The Existence of (s,t)-Monochromatic-rectangles in a 2-colored Checkerboard

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(1)國⽴政治⼤學應⽤數學系碩⼠學位論⽂. 政治大. (s,t) 單⾊矩形於⼆⾊棋盤之存在性立 ‧ 國. 學. ‧. The Existence of (s,t)-Monochromatic-rectangles in a 2-colored Checkerboard n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 碩⼠班學⽣：卓駿焰撰指導教授：張宜武博⼠中華民國 105 年 6 ⽉ 30 ⽇.

(2) 國⽴政治⼤學應⽤數學系卓駿焰君所撰之碩⼠學位論⽂ (s,t) 單⾊矩形於⼆⾊棋盤之存在性 The Existence of (s,t)-Monochromatic-rectangles in a 2-colored Checkerboard 政治大立. ‧ 國. 學 ‧. 業經本委員會審議通過論⽂考試委員會委員：. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 指導教授：系主任：. 中華民國 105 年 6 ⽉ 30 ⽇.

(3) 中⽂本⽂藉由矩形棋盤著⾊探討完全⼆分圖 Km,n 由兩種顏⾊任意塗邊，使得此兩⾊著邊之完全⼆分圖 Km,n 會包含單⾊⼦圖 Ks,2 、Ks,3 與 Ks,t （s ≥ 2），我們將討論參數 n 與 s 之間滿必須滿⾜何種關係。本⽂也將介紹處理棋盤著⾊問題的⼀般⽅法與技巧，以及透過棋盤如何將棋盤問題轉化為圖論問題，並且將它推廣。關鍵字：完全⼆分圖、單⾊⼦圖. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. ii. i n U. v.

(4) Abstract In this paper, we study the two edge-coloring of Km,n such that Km,n contains a monochromatic subgraph Ks,2 , Ks,3 or Ks,t . We find the relation between n, s by. 政治大 Keywords: Complete 立bipartite graph;Monochromatic subgraph. investigating a two coloring of a checkerboard.. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. iii. i n U. v.

(5) Contents 試委員會審. i. 中⽂. 立. Abstract. iii. ‧ 國. ‧. (2,2)-Monochromatic-rectangles in a Checkerboard. sit. 2. io. Introduction. al. n. 3. iv vi. y. Nat. 1. Ch. engchi U. er. List of Figures. ii. 學. Contents. 政治大. v ni. 1 3. 2.1. The Case of 2 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.2. The Case of 3 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 5. 2.3. The Case of 4 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 6. 2.4. The Case of 5 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 7. 2.5. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 9. (2,t)-monochromatic-rectangles in a Checkerboard. 13. 3.1. The Case of 2 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 13. 3.2. The Case of 3 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 13. 3.3. The Case of 4 × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . . .. 14. 3.4. The Case of 5 × n Checkerboard. . . . . . . . . . . . . . . . . . . . . . . . .. 14. 3.5. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 23. iv.

(6) 4.1. The Case of 3 × n Checkerboard. . . . . . . . . . . . . . . . . . . . . . . . .. 24. 4.2. The Case of 4 × n Checkerboard. . . . . . . . . . . . . . . . . . . . . . . . .. 24. 4.3. The Case of 5 × n Checkerboard. . . . . . . . . . . . . . . . . . . . . . . . .. 24. 4.4. The Case of 6 × n Checkerboard. . . . . . . . . . . . . . . . . . . . . . . . .. 25. 4.5. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 26. (3,t)-Monochromatic-rectangles in a Checkerboard 5.1. The Case of 5 × n Checkerboard. 5.2 5.3. . . . . . . . . . . . . . . . . . . . . . . . .. 27. The Case of 6 × n Checkerboard. 28. Summary. . . . . . . . . . . . . . . . . . . . . . . . 政 .治 . . . . . . . . . . . . . . . . . . .大 . . . . . . . . . . . . . . . . . . 立. 28. (s,2)-Monochromatic-rectangles in a Checkerboard. 29. The Case of (2s − 2) × n Checkerboard. . . . . . . . . . . . . . . . . . . . .. 6.2. The Case of (2s − 1) × n Checkerboard. . . . . . . . . . . . . . . . . . . . .. 29. 6.3. The Case of 2s × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . .. 30. 6.4. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ‧. 6.1. io. sit. y. Nat. al. n. 7. 27. 學. 6. 24. i n U. (s,t)-Monochromatic-rectangles in a Checkerboard. Ch. e n g c h i.. er. 5. (3,2)-Monochromatic-rectangles in a Checkerboard. ‧ 國. 4. v. 29. 31 32. 7.1. The Case of (2s − 1) × n Checkerboard. . . . . . . . . . . . . . . . . . . .. 32. 7.2. The Case of 2s × n Checkerboard . . . . . . . . . . . . . . . . . . . . . . . .. 33. 7.3. Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 33. Bibliography. 34. v.

(7) List of Figures m × n checkerboard . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2. 1.2. A 3 × 4 checkerboard and the correspond complete bipartite graph. . . . . . .. 政治大 A (2,2)-monochromatic-rectangle . . . . . . . . . . . . . . . . . . . . . . . . 立. 2. 2.2. A (3,4)-monochromatic-rectangle . . . . . . . . . . . . . . . . . . . . . . . .. 3. 2.3. There are two (2,1)-monochromatic-rectangles of the same color in the checker-. ‧ 國. 2.1. 學. 1.1. ‧. 4. y. board. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3. board. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.5. Every 2-colored 3 × 7 checkerboard contains a (2,2)-monochromatic-rectangle.. 6. 2.6. There is a 2-colored 3 × 6 checkerboard containing no (2,2)-monochromatic-. 2.4. There is no two (2,1)-monochromatic-rectangles of the same color in the checker-. n. er. io. sit. Nat. al. Ch. engchi. i n U. v. rectangles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 6. 2.7. Every 2-colored 4 × 7 checkerboard contains a (2,2)-monochromatic-rectangle.. 7. 2.8. There is a 2-colored 4 × 6 checkerboard containing no (2,2)-monochromaticrectangles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 8. Every 2-colored 5 × 5 checkerboard contains a (2,2)-monochromatic-rectangle.. 8. 2.10 The subgraph induced by {X1 , X3 , Y2 , Y7 } is a monochromatic copy of K2,2 . . .. 10. 2.11 The subgraph induced by {X1 , X3 , Y2 , Y7 } is a monochromatic copy of K2,2 .. .. 11. 2.12 The subgraph induced by {X4 , X5 , Y2 , Y5 } is a monochromatic copy of K2,2 .. .. 12. rectangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 19. 2.9. 3.1. There is a 2-colored 5 × 10 checkerboard containing no a (2,3)-monochromatic-. vi.

