# 最大，二分，外平面圖之容忍表示法 - 政大學術集成

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(2) Contents. Abstract. ii. 中文摘要. iii. 1 Introduction. 1. 1.1. History of Tolerance Graphs . . . . . . . . . . . . . . . . . . . . . .. 1.2. The Structure of Tolerance Graphs . . . . . . . . . . . . . . . . . .. 立. 政 治 大. ‧ 國. 學. 2 Tolerance Graphs. Bounded Tolerance Representations for Trees and Bipartite Graphs. 2.3. A Tolerance Representation of C4 . . . . . . . . . . . . . . . . . . .. 2.4. A Tolerance Representation of Concatenation of Two 4-cycles . . .. sit. 7. v i n C h of Concatenation A Tolerance Representation of Three 4-cycles . . . engchi U n. 2.5. 3 Some Results on Maximal Outerplanar Graphs 3.1. 4 6. er. io. al. 4. y. 2.2. 3. ‧. Definition and Theorem of Tolerance Graph . . . . . . . . . . . . .. Nat. 2.1. 1. 10 12. 20. A 2-connected Graph Which Is Maximal Outerplanar Graph and Bipartite Is Not Necessarily a Tolerance Graph. . . . . . . . . . . .. 20. 4 Open Problems and Further Directions of Studies. 30. References. 31. i.

(3) Abstract In this thesis, we prove a 2-connected graph G which is maximal outerplanar and bipartite is a tolerance graph if and only if there is no induced subgraphs H1 , H2 , H3 and H4 of G. keywords: Tolerance Graphs; Maximal Outerplanar Graphs; Bipartite graphs.. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. ii. i n U. v.

(4) 中文摘要 在這篇論文中，我們針對2-連通的最大外平面圖而且是二分圖的圖形，討論 其容忍表示法，並找到它的所有禁止子圖 H1 、H2 、H3 、H4 。 關鍵詞:最大外平面圖；二分圖；容忍表示法。. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. iii. i n U. v.

(5) 1 1.1. Introduction History of Tolerance Graphs In graph theory, interval graphs were introduced to study such problems of. intersecting intervals on the real line. Each vertex v in an interval graph G = (V, E) is associated with an interval Iv . If two vertices are connected by an edge in G, then the intersection of their associated intervals is nonempty. Those intervals could represent the durations of a set of events on a time line or sectors of consecutive. 治 政 as their having something important in common. 大 立. elements of a linearly ordered set. When two intervals intersect, we could explain. T olerance graphs which are a generalization of interval graph were introduced. ‧ 國. 學. by Golumbic and Monma [1]. In a tolerance graph, each vertex can be represented by an interval and a tolerance. If the size of the intersection of two intervals is. ‧. not at least as large as the tolerance associated with one of the vertices, then no. sit. y. Nat. edge is added between their vertices in the graph. Golumbic and Trenk [3] mention that their original motivation was the need to solve scheduling problems in which. io. n. al. er. resources such as vehicles, support personnel, rooms, etc. may be needed on an. i n U. v. exclusive basis, but where a measure of flexibility or tolerance would allow for. Ch. engchi. sharing or relinquishing the resource if a solution is not otherwise possible. We can think of the tolerance graph as a model of conflicts for events occurring in a block of time, in which intervals represent a set of events on a time line and their associated tolerances are flexibility of the event. Golumbic and Trenk [3] say that it is not even known how to get a tolerance representation when the graph is known to be a tolerance graphs. Similarly, the recognition problem for bounded tolerance graph is open when the graph is known to be a bounded tolerance graph. In spite of difficulty, we can also test whether it is not a bounded tolerance graph by some theorems. The complication of recognition. 1.

(6) problem for the class of tolerance graphs is not solved. Hayward and Shamir [4] show that recognition of tolerance graphs is NP. In our article, we discuss a 2-connected maximal outerplanar and bipartite graph G that is a tolerance graph and find all forbidden subgraphs of G. At first, in Chapter 1, we introduce a interval graph and a tolerance graph which is generalization of interval graph. In Chapter 2, we introduce the definitions, theorems, and propositions of tolerance graphs and discuss some special tolerance graphs and their tolerance representations. In Chapter 3, we show a tolerance representation of a 2-connected maximal outerplanar and bipartite graph and we. 政 治 大. find all forbidden subgraphs to test whether if 2-connected maximal outerplanar. 立. and bipartite graphs are tolerance graphs. The graph in Figure 1, appeared in. ‧ 國. 學. Golumbic and Trenk [3], shows the structure of tolerance graphs. It assist us to understand tolerance graph clearly. Finally, we bring up some open problems and. ‧. further directions of research.. n. er. io. sit. y. Nat. al. Ch. engchi. 2. i n U. v.

(7) 1.2. The Structure of Tolerance Graphs. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 1: Structure of the tolerance graph, appeared in Golumbic and Trenk, 2004.. 3.

