obvious contradiction −
12 = P v ∈ V ch( v ) = P v ∈ V ch ∗ ( v ) >
0. Weredistribute the charge via the following three discharging rules, which we apply simultaneouslyeverywhere they are applicable.(R1) Each 7 + -vertex gives charge to each 5-neighbor.(R2) Each 7 + -vertex gives charge to each 6-neighbor that has at least one 5-neighbor.(R3) Each 6-vertex gives charge to each 5-neighbor.We now verify that after discharging, each vertex v has nonnegative charge. We repeatedlyuse that G has no copy of configuration H . In particular, this implies that the number of5-neighbors for each vertex v is at most d ( v )2 . the electronic journal of combinatorics (2016), (2016),
0. Weredistribute the charge via the following three discharging rules, which we apply simultaneouslyeverywhere they are applicable.(R1) Each 7 + -vertex gives charge to each 5-neighbor.(R2) Each 7 + -vertex gives charge to each 6-neighbor that has at least one 5-neighbor.(R3) Each 6-vertex gives charge to each 5-neighbor.We now verify that after discharging, each vertex v has nonnegative charge. We repeatedlyuse that G has no copy of configuration H . In particular, this implies that the number of5-neighbors for each vertex v is at most d ( v )2 . the electronic journal of combinatorics (2016), (2016), ( v ) = : Each (5 , v has five 6 + -neighbors, so ch ∗ ( v ) > − (cid:0) (cid:1) >
0. Each(5 , v has four 6 + -neighbors, at least two of which are 7 + -neighbors; so ch ∗ ( v ) > − (cid:0) (cid:1) + 2 (cid:0) (cid:1) >
0. Each (5 , v has three 7 + -neighbors (otherwise G contains a copy of H ), so ch ∗ ( v ) = − (cid:0) (cid:1) = 0. d ( v ) = : Each (6 , v has ch ∗ ( v ) = ch( v ) = 0. Each (6 , v has at least two7 + -neighbors, so ch ∗ ( v ) > (cid:0) (cid:1) − (cid:0) (cid:1) = 0. Each (6 , v has four 7 + -neighbors, soch ∗ ( v ) = 0 + 4 (cid:0) (cid:1) − (cid:0) (cid:1) = 0. d ( v ) = : Each (7 , v has ch ∗ ( v ) > − (cid:0) (cid:1) = 0. Each (7 , v has six6 + -neighbors, at least two of which are 7 + -vertices (namely, the neighbors that are two furtherclockwise and two further counterclockwise around v from the 5-vertex; otherwise G has a copyof H ). So ch ∗ ( v ) > − (cid:0) (cid:1) − (cid:0) (cid:1) >
0. Each (7 , + -neighbors, at leastthree of which are 7 + -vertices; so ch ∗ ( v ) > − (cid:0) (cid:1) − (cid:0) (cid:1) >
0. Each (7 , + -neighbors, so ch ∗ ( v ) = 1 − (cid:0) (cid:1) = 0. d ( v ) = : Now v has at most four 5-neighbors, and gives each of these charge ; also v giveseach other neighbor charge at most . Thus ch ∗ ( v ) > − − ) − ) > d ( v ) > : Now v gives each neighbor charge at most , so ch ∗ ( v ) > d ( v ) − − d ( v )( ) = ( d ( v ) − > −
12 = P v ∈ V ch( v ) = P v ∈ V ch ∗ ( v ) >
0. This contradiction implies the result.
In this section we present the discharging argument for the proof of Theorem 1. It is convenientto collect all of the reducibitiy lemmas that we use to analyze the discharging (but prove later).
Lemma 8.
Every independent set J in a minimal G with | J | = 2 , satisfies | N ( J ) | > . Lemma 9.
A minimal G cannot have two nonadjacent -vertices with at least two commonneighbors. In particular, each vertex v in G has d ( v ) or more + -neighbors. Lemma 17.
Every minimal G has no -vertex v with − -neighbors u , u , and u that arepairwise nonadjacent. Lemma 18.
Every minimal G has no -vertex v with pairwise nonadjacent neighbors u , u ,and u , where d ( u ) = 5 , d ( u ) , and d ( u ) = 7 . Lemma 19.
Let u be a -vertex with nonadjacent vertices u and u each at distance two from u , where u is a -vertex and u is a − -vertex. A minimal G cannot have u and u with twocommon neighbors, and also u and u with two common neighbors. Lemma 20.
Every minimal G has no -vertex v with a -neighbor and two other − -neighbors, u , u , and u , that are pairwise nonadjacent. Lemma 21.
Let v , v , v be the corners of a -face, each a + -vertex. Let u , u , u be theother pairwise common neighbors of v , v , v , i.e., u is adjacent to v and v , u is adjacentto v and v , and u is adjacent to v and v . We cannot have | N ( { u , u , u } ) | . Inparticular, we cannot have d ( u ) = d ( u ) = 5 and d ( u ) . Lemma 22.
Let u be a -vertex with nonadjacent -vertices u and u each at distance twofrom u . A minimal G cannot have u and u with two common neighbors and also u and u with two common neighbors. the electronic journal of combinatorics (2016), (2016),
Let u be a -vertex with nonadjacent -vertices u and u each at distance twofrom u . A minimal G cannot have u and u with two common neighbors and also u and u with two common neighbors. the electronic journal of combinatorics (2016), (2016), emma 23. Suppose that a minimal G contains a -vertex v with no -neighbor. Now v cannothave at least five -neighbors, each of which has a -neighbor. Theorem 1.
Every planar graph G has independence ratio at least .Proof. We assume that the theorem is false, and let G be a minimal counterexample to thetheorem; by “minimal” we mean having the fewest vertices and, subject to that, the fewestnon-triangular faces (thus, G is a triangulation). We will use discharging with initial chargech( v ) = d ( v ) −
6. We use the following five discharging rules to guarantee that each vertexfinishes with nonnegative charge, which yields a contradiction.(R1) Each 6-vertex gives to each 5-neighbor unless either they share a common 6-neighbor andno common 5-neighbor or else the 5-neighbor receives charge from at least four vertices;in either of these cases, the 6-vertex gives the 5-neighbor .(R2) Each 8 + -vertex v gives + h w to each 6 − -neighbor w where h w is the number of 7 + -verticesin N ( v ) ∩ N ( w ).(R3) Each 7-vertex gives to each isolated 5-neighbor; gives 0 to each crowded 5-neighbor; gives to each other 5-neighbor; and gives to each 6-neighbor unless neither the 7-vertex northe 6-vertex has a 5-neighbor.(R4) After applying (R1)–(R3), each 5-vertex with positive charge splits it equally among its6-neighbors that gave it .(R5) After applying (R1)–(R4), each 6-vertex with positive charge splits it equally among its6-neighbors with negative charge.Now we show that after applying these five discharging rules, each vertex v finishes withnonnegative charge, i.e., ch ∗ ( v ) >
0. (It is worth noting that if some vertex v has nonnegativecharge after applying only (R1)–(R3), then v also has nonnegative charge after applying (R1)–(R5), i.e., ch ∗ ( v ) >
0. In fact, the analysis for most cases only needs (R1)–(R3). The final tworules are used only in Cases (iv)–(vi), near the end of the proof.) Since the sum of the initialcharges is −
12, this contradicts our assumption that G was a minimal counterexample. Subjectto proving the needed reducibility lemmas, this contradiction completes the proof of Theorem 1. d ( v ) > : We will show that v gives away charge at most d ( v )4 . To see that it does, let v first give charge to each neighbor. Now let each 6 − -neighbor w take from each 7 + -vertex in N ( v ) ∩ N ( w ). Since G [ N ( v )] is a cycle, each 7 + -neighbor gives away at most the it got from v . Each neighbor of v has received at least as much charge as by rule (R2) and v has given awaycharge d ( v )4 . Now ch ∗ ( v ) > ch( v ) − d ( v ) = d ( v ) − − d ( v ) = ( d ( v ) − > d ( v ) = : Let u , . . . , u denote the neighbors of v in clockwise order. First suppose that v has an isolated 5-neighbor. By Lemma 20, the subgraph induced by the remaining 6 − -neighborsmust have independence number at most 1. Hence v gives away charge at most either + or + 2( ); in either case, ch ∗ ( v ) > ch( v ) − v has no isolated5-neighbor. Suppose first that v has a (non-isolated) 5-neighbor. Now v has at most five total6 − -neighbors, again by Lemma 20. If v has at most four 6 − neighbors, then, since each 6 − -neighbor receives charge at most , we have ch ∗ ( v ) > ch( v ) − ) = 0. By Lemma 20, if v has the electronic journal of combinatorics (2016), (2016),
12, this contradicts our assumption that G was a minimal counterexample. Subjectto proving the needed reducibility lemmas, this contradiction completes the proof of Theorem 1. d ( v ) > : We will show that v gives away charge at most d ( v )4 . To see that it does, let v first give charge to each neighbor. Now let each 6 − -neighbor w take from each 7 + -vertex in N ( v ) ∩ N ( w ). Since G [ N ( v )] is a cycle, each 7 + -neighbor gives away at most the it got from v . Each neighbor of v has received at least as much charge as by rule (R2) and v has given awaycharge d ( v )4 . Now ch ∗ ( v ) > ch( v ) − d ( v ) = d ( v ) − − d ( v ) = ( d ( v ) − > d ( v ) = : Let u , . . . , u denote the neighbors of v in clockwise order. First suppose that v has an isolated 5-neighbor. By Lemma 20, the subgraph induced by the remaining 6 − -neighborsmust have independence number at most 1. Hence v gives away charge at most either + or + 2( ); in either case, ch ∗ ( v ) > ch( v ) − v has no isolated5-neighbor. Suppose first that v has a (non-isolated) 5-neighbor. Now v has at most five total6 − -neighbors, again by Lemma 20. If v has at most four 6 − neighbors, then, since each 6 − -neighbor receives charge at most , we have ch ∗ ( v ) > ch( v ) − ) = 0. By Lemma 20, if v has the electronic journal of combinatorics (2016), (2016), xactly five 6 − -neighbors, then one is a crowded 5-neighbor, which receives no charge from v .So, again, ch ∗ ( v ) > − ) = 0. Finally, suppose that v has only 6 + -neighbors. By Lemma 23, v gives charge to at most four 6-neighbors, so ch ∗ ( v ) > ch( v ) − ) = 0. d ( v ) = : Since ch( v ) = −