(8) 4.1. There is a 2-colored 5 × 20 checkerboard containing no a (3,2)-monochromaticrectangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . There is a 2-colored 6 × 20 checkerboard containing no a (3,2)-monochromaticrectangle. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 立. 政治大. 學 ‧. ‧ 國 io. sit. y. Nat. n. al. er. 4.2. 26. Ch. engchi. vii. i n U. v. 26.

(9) Chapter 1 Introduction 政治大 We often encounter problems related the Ramsey numbers [1] in many Mathematical Com立. ‧ 國. 學. petitions of High School Students. In this thesis, we use the ideas of the Ramsey numbers on the checkerboard problems. We follow [2] for the notations in graph theory and the definitions. ‧. of the complete bipartite graph Km,n , and follow [3] to construct the correspondence between the checkerboards and complete bipartite graphs. Jiong-Sheng Li provides the minimal sizes. y. Nat. sit. of k-colored square checkerboards which have monochromatic-rectangles in [3], we define the. n. al. er. io. generalized monochromatic-rectangles and discuss the existence of such rectangles in an m × n. i n U. v. checkerboard. In this thesis, we only consider the checkerboards which are arbitrarily colored. Ch. engchi. by two colors and we called it two-colored checkerboard.. In the second chapter, we discuss the minimal columns of the 2-colored checkerboard which has (2,2)-monochromatic-rectangles by fixing the rows. At the end, we convert the results to graphic problems. In the third chapter, we extend the second chapter to discuss the minimal columns of the 2-colored checkerboard which has (2,t)-monochromatic-rectangles by fixing the rows. At the end, we convert the results to graphic problems. In the forth chapter, we discuss the minimal columns of the 2-colored checkerboard which has (3,2)-monochromatic-rectangles by fixing the rows. At the end, we convert the results to graphic problems. In the fifth chapter, we extend the forth chapter to discuss the minimal columns of the 2-colored checkerboard which has (3,t)-monochromatic-rectangles by fixing the rows. At the end, we convert the results to graphic problems. 1.

(10) In the second chapter to the fifth chapter, all results have been proved in [4], but we improve the proof such that be more general and we also propose some amendments in the third chapter. In the last two chapter, we propose the generalized conclusions. We discuss the minimal columns of the 2-colored checkerboard which has (s,t)-monochromatic-rectangles by fixing the rows. At the end, we convert the results to graphic problems.. By [3] we convert the grids of a checkerboard into the edges of a complete bipartite graph, the number of rows and columns correspond to the number of vertices in complete partite sets, X and Y, respectively.. 立. 政治大. sit. y. ‧. ‧ 國. 學. Nat. Figure 1.1: m × n checkerboard. n. al. er. io. If the grid in the i-th row and the j-th column of the checkerboard is black, then the cor-. i n U. v. respond edge xi y j in the correspond complete bipartite graph is solid. And the white grid is. Ch. engchi. correspond the dashed edge. The following is a 2-colored m × n checkerboard correspond to a 2-coloring Km,n .. Figure 1.2: A 3 × 4 checkerboard and the correspond complete bipartite graph.. 2.

(11) Chapter 2 (2,2)-Monochromatic-rectangles in a Checkerboard 政治大立 ‧ 國. 學. Definition 2.1. An m × n rectangle is called a (s,t)-monochromatic-rectangle,. ‧. if in first column there are s grids including the first one and the last one that have the same color,. n. al. er. io. sit. y. Nat. and there are other t-1 columns including the last column that are copies of the first column.. Ch. engchi. i n U. v. Figure 2.1: A (2,2)-monochromatic-rectangle Figure 2.2: A (3,4)-monochromatic-rectangle. 3.

(12) 2.1. The Case of 2 × n Checkerboard If there are two (2,1)-monochromatic-rectangles of the same color, then the checker-. board has a (2,2)-monochromatic rectangle. Otherwise, there is no (2,2)-monochromatic rectangle.. 立. 政治大. ‧. ‧ 國. 學. Figure 2.3: There are two (2,1)-monochromatic-rectangles of the same color in the checkerboard.. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 2.4: There is no two (2,1)-monochromatic-rectangles of the same color in the checkerboard. Therefore, in a 2-colored 2 × n checkerboard a (2,2)-monochromatic-rectangle may not exist.. 4.

