Kuratowski-Pontrjagin theorem on planar graphs

Kuratowski-Pontrjagin theorem on planar graphs

JOURNAL OF COMBINATORIAL THEORY,Series B 24, 228-232 (1978) Kuratowski-Pontrjagin MICHAEL Mathematical fnstitute, Academy of Sciences Communicate...

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JOURNAL OF COMBINATORIAL THEORY,Series B 24, 228-232 (1978)

Kuratowski-Pontrjagin MICHAEL Mathematical

fnstitute,

Academy

of Sciences

Communicated

Theorem

on Planar Graphs

BURSTEIN of the Georgian

S.S.R.,

Tbilisi-380093,

U.S.S.R.

by A. A. Zykov

Received August 1975 FOR MY TEACHER, ACADEMICIAN

GEORGE CHOGOSHVILI

In- this paper we give a new proof of the well-known theorem of KuratowskiPontrjagin on planar graphs: A graph is planar iff it does not contain a subgraph homeomorphic to Kj or K,,, .

In this paper we give a new proof of the well-known theorem of Kuratowski-Pontrjagin. It was a difficult problem to characterize planar graphs before the appearance of Kuratowski’s paper [I] in 1930. Pontrjagin independently obtained the same result. In 1954 the theorem was reproved by Dirac and Shuster [2] in graph-theoretic terms. We follow the definitions of [4, 51, and we use a notion of planar hypergraph. A hypergraph is calledplanar if its vertices can be represented aspoints on the plane, its edgescan be represented as open 2-cells in the plane, so that edgesdo not intersect each other, and a vertex is incident to an edge iff the corresponding point belongs to the boundary of the corresponding 2-cell. Edges of degree 2 can be represented as usual by Jordan curves. A hypergraph is said to be Hamiltonian if it contains a cycle, consisting of the edgesof degree 2, that passesthrough all of its points (i.e., the usual Hamiltonian cycle). Let G be a hypergraph with fixed Hamiltonian cycle. Edges of G which do not belong to this cycle are called bridges. Imbed the Hamiltonian cycle of G on the plane. We say that two brideges E’ and E” overlap if they cannot be imbedded into the interior of this cycle as 2-cells containing their incident vertices on their boundaries without intersections. Thus, for a Hamiltonian hypergraph G with fixed Hamiltonian cycle one can define a graph of overlappings H(G) as follows: Vertices of H(G) are the bridges of G and two vertices are adjacent in H(G) iff they overlap. PROPOSITION.

Hamiltonian hypergraph G is planar 18 the graph H(G) is

bichromatic. 228 00958956/78/0242-0228$02.00/O Copyright 411 rights

0 1978 by Academic Press, Inc. of reproduction in any form reserved.

THEOREM ON PLANAR GRAPHS

229

Proof. The proof is trivial; edgeswhich must lie inside the cycle will be colored with first color and the edges which lie outside with the second color. Let r denote the class of Hamiltonian graphs which have a simple cycle of odd length for a graph of overlappings, and have no vertices of degree 2. COROLLARY.

Each graph from r is nonplanar.

Remark 1. KSand K3,3 belong to r (seeFig. 1).

FIGURE

LEMMA.

1

Any graph from r contains a subgraph homeomorphic to K5

or k3 , Proof. It is easy to see that any graph L from r contains another Hamiltonian cycle: The latter passesthrough all bridges of L and may be through some edgesof the original cycle (see Fig. 2). Two edgesof the first cycle that do not belong to the new one overlap (with respect to the new Hamiltonian cycle) iff they are not adjacent. But if we find three pairwise overlapped bridges (with respect to any cycle) we obtain a homeomorph of &,3 . So if a graph L from r has three pairwise disjoint edgeswhich do not belong to the new cycle, then it contains a subgraph homeomorphic to K 3,3. But if L has not such a triple, then L can be only KS .

6

7

2 FIGURE

THEOREM (Kuratowski-Pontrjagin). homeomorphof K5 or K3,3 .