1, we must show that v receives total charge at least 1. Let u , . . . , u be the neighbors of v . First suppose that v has five 6 + -neighbors. Now v willreceive charge at least 4( ) unless exactly two of these neighbors are 7-vertices for which v isa crowded 5-neighbor. However, in this case the other three neighbors are all 6-neighbors, soch ∗ ( v ) > − ) + = 0. Now suppose that v has exactly four 6 + -neighbors, say u , . . . , u .If v receives charge from each, then ch ∗ ( v ) > − ) = 0; so suppose that v receives chargefrom at most three neighbors. In total, v receives charge at least from u and u : at least2( ) if u is not a 6-vertex and at least + 0 if u is a 6-vertex. Similarly, v receives at least in total from u and u ; so, ch ∗ ( v ) > − ) = 0. Now suppose that v has exactly three6 + -neighbors, say u , u , u . Lemma 9 implies that u , u , u are consecutive neighbors of v . If u and u are both 6-vertices, then v receives charge from each. If both are 7 + -vertices, then v receives charge from each and charge from u . So assume that exactly one of u and u is a 6-vertex, say u . Now v receives charge from u and charge from each of u and u , fora total of + 2( ). In every case ch ∗ ( v ) > ch( v ) + 1 = 0. d ( v ) = : Note that (R5) will never cause a 6-vertex to have negative charge. Thus, inshowing that a 6-vertex has nonnegative charge, we need not consider it.Clearly, a 6-vertex with no 5-neighbor finishes (R1)–(R3) with nonnegative charge. Supposethat v is a 6-vertex with exactly one 5-neighbor. We will show that v finishes (R1)–(R3) withcharge at least . Let u , . . . , u denote the neighbors of v and assume that u is the only5-vertex. By Lemma 17, at least one of u , u , u is a 7 + -vertex, so it gives v charge . Ifone of u and u is a 6-vertex, then v gives charge only to u , finishing with charge at least2( ) − . Otherwise, v receives charge at least from each of u and u , so finishes with chargeat least 3( ) − . Similarly, if v has no 5-neighbor and at least one 8 + -neighbor, then v finishes(R1)–(R3) with charge at least .Now suppose that v has at least two 5-neighbors. By Lemma 9, At most one of u , u , u can be a 5-vertex. Similarly, for u , u , u ; hence, assume that v has exactly two 5-neighbors.These 5-neighbors can either be “across”, say u and u , or “adjacent”, say u and u .Suppose that v has 5-neighbors u and u . Note that all of its remaining neighbors mustbe 6 + -vertices. At least one of u , u , u must be a 7 + -vertex; similarly for u , u , u . Now weshow that the total net charge that v gives to u , u , u is 0. Similarly, the total net charge that v gives to u , u , u is 0. If both u and u are 7 + -vertices, then v gets from each and gives to u . Otherwise, one of u and u is a 6-vertex and the other is a 7 + -vertex; now v gets from the 7 + -vertex and gives only to u . The same is true for u , u , u . Thus, v finishes withcharge 0.Suppose instead that v has 5-neighbors u and u . By Lemmas 17 and 18 either both of u and u are 7 + -vertices or one is a 6-vertex and the other an 8 + -vertex. The same holdsfor u and u . Let w , . . . , w be the common neighbors of successive pairs of vertices in thelist u , u , u , u , u , u , u . Note that w ↔ w , since u is a 5-vertex and { v, u , w , w , u } ⊆ N ( u ). Similarly, w ↔ w . (See Figure 3.) By Lemma 9, since G has no separating 3-cycle, w and w are 6 + -vertices. Consider the possible degrees for u , u , u , u . Up to symmetry,they are (i) 7 + , + , + , + , (ii) 7 + , + , + ,
6, (iii) 8 + , + , , + , (iv) 8 + , , , + , (v) 6 , , + , + ,and (vi) 6 , + , + , Case (i) , v receives charge at least 4( ), so ch ∗ ( v ) >
0. In
Case (ii) , v receives charge the electronic journal of combinatorics (2016), (2016),
Case (ii) , v receives charge the electronic journal of combinatorics (2016), (2016), u v u u u u w w w w w w Figure 3: The closed neighborhood of v and some nearby vertices.at least + ( + + ) + , so ch ∗ ( v ) > − ( + ) >
0. In
Case (iii) , v receives charge atleast ( + ) + + = . Recall that w is a 6 + -vertex, by Lemma 9. If w is a 6 + -vertex, then v gives only to u , so ch ∗ ( v ) > − ( + ) = 0. So suppose that w is a 5-vertex. Now ineach case v gets charge at least back from u , via (R4). If w is a 6-vertex, then u receivescharge 2( ) + and sends back to each of v and w . Otherwise, w is a 7 + -vertex, so u sends u charge at least , and v gets back at least . Thus, in each instance of Case (iii), we havech ∗ ( v ) >
0. So we are in Cases (iv), (v), or (vi).
Case (iv): 8 + , , , + . If w is a 6 + -vertex, then both u and u are sent charge by fourvertices and hence v gives away at most 2( ). Since v gets at least from each of u and u ,we have ch ∗ ( v ) > ) − ) = 0. Hence, we assume that w is a 5-vertex.Now if w is a 6-vertex, then u receives charge , so gives back to v . If instead w is a7 + -vertex, then u receives charge at least from v and w together and then charge at least + from u for a total of . Since u has only one 6-neighbor, it gives the extra back to v by (R4). The same holds for u , so v gets back from each of u and u . So, the total chargethat v gets from u , u , u , u is at least + + + = .Suppose that u has at least two 5-neighbors. Now one of them, call it x , is a commonneighbor with either u or u , so we can apply Lemma 19 to { v, w , x } (again x w , since w has two other 5-neighbors; x cannot be identified with one of these other 5-neighbors, since G has no separating 3-cycle). Similarly, u has at most one 5-neighbor. Hence, by our argumentabove, both u and u finish (R1)–(R3) with charge at least . Now we show that u has atmost three 6-neighbors; similarly for u .Suppose that u has at least four 6-neighbors. Define y by N ( u ) = { v, u , w , y, w , u } .Recall that w ↔ w and w ↔ w , as noted before Case (i). If y is a 6-vertex, then we canapply Lemma 19 to { u , y, u } . (We cannot have y = w , since letting J = { u , u , w } gives | J | = 3 and | N ( J ) | − − − w and w must be 6-vertices. We can apply Lemma 19 to { v, w , w } unless w ↔ w ,so assume this. Also, we can apply Lemma 19 to { v, w , w } unless w ↔ w ; so assume this.Hence, N ( w ) ⊇ { u , u , w , w , w , w } , which is a contradiction since d ( w ) = 5.Thus, we conclude that u has at most two 6-neighbors other than u , so at most two 6-neighbors that finish (R1)–(R3) with negative charge. An analogous argument holds for u .Hence v gets at least from each of u and u via (R5), so ch ∗ ( v ) > − ) + + 2( ) = 0. the electronic journal of combinatorics (2016), (2016),
Case (iv): 8 + , , , + . If w is a 6 + -vertex, then both u and u are sent charge by fourvertices and hence v gives away at most 2( ). Since v gets at least from each of u and u ,we have ch ∗ ( v ) > ) − ) = 0. Hence, we assume that w is a 5-vertex.Now if w is a 6-vertex, then u receives charge , so gives back to v . If instead w is a7 + -vertex, then u receives charge at least from v and w together and then charge at least + from u for a total of . Since u has only one 6-neighbor, it gives the extra back to v by (R4). The same holds for u , so v gets back from each of u and u . So, the total chargethat v gets from u , u , u , u is at least + + + = .Suppose that u has at least two 5-neighbors. Now one of them, call it x , is a commonneighbor with either u or u , so we can apply Lemma 19 to { v, w , x } (again x w , since w has two other 5-neighbors; x cannot be identified with one of these other 5-neighbors, since G has no separating 3-cycle). Similarly, u has at most one 5-neighbor. Hence, by our argumentabove, both u and u finish (R1)–(R3) with charge at least . Now we show that u has atmost three 6-neighbors; similarly for u .Suppose that u has at least four 6-neighbors. Define y by N ( u ) = { v, u , w , y, w , u } .Recall that w ↔ w and w ↔ w , as noted before Case (i). If y is a 6-vertex, then we canapply Lemma 19 to { u , y, u } . (We cannot have y = w , since letting J = { u , u , w } gives | J | = 3 and | N ( J ) | − − − w and w must be 6-vertices. We can apply Lemma 19 to { v, w , w } unless w ↔ w ,so assume this. Also, we can apply Lemma 19 to { v, w , w } unless w ↔ w ; so assume this.Hence, N ( w ) ⊇ { u , u , w , w , w , w } , which is a contradiction since d ( w ) = 5.Thus, we conclude that u has at most two 6-neighbors other than u , so at most two 6-neighbors that finish (R1)–(R3) with negative charge. An analogous argument holds for u .Hence v gets at least from each of u and u via (R5), so ch ∗ ( v ) > − ) + + 2( ) = 0. the electronic journal of combinatorics (2016), (2016), ase (v): 6 , , + , + . Note that v receives charge at least 2( ) = from u and u . If w is a 6 + -vertex, then u receives charge from four neighbors, so v gives away charge at most + . Thus ch ∗ ( v ) >
0. So assume w is a 5-vertex. First, we show that v gets back at least from u . If d ( w ) = 6, then u gets charge + + = , so returns charge to each of v and w . Otherwise w is a 7 + -vertex, so u sends charge to u , and u returns at least to v . Thus, the total charge that v gets from u , u , and u is at least 2( ) + = .If w is a 6-vertex, then v gets back charge from u , via (R4), so ch ∗ ( v )0 − ) + + = 0.Instead, assume w is a 7 + -vertex. Now we show that v gets charge at least from u by (R5).Let y be the neighbor of u other than v, u , w , w , u . Applying Lemma 18 to { u , u , y } , showsthat y is an 8 + -vertex. If w is a 5-vertex, then we apply Lemma 19 to { v, w , w } to get acontradiction ( w cannot be adjacent to w , since w already has two other 5-neighbors, and w cannot be identified with u or w , since G has no separating 3-cycles). Hence w is a 6 + -vertex.So u receives charge at least from w and at least + from y . After u gives charge to u ,it has charge at least . So, by (R5), u gives each of its at most three 6-neighbors (including v ) charge at least ( ) = . Thus, ch ∗ ( v ) > − + = 0. Case (vi): 6 , + , + , First suppose that w is a 6 + -vertex. Note that v gets charge atleast 2( ) from u and u , so it suffices to show that v gives net charge at most to each of u and u . We consider u ; the case for u is symmetric. If w gives charge to u , then u receivescharge from four neighbors, so it gets charge only from v . Recall that w must be a 6 + -vertex,as noted before Case (i). Thus w fails to give charge to u only if u is a crowded 5-neighborof w ; suppose this is the case. So w is a 7-vertex and w is a 6-vertex. Now u gets charge + + = , so u returns charge to each of w and v , via (R4), as desired. By symmetry, u also returns to v . Thus ch ∗ ( v ) > − ) + 2( ) + 2( ) = 0. So instead, assume that w is a 5-vertex.Now we show that u returns to v via (R4). By symmetry the same is true of u . If w isa 6-vertex, then v gets back from u , since u receives + + and returns to each of w and v . So assume, that w is a 7 + -vertex. If w ↔ w , then we apply Lemma 8 to { w , u } ; so w w . If w is a 6 − -vertex, then we apply Lemma 19 to { v, w , w } to get a contradiction(as above, w cannot be identified with u or w , since G has no separating 3-cycle). Thus, w is a 7 + -vertex. So u has at least three 7 + -neighbors and at most two 6-neighbors. Thus, after u gives charge to u , by (R5) it gives charge ( ) = to v . So in each case, u gives at least to v via (R5). Since the same is true of u , we have ch ∗ ( v ) > − ) + 2( ) + 2( ) = 0. It is quite useful to know that a minimal counterexample has no separating 3-cycle; we provethis in Lemma 2. When proving coloring results, such a lemma is nearly trivial. However, forindependence results, it requires much more work. Albertson proved an analogous lemma whenshowing that planar graphs have independence ratio at least . Our proof generalizes his to thebroader context of showing that a minor-closed family of graphs has independence ratio at least c for some rational c . We will apply this lemma to planar graphs and will let c = . Lemma 2.