(8) 2 2.1. Tolerance Graphs Definition and Theorem of Tolerance Graph. Definition 2.1. A graph G = (V, E) is a tolerance graph if each vertex x ∈ V can be assigned a closed interval Ix and a tolerance tx > 0 and tx ∈ R so that xy ∈ E if and only if |Ix ∩ Iy | ≥ min {tx , ty }. Such a collection of intervals and tolerances is called a tolerance representation. Definition 2.2. A graph G = (V, E) is a bounded tolerance graph if G has a. 政 治 大. tolerance representation with tv ≤ |Iv |, for all v ∈ V .. 立. Example 2.3. Consider the graph in Figure 2 with tolerances. ⇐⇒ 5 = |Ia ∩ Ib | > min{ta , tb } = 1. bc ∈ E. ⇐⇒ 1 = |Ib ∩ Ic | ≥ min{tb , tc } = 1. ⇐⇒ 4 = |Ic ∩ Id | > min{tc , td } = 1. y. sit. cd ∈ E. ‧. ab ∈ E. Nat. ‧ 國. 學. ta = 1, tb = 6, tc = 1, td = 6. Since. er. io. ad ∈ E ⇐⇒ 1 = |Ia ∩ Id | ≥ min{ta , td } = 1. 0 = |I ∩ I | < min{t , t } = 1 a ⇐⇒ iv l C n bd ∈ / E ⇐⇒ 3h= |I ∩ I | < min{t e n g c h i U , t } = 6,. n. ac ∈ /E. a. c. a. b. d. b. c. d. G is a tolerance graph with tolerances ta = 1, tb = 6, tc = 1, td = 6.. Figure 2: an example of tolerance graph and its tolerance representation. 4.

(9) Definition 2.4. A transitive orientation F of graph G = (V, E) is an assignment → ∈ F and − → ∈ F then − → ∈ F. of a direction to each edge in E such that if − xy yz xz Definition 2.5. A graph G is called a comparability graph if it has a transitive orientation. Example 2.6. C4 is a comparability graph but C5 is not a comparability graph by the following graphs. Let G1 = (V1 , E1 ) where V1 = {a, b, c, d}, E1 = {ab, bc, cd, ad} and G2 = (V2 , E2 ) where V2 = {a, b, c, d, e}, E2 = {ab, bc, cd, ad, ea}. In any F1 be an orientation to → − → − → − − → each edge of G1 = (V1 , E1 ). We let ab ∈ F1 , cb ∈ F1 , cd ∈ F1 and ad ∈ F1 . Hence,. 政 治 大 be any transitive 立 orientation of G = (V , E ). Without loss of generality, → −. we can easily check F1 is a transitive orientation of G1 . Let F2. 2. 2. 2. 2. 2. ‧. ‧ 國. 學. − − we let → ea ∈ F2 . If ab ∈ F2 , then eb ∈ E2 , but eb ∈ / E2 . Hence, if → ea ∈ F2 , then → − → − → − → − − ba ∈ F2 . Similarly, we know that bc ∈ F2 , dc ∈ F2 , de ∈ F2 , and → ae ∈ F2 . We − reach a contradiction to → ea ∈ F . Hence, G has no transitive orientation.. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 3: Graphs show that C4 has a transitive orientation and C5 has no transitive orientation.. Definition 2.7. A cocomparability graph is a graph whose complement is a comparability graph. Definition 2.8. Three vertices v1 , v2 , v3 ∈ V (G) form an asteroidal triple(AT) of G if there is a path from vi to vj , for all i, j ∈ {1, 2, 3} which avoids using any vertex in the closed neighborhood N [vk ] = {vk } ∪ N (vk ).. 5.

(10) Figure 4: The vertices a, b, c form an asteroidal triple(AT) in both graphs.. Definition 2.9. A graph G is called asteroidal triple f ree (AT −f ree) if it contains no asteroidal triple(AT).. 政 治 大. Lemma 2.10. [2] All cocomparability graphs are AT-free.. 立. Theorem 2.11. [3] If G is a bounded tolerance graph, then G is a cocomparability. ‧ 國. 學. graph.. y. Nat. io. sit. Bounded Tolerance Representations for Trees and Bipartite Graphs. al. er. 2.2. ‧. Corollary 2.12. If G is a bounded tolerance graph, G is AT-free.. v. n. Theorem 2.13. [3] If T is a tree, the following are equivalent.. Ch. engchi. (i) T is a bounded tolerance graph.. (ii) T has no subtree isomorphic to the graph T2 . (iii) T is a caterpillar. (iv) T is an interval graph. (v) T is a permutation graph. (vi) T is a cocomparability graph. (vii) T has no asteroidal triple(AT).. 6. i n U.

(11) Theorem 2.14. [3] Let G = (X, Y, E) be a bipartite graph. The following conditions are equivalent. (i) G is a bounded tolerance graph. (ii) G is a trapezoid graph. (iii) G is a cocomparability graph. (iv) G is AT-free. (v) G is a permutation graph.. 4. 學 ‧. ‧ 國. 2.3. 政 治 大 A Tolerance Representation of C 立. er. io. sit. y. Nat. al. n. Figure 5: The graph G1 and its tolerance representation.. Ch. engchi. i n U. v. Let G1 = (V1 , E1 ), where V1 = {a, b, c, d}, E1 = {ab, bc, cd, ad}, < I, t > be a tolerance representation of G1 where I = {Ix |x ∈ V1 }, t = {tx |x ∈ V1 }, kxy = |Ix ∩ Iy | x, y ∈ V1 and Ix = (lx , rx ). Lemma 2.15. If xy ∈ E1 then kxy > kpq for some pq ∈ / E1 . Proof. Because ab ∈ E1 , we have kab = |Ia ∩ Ib | ≥ min{ta , tb }. If ta ≥ tb , then kab = |Ia ∩ Ib | ≥ tb > |Ib ∩ Id | = kbd . If ta < tb , then kab = |Ia ∩ Ib | ≥ ta > |Ia ∩ Ic | = kac . Therefore, kab > kbd or kab > kac . Similarly, kxy > kbd or kxy > kac , for all x, y ∈ E1 and we can know that if xy ∈ E1 then kxy > kpq , for some pq ∈ / E1 .. 7.

