Any Maximal Planar Graph with Only One
Separating Triangle is Hamiltonian
∗
CHIUYUAN CHEN cychen@mail.nctu.edu.tw
Department of Applied Mathematics, National Chiao Tung University, Hsinchu 300, Taiwan, ROC Received December 27, 2000; Revised December 27, 2000; Accepted December 28, 2000
Abstract. A graph is hamiltonian if it has a hamiltonian cycle. It is well-known that Tutte proved that any 4-connected planar graph is hamiltonian. It is also well-known that the problem of determining whether a 3-4-connected planar graph is hamiltonian is NP-complete. In particular, Chv´atal and Wigderson had independently shown that the problem of determining whether a maximal planar graph is hamiltonian is NP-complete. A classical theorem of Whitney says that any maximal planar graph with no separating triangles is hamiltonian, where a separating triangle is a triangle whose removal separates the graph. Note that if a planar graph has separating triangles, then it can not be 4-connected and therefore Tutte’s result can not be applied. In this paper, we shall prove that any maximal planar graph with only one separating triangle is still hamiltonian.
Keywords: planar graph, maximal planar graph, hamiltonian cycle, separating triangle, NP-complete
1. Introduction
Our terminology and notation in graphs are standard; see Chartrand and Lensniak (1981) and West (1996), except as indicated. Graphs discussed in this paper are assumed simple and finite. A graph is hamiltonian if it has a hamiltonian cycle. A graph is planar if it can be drawn in the plane with no two edges crossing. A plane graph is a graph drawn in the plane with no two edges crossing. Unless otherwise specified, a planar graph means the plane embedding of the graph. A planar graph divides the plane into regions, which are called faces. The unbounded region is called the exterior face; the other faces are called
interior faces. An edge is a boundary edge if it is on the exterior face. A graph is maximal planar if it is planar and no edge can be added without losing planarity. Note that any face of
a maximal planar graph is a triangle. A triangulation is a 2-connected planar graph in which all faces (except possibly the exterior face) are triangles. A triangle of a planar graph is a
separating triangle if it does not form the boundary of a face. That is, a separating triangle
has vertices both inside it and outside it; therefore its removal separates the graph. For example, the planar graph in figure 1 has three separating triangles: ACE, BDC, and FBA.
It is well-known that the problem of determining whether a 3-connected planar graph is hamiltonian is NP-complete (Garey et al., 1976). In particular, Chv´atal and Wigderson had independently shown that the problem of determining whether a maximal planar graph is
∗This research was partially supported by the National Science Council of the Republic of China under grant
s E s C sD s s s A sB s s F ✏✏✏✏✏ PPPPP ✚✚ ✚✚ ✚✚❅ ❅ ❅ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ✭ ❤❤❤❤❤❤❤❤❤ ✟✟✟❍❍❍ L L L L L L LL ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ✁✁ ✁ ❆ ❆ ❆
Figure 1. An illustration of separating triangles.
hamiltonian is NP-complete (Chv´atal, 1985). On the other hand, Whitney (1931) proved that any maximal planar graph with no separating triangles is hamiltonian; Asano et al. (1984) proposed a linear-time algorithm for finding hamiltonian cycles in such planar graphs. Tutte (1956) proved that any 4-connected planar graph is hamiltonian; Chiba and Nishizeki (1989) proposed a linear-time algorithm for finding hamiltonian cycles in these planar graphs. Note that Tutte’s result can be viewed as a generalization of Whitney’s result since any maximal planar graph with at least five vertices and with no separating triangles is 4-connected (we prove this in Section 4). Dillencourt (1990) also general-ized Whitney’s result, but in a different way: he relaxed the requirement that the triangu-lation must be maximal planar but still insisted that the triangutriangu-lation has no separating triangles.
All the results of Whitney (1931), Tutte (1956), and Dillencourt (1990) insist that the given planar graph has no separating triangles. Note that if a planar graph has separating triangles, then it can not be 4-connected and therefore Tutte’s result can not be applied. In this paper, we shall prove that any maximal planar graph with only one separating triangle is still hamiltonian. Note that there exist non-hamiltonian maximal planar graphs with 7 separating triangles.