Let c > be rational. Let G be a minor-closed family of graphs. If G is a minimalcounterexample to the statement that every n -vertex graph in G has an independent set of sizeat least cn , then G has no separating 3-cycle. the electronic journal of combinatorics (2016), (2016),
Let c > be rational. Let G be a minor-closed family of graphs. If G is a minimalcounterexample to the statement that every n -vertex graph in G has an independent set of sizeat least cn , then G has no separating 3-cycle. the electronic journal of combinatorics (2016), (2016), roof. Suppose to the contrary that G has a separating 3-cycle X . Let A and A be inducedsubgraphs of G with V ( A ) ∩ V ( A ) = X and A ∪ A = G .Our plan is to find big independent sets in two smaller graphs in G (by minimality) and piecethose independent sets together to get an independent set in G of size at least c | G | (for brevity,we write | G | for | V ( G ) | ). More precisely, we consider independent sets in each A i , either with X deleted, or with some pair of vertices in X identified. In Claims 1–3, we prove lower boundson α ( G ) in terms of | A | and | A | . In Claim 4, we examine | A | and | A | modulo b , where c = ab in lowest terms. In each case, we show that one of the independent sets constructed in Claims1–3 has size at least c | G | . Our proof relies heavily on the fact that α ( H ) is an integer (for everygraph H ), which often allows us to gain slightly over c | H | . Claim 1. α ( G ) > ⌈ c ( | A | − ⌉ + ⌈ c ( | A | − ⌉ . The union of the independent sets obtained by applying minimality of G to A \ X and A \ X is independent in G . Claim 2. α ( G ) > ⌈ c ( | A i | − ⌉ + ⌈ c | A j |⌉ − whenever { i, j } = { , } . For concreteness, let i = 1 and j = 2; the other case is analogous. Apply minimality to A to get an independent set I in A with | I | > ⌈ c | A |⌉ . Form A ′ from A by contracting X to a single vertex u . Apply minimality to A ′ to get an independent set I in A ′ with | I | > ⌈ c ( | A | − ⌉ . If u ∈ I , then I ∪ I \ { u } is independent in G and has the desired size.Otherwise, I ∪ I \ X is an independent set of the desired size in G . Claim 3. α ( G ) > ⌈ c ( | A | − ⌉ + ⌈ c ( | A | − ⌉ − . Let X = { x , x , x } . For each k ∈ { , } and t ∈ { , } , form A k,t from A k by contracting x x t to a vertex x k,t . Applying minimality to A k,t gives an independent set I k,t in A k,t with | I k,t | > ⌈ c ( | A k | − ⌉ .If at most one of I ,t and I ,t contains a vertex of X (or a contraction of two vertices in X ), then to get a big independent set, we take their union, discarding this at most one vertex.Formally, if { x k,t , x − t } ∩ I k,t = ∅ , then ( I ,t ∪ I ,t ) \ X is an independent set in G of the desiredsize. So assume that each of I ,t and I ,t contains a vertex (or a contraction of an edge) of X .Now we look for a vertex x ℓ of X such that each of I ,t and I ,t contains x ℓ or a contractionof x ℓ . Formally, if x − t ∈ I ,t ∩ I ,t , then ( I ,t ∪ I ,t ) \ X is an independent set in G of the desiredsize. Similarly, if x ,t ∈ I ,t and x ,t ∈ I ,t , then ( I ,t ∪ I ,t ∪ { x } ) \ { x ,t , x ,t } is an independentset in G of the desired size.So, by symmetry, we may assume that x , ∈ I , and x ∈ I , . Also, either x , ∈ I , or x , ∈ I , . If x , ∈ I , , then ( I , ∪ I , ) \ { x , } is an independent set in G of the desiredsize. Otherwise, x , ∈ I , and ( I , ∪ I , ∪ { x } ) \ { x , , x , } is an independent set in G of thedesired size. Claim 4.
The lemma holds.
Let a and b be positive integers such that c = ab and gcd( a, b ) = 1. For each i ∈ { , } , let N i = | A i | − j ∈ { , , , } , choose k ji such that 1 k ji b and k ji ≡ a ( N i + j )(mod b ). In other words, ⌈ c ( N i + j ) ⌉ = ab ( N i + j ) + b − k ji b . Intuitively, if there exist i and j suchthat k ji is small compared to b , then we improve our lower bound on the independence number(in some smaller graph) by the fact that the independence number is always an integer. In thepresent claim, we show that if some k ji is small, then G has an independent set of the desiredsize. In contrast, if all k ji are big, then we get a contradiction.By symmetry, we may assume that k k . Subclaim 4a. k + k > b + 1 − a and k + k > b + a + 1 and k + k > b + a + 1 and the electronic journal of combinatorics (2016), + k > b + a + 1 . If any independent set constructed in Claims 1–3 has size at least c | G | , then we are done. Sowe assume not; more precisely, we assume that each of these independent sets has size at most a | G |− b . Each of the four desired bounds follow from simplifying the inequalities in Claims 1–3.Note that | G | = N + N + 3.By Claim 1, we have α ( G ) > ⌈ c ( | A | − ⌉ + ⌈ c ( | A | − ⌉ = ab ( N + N ) + b − k b + b − k b = ab | G | + b − a − k − k b . Hence k + k > b + 1 − a .By Claim 2, we have α ( G ) > ⌈ c ( | A | − ⌉ + ⌈ c | A |⌉− ab ( N +1+ N +3)+ b − k b + b − k b − ab | G | + b + a − k − k b −
1. Hence k + k > b + a + 1. Similarly, k + k > b + a + 1.By Claim 3, we have α ( G ) > ⌈ c ( | A | − ⌉ + ⌈ c ( | A | − ⌉ − > ab ( N + 2 + N + 2) + b − k b + b − k b − ab | G | + b + a − k − k b −
1. Hence k + k > b + a + 1.Now to get a contradiction, it suffices to show that k ji a for some i ∈ { , } and some j ∈ { , , } ; since k ji b for all i and j , this will contradict one of the equalities above. Subclaim 4b.
Either k a or k a . In each case we get a contradiction, so the claim istrue, and the lemma holds. By Subclaim 4a, we have k + k > b + 1 − a . By symmetry, we assumed k > k , so wehave k > b +1 − a . Since, k ≡ k + 2 a (mod b ) and b +1 − a + 2 a > b , we have k k + 2 a − b .Now we consider two cases, depending on whether k b − a or k > b − a + 1. If k b − a ,then k k + 2 a − b ( b − a ) + 2 a − b = a , a contradiction. Suppose instead that k > b − a + 1.Now k ≡ k + a (mod b ). Since k > b − a + 1, we see that k + a > b + 1, so k k + a − b a ,a contradiction.Now we turn to proving a series of lemmas showing that G cannot have too many 6 − -verticesnear each other. Many of these lemmas will rely on applications of the following result, whichwe think may be of independent interest. The idea for the proof is to find big independent setsfor two smaller graphs, and piece them together to get a big independent set in G .For S ⊆ V ( G ), let the interior of S be I ( S ) = { x ∈ S | N ( x ) ⊆ S } . For vertex sets V , V ⊂ V ( G ) we write V ↔ V if there exists an edge v v ∈ E ( G ) with v ∈ V and v ∈ V ;otherwise, we write V i V j . Lemma 3.