(12) By Lemma 2.15, we have the following lemma. Lemma 2.16. One of kac , kbd is the smallest value in kxy , for all x, y ∈ V1 .. Without loss of generality, we assume that kac is the smallest value of kxy , for all x, y ∈ V1 . Proposition 2.17. We have the following properties about G1 = (V1 , E1 ) in Figure 5.. (i) ta ≤ kab and tc ≤ kbc ; (ii) ta ≤ kad and tc ≤ kcd ;. 立. 政 治 大. (iii) If rb > rd then tb > kcd ;. ‧ 國. 學. (iv) If rd > rb then td > kbc .. ‧. Proof.. (i) At first, we discuss the position of Ia and Ic . If Ia ⊆ Ic , then kac =. Nat. sit. y. |Ia ∩ Ic | = |Ia |. Because ab ∈ E, we have kab > kac = |Ia |. We reach a. al. er. io. contradiction to kab ≤ |Ia |. Hence, Ia * Ic . Similarly, Ic * Ia . Without loss. n. of generality, we let la < lc and ra < rc . Next, we know that kab > kac and. Ch. i n U. v. kbc > kac . Therefore, lb < lc and rb > ra . The tolerance representation of G1. engchi. is shown in Figure 6. Hence, we have the following four cases about ta and tc .. Figure 6: The tolerance representation of the vertices a, b and c.. 8.

(13) case1: Suppose that the tolerance ta > kab and the tolerance tc > kbc . Since ab ∈ E1 and bc ∈ E1 , we have the result tb ≤ kab and tb ≤ kbc . Because kad > kac and kcd > kac , ld < lc and rd > ra . Suppose that ld < lb . We obtain that kab ≤ kbd , but kbd = |Ib ∩ Id | < min{tb , td } ≤ tb ≤ kab which reaches a contradiction (see Figure 7). Hence, we know that ld ≥ lb . The graph is shown in Figure 8. Because ad ∈ E1 , kad = |Ia ∩ Id | ≥ min{ta , td }. We know that ta > kab ≥ kad (see Figure 8). Therefore, we obtain that td ≤ kad . Hence, |Ib ∩ Id | = kbd ≥ kad ≥ td , it implies that bd ∈ E1 , again, we reach a contradiction.. 立. 政 治 大. ‧. ‧ 國. 學. Nat. Figure 8: The tolerance representa-. tion of G1 with ld < lb .. tion of G1 with ld ≥ lb .. n. al. er. io. sit. y. Figure 7: The tolerance representa-. i n U. v. case2: Suppose that the tolerance ta > kab and the tolerance tc ≤ kbc .. Ch. engchi. Since ab ∈ E1 , we get that tb ≤ kab . Because kad > kac and kcd > kac , ld < lc and rd > ra . Suppose that ld < lb . We obtain that kab ≤ kbd , but kbd = |Ib ∩ Id | < min{tb , td } ≤ tb ≤ kab which reaches a contradiction (see Figure 7). Hence, we know that ld ≥ lb . The graph is shown in Figure 8. Because ad ∈ E1 , kad = |Ia ∩ Id | ≥ min{ta , td }. We know that ta > kab ≥ kad (see Figure 8). Therefore, we obtain that td ≤ kad . Hence, |Ib ∩ Id | = kbd ≥ kad ≥ td , it implies that bd ∈ E1 , again, we reach a contradiction. case3: Suppose that the tolerance ta ≤ kab and tc > kbc . Since bc ∈ E1 , we obtain that tb ≤ kbc . Similarly, we also reach a contra-. 9.

(14) diction. Therefore, we have ta ≤ kab and tc ≤ kbc . (ii) We can exchange two points b and d and use the result of (i) to get that ta ≤ kad and tc ≤ kcd . (iii) If rb > rd , we can easily obtain that tb > kbd = |Ib ∩ Id | ≥ |Ic ∩ Id | = kcd from Figure 9. (iv) If rb < rd , we can easily obtain that td > kbd = |Ib ∩ Id | ≥ |Ib ∩ Ic | = kbc from Figure 10.. 立. 政 治 大. ‧. ‧ 國. 學. Nat. tion of G1 with rb > rd .. tion of G1 with rb < rd .. n. al. er. sit. y. Figure 10: The tolerance representa-. io. Figure 9: The tolerance representa-. 2.4. Ch. engchi. i n U. v. A Tolerance Representation of Concatenation of Two 4-cycles Let G2 be the concatenation of two 4-cycles (see Figure 11).. Because G2 is AT-free and by Theorem 2.14, it is a bounded tolerance graph. Hence it has a tolerance representation. Let G2 = (V2 , E2 ), V2 = {a, b, c, d, e, f }, E2 = {ab, bc, cd, ad, be, ef, cf } and < I, t > be a tolerance representation of G2 where I = {Ix |x ∈ V2 }, t = {tx |x ∈ V2 }, kxy = |Ix ∩ Iy |, for all x, y ∈ V2 and Ix = (lx , rx ).. 10.