2. Preliminaries
Since Whitney’s result (Whitney, 1931) plays a crucial role in (Asano et al., 1984; Chiba and Nishizeki, 1989; Dillencourt, 1990) and this paper, we state this result first. Let G be a triangulation, let R be the exterior face of G, and let A and B be two vertices on R. We say that (G, R, A, B) satisfies Whitney’s condition (Condition (W ) for short) if (G, R, A, B) satisfies Conditions (W1) and (W2) described below. We say that (G, R, A, B) satisfies Condition (W1 ) if G has no separating triangles. We say that (G, R, A, B) satisfies Condition (W2) if either
(W2a) a0, a1, . . . , am is the path from A to B and b0, b1, . . . , bn is the path from B to
A (a0 = bn = A, b0 = am = B), and there is no chord (a chord is an edge joining two
(a) s bn−1 s A s a1 s a2 s s sB s b1 s b2 s s s ✟✟✟ ❍❍❍ ❅ ❅ ❅ ✟ ✟ ✟ ❍ ❍ ❍ ❅ ❅ ❅ ✁✁ ✁✁ ✁✁ ❧ ❧ ❧ ❧ ❧ ❧ ❧ ❧❧ (b) s s A s a1 s a2 s s sB s b1 s b2 s s C s c1 c2 ✟✟✟ ❍❍❍ ❅ ❅ ❅ ✟ ✟ ✟ ❍ ❍ ❍ ❅ ❅ ❅ ✁✁ ✁✁ ✁✁ ❧ ❧ ❧ ❧ ❧ ❧ ❧ ❧❧
Figure 2. An illustration of Conditions (W2a) and (W2b). Note that only vertices on the exterior faces are shown. (W2b) a0, a1, . . . , am is the path from A to B, b0, b1, . . . , bn is the path from B to C,
c0, c1, . . . , ckis the path from C to A for some vertex C on R (a0= ck= A, b0= am = B,
c0= bn = C), and there is no chord of the form aiaj, bibj, or cicj.
For example, in figure 2(a), the chords of the exterior face are a1bn−1and a2b1; since there
is no chord of the form aiajor bibj, (G, R, A, B) satisfies Condition (W2a). In figure 2(b),
the chords of the exterior face are a1c2, a2b1, and b2c1; since there is no chord of the form
aiaj, bibj, or cicj, (G, R, A, B) satisfies Condition (W2b).
Whitney (1931) proved that
Lemma 1 (Whitney’s Lemma). Let G be a triangulation, let R be the exterior face of G, and let A and B be two vertices on R. If (G, R, A, B) satisfies Condition (W ), then G has a hamiltonian path from A to B.
Theorem 2 (Whitney’s Theorem). If G is a maximal planar graph with no separating triangles, then G is hamiltonian.
The following lemma will be used heavily in this paper.
Lemma 3. Let G be a maximal planar graph with no separating triangles, let R be the
exterior face of G, and let A and B be two vertices on R. Then (G, R, A, B) satisfies Condition (W ).
Proof: Since G has no separating triangles, (G, R, A, B) satisfies Condition (W1). Since
R is a triangle, it has no chords; therefore (G, R, A, B) satisfies Condition (W2). Thus,
(G, R, A, B) satisfies Condition (W). 3. What we are afraid of?
Let G be a maximal planar graph with only one separating triangle A BC. Let Gin(Gout)
triangle A BC. Then, both Gin and Gout are maximal planar graphs with no separating
triangles.
Consider Gin. A, B, C form the exterior face, say R, of Gin. By Lemma 3, (Gin, R, A, B)
satisfies Condition (W). By Whitney’s Lemma, Ginhas a hamiltonian path Pinfrom A to
B. Then
Pin= A, Pin( A, C), C, Pin(C, B), B,
where Pin( A, C) (Pin(C, B)) is the subpath of Pinbetween A and C (C and B).
Next consider Gout. A, B, C form an interior face, say R, of Gout. Note that a plane
graph can always be embedded in the plane so that a given face of the graph becomes the exterior face; see Nishizeki and Chiba (1988). Therefore, we can embed Goutin the plane so
that Rbecomes the exterior face of Gout. By Lemma 3, (Gout, R, B, C) satisfies Condition
(W). By Whitney’s Lemma, Gouthas a hamiltonian path Poutfrom B to C. Then
Pout= B, Pout(B, A), A, Pout( A, C), C,
where Pout(B, A) (Pout( A, C)) is the subpath of Poutbetween B and A ( A and C).