Let G be a minor-closed family of graphs. Let G be a minimal counterexample tothe statement that every n -vertex graph in G has an independent set of size at least cn (for somefixed c > ). Let S , . . . , S t be pairwise disjoint subsets of a nonempty set S ⊆ V ( G ) such that t < | S | and G [ S i ] is connected for all i ∈ { , . . . , t } . Now there exists X ⊆ { , ..., t } such that S i S j for all distinct i, j ∈ X and α (cid:0) G (cid:2) I ( S ) ∪ S i ∈ X S i (cid:3)(cid:1) < | X | + ⌈ c ( | S | − t ) ⌉ .Proof. Suppose to the contrary that α (cid:0) G (cid:2) I ( S ) ∪ S i ∈ X S i (cid:3)(cid:1) > | X | + ⌈ c ( | S | − t ) ⌉ for all X ⊆{ , ..., t } such that S i S j for all distinct i, j ∈ X . Create G ′ from G by contracting S i toa single vertex w i for each i ∈ { , . . . , t } and removing the rest of S . (Note that we allow t = 0.) Since t < | S | , we have | G ′ | < | G | and hence minimality of G gives an independentset I in G ′ with | I | > c | G ′ | = c ( | G | − | S | + t ). Let W = I ∩ { w , . . . , w t } . By assumption,we have α (cid:0) G (cid:2) I ( S ) ∪ S w i ∈ W S i (cid:3)(cid:1) > | W | + ⌈ c ( | S | − t ) ⌉ . If T is a maximum independent set in G (cid:2) I ( S ) ∪ S w i ∈ W S i (cid:3) , then ( I \ W ) ∪ T is an independent set in G of size at least | I |− | W | + | T | > c ( | G | − | S | + t ) − | W | + ( | W | + ⌈ c ( | S | − t ) ⌉ ) > c | G | , a contradiction. the electronic journal of combinatorics (2016), e will often apply Lemma 3 with S = J ∪ N ( J ) for an independent set J . In this case, wealways have J ⊆ I ( S ). We state this case explicitly in Lemma 4 Lemma 4.
Let G be a minor-closed family of graphs. Let G be a minimal counterexample tothe statement that every n -vertex graph in G has an independent set of size at least cn (for somefixed c > ). No independent set J of G and nonnegative integer k simultaneously satisfy thefollowing conditions.1. | J | > c ( | N ( J ) | + k ) .2. For at least | J | − k vertices x ∈ J , there is an independent set { u x , v x } of size 2 in N ( x ) \ [ y ∈ J \{ x } N ( y ) .Proof. Suppose the lemma is false. Let S = J ∪ N ( J ) and t = | J | − k . Pick x , . . . , x t ∈ J satisfying condition (2). For i ∈ { , . . . , t } , let S i = { x i , u x i , v x i } . Applying Lemma 3, weget X ⊆ { , . . . , t } such that S i S j for all distinct i, j ∈ X and α (cid:0) G (cid:2) J ∪ S i ∈ X S i (cid:3)(cid:1) < | X | + ⌈ c ( | S | − t ) ⌉ . By (2), we have α (cid:0) G (cid:2) J ∪ S i ∈ X S i (cid:3)(cid:1) > | ( J \ X ) ∪ S x ∈ X { u x , v x } | > ( | J |−| X | )+2 | X | = | X | + | J | . Hence | X | + ⌈ c ( | S | − t ) ⌉ > | X | + | J | , giving ⌈ c ( | S | − t ) ⌉ > | J | > ⌈ c ( | N ( J ) | + k ) ⌉ by (1). But | S | − t = ( | J | + | N ( J ) | ) − ( | J | − k ) = | N ( J ) | + k ; so ⌈ c ( | S | − t ) ⌉ = ⌈ c ( | N ( J ) | + k ) ⌉ ,contradicting the previous inequality. This contradiction finishes the proof.As a simple example of how to apply Lemma 4, we note that it immediately implies thatevery planar graph G has independence ratio at least . By Euler’s theorem, G has a 5 − -vertex v . If d ( v )
4, then let G ′ = G \ ( v ∪ N ( v )). Let I ′ be an independent set in G ′ of size at least( n − /
5, and let I = I ′ ∪ { v } . If instead d ( v ) = 5, then apply Lemma 4, with c = , J = { v } ,and k = 0; since K is nonplanar, v has some pair of nonadjacent neighbors. This completesthe proof. Lemma 5.
Let G be a minor-closed family of graphs. Let G be a minimal counterexample tothe statement that every n -vertex graph in G has an independent set of size at least cn (for somefixed c > ). For any non-maximal independent set J in G , we have | N ( J ) | > (cid:22) − cc | J | (cid:23) + 2 . Proof.
Assume the lemma is false and choose a counterexample J minimizing | J | . Suppose G [ J ∪ N ( J )] is not connected. Now we choose a partition { J , . . . , J k } of J , minimizing k , suchthat k > G [ J i ∪ N ( J i )] is connected for each i ∈ { , . . . , k } . Applying the minimality of | J | to each J i we conclude that | N ( J i ) | > (cid:4) − cc | J i | (cid:5) + 2 for each i ∈ { , . . . , k } . The minimalityof k gives | N ( J ) | = (cid:12)(cid:12)(cid:12)S ki =1 N ( J i ) (cid:12)(cid:12)(cid:12) = P ki =1 | N ( J i ) | , so | N ( J ) | > k + P ki =1 (cid:4) − cc | J i | (cid:5) > k + P ki =1 1 − cc | J i | > − cc | J | , a contradiction. Hence, G [ J ∪ N ( J )] is connected.Let S = J ∪ N ( J ). Apply Lemma 3 with t = 1 and S = S . This shows that either | J | α ( G [ I ( S )]) < ⌈ c ( | S | − ⌉ or α ( G [ S ]) < ⌈ c ( | S | − ⌉ , since the only possibilitiesare X = ∅ and X = { } . By assumption J is a counterexample, so | N ( J ) | (cid:4) − cc | J | (cid:5) + 1,which implies that | S | = | J | + | N ( J ) | | J | + (cid:4) − cc | J | (cid:5) + 1 = j | J | c k + 1. Now ⌈ c ( | S | − ⌉ l c (( j | J | c k + 1) − m = l c j | J | c km ⌈| J |⌉ = | J | . Hence, we cannot have X = ∅ in Lemma 3. the electronic journal of combinatorics (2016), (2016),
Assume the lemma is false and choose a counterexample J minimizing | J | . Suppose G [ J ∪ N ( J )] is not connected. Now we choose a partition { J , . . . , J k } of J , minimizing k , suchthat k > G [ J i ∪ N ( J i )] is connected for each i ∈ { , . . . , k } . Applying the minimality of | J | to each J i we conclude that | N ( J i ) | > (cid:4) − cc | J i | (cid:5) + 2 for each i ∈ { , . . . , k } . The minimalityof k gives | N ( J ) | = (cid:12)(cid:12)(cid:12)S ki =1 N ( J i ) (cid:12)(cid:12)(cid:12) = P ki =1 | N ( J i ) | , so | N ( J ) | > k + P ki =1 (cid:4) − cc | J i | (cid:5) > k + P ki =1 1 − cc | J i | > − cc | J | , a contradiction. Hence, G [ J ∪ N ( J )] is connected.Let S = J ∪ N ( J ). Apply Lemma 3 with t = 1 and S = S . This shows that either | J | α ( G [ I ( S )]) < ⌈ c ( | S | − ⌉ or α ( G [ S ]) < ⌈ c ( | S | − ⌉ , since the only possibilitiesare X = ∅ and X = { } . By assumption J is a counterexample, so | N ( J ) | (cid:4) − cc | J | (cid:5) + 1,which implies that | S | = | J | + | N ( J ) | | J | + (cid:4) − cc | J | (cid:5) + 1 = j | J | c k + 1. Now ⌈ c ( | S | − ⌉ l c (( j | J | c k + 1) − m = l c j | J | c km ⌈| J |⌉ = | J | . Hence, we cannot have X = ∅ in Lemma 3. the electronic journal of combinatorics (2016), (2016), nstead, we must have X = { } , which implies that α ( G [ S ]) < ⌈ c ( | S | − ⌉ . Since J is non-maximal, we have S = V ( G ), so we may apply minimality of G to G [ S ] to concludethat α ( G [ S ]) > ⌈ c | S |⌉ . Combining this inequality with the previous one, we have ⌈ c | S |⌉ = ⌈ c ( | S | − ⌉ . Now the upper bound on ⌈ c ( | S | − ⌉ from the previous paragraph gives ⌈ c | S |⌉ = ⌈ c ( | S | − ⌉ | J | . Finally, applying Lemma 3 with t = 0 (simply deleting J ∪ N ( J )) shows that | J | < ⌈ c ( | S | ) ⌉ . These two final inequalities contradict each other, which finishes the proof.Lemmas 2–5 hold in a more general setting than just c = , as we showed. In the rest of thissection, we consider only a planar graph G that is minimal among those with independence ratioless than . To remind the reader of this, we often call it a minimal G . Applying Lemma 5with c = gives the following corollary. Lemma 6.
For any non-maximal independent set J in a minimal G , we have | N ( J ) | > (cid:22) | J | (cid:23) + 2 . In particular, if | J | = 1 , then | N ( J ) | > ; if | J | = 2 , then | N ( J ) | > ; and if | J | = 3 , then | N ( J ) | > . The case | J | = 1 shows that G has minimum degree 5, and this is the best we can hope forwhen | J | = 1. Recall that G is a planar triangulation, since we chose it to have as few non-triangular faces as possible. As a result, we can improve the bound when | J | = 2 to | N ( J ) | > | J | = 3 to | N ( J ) | >
13. Theseimprovements are the focus of the next ten lemmas. In many instances, the proofs are easyapplications of Lemma 3. First, we need a few basic facts about planar graphs.