(13) 2.2. The Case of 3 × n Checkerboard. Definition 2.2. In a checkerboard, two (s,1)-monochromatic-rectangles of the same color are the same, if one is a copy of the other one. Definition 2.3. An n × 1 column contains a (s,1)-monochromatic-rectangle means that there are s grids of the same color in the column. Note: An n × 1 column contains at most Csn distinct (s,1)-monochromatic-rectangles. We consider. 政治大 there exists a (2,2)-monochromatic-rectangle. 立. Lemma 2.4. In every 2-colored 3 × n checkerboard, n = 7 is the smallest number such that. ‧ 國. 學. Proof. To prove that we need to exhibit a 2-colored 3 × 6 checkerboard that has no (2,2)monochromatic-rectangles. In a column, there are at most C23 distinct black (2,1)-monochromatic-. ‧. rectangles. So we can distribute the C23 distinct black (2,1)-monochromatic-rectangles and the. sit. y. Nat. C23 distinct white (2,1)-monochromatic-rectangles to the 6 columns, then the 2-colored 3 × 6. io. er. checkerboards have no (2,2)-monochromatic-rectangles.. By pigeonhole principle, there are at least ⌈ 3×2 7 ⌉ = 11 grids of the same color. Without loss. n. al. Ch. of generality, let the color be black. Then we have. iv n ∑ Udi ≥ 7. e n g c h i i =1. 11, where di is the number of. black grids of the ith column of the checkerboard. Assume 2-colored 3 × 7 checkerboard has a coloring such that there is no (2,2)-monochromatic-rectangle, then any two columns don’t cond. tain the same black (2,1)-monochromatic-rectangles, each column contains C2 i distinct black (2,1)-monochromatic-rectangles, and the total number of distinct black (2,1)-monochromaticrectangles is not more than C23 . So we have C2d1 + C2d2 + · · · + C2d7 ≤ C23. (2.2.1). Let d1 + d2 + · · · + d7 = 11 + t, where t ∈ N. Then we can transform (2.2.1) to d21 + d22 + · · · + d27 ≤ 17 + t 5. (2.2.2).

(14) By Cauchy–Schwarz inequality, we get. (d21 + d22 + · · · + d27 )(12 + 12 + · · · + 12 ) ≥ (d1 + d2 + · · · + d7 )2 ⇒ (17 + t) × 7 ≥ (11 + t)2 So we have t2 + 15t + 2 ≤ 0 But t ∈ N, we reach a contradiction in the last inequality. Therefore, every 2-coloring of 3 × 7 checkerboard yields a (2,2)-monochromatic-rectangle. □. 立. 政治大. ‧. ‧ 國. 學. n. al. er. io. sit. y. Nat. Figure 2.5: Every 2-colored 3 × 7 checkerboard contains a (2,2)-monochromatic-rectangle.. Ch. engchi. i n U. v. Figure 2.6: There is a 2-colored 3 × 6 checkerboard containing no (2,2)-monochromaticrectangles.. 2.3. The Case of 4 × n Checkerboard. Lemma 2.5. In every 2-colored 4 × n checkerboard, n = 7 is the smallest number such that there exists a (2,2)-monochromatic-rectangle. Proof. By Lemma2.4, in every 2-colored 3 × 7 checkerboard, there is a (2,2)-monochromatic-. 6.

(15) rectangle. Therefore, in every 2-colored 4 × 7 checkerboard, there is a (2,2)-monochromaticrectangle. □. Figure 2.7: Every 2-colored 4 × 7 checkerboard contains a (2,2)-monochromatic-rectangle.. The Case of 5立 × n Checkerboard. 學. ‧ 國. 2.4. 政治大. Lemma 2.6. In every 2-colored 5 × n checkerboard, n = 5 is the smallest number such that. ‧. there exists a (2,2)-monochromatic-rectangle.. y. Nat. Proof.To prove that we need to exhibit a 2-colored 5 × 4 checkerboard that has no (2,2)-. er. io. sit. monochromatic-rectangles. By the Lemma2.5, we have a 2-colored 4 × 5 checkerboard which doesn’t have (2,2)-monochromatic-rectangles. We rotate the checkerboard, so we have the 2-. n. al. i n U. v. colored 5 × 4 checkerboard that has no (2,2)-monochromatic-rectangles.. Ch. engchi. By pigeonhole principle, there are at least ⌈ 5×2 5 ⌉ = 13 grids of the same color. Without loss 5. of generality, let the color be black. Then we have ∑ di ≥ 13, where di is the number of black grids of the. ith. i =1. column of the checkerboard. Assume two colored 5 × 5 checkerboard has. a coloring such that there is no (2,2)-monochromatic-rectangles, then any two columns don’t d. contain the same black (2,1)-monochromatic-rectangles, each column contains C2 i distinct black (2,1)-monochromatic-rectangles, and the total number of distinct black (2,1)-monochromaticrectangles is not more than C25 . So we have C2d1 + C2d2 + · · · + C2d5 ≤ C25. 7. (2.4.1).

(16) Figure 2.8: There is a 2-colored 4 × 6 checkerboard containing no (2,2)-monochromaticrectangles. Let d1 + d2 + · · · + d5 = 13 + t, where t ∈ N. Then we can transform (2.4.1) to. 政治大. d21 + d22 + · · · + d25 ≤ 23 + t. 立. (2.4.2). By Cauchy–Schwarz inequality, we get. ‧ 國. 學 ⇒ (23 + t) × 5 ≥ (13 + t)2. n. al. t2 + 21t + 54 ≤ 0. Ch. engchi. er. io. sit. y. Nat. So we have. ‧. (d21 + d22 + · · · + d25 )(12 + 12 + · · · + 12 ) ≥ (d1 + d2 + · · · + d5 )2. i n U. v. But t ∈ N, we reach a contradiction in the last inequality. Therefore, every 2-colored 5 × 5 checkerboard yields a (2,2)-monochromatic-rectangle. □. Figure 2.9: Every 2-colored 5 × 5 checkerboard contains a (2,2)-monochromatic-rectangle.. 8.

(17) 2.5. Summary The case of m × n checkerboard, where m ≥ 6, can be obtained by rotation of the. rectangles. For example, 6 × 5 checkerboard can be considered to 5 × 6 checkerboard, so it has a (2,2)-monochromatic-rectangle.. We can convert the above theorems to graphic problems. We have the following proposition. • If n > 6, every 2-coloring of K3,n contains a monochromatic K2,2 .. 政治大 • If n > 4, every 2-coloring 立 of K contains a monochromatic K. • If n > 6, every 2-coloring of K4,n contains a monochromatic K2,2 . 5,n. 2,2 .. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 9. i n U. v.