(15) Figure 11: The graph G2 .. Proposition 2.18. If kac is the smallest value in all of kxy , for all x, y ∈ {a, b, c, d}, then kce is the smallest value in all of kij , for all i, j ∈ {b, c, e, f }.. 政 治 大. Proof. Let kac is the smallest value in all of kxy s, for all x, y ∈ {a, b, c, d}. Because the subgraph G{b,e,f,c} of G2 is also a C4 , one of kce and kbf is the smallest value. 立. in all of kij , i, j ∈ {b, c, e, f } by Lemma 2.16. If kbf is the smallest value in all. ‧ 國. 學. of kij , i, j ∈ {b, c, e, f }. Because the subgraph G{a,b,c,d} of G2 is a C4 , we can draw a tolerance representation of a subgraph G1 of G2 , where G1 = (V1 , E1 ), V1 =. ‧. {a, b, c, d}, E1 = {ab, bc, cd, ad}. The tolerance representation is shown in Figure 12. Next, we discuss the right endpoints of the interval Ib and Id . If rb ≤ rd , we get that. y. Nat. sit. kbd ≥ kbc ≥ tb (see Figure 13). Therefore, we obtain that bd ∈ E2 . It contradicts to. al. er. io. the fact that bd ∈ / E2 . If rb > rd , by Proposition 2.17 (ii) and (iii), we have tb > kcd and tc ≤ kcd . Therefore, tb > kcd ≥ tc . Now, we discuss the left endpoints of Ic. n. v i n and I . If l ≥ l , we get that C k ≥ k ≥ t (see Figure 14 ). Therefore, we obtain hengchi U ce ∈ E , but ce ∈ / E . We get a contradiction. If l < l , we can easily see that e. c. 2. e. ce. cd. c. 2. c. e. kce ≥ kbe ≥ tb > tc (see Figure 15). Hence, we also get a contradiction. Thus, kce is the smallest value in all kij , i, j ∈ {b, c, e, f }.. Figure 12: A tolerance representation of a subgraph G1 of G2 .. 11.

(16) Figure 13: A tolerance representation of G2 with rb ≤ rd .. 政 治 大 Figure 14: A tolerance representation Figure 15: A tolerance representation 立 of G with r > r and l ≥ l . of G with r > r and l < l . d. c. e. 2. b. d. c. e. 學 ‧. 2.5. b. ‧ 國. 2. A Tolerance Representation of Concatenation of Three 4-cycles. sit. y. Nat. al. er. io. There are two types of the concatenation of three 4-cycles. The first type is. v. n. shown in Figure 16 which will be mentioned later. Now, we discuss the second type, say G3 (see Figure 17).. Ch. engchi. i n U. Figure 16: The first type of graph G3 .. By Figure 17, we know that G3 is a tolerance graph. Hence, it has a tolerance representation. Let G3 = (V3 , E3 ), V3 = {a, b, c, d, e, f, g, h}, E3 = {ab, bc, cd, ad, be, ef,. 12.

(17) cf, ch, f g, gh}, < I, t > be a tolerance representation of G3 , where I = {Ix |x ∈ V3 }, t = {tx |x ∈ V3 }, kxy = |Ix ∩ Iy |x, y ∈ V3 and Ix = (lx , rx ). By Proposition 2.18, we know that if kac is the smallest value in all kxy , for all x, y ∈ {a, b, c, d}, then kce is the smallest value in all kij , for all i, j ∈ {b, c, e, f }. We can use Proposition 2.18 again to get that kcg is the smallest value in all kpq , for all p, q ∈ {c, f, g, h}. Now, we want to prove that it is impossible that kac is the smallest value of kxy , for all x, y ∈ {a, b, c, d} of G3 .. 立. 政 治 大. ‧ 國. 學 ‧. Figure 17: The second type of graph G3 and its tolerance representation.. y. Nat. io. sit. Proposition 2.19. If < I, t > is a tolerance representation of G3 and kxy is defined. er. by above, then kac is not the smallest value of kxy , for all x, y ∈ {a, b, c, d} in G3 .. al. n. v i n C h that k is theU smallest value of k Proof. For a contradiction, suppose engchi ac. xy ,. for all. x, y ∈ {a, b, c, d}. By Proposition 2.18, we obtain that kce is the smallest value of kij , for all i, j ∈ {b, c, e, f } and kcg is the smallest value of kpq , for all p, q ∈ {c, f, g, h}. Therefore, we can draw a tolerance representation of subgraph G2 of G3 , where G2 = (V2 , E2 ), V2 = {a, b, c, d, e, f }, E2 = {ab, bc, cd, ad, be, ef, cf }. The tolerance representation of subgraph G2 of G3 is shown in Figure 18. Now, consider the interval Ig , it has two conditions in the tolerance representation of G3 . One is that Ig ⊆ Ic and the other is that Ig * Ic . At first, let Ig ⊆ Ic , it implies that kcg = |Ic ∩ Ig | = |Ig | ≥ |If ∩ Ig | = kf g . We get a contraction to the fact that kcg is the smallest value of kpq , for all p, q ∈ {c, f, g, h}. So we obtain. 13.