Lemma 7. If G is a plane triangulation with no separating 3-cycle and δ ( G ) = 5 , then(a) If v ∈ V ( G ) , then G [ N ( v )] is a cycle; and(b) G is -connected with | V ( G ) | > ; and(c) If v, w ∈ V ( G ) are distinct, then G [ N ( v ) ∩ N ( w )] is the disjoint union of copies of K and K .Proof. Plane triangulations are well-known to be 3-connected. Property (a) follows by notingthat G \{ v } is 2-connected and hence each face boundary is a cycle; so G [ N ( v )] has a hamiltoniancycle. This cycle must be induced since G has no separating 3-cycle.For (b), suppose that G has a separating set { x, y, z } . Since G has no separating 3-cycle,we assume that xy E ( G ). By (a), N ( x ) induces a cycle C . Since G is 3-connected, x musthave a neighbor in each component of G \ { x, y, z } . So C has a vertex in each component of G \ { x, y, z } and hence C \ { x, y, z } is disconnected. But x V ( C ) and since xy E ( G ), also y V ( C ). So, C \ { z } is disconnected, which is impossible. Since G is a plane triangulationand δ ( G ) = 5, we have 5 | G | | E ( G ) | = 6 | G | −
12, so | G | > δ ( G ) = 5, it follows that no neighborhood contains K or C . If G [ N ( v ) ∩ N ( w )]had an induced P (path on 3 vertices), then the neighborhood of the center of this P wouldcontain K or C . This proves (c). Lemma 8.
Every independent set J in a minimal G with | J | = 2 , satisfies | N ( J ) | > . the electronic journal of combinatorics (2016), (2016),
Every independent set J in a minimal G with | J | = 2 , satisfies | N ( J ) | > . the electronic journal of combinatorics (2016), (2016), roof. By Lemma 7(b), | G | >
12; so J cannot be a maximal independent set when | N ( J ) | | N ( J ) | = 8. Let J = { x, y } . If we can apply Lemma 4with k = 0, then we are done. If we cannot, then by symmetry we may assume that there isno independent 2-set in N ( x ) \ N ( y ). So N ( x ) \ N ( y ) is a clique. Since d ( x ) > N ( x )induces a cycle, | N ( x ) \ N ( y ) |
2. Now, since x is a 5 + -vertex, G [ N ( x ) ∩ N ( y )] induces P ; thiscontradicts Lemma 7(c).A direct consequence of Lemma 8 is the following useful fact. Lemma 9.
A minimal G cannot have two nonadjacent 5-vertices with at least two commonneighbors. In particular, each vertex v in G has d ( v )2 or more + -neighbors.Proof. The first statement follows immediately from Lemma 8. Now we consider the second.Let v be a vertex with d ( v ) = k and neighbors u , . . . , u k in clockwise order. If more than k/ v are 5-vertices, then (by Pigeonhole) there exists an integer i such that u i and u i +2 are 5-vertices (subscripts are modulo k ). Now we apply Lemma 8 to u i and u i +2 . Recallthat u i and u i +2 are nonadjacent, since G has no separating 3-cycle, as shown in Lemma 2.Now we consider the case when | J | = 3. Lemma 6 gives | N ( J ) | >
12. Our next few lemmasshow certain conditions under which we can conclude that | N ( J ) | > Lemma 10.
Let J be an independent set in a minimal G with | J | = 3 and | N ( J ) | > .Choose S , S ⊆ J ∪ N ( J ) such that S ∩ S = ∅ and both G [ S ] and G [ S ] are connected. If α ( G [ S i ∪ J ]) > for each i ∈ { , } , then | N ( J ) | > .Proof. Suppose not and choose a counterexample minimizing | J ∪ N ( J ) | − | S ∪ S | . Clearly | N ( J ) | = 12. First we show that S ∪ S = J ∪ N ( J ). It suffices to show that G [ J ∪ N ( J )]is connected, since then we can add to either S or S any vertex in N ( S ∪ S ) \ ( S ∪ S ).In particular, we show that every x ∈ J satisfies ( x ∪ N ( x )) ∩ ( ∪ y ∈ ( J \{ x } ) ( y ∪ N ( y ))) = ∅ .Suppose not. By Lemma 8, we have | ∪ y ∈ J \{ x } N ( y ) | >
9. Now | ∪ y ∈ J N ( y ) | > d ( x ) >
14, acontradiction. Now we must have G [ J ∪ N ( J )] connected, so we can assume S ∪ S = J ∪ N ( J ).Similarly, we assume S ↔ S .Now we apply Lemma 3 with S = J ∪ N ( J ), t = 2, and S and S as above. Since S ↔ S ,we have | X |
1. We cannot have | X | = 1 since, by hypothesis, α ( G [ S i ∪ J ]) > i ∈ { , } . So suppose that X = ∅ . Now we have α ( G [ J ]) > | J | = 3 = (cid:6) ( | J ∪ N ( J ) | − (cid:7) = (cid:6) (3 + 12 − (cid:7) . This contradiction completes the proof. Lemma 11.
Let J = { u , u , u } . If J is an independent set in a minimal G where1. N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set; and2. α ( G [ J ∪ N ( u ) ∪ N ( u )]) > ,then | N ( J ) | > .Proof. Since G is a planar triangulation with minimum degree 5 and at least three 6 + -verticesby Lemma 9, we have 5 | G | + 3 | E ( G ) | = 6 | G | −
12 and hence | G | >
15. Thus J cannot be amaximal independent set when | N ( J ) |
11. So, by Lemma 6, we know that | N ( J ) | >
12. Let I be an independent set of size 2 in N ( u ) \ ( N ( u ) ∪ N ( u )).First, suppose N ( u ) ∩ N ( u ) = ∅ . We apply Lemma 10 with S = { u } ∪ I and S = { u , u } ∪ N ( u ) ∪ N ( u ). Clearly, G [ S ] is connected. Also, G [ S ] is connected since N ( u ) ∩ the electronic journal of combinatorics (2016), (2016),
12. Let I be an independent set of size 2 in N ( u ) \ ( N ( u ) ∪ N ( u )).First, suppose N ( u ) ∩ N ( u ) = ∅ . We apply Lemma 10 with S = { u } ∪ I and S = { u , u } ∪ N ( u ) ∪ N ( u ). Clearly, G [ S ] is connected. Also, G [ S ] is connected since N ( u ) ∩ the electronic journal of combinatorics (2016), (2016), ( u ) = ∅ , by assumption. The set I ∪ { u , u } shows that α ( G [ S ∪ J ]) > α ( G [ S ∪ J ]) >
4. So the hypotheses of Lemma 10 are satisfied, giving | N ( J ) | > N ( u ) ∩ N ( u ) = ∅ . This implies N ( u ) \ ( N ( u ) ∪ N ( u )) = N ( u ) \ N ( u ).If N ( u ) \ N ( u ) contains an independent 2-set as well, then applying Lemma 4 with k = 1 gives | N ( J ) | >
13, as desired. Otherwise, | N ( u ) \ N ( u ) |
2, so G [ N ( u ) ∩ N ( u )] contains P ,contradicting Lemma 7(c).One particular case of Lemma 11 is easy to verify in our applications, so we state it separately,as Lemma 13. First, we need the following lemma. Lemma 12.
Let v be a + -vertex in G . If S ⊆ V ( G ) with { v } ∪ N ( v ) ⊆ S and | S | > , then α ( G [ S ]) > .Proof. If d ( v ) >
8, then the neighbors of v induce an 8 + -cycle (by Lemma 7(a)), which hasindependence number at least 4; so we are done. So suppose d ( v ) = 7. Let u , . . . , u denote theneighbors of v in clockwise order; note that G [ N ( v )] is a 7-cycle, again by Lemma 7(a). Pick w , w ∈ S \ ( { v } ∪ N ( v )). Let H i = G [ N ( v ) \ N ( w i )] for each i ∈ { , } . If H i contains anindependent 3-set J for some i ∈ { , } , then J ∪ { w i } is the desired independent 4-set, so weare done. Therefore, we must have | H i | i ∈ { , } . So, | N ( v ) ∩ N ( w i ) | > N ( v ) ∩ N ( w i ) has at least two components; therefore, so does H i . It must have exactly two components or we get an independent 3-set in H i . Similarly, if | H i | = 4, then H i has no isolated vertex. So, either H i is 2 K or | H i |
3. Now in each case weget a subdivision of K , ; the branch vertices of one part are v, w , w and the branch verticesof the other are three of the u i . This contradiction finishes the proof. Lemma 13.