(18) Example 2.7. Every 2-coloring of K3,7 exists a monochromatic K2,2 subgraph.. 治政 Figure 2.10: The subgraph induced by {X , X , Y , Y大 } is a monochromatic copy of K 立 3. 1. 2. 7. ‧ 國. 學. Example 2.8. There is a 2-coloring of K3,6 no a monochromatic K2,2 subgraph.. ‧. n. er. io. sit. y. Nat. al. Ch. engchi. 10. i n U. v. 2,2 ..

(19) Example 2.9. Every 2-coloring of K4,7 exists a monochromatic K2,2 subgraph.. 治政 Figure 2.11: The subgraph induced by {X , X , Y , Y大 } is a monochromatic copy of K 立 1. 3. 2. 7. ‧ 國. 學. Example 2.10. There is a 2-coloring of K4,6 no a monochromatic K2,2 subgraph.. ‧. n. er. io. sit. y. Nat. al. Ch. engchi. 11. i n U. v. 2,2 ..

(20) Example 2.11. Every 2-coloring of K5,5 exists a monochromatic K2,2 subgraph.. 治政 Figure 2.12: The subgraph induced by {X , X , Y , Y大 } is a monochromatic copy of K 立 4. 5. 2. 5. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 12. i n U. v. 2,2 ..

(21) Chapter 3 (2,t)-monochromatic-rectangles in a Checkerboard 政治大立 ‧ 國. 學. 3.1. ‧. The Case of 2 × n Checkerboard. y. Nat. If there are t (2,1)-monochromatic-rectangles of the same color, then the checkerboard. er. io. sit. has a (2,t)-monochromatic-rectangle. Otherwise, there is no (2,t)-monochromatic-rectangle.. n. al. i n 3.2 The Case of 3 × C nh Checkerboard engchi U. v. Theorem 3.1. If n > (6t − 6), where t ≥ 2, then in every 2-colored 3 × n checkerboard. There is a (2,t)-monochromatic-rectangle. Proof. If n = (6t − 6) + 1 = 6t − 5 (Only prove that every coloring of two colors 3 × (6t − 5) checkerboard, there is a (2,t)-monochromatic-rectangle.) By pigeonhole principle, there are at least ⌈ 6t2−5 ⌉ = 3t − 2 columns that have at least two of same color grids. Without loss of generality, let the color be black. Then di ≥ 2. i = 1, 2, . . . , (3t − 2) , where di is the. number of black grids of the ith column of the checkerboard. Assume two colored 3 × (6t − 5) checkerboard has a coloring such that there is no (2,t)-monochromatic-rectangles, then any t columns don’t contain the same black (2,1)-monochromatic-rectangles, each column contains d. C2 i distinct black (2,1)-monochromatic-rectangles, and the total number of distinct black (2,1)13.

(22) monochromatic-rectangles is not more than (t − 1) · C23 . So we have, d. C2d1 + C2d2 + · · · + C2 3t−2 ≤ (t − 1) · C23 Because di ≥ 2 d. C22 + C22 + · · · + C22 ≤ C2d1 + C2d2 + · · · + C2 3t−2 combining the two results shows C22 + C22 + · · · + C22 ≤(t − 1) · C23. 政⇒ 3t 治 − 2 ≤3t − 3 大. 立. ‧ 國. 學. 1 ≤ 0 we reach a contradiction. So, If n > (6t − 6), where t ≥ 2, then in every 2-coloring of 3 × n checkerboard. There is a (2,t)-monochromatic-rectangle. □. ‧. The Case of 4 × n Checkerboard. sit. y. Nat. 3.3. n. al. er. io. Theorem 3.2. If n > (6t − 6), where s ≥ 2, then in every 2-colored 4 × n checkerboard. There is a (2,t)-monochromatic-rectangle.. Ch. engchi. i n U. v. Proof. By Theorem 3.1, in every 2-colored 3 × (6t − 6) checkerboard, there is a (2,t)monochromatic-rectangle. Therefore, in every 2-colored 4 × (6t − 6) checkerboard, there is a (2,t)-monochromatic-rectangle. □. 3.4. The Case of 5 × n Checkerboard. Theorem 3.3. If n > (5t − 6), where t ≥ 2, then in every 2-colored 5 × n checkerboard. There is a (2,t)-monochromatic-rectangle, where t is even. And C2d1 + C2d2 + · · · + C2dn > (2t − 3) · C25 where di is the number of black grids of the ith column of the checkerboard. Proof. Suppose t = 2k − 2, where k is integer greater than two, we use induction on k, If n = (10k − 16) + 1 = 10k − 15 (Only prove that every coloring of two colors 5 × (10k − 15) 14.

(23) checkerboard, there is a (2k − 2, 2)-monochromatic-rectangle. Basis step: When k = 2, by the Lemma 2.6 we have in every 2-coloring of 5 × 5 checkerboard, there is a (2,2)-monochromatic-rectangle. Suppose k = s is true, s ≥ 2 and s ∈ N for all 2-coloring of 5 × (10s − 15) checkerboard, there is a (2, 2s − 2)-monochromatic-rectangle. By pigeonhole principle, there are at least ⌈. 5×(10s−15) ⌉ 2. = 25s − 37 grids of the same color.. Without loss of generality, let the color be black. So, we have. 10s−15. ∑. i =1. d. di = 25s − 37, and. C2d1 + C2d2 + · · · + C2 10s−15 > (2s − 3) · C25 Induction step: When k = s + 1, n = 10(s + 1) − 15 = 10s − 5, by pigeonhole principle, there are at least ⌈. 5×(10s−5) ⌉ 2. = 25s − 12 grids of the same color in 5 × (10s − 5) checker-. 政治大. 10s−5. board. Without loss of generality, let the color be black. We have ∑ di = 25s − 12. By induction hypothesis,. 10s−5. ∑. 10. i. ‧ 國. i =1. i. i =1. = 25 and. d C2 i. is the number of black. 學. i =11. 立 d = 25s − 37, so ∑ d. (2,1)-monochromatic-rectangles in the ith column, i = 1, 2, · · · , (10s − 5). Assume two col-. ‧. ored 5 × (10s − 5) checkerboard has a coloring such that there is no (2, 2s)-monochromaticrectangles, then any 2s columns don’t contain the same black (2,1)-monochromatic-rectangles,. Nat. d. sit. y. each column contains C2 i distinct black (2,1)-monochromatic-rectangles, and the total number. n. al. er. io. of distinct black (2,1)-monochromatic-rectangles is not more than (2s − 1) · C25 . So we have C2d1. Ch. + C2d2. d + · · · + C2 10s−5. engchi. iv 5 n ≤U (2s − 1) · C2. By induction hypothesis d. C2d11 + C2d12 + · · · + C2 10s−5 > (2s − 3) · C25 So, we have d. C2d1 + C2d2 + · · · + C2 10 < (2s − 1) · C25 − (2s − 3) · C25 d. C2d1 + C2d2 + · · · + C2 10 < 20. 15. (3.4.1).