(18) Figure 18: The tolerance representation of subgraph G2 of G3 .. that Ig * Ic . Because f g ∈ E3 , that is, If ∩ Ig is not empty. Next, we continue to consider the size of kcg and kce . If kcg ≥ kce , then lg ≤ le . We can add an interval Ig in Figure 18 and is shown in the following figures. From Figure 19, we get that if rg ≤ rb then |Ig ∩ Ib | =. 政 治 大 and r ≤ 立r . Hence, we can get that r ≤ r. |Ig | ≥ |Ig ∩ If | ≥ tg . It implies that gb ∈ E3 , but gb ∈ / E3 . Therefore, rg > rb . Similarly, rg < re. b. e. b. g. ≤ re . Next, consider. ‧ 國. 學. the interval If , then we have some cases in the right endpoints of Ib , Ie , If and Ig . We use figures to explain the contradictions of the following cases.. ‧. n. er. io. sit. y. Nat. al. Ch. i n U. e n g c Figure h i 20:. Figure 19: The tolerance representation of G3 with rg ≤ rb .. v. The tolerance representa-. tion of G3 with rg ≥ re .. Case1: Suppose that rf ≤ rb ≤ rg ≤ re or rb ≤ rf ≤ rg ≤ re . From the following figures, we can easily get that |Ig ∩ Ie | ≥ |Ie ∩ If | ≥ te . Hence, we obtain that ge ∈ E3 . But, it contradicts to the fact that ge ∈ / E3 . Case2: Suppose that rb ≤ rg ≤ rf ≤ re or rb ≤ rg ≤ re ≤ rf . From the following figures, we can easily get that |Ig ∩ Ie | ≥ |Ie ∩ Ib | ≥ te . Hence, we obtain that ge ∈ E3 . But, it contradicts to the fact that ge ∈ / E3 .. 14.

(19) Figure 21: The tolerance representa-. Figure 22: The tolerance representa-. tion of G3 with rf ≤ rb ≤ rg ≤ re .. tion of G3 with rb ≤ rf ≤ rg ≤ re .. 學. ‧ 國. 立. 政 治 大 Figure 24: The tolerance representa-. tion of G3 with rb ≤ rg ≤ rf ≤ re .. tion of G3 with rb ≤ rg ≤ re ≤ rf .. ‧. Figure 23: The tolerance representa-. sit. y. Nat. If kcg < kce , then lg > le . We add an interval Ig in Figure 18 and is shown in. io. n. al. er. the following figures. In the Figure 25, we can prove that rg > rb and rg > re. i n U. v. by similar way in the case kcg ≥ kce . Next, we also discuss the condition of. Ch. engchi. the right endpoints of Ib , Ie , If and Ig .. Figure 25: The tolerance representation of G3 with rg ≤ rb and rg ≤ re .. Case1: Suppose that rf ≤ rb ≤ re < rg , rf ≤ re < rb < rg or rb ≤ rf ≤ re < rg .. 15.

(20) From Figure 26, Figure 27 and Figure 28, we see that |Ig ∩Ie | > |If ∩Ig | ≥ tg . Therefore, we get that ge ∈ E3 , but in fact ge ∈ / E3 . Hence, we obtain a contradiction. Case2: Suppose that rb ≤ re ≤ rf < rg . We can draw a graph shown in Figure 29 and in the same figure the interval If will not cause any contradiction. Next, we continue to discuss left endpoint and right endpoint of the interval Ih . If le < lh , then |Ic ∩ Ih | < |Ic ∩ Ie | < tc from Figure 29. This contradicts to the fact that tc ≤ |Ic ∩ Ih |.Therefore, we get that le ≥ lh . There are three cases about. 政 治 大. the position of the right endpoint rh of the interval Ih .. 立. ‧. ‧ 國. 學 Figure 27: The tolerance representa-. tion of G3 with rf ≤ rb ≤ re < rg .. tion of G3 with rf ≤ re < rb < rg .. n. al. er. io. sit. y. Nat. Figure 26: The tolerance representa-. Ch. engchi. i n U. v. Figure 28: The tolerance representation of G3 with rb ≤ rf ≤ re < rg .. (i) If rh ≥ re , we can easily get that |Ih ∩ Ie | = |Ie | ≥ |Ie ∩ If | ≥ te from the following figure. It implies that he ∈ E3 , but he ∈ / E3 . We obtain a contradiction.. (ii) If rb < rh < re , we can easily see that |Ih ∩ Ie | ≥ |Ib ∩ Ie | ≥ te from. 16.

(21) Figure 29: The tolerance representation of G3 with rb ≤ re ≤ rf < rg and le < lh .. 政 治 大. 立. Figure 30: The tolerance representation of G3 with rh ≥ re .. ‧ 國. 學. the following figure. It implies that he ∈ E3 , but he ∈ / E3 . We reach. ‧. a contradiction.. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 31: The tolerance representation of G3 with rb < rh < re .. (iii) If rh ≤ rb , we can easily see that |Ig ∩ Ie | ≥ |Ih ∩ Ig | ≥ tg from the following figure. It implies that ge ∈ E3 , but ge ∈ / E3 . We obtain a contradiction. Case3: Suppose that rb ≤ re < rg ≤ rf . A contradiction of this case is the same as Case2. Case4: Suppose that re ≤ rf < rb < rg . From Figure 33, we get that |Ig ∩ Ib | ≥ |Ig ∩ If | ≥ tg . Therefore, we get that. 17.