Let J = { u , u , u } . If J is an independent set in a minimal G where1. N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set; and2. G [ J ∪ N ( u ) ∪ N ( u )] contains a + -vertex and its neighborhood,then | N ( J ) | > .Proof. We apply Lemma 11 using Lemma 12 to verify hypothesis (2). To do so, we let S = { u , u , u } ∪ N ( u ) ∪ N ( u ), and we need that |{ u , u , u } ∪ N ( u ) ∪ N ( u ) | >
10. This isimmediate from Lemma 8, since | S | > |{ u , u , u }| + | N ( u ) ∪ N ( u ) | > Lemma 14.
Let J be an independent 3-set in G . Choose S , S , S ⊆ J ∪ N ( J ) such that G [ S i ] is connected and S i ∩ S j = ∅ for all distinct i, j ∈ { , , } . If | N ( J ) | , then either1. S i S j for some { i, j } ⊆ { , , } ; or2. α ( G [ S i ∪ J ]) for some i ∈ { , , } .Proof. This is an immediate corollary of Lemma 3 with S = J ∪ N ( J ) and t = 3. If S i ↔ S j forall { i, j } ∈ { , , } , then in Lemma 3 either | X | = 1 or | X | = 0. We cannot have | X | = 0, since α ( G [ I ( S )]) > α ( G [ J ]) > | J | = 3 = (cid:6) (13 + 3 − (cid:7) . Hence | X | = 1, which implies (2).The next lemma can be viewed as a variant on the result we get by applying Lemma 4 with | J | = 3 and k = 0 (and c = ). As in that case, we require that each of N ( u ) \ ( N ( u ) ∪ N ( u ))and N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set. However, here we do not requirethat N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set. Instead, we have hypothesis (2)below. Not surprisingly, the proof is similar to that of Lemma 4. the electronic journal of combinatorics (2016), (2016),
Let J be an independent 3-set in G . Choose S , S , S ⊆ J ∪ N ( J ) such that G [ S i ] is connected and S i ∩ S j = ∅ for all distinct i, j ∈ { , , } . If | N ( J ) | , then either1. S i S j for some { i, j } ⊆ { , , } ; or2. α ( G [ S i ∪ J ]) for some i ∈ { , , } .Proof. This is an immediate corollary of Lemma 3 with S = J ∪ N ( J ) and t = 3. If S i ↔ S j forall { i, j } ∈ { , , } , then in Lemma 3 either | X | = 1 or | X | = 0. We cannot have | X | = 0, since α ( G [ I ( S )]) > α ( G [ J ]) > | J | = 3 = (cid:6) (13 + 3 − (cid:7) . Hence | X | = 1, which implies (2).The next lemma can be viewed as a variant on the result we get by applying Lemma 4 with | J | = 3 and k = 0 (and c = ). As in that case, we require that each of N ( u ) \ ( N ( u ) ∪ N ( u ))and N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set. However, here we do not requirethat N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set. Instead, we have hypothesis (2)below. Not surprisingly, the proof is similar to that of Lemma 4. the electronic journal of combinatorics (2016), (2016), emma 15. Let J = { u , u , u } . If J is an independent set in a minimal G such that1. N ( u i ) \ ( N ( u j ) ∪ N ( u )) contains an independent 2-set M i for all { i, j } = { , } ; and2. α ( G [ J ∪ V ( H )]) > , where H is u ’s component in G [ { u } ∪ N ( J )] \ ( M ∪ M ) ,then | N ( J ) | > .Proof. First, we show that u is distance two from each of u and u . Suppose not; by symmetry,assume that u is distance at least three from u . Now N ( u ) \ ( N ( u ) ∪ N ( u )) = N ( u ) \ N ( u ).By Lemma 7, N ( u ) ∩ N ( u ) consists of disjoint copies of K and K . Thus, since d ( u ) > N ( u ) \ ( N ( u ) ∪ N ( u )) contains an independent 2-set. Now, if | N ( J ) |
13, thenapplying Lemma 4 with k = 0 gives a contradiction. Hence, u is distance two from each of u and u .Choose disjoint subsets S , S , S ⊂ J ∪ N ( J ) where G [ S i ] is connected for all i ∈ { , , } and { u i } ∪ M i ⊆ S i for each i ∈ { , } and u ∈ S , first maximizing | S | and subject to thatmaximizing | S | + | S | + | S | . Since J ⊆ S ∪ S ∪ S , maximality of | S | + | S | + | S | gives S ∪ S ∪ S = J ∪ N ( J ).Now we apply Lemma 3, with S = S ∪ S ∪ S . To get a contradiction, we need only verify,for each possible X , that α ( G [ I ( S ) ∪ S i ∈ X S i ]) > | X | + (cid:6) ( | S | − | J | ) (cid:7) = | X | + 3. Since S ↔ S and S ↔ S , either | X | X = { , } . In the latter case, M ∪ M ∪ { u } is the desiredindependent 5-set. If instead X = ∅ , then J is the desired independent 3-set.So we must have X = { i } for some i ∈ { , , } . If i ∈ { , } , then M i ∪ { u , u − i } ) is thedesired independent set. So instead assume that X = { } . But, by the maximality of | S | , G [ J ∪ S ] contains u ’s component in G [ { u } ∪ N ( J )] \ M \ M . So by (2), G [ J ∪ S ] has anindependent 4-set, as desired.Again, one particular case of Lemma 15 is easy to verify, so we state it separately. Lemma 16.
Let J = { u , u , u } . If J is an independent set in a minimal G such that1. N ( u i ) \ ( N ( u j ) ∪ N ( u )) contains an independent 2-set M i for all { i, j } = { , } ; and2. u ’s component H in G [ { u }∪ N ( J )] \ ( M ∪ M ) satisfies | J ∪ V ( H ) | > and G [ J ∪ V ( H )] contains a + vertex and its neighborhood,then | N ( J ) | > .Proof. We apply Lemma 15, using Lemma 12 to verify hypothesis (2).Thus far, our lemmas have not focused much on the actual planar embedding of G . At thispoint we transition and start analyzing the embedding, as well. Lemma 17.
Every minimal G has no 6-vertex v with − -neighbors u , u , and u that arepairwise nonadjacent.Proof. Lemma 6, applied with J = { u , u , u } , yields 12 | N ( { u , u , u } ) | d ( u ) + d ( u ) + d ( u ) −
5. Hence, by symmetry, assume that the vertices are arranged as in Figure 4(a) with allvertices distinct as drawn or as in Figure 4(b) with at most one pair of vertices identified.The first case is impossible by Lemma 4 with k = 1, using the vertices labeled 2 for u andthose labeled 3 for u . When the vertices in Figure 4(b) are distinct as drawn, we apply Lemma 4 the electronic journal of combinatorics (2016), (2016),
5. Hence, by symmetry, assume that the vertices are arranged as in Figure 4(a) with allvertices distinct as drawn or as in Figure 4(b) with at most one pair of vertices identified.The first case is impossible by Lemma 4 with k = 1, using the vertices labeled 2 for u andthose labeled 3 for u . When the vertices in Figure 4(b) are distinct as drawn, we apply Lemma 4 the electronic journal of combinatorics (2016), (2016), u u u
33 212 (a) A 6-vertex, v , with non-adjacentneighbors u , u , and u such that d ( u ) = 5 and d ( u ) = d ( u ) = 6. v u u u
33 21244 (b) A 6-vertex, v , with non-adjacent6-neighbors u , u , and u . Figure 4: The two cases of Lemma 17.with k = 0, using the vertices labeled 2 for u , the vertices labeled 3 for u , and those labeled 4for u . Instead, by symmetry and the fact that G contains no separating 3-cycle, assume that thevertices labeled 2 and 3 that are drawn at distance four are identified; so | N ( { u , u , u } ) | = 12.Now the pairs of vertices labeled 1 each have a common neighbor, so the vertices labeled 1must be an independent set, to avoid a separating 3-cycle. Now, we apply Lemma 11, usingthe vertices labeled 4 for the independent 2-set. This implies that | N ( { u , u , u } ) | >
13, whichcontradicts our conclusion above that | N ( { u , u , u } ) | = 12. Lemma 18.
Every minimal G has no 6-vertex v with pairwise nonadjacent neighbors u , u ,and u , where d ( u ) = 5 , d ( u ) , and d ( u ) = 7 .Proof. Let J = { u , u , u } . By Lemma 6, 12 | N ( J ) | − x y , as in Fig-ure 5(b). For each i ∈ { , , } , let S i consist of the vertices labeled i . Now for each i ∈ { , , } , G [ S i ] is connected. Clearly, for each i ∈ { , } the vertices labeled I i form an independent 4-set.Since x y , the vertices labeled I also form an independent 4-set. Note that S ↔ S and S ↔ S ; however, possibly S S . If S ↔ S , then we can apply Lemma 14 to get a contra-diction. So, we assume that S S . But now we have an independent 5-set consisting of u ,the two vertices labeled { , I } and the two vertices labeled { , I } ; hence α ( G [ S ∪ S ∪ J ]) > x ↔ y .Suppose w z , as in Figure 5(c). For each i ∈ { , , } , let S i consist of the vertices labeled i . Clearly G [ S i ] is connected for each i ∈ { , } . Also, G [ S ] is connected because x ↔ y . Notethat for each i ∈ { , } , the vertices labeled I i form an independent 4-set. Since x ↔ y and w z , the vertices labeled I also form an independent 4-set. Note that S ↔ S and S ↔ S ;however, possibly S S . If S ↔ S , then we apply Lemma 14 to get a contradiction. Soinstead we assume that S S . But now we again have an independent 5-set, consisting of u , the electronic journal of combinatorics (2016), (2016),
Every minimal G has no 6-vertex v with pairwise nonadjacent neighbors u , u ,and u , where d ( u ) = 5 , d ( u ) , and d ( u ) = 7 .Proof. Let J = { u , u , u } . By Lemma 6, 12 | N ( J ) | − x y , as in Fig-ure 5(b). For each i ∈ { , , } , let S i consist of the vertices labeled i . Now for each i ∈ { , , } , G [ S i ] is connected. Clearly, for each i ∈ { , } the vertices labeled I i form an independent 4-set.Since x y , the vertices labeled I also form an independent 4-set. Note that S ↔ S and S ↔ S ; however, possibly S S . If S ↔ S , then we can apply Lemma 14 to get a contra-diction. So, we assume that S S . But now we have an independent 5-set consisting of u ,the two vertices labeled { , I } and the two vertices labeled { , I } ; hence α ( G [ S ∪ S ∪ J ]) > x ↔ y .Suppose w z , as in Figure 5(c). For each i ∈ { , , } , let S i consist of the vertices labeled i . Clearly G [ S i ] is connected for each i ∈ { , } . Also, G [ S ] is connected because x ↔ y . Notethat for each i ∈ { , } , the vertices labeled I i form an independent 4-set. Since x ↔ y and w z , the vertices labeled I also form an independent 4-set. Note that S ↔ S and S ↔ S ;however, possibly S S . If S ↔ S , then we apply Lemma 14 to get a contradiction. Soinstead we assume that S S . But now we again have an independent 5-set, consisting of u , the electronic journal of combinatorics (2016), (2016), u u z x w y (a) The forbidden configuration. , I , I , I , I , I , I , I , I , I , I , I , I (b) The x y case. , I , I , I , I , I , I , I , I , I , I , I , I (c) The x ↔ y and w z case. , I , I , I , I , I , I , I , I , I , I , I , I (d) The x ↔ y and w ↔ z case. u u u Q Q Q a b a b (e) The case of one identified pair. Figure 5: The case of Lemma 18. the electronic journal of combinatorics (2016), he two vertices labeled { , I } , and the two vertices labeled { , I } ; hence α ( G [ S ∪ S ∪ J ]) > w ↔ z .Now consider Figure 5(d). For each i ∈ { , , } , let S i consist of the vertices labeled i . Notethat G [ S i ] is connected for each i ∈ { , } . Also, G [ S ] is connected because x ↔ y and w ↔ z .Clearly, the vertices labeled I i form an independent 4-set for each i ∈ { , } . Since x ↔ y , thevertices labeled I also form an independent 4-set. Note that S ↔ S and S ↔ S ; however,possibly S S . If S ↔ S , then we apply Lemma 14 to get a contradiction. So, instead weassume that S S . But now we have an independent 5-set, consisting of u , the two verticeslabeled { , I } , and the two vertices labeled { , I } ; hence α ( G [ S ∪ S ∪ J ]) >
5. So, we applyLemma 3 to get a contradiction.Hence, we may assume that exactly one pair of vertices in Figure 5(a) is identified. Noneighbor of u can be identified with a neighbor of u , since then u and u would have threecommon neighbors, violating Lemma 8. Hence, to avoid separating 3-cycles, we assume thata vertex labeled 2 a is identified with a vertex labeled Q (the case where a vertex labeled 2 b is identified with a vertex lableled Q is nearly identical, so we omit the details). But now therightmost vertex labeled 1 and the leftmost vertex labeled 1 are on opposite sides of a separatingcycle and hence nonadjacent. Therefore, u together with the vertices labeled 1 is an independent4-set. So, now we apply Lemma 11 to get a contradiction, using the vertices labeled 2 b for theindependent 2-set. Lemma 19.