(24) By the definition of combination we get d ( d − 1) d1 ( d1 − 1) d2 ( d2 − 1) + + · · · + 10 10 < 20 2 2 2 multiple 2 in both side. (d21 + d22 + · · · + d210 ) − (d1 + d2 + · · · + d10 ) < 40. d21 + d22 + · · · + d210 < 65. 政治大. 立. By Cauchy–Schwarz inequality. ‧ 國. 學. (d21 + d22 + · · · + d210 ) · 10 ≥ (d1 + d2 + · · · + d10 )2. 252 ≥ 10. y. + · · · + d210 ). er. io. sit. + d22. ‧. Nat. (d21. n. a l 62.5 ≤ (d21 + d22 + · · · + d210i v) ≤ 64 =⇒ n U i engch 2. Ch. ∵ di are integer ∴ (d21 + d22 + · · · + d10 ) must be 63 or 64 Case1 : d21 + d22 + · · · + d210 = 63. d22 + d23 + · · · + d210 = 63 − d21 and. d2 + d3 + · · · + d10 = 25 − d1 By Cauchy–Schwarz inequality. 16.

(25) (d22 + d23 + · · · + d210 ) · 9 ≥ (d2 + d3 + · · · + d10 )2. (63 − d21 ) · 9 ≥ (25 − d1 )2. 10d21 − 50d1 + 58 ≤ 0. 1.83 ≤ d ≤ 3.17 政治大 must be 立2 or 3. Similarly, d , d , · · · , d 1. and d1 is integer, so d1. 2. 3. 10. must be 2 or 3 But d1 + d2 +. ‧ 國. 學. · · · + d10 = 25, so d1 , d2 , · · · , d10 consists of five 2′ s and five 3′ s Therefore, d21 + d22 + · · · + d210 = 65 ̸= 63, we reach a contradiction.. n. al. er. io. sit. d22 + d23 + · · · + d210 = 64 − d21. y. Nat. and. ‧. Case2 : d21 + d22 + · · · + d210 = 64. Ch. engchi. i n U. v. d2 + d3 + · · · + d10 = 25 − d1 By Cauchy–Schwarz inequality. (d22 + d23 + · · · + d210 ) · 9 ≥ (d2 + d3 + · · · + d10 )2. (64 − d21 ) · 9 ≥ (25 − d1 )2. 10d21 − 50d1 + 49 ≤ 0. 17.

(26) 1.34 ≤ d1 ≤ 3.66 and d1 is integer, so d1 must be 2 or 3. Similarly, d2 , d3 , · · · , d10 must be 2 or 3. But d1 + d2 + · · · + d10 = 25, so d1 , d2 , · · · , d10 consists of five 2′ s and five 3′ s. Therefore, d21 + d22 +. · · · + d210 = 65 ̸= 64, we reach a contradiction. So for all 2-colored 5 × (10s − 5) checkerboard, there is a (2,2s)-monochromatic rectangle. By induction, ∀k ≥ 2 and k ∈ N, in every 2-colored 5 × (5t − 6) checkerboard, there is a. (2, t)-monochromatic rectangle, where t = 2k − 2.. 政治大. Lemma 3.4. In every 2-colored 5 × 11 checkerboard, there is a (2,3)-monochromatic-rectangle.. 立. Proof. By pigeonhole principle, there are at least ⌈ 5×211 ⌉ = 28 grids of the same color.. ‧ 國. 學. 11. Without loss of generality, let the color be black. Then we have ∑ di ≥ 28, where di is the i =1. number of black grids of the ith column of the checkerboard. Assume two colored 5 × 11. ‧. checkerboard has a coloring such that there is no monochromatic-rectangles, then any three. Nat. sit. d. y. columns don’t contain the same black (2,1)-monochromatic-rectangles, each column contains. al. er. io. C2 i distinct black (2,1)-monochromatic-rectangles, and the total number of distinct black (2,1)-. n. monochromatic-rectangles is not more than 2 · C25 . So we have. Ch. C2d1. e n g c hdi. + C2d2. i n U. v. + · · · + C2 11 ≤ 2 × C25. (3.4.2). Let d1 + d2 + · · · + d11 = 28 + t, where t ∈ N. Then we can transform (3.4.2) to d21 + d22 + · · · + d211 ≤ 68 + t By Cauchy–Schwarz inequality, we get. (d21 + d22 + · · · + d211 )(12 + 12 + · · · + 12 ) ≥ (d1 + d2 + · · · + d11 )2 ⇒ (68 + t) × 11 ≥ (28 + t)2. 18. (3.4.3).