If Pout( A, C) = ∅, then
A, Pin( A, C), C, Pin(C, B), B, Pout(B, A), A
is a hamiltonian cycle of G. Similarly, if Pin( A, C) = ∅, then
A, Pout( A, C), C, Pin(C, B), B, Pout(B, A), A
is a hamiltonian cycle of G. What we are afraid of is that
Pin( A, C) = ∅ and Pout( A, C) = ∅.
Then it is impossible to use Pinand Poutto derive a hamiltonian cycle of G.
4. The main result
We first prove a lemma mentioned in the introduction. A vertex cut of a connected graph
G= (V, E) is a subset Vof V such that its removal disconnects G. A minimal vertex cut is a vertex cut such that no proper subset of it is also a vertex cut.
Lemma 4. Any maximal planar graph with at least five vertices and with no separating triangles is 4-connected.
Proof: Whitney (1932) proved that a maximal planar graph G with at least four vertices is 3-connected. Hakimi and Schmeichel (1978) proved that if Vis a minimal vertex cut of a maximal planar graph G, then the subgraph of G induced by V is a cycle without
chords. (Recall that a chord is an edge joining two non-consecutive vertices of a cycle.) Let Vbe a minimal vertex cut of G. Since G is 3-connected,|V| ≥ 3. Suppose |V| = 3. Then by Hakimi and Schmeichel (1978), the subgraph of G induced by Vis a triangle; this contradicts the assumption that G has no separating triangles. Therefore,|V| ≥ 4. This proves that G is 4-connected.
We now prove a variation of Whitney’s Lemma.
Lemma 5. Let G be a maximal planar graph with no separating triangles, let R be the
exterior face of G, and let A, B, C be the three vertices on R. Then G has a hamiltonian path from A to B passing through the edge AC.
Proof: If G has only three vertices, then this lemma is clearly true. In the following, assume that G has at least four vertices. Let P0, P1, . . . , Pr (P0 = B, Pr = C) be the
sequence of vertices adjacent to A such that each APi is the immediate counter-clockwise
edge of APi−1around A; see figure 3. Note that PiPi+1is an edge of G for all i , 0≤ i ≤ r −1.
Since G has no separating triangles, the following three properties hold:
1. G has no edge of the form PiPj (0≤ i < i + 2 ≤ j ≤ r), since otherwise PiPjA is a
separating triangle.
2. G has no edge of the form BPi (2 ≤ i ≤ r − 1), since otherwise BPiA is a separating
triangle.
3. G has no edge of the form CPi (1≤ i ≤ r − 2), since otherwise PiCA is a separating
triangle.
See figure 3. Let Gbe the subgraph of G derived by deleting A. Then Gis a triangulation. Let Rbe the exterior face of G. Then R= B, P1, . . . , Pr−1, C. Since G has no separating
triangles, Galso has no separating triangles; thus (G, R, C, B) satisfies Condition (W1). By (1)–(3), Rhas no chords and therefore (G, R, C, B) satisfies Condition (W2). Thus, (G, R, C, B) satisfies Condition (W). By Lemma 1, Ghas a hamiltonian path P from C
s P0 = B s P1 sA s P2 s P3 Pr = C G ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ❙ ✭✭✭✭✭✭✭✭✭❤❤❤❤❤❤❤❤❤❛❛❛❛❛❛❛❛ L L L L L L LL
to B. The path P together with the edges AC form a hamiltonian path of G from A to B passing through the edge AC.
We classify maximal planar graphs here. Let G be a maximal planar graph. Note that unless G is a triangle, then it is impossible for G to have a hamiltonian cycle passing through all of its three boundary edges. We say that G is hamiltonian (non-hamiltonian)
for any two boundary edges if for any two boundary edges, G has (does not have) a
hamiltonian cycle passing through them. We say that G is hamiltonian (non-hamiltonian)
for any boundary edge if for any boundary edge, G has (does not have) a hamiltonian
cycle passing through it. A K4 (the complete graph with four vertices) is hamiltonian
for any two boundary edges. Note that not every hamiltonian maximal planar graph is hamiltonian for any two boundary edges. For example, the graph in figure 1 has a hamil-tonian cycle passing through both DF and EF, but it does not have a hamilhamil-tonian cycle passing through both DE and DF (or DE and EF). The two graphs in figure 4 are even worse: although they are hamiltonian, they are non-hamiltonian for any two boundary edges.