Let u be a -vertex with nonadjacent vertices u and u each at distance two from u , where u is a -vertex and u is a − -vertex. A minimal G cannot have u and u with twocommon neighbors, and also u and u with two common neighbors.Proof. Figure 6 shows the possible arrangements when u is a 6-vertex. The case when u is a 5-vertex is similar, but easier. In particular, when u is a 5-vertex, we already know | N ( { u , u , u } ) |
12, so all vertices in the corresponding figures must be distinct as drawn.Furthermore, it now suffices to apply Lemma 4 with k = 1. We omit further details. So supposeinstead that d ( u ) = 6.First, suppose all vertices in the figures are distinct as drawn. Now Figures 6(a,c) areimpossible by Lemma 4 with k = 0; for each i ∈ { , , } , we use the vertices labeled i as theindependent 2-set for u i . For Figure 6(b), let I be the vertices labeled u or 1 a and let I be thevertices labeled u or 1 b . To avoid a separating 3-cycle, at least one of I or I is independent.Hence Figure 6(b) is impossible by Lemma 15; for the independent 4-set, use I or I and foreach i ∈ { , } , use the vertices labeled i as the independent 2-set for u i .By Lemma 6, | N ( J ) | >
12, so exactly one pair of vertices is identified in one of Fig-ures 6(a,b,c). First, consider Figures 6(a,c) simultaneously. Since G has no separating 3-cycle,the identified pair must contain a vertex labeled 3. Now we apply Lemma 4 with k = 1, usingthe vertices labeled 3 b in place of those labeled 3.Finally, for Figure 6(b), we apply Lemma 11. For the independent 2-set we use either thevertices labeled 3 or the vertices labeled 4; at least one of these pairs contains no identifiedvertex. For the independent 4-set, we use either u and the vertices labeled 5 a or else u andthe vertices labeled 5 b . Since G has no separating 3-cycle, at least one of these 4-sets will beindependent. Lemma 20.
Every minimal G has no 7-vertex v with a 5-neighbor and two other − -neighbors, u , u , and u , that are pairwise nonadjacent. In other words, Figures 7(a–e) are forbidden. the electronic journal of combinatorics (2016), (2016),
Every minimal G has no 7-vertex v with a 5-neighbor and two other − -neighbors, u , u , and u , that are pairwise nonadjacent. In other words, Figures 7(a–e) are forbidden. the electronic journal of combinatorics (2016), (2016), u u b b (a) Here u and u have a com-mon neighbor in N ( u ). u a, bu b, a u , a, b a, b, a, b a, b (b) Here u and u have adjacentneighbors in N ( u ). u u u
22 3 3 b b (c) Here u and u have neighbors at dis-tance 2 in N ( u ). Figure 6: The cases of Lemma 19. The three possibilities for an independent 3-set { u , u , u } where d ( u ) = 6, d ( u ) d ( u ) = 5, and each of u and u has twoneighbors in common with u . the electronic journal of combinatorics (2016), (2016),
22 3 3 b b (c) Here u and u have neighbors at dis-tance 2 in N ( u ). Figure 6: The cases of Lemma 19. The three possibilities for an independent 3-set { u , u , u } where d ( u ) = 6, d ( u ) d ( u ) = 5, and each of u and u has twoneighbors in common with u . the electronic journal of combinatorics (2016), (2016), u u u (a) A 7-vertex, v , with non-adjacent5-neighbors, u , u , and u . v u u u (b) A 7-vertex, v , with a 6-neighbor, u , and two 5-neighbors, u and u . vu u u (c) A 7-vertex, v , with a 6-neighbor, u , and two 5-neighbors, u and u . v u
41 4 u u (d) A 7-vertex, v , with a 5-neighbor, u , and two 6-neighbors, u and u . vu u u
11 22 (e) A 7-vertex, v , with a 5-neighbor, u , and two 6-neighbors, u and u . Figure 7: The five cases of Lemma 20. the electronic journal of combinatorics (2016), (2016),
11 22 (e) A 7-vertex, v , with a 5-neighbor, u , and two 6-neighbors, u and u . Figure 7: The five cases of Lemma 20. the electronic journal of combinatorics (2016), (2016), + + + u u
221 1 33 (a) A 3-face v v v , such that thepairwise common neighbors of v , v , v have degrees 5, 5, and at most 6. Figure 8: The key case of Lemma 21.
Proof.
Lemma 6 yields 12 | N ( { u , u , u } ) | d ( u ) + d ( u ) + d ( u ) − −
13. InFigure 7(a), | N ( { u , u , u } ) | = 11. So, by symmetry, we assume that the vertices are arrangedas in Figures 7(b,c) with all vertices distinct as drawn or as in Figures 7(d,e) with at most onepair of vertices identified.First suppose the vertices are disinct as drawn. For Figures 7(b,c,d), we apply Lemma 4; for(b) and (c) we use k = 1, and for (d) we use k = 0. For Figure 7(e), we apply Lemma 16, usingthe vertices labeled 1 for M and the those labeled 2 for M . Now | N ( { u , u , u } ) | >
14 is acontradiction.So, instead suppose that a single pair of vertices is identified in one of Figures 7(d,e). Firstconsider (d). If a vertex labeled 1 is identified with another vertex, then we apply Lemma 13using the vertices labeled 2 for the independent 2-set (vertices labeled 1 and 2 cannot be iden-tified, since they are drawn at distance at most 3). Otherwise, the identified vertices must bethose labeled 2 and 3 that are drawn at distance four. Now the vertices labeled 4 are pairwise atdistance two, so must be an independent 4-set. Now we get a contradiction, by applying Lemma11 using the vertices labeled 1 for the independent 2-set.Finally, consider Figure 7(e), with a single pair of vertices identified. Again we applyLemma 4, with k = 1. Since u has three possibilities for its pair of nonadjacent neighbors,and no neighbor of u appears in all three of these pairs, u satisfies condition (2). Similarly, u also satisfies condition (2). Lemma 21.
Let v , v , v be the corners of a 3-face, each a + -vertex. Let u , u , u be theother pairwise common neighbors of v , v , v , i.e., u is adjacent to v and v , u is adjacentto v and v , and u is adjacent to v and v . We cannot have | N ( { u , u , u } ) | . Inparticular, we cannot have d ( u ) = d ( u ) = 5 and d ( u ) .Proof. If the only pairwise common neighbors of the u i are the v i , then two u i are 5-vertices andthe third is a 6 − -vertex. The case where the u i have more pairwise common neighbors is nearly the electronic journal of combinatorics (2016), dentical, and we remark on it briefly at the end of the proof. So suppose that d ( u ) = d ( u ) = 5and d ( u ) = 6, as shown in Figure 8; the case where d ( u ) = 5 is nearly identical. We will applyLemma 4 with J = { u , u , u } and k = 0. Clearly, J is an independent set. Now we verifythat each vertex of J satisfies condition (2). Since G has no separating 3-cycle, the two verticesin each pair with a common label (among { , , } ) are distinct and nonadjacent. Similarly, thevertices with labels in { , , } are distinct, since they are drawn at pairwise distance at mostthree, and G has no separating 3-cycle. Thus, we can apply Lemma 4, as desired.In the more general case where the u i have pairwise common neighbors in addition to the v i ,the argument above still shows that the vertices with labels in { , , } are distinct. So again,we can apply Lemma 4 with k = 0. Lemma 22.