(27) So we have t2 + 45t + 36 ≤ 0 But t ∈ N, the last inequality is contradiction. Therefore, every 2-colored 5 × 11 checkerboard yields a (2,3)-monochromatic-rectangle. □. 立. 政治大. ‧ 國. 學 ‧. Figure 3.1: There is a 2-colored 5 × 10 checkerboard containing no a (2,3)-monochromaticrectangle.. n. er. io. sit. y. Nat. al. Ch. engchi. 19. i n U. v.

(28) Theorem 3.5. If n > (5t − 5), where t ≥ 2, then in every 2-colored 5 × n checkerboard. There is a (2,t)-monochromatic-rectangle, where s is odd. And C2d1 + C2d2 + · · · + C2dn > (2t − 2) · C25 where di is the number of black grids of the ith column of the checkerboard. Proof. Suppose t = 2k − 1, where k is integer greater than two, we use induction on k, If n = (10k − 10) + 1 = 10k − 9 (Only prove that every 2-colored 5 × (10k − 9) checkerboard, there is a (2, 2k − 1)-monochromatic-rectangle. Basis step: When k = 2, by the Lemma 3.4 we have in every 2-colored 5 × 11 checkerboard, there is a (2,3)-monochromatic-rectangle. Suppose k = s is true, s ≥ 2 and s ∈ N. 治政 By pigeonhole principle, there are at least ⌈ ⌉大 = 25s − 22 grids of the same color. 立 Without loss of generality, let the color be black. So, we have ∑ d = 25s − 22, and. for all 2-colored 5 × (10s − 9) checkerboard, there is a (2, 2s − 1)-monochromatic-rectangle. 10t−9. 學. ‧ 國. 5×(10s−9) 2. d. C2d1 + C2d2 + · · · + C2 10s−9 > (2s − 2) · C25. i =1. i. 5×(10s+1) ⌉ 2. = 25s + 3 grids of the same color in 5 × (10s + 1) checkerboard.. Nat. 10s+1. y. there are at least ⌈. ‧. Induction step: When k = s + 1, n = 10(s + 1) − 9 = 10s + 1, by pigeonhole principle,. 10s−5. sit. Without loss of generality, let the color be black. We have ∑ di = 25s + 3. By induction i =1. 10. d. al. n. i =11. i =1. er. io. hypothesis, ∑ di = 25s − 22, so ∑ di = 25 and C2 i is the number of black-bars in the ith. i n U. v. column, i = 1, 2, · · · , (10s + 1). Assume two colored 5 × (10s + 1) checkerboard has a color-. Ch. engchi. ing such that there is no (2, 2s + 1)-monochromatic-rectangles, then any 2s + 1 columns don’t d. contain the same black (2,1)-monochromatic-rectangles, each column contains C2 i distinct black (2,1)-monochromatic-rectangles, and the total number of distinct black (2,1)-monochromaticrectangles is not more than 2s · C25 . So we have d. C2d1 + C2d2 + · · · + C2 10s+1 ≤ 2s · C25 By induction hypothesis d. C2d11 + C2d12 + · · · + C2 10s+1 > (2s − 2) · C25. 20. (3.4.4).

(29) So, we have d. C2d1 + C2d2 + · · · + C2 10 < 2s · C25 − (2s − 2) · C25 d. C2d1 + C2d2 + · · · + C2 10 < 20 By the definition of combination we get d1 ( d1 − 1) d2 ( d2 − 1) d ( d − 1) + + · · · + 10 10 < 20 2 2 2 multiple 2 in both side. 治政大· · · + d + · · · + d ) − (d + d + 立. + d22. 2 10. 1. 2. 10 ). < 40. 學. ‧ 國. (d21. d21 + d22 + · · · + d210 < 65. ‧. By Cauchy–Schwarz inequality. sit. y. Nat. n. al. er. io. (d21 + d22 + · · · + d210 ) · 10 ≥ (d1 + d2 + · · · + d10 )2. Ch (d21. e2 n g c h 2i. i n U. + d2 + · · · + d10 ) ≥. v. 252 10. =⇒ 62.5 ≤ (d21 + d22 + · · · + d210 ) ≤ 64 ∵ di are integer ∴ (d21 + d22 + · · · + d210 ) must be 63 or 64 Case1 : d21 + d22 + · · · + d210 = 63 d22 + d23 + · · · + d210 = 63 − d21 and. 21.

(30) d2 + d3 + · · · + d10 = 25 − d1 By Cauchy–Schwarz inequality. (d22 + d23 + · · · + d210 ) · 9 ≥ (d2 + d3 + · · · + d10 )2. (63 − d21 ) · 9 ≥ (25 − d1 )2. 治政 10d − 50d + 58 ≤ 大0 2 1. 1. 學. ‧ 國. 立. 1.83 ≤ d1 ≤ 3.17. ‧. and d1 is integer, so d1 must be 2 or 3. Similarly, d2 , d3 , · · · , d10 must be 2 or 3 But d1 + d2 +. sit. y. Nat. · · · + d10 = 25, so d1 , d2 , · · · , d10 consists of five 2′ s and five 3′ s Therefore, d21 + d22 + · · · +. n. al. er. io. d210 = 65 ̸= 63, we reach a contradiction. Case2 : d21 + d22 + · · · + d210 = 64. v i n Ch =U 64 − d21 i engch. d22. + d23. + · · · + d210. and. d2 + d3 + · · · + d10 = 25 − d1 By Cauchy–Schwarz inequality. (d22 + d23 + · · · + d210 ) · 9 ≥ (d2 + d3 + · · · + d10 )2. (64 − d21 ) · 9 ≥ (25 − d1 )2. 22.