3

2

Any

nonplanar graph contains a

230

MICHAEL

BURSTEIN

ProoJ: Let L be critical nonplanar graph; it means that after deleting any edge of L (without its end points) we obtain a planar graph. To prove the theorem it therefore suffices to prove that L contains a subgraph homeomorphic to a graph from the class I’. We assume that L has no vertices of degree 2. Evidently, L has no multiple edges and L is a block, because if L is not a block, then all its blocks are planar, so L is also planar. Let C be a cycle in L of maximum length and X a set of vertices lying on this cycle. Consider a subgraph of L induced by vertices not from X. We call a piece of L (with respect to C) any connected component of this subgraph with the addition of all edges incident to vertices of the component and their end points. An edge of L whose end points are in X is also a piece. Let L, , L, ,..., LI, be the pieces. For a piece Li consider a set of vertices Ei _CX of this piece lying on C. Consider a hypergraph G, whose vertex set is X and whose edges are all the edges of the cycle C and the sets Ei , i = 1, 2 ,..., k. G is a Hamiltonian hypergraph with a Hamiltonian cycle C. Clear, that for any i = 1, 2 ,..., k, / Ei ( 2 2 and vi Ei = X. Two consecutive vertices of a cycle C do not belong to the same Ei , because if they do, then they can be connected by a path with all edges from Li, and L would have a longer cycle. So k > 2. Let us note that for any i = 1, 2,..., k a graph C v Li is planar; really, C U Li is obtained from L after deleting the edges of other pieces. A hypergraph G is nonplanar because if it were planar then we could represent it on the plane and imbed each piece Li into the 2-cell representing an edge Ei ; this is impossible, for L is nonplanar. Therefore, H(G) is not bichromatic. After deleting any edge Es we obtain a planar hypergraph. So H(G) is critical not bichromatic, i.e., H(G) is simply an odd cycle. If for all i, 1Ei j = 2 we are finished. In fact, in each Li we can take a path connecting the vertices from Ei , and these paths together with C constitue the homeomorph of the graph from r. We can assume that the pieces are numbered so that Ei overlaps Etml and &I (6, = & > Et,, = -4). Orient the edges of C in a cyclic fashion, and let [x, ~1 denote the set of vertices of the oriented path from x to y along C. Denote also (x, y] =

[x, vl\C4 [x3JJ>= k YI\~J~~and (x3Y>= (x9YIKYI. Suppose that / Ei j > 3 for some i; with no loss of generality we can assume that 1El j > 3. Let El = (x1 , x2 ,..., xt> and this order of the vertices of El coinsides with the order of these vertices on C. For any vertex x E Ei the subset Ei\{x} does not overlap one of the edges K1 or Ei+l , for if it does, then we can delete some edges of Li incident to x from L and the remaining graph will be nonplanar. So any proper subset of EC does not overlap Expl or E,+l . That is why the set [xiP1, xi+,] contains Ez or Ek for any i = 1, 2,..., t

THEOREM ON PLANAR GRAPHS

and t = t = one

231

one of them touches both [xiP1, xi) and (xi, xi+J. It is possible only if 3 because the consecutive vertices of C do not belong to E1 . But even if 3 each set [x1 , x3], [x2 , x,], [x, , xz] contains E, or El, (see Fig. 3). So, of them, say E, , is contained in two of these sets, say

FIGURE 3

It is clear that x1 E E2 ; otherwise E2 would not overlap El . If x2, xQ E ES then El C E2 and EI, overlapps E, as well and it means that k = 3. If / E3 1> 3, then a 2-element subset of E3 overlaps both El and E, and this contradicts our above remark. If / ES 1 = 2, then E2 contains consecutive vertices of C. So El $ E, In this case E2 is not contained in the third set [x3 , x.J and it means that Ek C [x, , xJ. It is evident that / E, / = 2 for consecutive vertices of C do not belong to E2 . So is El, . Clearly, E, = {x, , x’} where x’ E (x2, x3). It is easy to see that E, and Ek overlap because Ek touches both sets [x3 , x1) and (x1 , x2]. So k = 3 and we obtain the hypergraph depicted in Fig. 3. In this case we can easily find a homeomorph of K3,3 in L: Take a vertex in L, connected with x1 , x, and x, by three disjoint paths. This completes the proof. One can observe that in the latter case (Fig. 3) the length of the cycle C in L can be increased. Remark 2. In comparison with the classical proofs of [I] or [2] of the theorem our proof is constructive. It shows how to find K5 or K3,3 in a nonplanar graph. As in [3], a simple algorithm for planarity testing can be based on this proof. It is unnecessary for the cycle C to be of maximum length; only the consecutive vertices of this cycle must not lie in the same piece.

232

MICHAEL

BURSTEIN

ACKNOWLEDGMENT I am grateful to Professor D. Baladze for useful conversations with regard to this paper. I also wish to thank the referee, who pointed out that the proof in the original version was insufficiently detailed.

REFERENCES 1. K. KIJRATOWSKI, Sur le probleme 15-16 (1930), 271-283. 2. G. A. DIRAC AND S. SHUSTER, A

des courbes gauches en topologie,

Fund. Math.

theorem of Kuratowski, Ned&. Acad. Wetensch. Proc. Ser. A 57 (1954), 343-348. 3. W. T. TUTTE, How to draw a graph, Proc. London Math. Sot. 13 (1963) 743-767. 4. F. HARARY, “Graph Theory,” Addison-Wesley, Reading Mass., 1969. 5. A. A. ZYKOV, Hypergraphs, Uspehi Mat. Nauk, 6 (1974), 89-154 [in Russian].