Let u be a 7-vertex with nonadjacent 5-vertices u and u each at distance twofrom u . A minimal G cannot have u and u with two common neighbors and also u and u with two common neighbors.Proof. This situation is shown in Figures 9(a,b,c), possibly with some vertices identified. Let J = { u , u , u } . Suppose that more than a single pair of vertices is identified, which implies | N ( J ) |
11. If J is a non-maximal independent set, then this contradicts Lemma 6. Sosuppose that J is a maximal independent set. If | N ( J ) |
10, then | G |
13, so J is the desiredindependent set of size | G | . Otherwise, | G | = 14, so exactly three vertices are identified. Nowwe find an independent 4-set. Either we can take the four vertices labeled 4, or the two labeled i , together with J \ { u i } , for some i ∈ { , , } . Thus, at most one pair of vertices drawn asdistinct are identified.If all vertices labeled 2 or 3 are distinct as drawn, then we apply Lemma 16 and get acontradiction. By Lemma 6, the only other possibility is that exactly one pair of vertices isidentified. Such a pair must consist of vertices labeled 2 and 3 that are drawn at distance four(otherwise we apply Lemma 4, with k = 1). In Figure 9(a), this is impossible, since the two5-vertices u and u would have two neighbors in common, violating Lemma 9.Now we consider the cases shown in Figures 9(b,c) simultaneously. We apply Lemma 11using the vertices labeled 1 for the independent 2-set. Let I be the set of vertices labeled 4. If I is independent, then we are done; so assume not. Recall that a vertex labeled 2 is identifiedwith a vertex labeled 3.Suppose the vertices labeled 4 in N ( u ) \ N ( u ) and N ( u ) \ N ( u ) are not adjacent. Now bysymmetry, we may assume that the vertex labeled 4 in N ( u ) ∩ N ( u ) is adjacent to the vertexlabeled 4 in N ( u ) \ N ( u ). Let I be the set made from I by replacing the vertex labeled 4in N ( u ) ∩ N ( u ) with the vertex labeled 4 b . If I is independent, then we are done; so assumenot. Now the vertex labeled 4 b must be adjacent to the vertex labeled 4 in N ( u ) \ N ( u ),but this makes a separating 3-cycle (consisting of two vertices labeled 4 and one labeled 4 b ), acontradiction.So, we may assume that the vertices labeled 4 in N ( u ) \ N ( u ) and N ( u ) \ N ( u ) areadjacent. Suppose the topmost vertex labeled 2 is identified with the topmost vertex labeled3. Now again we are done; our independent 4-set consists of the two neighbors of u labeled 4,together with an independent 2-set from among the two leftmost and two rightmost vertices (byplanarity, they cannot all four be pairwise adjacent).The only remaining possibility is that the bottommost vertex labeled 2 is identified withthe bottommost vertex labeled 3 (since the two topmost vertices labeled 4 are adjacent). If the electronic journal of combinatorics (2016), u u (a) Here u and u have a commonneighbor in N ( u ). u b u u (b) Here u and u have adjacentneighbors in N ( u ). u
14 4 4 c b u u (c) Here u and u have neighbors atdistance 2 in N ( u ). Figure 9: The cases of Lemma 22. The three possibilities for an independent 3-set { u , u , u } where d ( u ) = 7, d ( u ) = d ( u ) = 5, and each of u and u has twoneighbors in common with u . the electronic journal of combinatorics (2016), (2016),
14 4 4 c b u u (c) Here u and u have neighbors atdistance 2 in N ( u ). Figure 9: The cases of Lemma 22. The three possibilities for an independent 3-set { u , u , u } where d ( u ) = 7, d ( u ) = d ( u ) = 5, and each of u and u has twoneighbors in common with u . the electronic journal of combinatorics (2016), (2016), e are in Figure 9(b), then the vertex labeled 4 b is a 5-vertex; since it shares two neighborswith u , another 5-vertex, we contradict Lemma 9. Hence, we must be in Figure 9(c). Nowour independent 4-set consists of the two neighbors of u labeled 4b and 4c, together with anindpendent 2-set from among the four topmost vertices (again, by planarity, they cannot all bepairwise adjacent). Lemma 23.
Suppose that a minimal G contains a 7-vertex v with no 5-neighbor. Now v cannothave at least five 6-neighbors, each of which has a 5-neighbor.Proof. Suppose to the contrary. Denote the neighbors of v in clockwise order by u , . . . , u . Case 1:
Vertices u , u , u , u are 6-vertices, each with a 5-neighbor.First, suppose that u and u have a common 5-neighbor, w . Consider the 5-neighbor w of u . By Lemma 9, it cannot be common with u ; similarly, the 5-neighbor w of u cannot becommon with u . (We must have w and w distinct, since otherwise we apply Lemma 21 to { u , u , w } . Also, we must have w and w each distinct from w , since G has no separating3-cycles.)First, suppose that w has two common neighbors with u . If w u , then we applyLemma 19 to { w , u , u } ; so assume w ↔ u . Now let J = { u , u , w } . Clearly, J is anindependent 3-set. Also | N ( J ) | − w cannot have two common neighbors with u . Similarly, w cannot have two common neighborswith u . Hence, w ↔ u and also w ↔ u . Now we must have w ↔ w ; otherwise we applyLemma 22 to { v, w , w } . Similarly, we must have w ↔ w and w ↔ w ; these edges cut off w from u , so u w . Since u and w are nonadjacent, but have a 5-neighbor in common,they must have two neighbors in common. So we apply Lemma 19 to { u , u , w } . Hence, weconclude that the common neighbor of u and u is not a 5-neighbor.Since u and u are 6 − -vertices, by Lemma 17, vertex u cannot have another 6 − -vertex thatis nonadjacent to u and u . Thus, a 5-neighbor of u must be a common neighbor of u ; callthis 5-neighbor w . Similarly, the common neighbor w of u and u is a 5-vertex. We musthave w ↔ w , for otherwise we apply Lemma 22. We may assume that u is a 6-vertex. If not,then v ’s five 6-neighbors, each with a 5-neighbor, are successive ; so, by symmetry, we are in thecase above, where u and u have a common 5-neighbor.By planarity, either u w or else u w ; by symmetry, assume the former. Since u and w share a 5-neighbor (and are nonadjacent), they have two common neighbors. Now if u w ,then we apply Lemma 19 to { u , u , w } . Hence, assume u ↔ w . This implies that u w .Now, the same argument implies that u ↔ w . Now let J = { u , u , u } . Lemma 6 gives12 | N ( J ) | − J have no additional pairwise commonneighbors. Hence, we have an independent 2-set M in N ( u ) \ ( N ( u ) ∪ N ( u )). Similarly,we have an independent 2-set M in N ( u ) \ ( N ( u ) ∪ N ( u )). Now we apply Lemma 10with J = { u , u , u } and S = M ∪ { u } and S = M ∪ { u } . In each case, we have α ( G [ S i ∪ J ]) > | M i ∪ { u − i , u } | = 4. This implies that | N ( J ) | >
13, a contradiction. Hence, v cannot have four successive 6-neighbors, each with a 5-neighbor. Case 2:
Vertices u , u , u , u , u are 6-vertices, each with a 5-neighbor.Suppose that the common neighbor w of u and u is a 5-vertex. By symmetry (between u and u ) and Lemma 17, assume that the common neighbor w of u and u is a 5-vertex. If w w , then we apply Lemma 22; so assume that w ↔ w . If u w , then apply Lemma 19to { u , u , w } (note that u and w have two common neighbors, since they have a common5-neighbor). So assume that u ↔ w . Similarly, we assume that u ↔ w , since otherwise we the electronic journal of combinatorics (2016), pply Lemma 19 to { u , u , w } . Now consider the 5-neighbor w of u . By Lemma 9, it cannotbe a common neighbor of u (because of w ). If it is a common neighbor of u , then we applyLemma 22 to { w , w , v } ; note that w w , since they are cut off by edge w u . Hence, w isneither a common neighbor of u nor of u . Now we apply Lemma 19 to { u , w , w } . Thus, weconclude that the common neighbor of u and u is not a 5-vertex.Let x denote the common neighbor of u and u ; as shown in the previous paragraph, x must be a 6 + -vertex. Suppose that the 5-neighbor w of u is also a neighbor of x . If w u ,then we apply Lemma 19 to { u , u , w } ; so assume that w ↔ u . Now if the 5-neighbor w of u is also adjacent to x , then we apply Lemma 19 to { u , w , u } (we must have w u due to edge w u ). So, by symmetry (between u and u ), we may assume that w ↔ u . Now,by Lemma 17, the 5-neighbor w of u has is adjacent to either u or u . In either case, wemust have w ↔ w ; otherwise, we apply Lemma 22 to { v, w , w } . If w ↔ u , then w ↔ w and w ↔ w ; otherwise, we apply Lemma 22 to { v, w , w } or { v, w , w } . Now we applyLemma 19 to { u , u , w } . So instead w u . Finally, we apply Lemma 19 to { u , w , u } .This completes the proof. Acknowledgments
As we mentioned in the introduction, the ideas in this paper come largely from Albertson’sproof [1] that planar graphs have independence ratio at least . In fact, many of the reducibleconfigurations that we use here are special cases of the reducible configurations in that proof.We very much like that paper, and so it was a pleasure to be able to extend Albertson’s work.It seems that the part of his own proof that Albertson was least pleased with was verifying“unavoidability”, i.e., showing that every planar graph contains a reducible configuration. Inthe introduction to Albertson [1], he wrote: “Finally Section 4 is devoted to a massive, uglyedge counting which demonstrates that every triangulation of the plane must contain someforbidden subgraph.” This is one reason that we included in this paper a short proof of thissame unavoidability statement, via discharging. We think Mike might have liked it.The first author thanks his Lord and Savior, Jesus Christ. References [1] M. O. Albertson. A lower bound for the independence number of a planar graph.
J. Com-binatorial Theory Ser. B , 20(1):84–93, 1976.[2] K. Appel and W. Haken. Every planar map is four colorable. I. Discharging.
Illinois J.Math. , 21(3):429–490, 1977.[3] K. Appel and W. Haken. Every planar map is four colorable. II. Reducibility.
Illinois J.Math. , 21(3):491–567, 1977.[4] C. Berge.
Graphes et hypergraphes . Dunod, Paris, 1970. Monographies Universitaires deMath´ematiques, No. 37.[5] M. R. Garey and D. S. Johnson. The rectilinear Steiner tree problem is NP-complete.
SIAMJ. Appl. Math. , 32(4):826–834, 1977. the electronic journal of combinatorics (2016), (2016),