(31) 10d21 − 50d1 + 49 ≤ 0. 1.34 ≤ d1 ≤ 3.66 and d1 is integer, so d1 must be 2 or 3. Similarly, d2 , d3 , · · · , d10 must be 2 or 3. But d1 + d2 + · · · + d10 = 25, so d1 , d2 , · · · , d10 consists of five 2′ s and five 3′ s. Therefore, d21 + d22 +. · · · + d210 = 65 ̸= 64, we reach a contradiction. So for all 2-colored 5 × (10s − 5) checkerboard, there is a (2, 2s + 1)-monochromatic rectangle.. 立. 政治大. By induction, ∀k ≥ 2 and k ∈ N, in every 2-colored 5 × (5t − 5) checkerboard, there is a. ‧ 國. ‧. Summary. Nat. y. 3.5. 學. (2, t)-monochromatic rectangle,where t = 2k − 1. □. er. io. Let s ≥ 2. sit. We can convert the above theorems to graphic problems. We have the following proposition.. al. n. v i n C h of K3,n exists aUmonochromatic K2,t subgraph. • If n > 6(t − 1), every 2-coloring engchi • If n > 6(t − 1), every 2-coloring of K4,n exists a monochromatic K2,t subgraph. • If n > (10t − 16), every 2-coloring of K5,n exists a monochromatic K2,(2t−2) subgraph. • If n > (10t − 10), every 2-coloring of K5,n exists a monochromatic K2,(2t−1) subgraph.. 23.

(32) Chapter 4 (3,2)-Monochromatic-rectangles in a Checkerboard 政治大立 ‧ 國. 學. 4.1. ‧. The Case of 3 × n Checkerboard. y. Nat. If there are two columns of all grids are of the same color, then the checkerboard has a. er. io. sit. (3,2)-monochromatic-rectangle. Otherwise, there is no (3,2)-monochromatic-rectangle.. n. al. i n 4.2 The Case of 4 × C nh Checkerboard engchi U. v. If every column of a 4 × n checkerboard has two black grids and two white grids, then it doesn’t have s a (3,2)monochromatic-rectangle.Therefore, for every two color 4 × n checkerboard exists a coloring such that has no (3,2)-monochromatic-rectangles in the 4 × n checkerboard.. 4.3. The Case of 5 × n Checkerboard. Lemma 4.1. In every 2-colored 5 × n checkerboard, n = 21 is the smallest number such that there exists a (3,2)-monochromatic-rectangle. Proof.To prove that we need to exhibit a two colored 5 × 20 checkerboard has no (3,2)24.

(33) monochromatic-rectangles. In a column, there are at most C35 distinct black (3,1)-monochromaticrectangles. So we can distribute the C35 distinct black (3,1)-monochromatic-rectangles and the C35 distinct white (3,1)-monochromatic-rectangles to the 20 columns, then the two colored 5 × 20 checkerboards have no (3,2)-monochromatic-rectangles. By pigeonhole principle, there are at least ⌈ 21 2 ⌉ = 11 columns with at least three grids of the same color. Without loss of generality, let the color be black. Then di ≥ 3 i = 1, 2, . . . , 11 , where di is the number of black grids of the ith column of the checkerboard. Assume 2-colored 5 × 21 checkerboard has a coloring such that there is no (3,2)monochromatic-rectangles, then any two columns don’t contain the same black (3,1)-monochromatic-rectangles, each column d. 政治大. contains C3 i distinct black (3,1)-monochromatic-rectangles, and the total number of distinct. 立. black (3,1)-monochromatic-rectangles is not more than C35 . So we have. ‧ 國. 學 ‧. Because di ≥ 3. C3d1 + C3d2 + · · · + C3d11 ≤ C35. y. Nat. al. er. io. sit. C33 + C33 + · · · + C33 ≤ C3d1 + C3d2 + · · · + C3d11. n. combining the two results shows. Ch. engchi. i n U. v. C33 + C33 + · · · + C33 ≤C35. ⇒ 11 ≤10 11 ≤ 10 iwe reach a contradiction in the last inequality. So, If n > 21, then in every 2-coloring of 5 × n checkerboard. There is a (3,2)monochromatic-rectangle. □. 4.4. The Case of 6 × n Checkerboard. Lemma 4.2. In every 2-colored 6 × n checkerboard, n = 21 is the smallest number such that there exists a (3,2)-monochromatic-rectangle. Proof. By Lemma4.1, in every 2-colored 5 × 21 checkerboard, there is a (3,2)monochromatic25.

(34) Figure 4.1: There is a 2-colored 5 × 20 checkerboard containing no a (3,2)-monochromaticrectangle.. 政治大 rectangle. Therefore, in every 立2-colored 6 × 21 checkerboard, there is a (3,2)monochromatic‧. ‧ 國. 學. rectangle. □. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. Figure 4.2: There is a 2-colored 6 × 20 checkerboard containing no a (3,2)-monochromaticrectangle.. 4.5. Summary. We can convert the above theorems to graphic problems. We have the following proposition. • If n > 20, every 2-coloring of K5,n contains a monochromatic K3,2 . • If n > 20, every 2-coloring of K6,n contains a monochromatic K3,2 . 26.

(35) Chapter 5 (3,t)-Monochromatic-rectangles in a. ‧. The Case of 5 × n Checkerboard. 學. 5.1. ‧ 國. Checkerboard 政治大立. y. Nat. Theorem 5.1. If n > 20(t − 1), where t ≥ 2, then in every 2-colored 5 × n checkerboard,. er. io. sit. there is an (3,t)-Monochromatic-rectangle.. Proof. If n = (20t − 20) + 1 = 20t − 19 (Only prove that every 2-colored 5 × (20t − 19). al. n. v i n Ch checkerboard, there is a s-monochromatic-rectangle.) e n g c h i ByUpigeonhole principle, there are at least 20t−19 ⌈. 2. ⌉ = 10t − 9 columns that have at least three of same color grids. Without loss of. generality, let the color be black. Then di ≥ 3. i = 1, 2, . . . , (10t − 9) , where di is the. number of black grids of the ith column of the checkerboard. Assume 2-colored 3 × (20t − 19) checkerboard has a coloring such that there is no (3,t)-Monochromatic-rectangles, then any s columns don’t contain the same black (3,1)-monochromatic-rectangles, each column contains d. C3 i distinct black (3,1)-monochromatic-rectangles, and the total number of distinct black (3,1)monochromatic-rectangles is not more than (t − 1) · C23 . So we have, d. C3d1 + C3d2 + · · · + C3 10t−9 ≤ (t − 1) · C35. 27.