(22) Figure 32: The tolerance representation of G3 with rh ≤ rb . gb ∈ E3 , but in fact gb ∈ / E3 . We reach a contradiction.. 立. 政 治 大. Figure 33: The tolerance representation of G3 with re ≤ rf < rb < rg .. ‧. ‧ 國. 學 y. Nat. Case5: Suppose that re < rb ≤ rf < rg .. io. sit. This case and Case2 are the same. We also discuss the left endpoint and. n. al. er. the right endpoint of the interval Ih . If le < lh , then we get that |Ic ∩ Ih | <. i n U. v. kce = |Ic ∩ Ie | < tc from the Figure 34. This contradicts to the fact that. Ch. engchi. tc ≤ |Ic ∩ Ih |. Therefore, we know that lh ≤ le . Next, consider the right endpoint of Ih . There are two cases in the right endpoint rh of the interval Ih .. Figure 34: The tolerance representation of G3 with le < lh .. 18.

(23) (i) If rh ≥ re , we can easily get that |Ih ∩ Ie | = |Ie | ≥ |Ie ∩ If | ≥ te from the following figure. It implies that he ∈ E3 , but he ∈ / E3 . We obtain a contradiction.. Figure 35: The tolerance representation of G3 with rh ≥ re .. 政 治 大. (ii) If rb < rh < re , we can easily get that |Ih ∩ Ie | ≥ |Ib ∩ Ie | ≥ te from. 立. the following figure. It implies that he ∈ E3 , but he ∈ / E3 . We reach a. ‧. ‧ 國. 學. contradiction.. er. io. sit. y. Nat. al. n. Figure 36: The tolerance representation of G3 with rh < re .. Ch. engchi. i n U. v. Case6: Suppose that re < rb < rg ≤ rf . A contradiction of this case is the same as Case5.. 19.

(24) 3 3.1. Some Results on Maximal Outerplanar Graphs A 2-connected Graph Which Is Maximal Outerplanar Graph and Bipartite Is Not Necessarily a Tolerance Graph.. Definition 3.1. A graph G is an outerplanar graph if it has an embedding in the plane with every vertex on the boundary of the unbounded face. A maximal outerplanar graph is a simple outplanar graph that is not a spanning subgraph of a large simple outplanar graph.. 立. 政 治 大. ‧ 國. 學 ‧. Figure 37: The left is an example of outerplanar graph and the right is an example of nonouterplanar graph.. io. sit. y. Nat. n. al. er. In this article, we discuss a graph that is maximal outerplanar and bipartite.. i n U. v. Furthermore, we want to know that whether this graph is a tolerance graph or not.. Ch. engchi. We adopt that every maximal outerplanar graph in this article is 2-connected.. Figure 38: Some graphs are maximal outerplanar and bipartite.. 20.

(25) Figure 39: The graph H1 .. 政 治 大. 學. ‧ 國. 立. Figure 40: The graph H2 .. Figure 41: The graph H3 .. Figure 42: The graph H4 .. ‧. Theorem 3.2. Let G be a maximal outerplanar and bipartite graph with vertices. y. Nat. number n(G) ≥ 4. G is a tolerance graph if and only if G has no induced subgraphs. er. io. sit. H1 , H2 , H3 and H4 .. al. n. v i n we have the following cases ofC thehdegree i Uv ∈ V . e n gofcvhwhere. Proof. Let G=(V, E) be a graph that is maximal outerplanar and bipartite. Then. Case1: Suppose that the degree of v is not more than 3 for all v ∈ V . In this case, it is easy to know that G is AT-free. Therefore, by Theorem 2.14, we obtain that G is a bounded tolerance graph. The graph G is shown in Figure 43.. 21.

(26) Figure 43: The graph is maximal outerplanar and bipartite with deg(v) ≤ 3 for all v ∈V.. Case2: Suppose that there is the only one vertex u∈ V whose degree is greater than 3. Let u be the common neighbor of v1 , v2 , ......., vk . Let t1 , t2 , ......, tk−1 be. 政 治 大. vertices between vi and vi+1 , i = 1, 2, ......, k − 1, respectively. The graph G. 立. is shown in Figure 44. We have a tolerance representation of G that is shown. ‧ 國. 學. in Figure 45. Hence, G is a tolerance graph with tolerances tu = ∞, tvi = 1, for all i = 1, 2, ......, k and ttj = ∞, for all j = 1, 2, ......, k − 1.. ‧. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 44: The graph is maximal outerplanar and bipartite with deg(u)≥ 4. for the only vertex u ∈ V .. Figure 45: The tolerance representation of G. 22.

(27) Case3: Suppose that there are more than two vertices whose degree are greater than 3. Let u1 and u2 be two vertices of G with deg( u1 )≥ 4 and deg( u2 )≥ 4. We have the following relation in u1 and u2 .. (i) If the edge u1 u2 ∈ E and u1 u2 on the boundary of the unbounded face. Let u1 be the common neighbor of v1 , v2 , ..., vk , and u2 be the common neighbor of s1 , s2 , ..., si . Let t1 , t2 , ..., tk−1 be vertices between vi and vi+1 , i = 1, 2, ..., k − 1, respectively and r1 , r2 , ..., ri−1 be vertices. 政 治 大. between sj and sj+1 , j = 1, 2, ...., i − 1, respectively. The graph G is. 立. shown in Figure 46.. ‧. ‧ 國. 學 er. io. sit. y. Nat. n. a l Figure 46: The graph G. i v n Ch U e n that g c Gh isi a tolerance graph. For a contradiction, suppose. Therefore, G. has a tolerance representation < I, t >, where I = {Ix |x ∈ V }, t = {tx |x ∈ V }, kxy = |Ix ∩ Iy |, x, y ∈ V . We consider the subgraph H1 of G that is shown in Figure 47 and Figure 48. Because G is a tolerance graph, the subgraph H1 of G also has a tolerance representation. By Lemma 2.16, we know that one of kt2 u1 and kv2 v3 is the smallest value of kxy s, for all x, y ∈ {u1 , v2 , t2 , v3 } in G. By Proposition 2.19, we know that kt2 u1 is not the smallest value of kxy , for all x, y ∈ {u1 , v2 , t2 , v3 } in G. Hence, we get that kv2 v3 is the smallest value of kxy s, for all x, y ∈ {u1 , v2 , t2 , v3 } in G. By Proposition 2.18, we can obtain that kv1 u2. 23.