What makes a maximal planar graph with separating triangles non-hamiltonian? We in-troduce the definitions of “hamiltonian for any two boundary edges”, “non-hamiltonian for any two boundary edges”, “hamiltonian for any boundary edge”, and “non-hamiltonian for any boundary edge” because we suspect they affect the hamiltonicity of maximal planar graphs with separating triangles. Suppose ABC is a separating triangle of a maximal planar graph G. Then, the graph Gin derived by deleting all the vertices outside the separating
triangle ABC is still a maximal planar graph. If Gin is “non-hamiltonian for any
bound-ary edge,” then G is certainly non-hamiltonian. “Non-hamiltonian for any two boundbound-ary edges” maximal planar graphs had been used as building blocks for constructing non-hamiltonian maximal planar graphs. For example, in Dillencourt (1996), Dillencourt used the two graphs in figure 4 as building blocks to construct non-hamiltonian maximal planar graphs. (This approach was also used in Nishizeki (1980).) In Dillencourt (1996), A, B,
Cin figure 4(a) (or figure 4(b)) were identified with A, B, C in figure 1 (that is, the interior face ABC of figure 1 was replaced with figure 4(a) (or figure 4(b))); the resultant graph
(a) s A sC s s s s s B ✘✘✘✘✘✘ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍ ❍❍ ❅ ❅ ✁✁ ✁✁ ✁✁ ✁✁ ✁✁ ✁✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❉ ❉ ❉ ❉ ❉ ❉ ❉❉ ❅ ❅ ❅ ❅❅ (b) s A sC s s s s s B s s ✘✘✘✘✘✘ ✟ ✟ ✟ ✟ ✟ ✟ ❍❍❍❍ ❍❍ ❅ ❅ ✁✁ ✁✁ ✁✁ ✁✁ ✁✁ ✁✁ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❆ ❉ ❉ ❉ ❉ ❉ ❉ ❉❉ ❅ ❅ ❅ ❅❅ ✡ ✡ ✡ ✡ ✡ ❏ ❏ ❏ ❏❏ L L L L L L LL
is a non-hamiltonian maximal planar graph. It is not difficult to verify that the above two non-hamiltonian maximal planar graphs each has 7 separating triangles.
We now prove a stronger version of Whitney’s Theorem.
Theorem 6. If G is a maximal planar graph with no separating triangles, then G is
hamiltonian for any two boundary edges.
Proof: Let R be the exterior face of G and let A, B, C be the three vertices on R. By Lemma 5, G has a hamiltonian path P from A to B with AC being an edge of it. The path
P together with the edge BA form a hamiltonian cycle of G passing through both AB and AC. By similar arguments, G has a hamiltonian cycle passing through both BA and CB and
a hamiltonian cycle passing through both CB and BA. Thus, G is hamiltonian for any two boundary edges.
We now prove the main result.
Theorem 7. If G is a maximal planar graph with only one separating triangle, then G is
hamiltonian.
Proof: Let A, B, C be the vertices form this unique separating triangle. Let Gin(Gout)
be the subgraph of G derived by deleting all the vertices outside (inside) the separating triangle ABC. Then, both Gin and Gout are maximal planar graphs with no separating
triangles. Consider Gin. A, B, C form the exterior face of Gin. By Theorem 6, Ginhas a
hamiltonian cycle Cinpassing through both BA and AC. Then
Cin = B, A, C, Pin(C, B), B,
where Pin(C, B) is the subpath of Cinbetween C and B. Consider Gout. A, B, C form an
interior face, say R, of Gout. Embed Goutin the plane so that Rbecomes the exterior face
of Gout. By Theorem 6, Gouthas a hamiltonian cycle Coutpassing through both AC and CB.
Then
Cout = A, C, B, Pout(B, A), A,
where Pout(B, A) is the subpath of Coutbetween B and A. Then
A, C, Pin(C, B), B, Pout(B, A), A
a hamiltonian cycle of G. References
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