(28) is the smallest value of kxy s, for all x, y ∈ {u1 , v1 , s1 , u2 } in G. Since kv1 u2 is the smallest value of kxy s, for all x, y ∈ {u1 , v1 , s1 , u2 } in G, but kv1 u2 is not the smallest value of kxy s, for all x, y ∈ {u1 , v1 , s1 , u2 } in G by Proposition 2.19. We reach a contradiction. Therefore, G is not a tolerance graph.. 48: The subgraph H of G that 政 治Figure大 s, for k is the smallest value of k s, for. Figure 47: The subgraph H1 of G that. 立. kt2 u1 is the smallest value of kxy. 1. v2 v3. all x, y ∈ {u1 , v2 , t2 , v3 } in G.. ‧ 國. 學. all x, y ∈ {u1 , v2 , t2 , v3 } in G.. xy. (ii) If the edge u1 u2 ∈ E and u1 u2 is not on the boundary of the unbounded. ‧. face.. y. Nat. Let u1 be the common neighbor of v1 , v2 , ..., vk , and u2 be the common. io. sit. neighbor of s1 , s2 , ..., si . Let t1 , t2 , ..., tk be vertices between vi−1 and. n. al. er. vi , i = 1, 2, ..., k, respectively and r1 , r2 , ..., ri be vertices between sj−1. i n U. v. and sj , j = 1, 2, ..., i, respectively. The graph G is shown in Figure 49.. Ch. engchi. Figure 49: The graph G.. Similarly, we consider the subgraph H2 of G that is shown in the following figures. By Proposition 2.18 and Proposition 2.19, we also prove that G is not a tolerance graph.. 24.

(29) Figure 50: The subgraph H2 of G that. Figure 51: The subgraph H2 of G that. kt2 u1 is the smallest value of kxy , for. kv1 v2 is the smallest value of kxy , for. all x, y ∈ {u1 , v1 , t2 , v2 } in G.. all x, y ∈ {u1 , v1 , t2 , v2 } in G.. 治 政 大 is on the boundary of the unbounded face. 立 Let u be the common neighbor of v , v , ..., v , and u. (iii) If the edge u1 u2 ∈ / E and the shortest path from u1 to u2 with odd length. 1. 1. 2. k. 2. be the common. ‧ 國. 學. neighbor of s1 , s2 , ..., si . Let t1 , t2 , ..., tk−1 be vertices between vi−1 and vi , i = 1, 2, ..., k, respectively and r1 , r2 , ..., ri−1 be vertices between. ‧. sj−1 and sj , j = 1, 2, ..., i, respectively. Let p1 , p2 , ..., p2n be vertices. y. Nat. of the shortest path from u1 to u2 and q1 , q2 , ..., q2n be vertices of the. graph G is shown in Figure 52.. n. al. Ch. engchi. er. io. sit. shortest path from v1 to s1 where pj qj ∈ E, for all j = 1, ..., 2n. The. i n U. v. Figure 52: The graph G.. Similarly, we consider the subgraph H3 of G that is shown in the following figures. By Proposition 2.18 and Proposition 2.19, we also prove that G is not a tolerance graph.. 25.

(30) Figure 53: The subgraph H3 of G that. Figure 54: The subgraph H3 of G that. kt2 u1 is the smallest value of kxy s, for. kv2 v3 is the smallest value of kxy s, for. all x, y ∈ {u1 , v2 , t2 , v3 } in G.. all x, y ∈ {u1 , v2 , t2 , v3 } in G.. (iv) If the edge u1 u2 ∈ / E and the shortest path from u1 to u2 with even length is on the boundary of the unbounded face.. 政 治 大 common neighbor of s , s , ..., s . Let t , t , ..., t be vertices between 立 v and v , i = 1, 2, ..., k, respectively and r , r , ..., r be vertices. Let u1 be the common neighbor of v1 , v2 , ......., vk , and u2 be the 1. i. 1. 2. k−1. i. 1. 2. i−1. 學. ‧ 國. i−1. 2. between sj−1 and sj , j = 1, 2, ..., i, respectively. Let p1 , p2 , ..., p2n+1 be vertices of the shortest path from u1 to u2 and q1 , q2 , ..., q2n+1 be vertices. ‧. of the shortest path from v1 to s1 where pj qj ∈ E, for all j = 1, ..., 2n + 1.. n. al. er. io. sit. y. Nat. The graph G is shown in Figure 55.. Ch. engchi. i n U. v. Figure 55: The graph G.. We have a tolerance representation of G that is shown in Figure 56. Hence, G is a tolerance graph with tolerances tu1 = ∞, tu2 = ∞, tvj = 1, for all j = 1, 2, ..., k, tsj = 1, for all j = 1, 2, ..., i, ttj = ∞, for all j = 1, 2, ..., k − 1, trj = ∞, for all j = 1, 2, ..., i − 1, tpj = 1, for all j = 1, 3, ..., 2n + 1, tpj = ∞, for all j = 2, 4, ..., 2n, tqj = 1, for all j = 2, 4, ..., 2n and tpj = ∞, for all j = 1, 3, ..., 2n + 1.. 26.

(31) Figure 56: The tolerance representation of graph G.. (v) If the edge u1 u2 ∈ / E and the shortest path from u1 to u2 with odd length is not on the boundary of the unbounded face. Let u1 be the common neighbor of v1 , v2 , ..., vk , and u2 be the common. 政 治 大 and v , i = 1, 2,立 ..., k, respectively and r , r , ..., r. neighbor of s1 , s2 , ..., si . Let t1 , t2 , ..., tk−1 be vertices between vi−1 i. 1. 2. i−1. be vertices between. ‧ 國. 學. sj−1 and sj , j = 1, 2, ..., i, respectively. Let p1 , p2 , ..., p2n+1 be vertices of the shortest path from u1 to s1 and q1 , q2 , ..., q2n+1 be vertices of the. ‧. shortest path from v1 to u2 where pj qj ∈ E, for all j = 1, ..., 2n + 1. The graph G is shown in Figure 57.. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 57: The graph G.. Similarly, we consider the subgraph H4 of G that is shown in the following figures. By Proposition 2.18 and Proposition 2.19, we also prove that G is not a tolerance graph.. 27.

(32) Figure 58: The subgraph H4 of G that. Figure 59: The subgraph H4 of G that. kt2 u1 is the smallest value of kxy s, for. kv2 v3 is the smallest value of kxy s, for. all x, y ∈ {u1 , v2 , t2 , v3 } in G.. all x, y ∈ {u1 , v2 , t2 , v3 } in G.. (vi) If the edge u1 u2 ∈ / E and the shortest path from u1 to u2 with even. 政 治 大 Let u be the common neighbor of v , v , ..., v and u be the common 立 neighbor of s , s , ......., s . Let t , t , ......, t be vertices between length is not on the boundary of the unbounded face. 1. 1. 1. 2. i. 1. 2. k. 2. 2. k−1. ‧ 國. 學. vi−1 and vi , i = 1, 2, ..., k, respectively and r1 , r2 , ..., ri−1 be vertices between sj−1 and sj , j = 1, 2, ..., i, respectively. Let p1 , p2 , ..., p2n be. ‧. vertices of the shortest path from u1 to s1 and q1 , q2 , ..., q2n be vertices. y. Nat. of the shortest path from v1 to u2 where pj qj ∈ E, for all j = 1, ..., 2n.. n. er. io. al. sit. The graph G is shown in Figure 60.. Ch. engchi. i n U. v. Figure 60: The graph G.. We have a tolerance representation of G as Figure 61 shown. Hence, G is a tolerance graph with tolerances tu1 = ∞, tu2 = ∞, tvj = 1, for all i = 1, 2, ..., k , tsj = 1, for all j = 1, 2, ..., i − 1, ttj = ∞, for all j = 1, 2, ..., k − 1, trj = ∞, for all j = 1, 2, ..., i − 1,. 28.

(33) tpj = 1, for all j = 1, 3, ..., 2n − 1, tpj = ∞, for all j = 2, 4, ..., 2n, tqj = 1, for all j = 2, 4, ..., 2n and tpj = ∞, for all j = 1, 3, ..., 2n − 1.. Figure 61: The tolerance representation of graph G.. 立. 政 治 大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 29. i n U. v.

(34) 4. Open Problems and Further Directions of Studies In this article, we have presented the tolerance representations and all forbid-. den subgraphs of 2-connected graphs which are maximal outerplanar graphs and bipartite. There are still some open problems for future studies.. 1. In Figure 1, we have known a hierarchy of classes of tolerance graph. Furthermore, a. We would like to characterize the graphs which are both tolerance and. 政 治 大. cocomparability.. 立. b. We would like to characterize the graphs which are both tolerance and. ‧ 國. 學. trapezoid.. 2. The general question of characterizing bipartite bounded tolerance is proved by. ‧. the following theorem. [3]. Let G = (X, Y, E)be a bipartite graph. The following conditions are equiva-. (i) G is a bounded tolerance graph.. n. al. Ch. (ii) G is a trapezoid graph.. engchi. er. io. sit. y. Nat. lent.. i n U. v. (iii) G is a cocomparability graph. (iv) G is AT-free. (v) G is a permutation graph.. Furthermore, we would like to prove the theorem for tolerance graph instead of bounded tolerance graph.. 30.

(35) References [1] M. Golumbic and C. Monma, A generalization of interval graphs with tolerances, Congressus Numerantium, 35 (1982), pp. 321–331. [2] M. Golumbic, D. Rotem, and J. Urrutia, Comparability graphs and intersection graphs, Discrete Math., 43 (1983), pp. 37–46. [3] M. Golumbic and A. Trenk, Tolerance graphs, Cambridge Univ Pr, 2004. [4] R. Hayward and R. Shamir, A note on tolerance graph recognition, Discrete. 政 治 大. Applied Mathematics, 143 (2004), pp. 307–311.. 立. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 31. i n U. v.

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