(36) Because di ≥ 3 d. C33 + C33 + · · · + C33 ≤ C3d1 + C3d2 + · · · + C3 10t−9 combining the two results shows C33 + C33 + · · · + C33 ≤(t − 1) · C35. ⇒ 10t − 9 ≤10t − 10 1 ≤ 0 we reach a contradiction in the last inequality. So, If n > (20t − 20), where t ≥ 2, then in every 2-colored 5 × n checkerboard. There is an (3,t)-Monochromatic-rectangle. □. 學. The Case of 6 × n Checkerboard. ‧ 國. 5.2. 立. 政治大. Theorem 5.2. If n > 20(t − 1), where s ≥ 2, then in every 2-colored 6 × n checkerboard,. ‧. there is a (3,t)-Monochromatic-rectangle.. Nat. sit. y. Proof. By Theorem 5.1, in every 2-colored 5 × (20t − 20) checkerboard, there is a (3,t)-. al. er. io. Monochromatic-rectangle . Therefore, in every 2-colored 6 × (20t − 20) checkerboard, there. n. is a (3,t)-Monochromatic-rectangle. □. 5.3. Ch. engchi. i n U. v. Summary. We can convert the above theorems to graphic problems. We have the following proposition. Let s ≥ 2 • If n > 20(t − 1), every 2-coloring of K5,n exists a monochromatic K3,t subgraph. • If n > 20(t − 1), every 2-coloring of K6,n exists a monochromatic K3,t subgraph.. 28.

(37) Chapter 6 (s,2)-Monochromatic-rectangles in a Checkerboard 政治大立 ‧ 國. 學. 6.1. ‧. The Case of (2s − 2) × n Checkerboard. y. Nat. If every column of a (2s − 2) × n checkerboard has s-1 black grids and s-1 white grids, then. io. sit. it doesn’t have s a (s,2)-monochromatic-rectangle.Therefore, for every two color 2s − 2 × n. 6.2. al. n. 2) × n checkerboard.. er. checkerboard exists a coloring such that has no (s,2)-monochromatic-rectangle in the (2s −. Ch. engchi. i n U. v. The Case of (2s − 1) × n Checkerboard. Lemma 6.1. In every 2-colored (2s − 1) × n checkerboard, n = (2Cs2s−1 + 1) is the smallest number such that there exists a (2,2)-monochromatic-rectangle. Proof.To prove that we need to exhibit a 2-colored (2s − 1) × 2Cs2s−1 checkerboard that has no (2,2)-monochromatic-rectangles. In a column, there are at most Cs2s−1 distinct black (s,1)monochromatic-rectangles. So we can distribute the Cs2s−1 distinct black (s,1)-monochromaticrectangles and the Cs2s−1 distinct white (s,1)-monochromatic-rectangles to the 2Cs2s−1 columns, then the 2-colored (2s − 1) × 2Cs2s−1 checkerboards have no (s,2)-monochromatic-rectangles. By pigeonhole principle, there are at least ⌈. 2Cs2s−1 +1 ⌉ 2. 29. = Cs2s−1 + 1 columns with at least s.

(38) grids are of the same color. Without loss of generality, let the color be black. Then di ≥ s. i = 1, 2, . . . , Cs2s−1 + 1 , where di is the number of black grids of the ith column of the. checkerboard. Assume 2-colored 2s − 1 × 2Cs2s−1 + 1 checkerboard has a coloring such that there is no (s,2)monochromatic-rectangles, then any two columns don’t contain the same black d. (s,1)-monochromatic-rectangles, each column contains Cs i distinct black (s,1)-monochromaticrectangles, and the total number of distinct black (s,1)-monochromatic-rectangles is not more than Cs2s−1 . So we have. Cs2s−1 +1. ∑. Cs k ≤ Cs2s−1 d. k =1. 政治大立C + C + · · · + C ≤ ∑. Because di ≥ s. s s. s s. s s. Cs2s−1 +1. d. Cs k. 學. ‧ 國. k =1. combining the two results shows. ‧. Css + Css + · · · + Css ≤Cs2s−1. io. y. sit. Nat. ⇒ Cs2s−1 + 1 ≤Cs2s−1. n. al. er. 1 ≤ 0 we reach a contradiction in the last inequality. So, If n > 2Cs2s−1 + 1, then in every. i n U. v. 2-colored (2s − 1) × (2Cs2s−1 + 1) checkerboard. There is a (s,2)-monochromatic-rectangle.. □. 6.3. Ch. engchi. The Case of 2s × n Checkerboard. Lemma 6.2. In every 2-colored (2s) × (2Cs2s−1 + 1) checkerboard, there is a (s,2)monochromaticrectangle. Proof. By Lemma6.1, in every 2-colored (2s − 1) × (2Cs2s−1 + 1) checkerboard, there is a (s,2)monochromatic-rectangle. Therefore, in every 2-colored (2s) × (2Cs2s−1 + 1) checkerboard, there is a (s,2)monochromatic-rectangle. □. 30.

(39) 6.4. Summary. We can convert the above theorems to graphic problems. We have the following proposition. • If n > (2Cs2s−1 + 1), every 2-coloring of K(2s−1),n contains a monochromatic Ks,2 . • If n > (2Cs2s−1 + 1), every 2-coloring of K2s,n contains a monochromatic Ks,2 .. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 31. i n U. v.

(40) Chapter 7 (s,t)-Monochromatic-rectangles in a Checkerboard 政治大立 ‧ 國. 學. 7.1. ‧. The Case of (2s − 1) × n Checkerboard. y. Nat. Theorem 7.1. In every 2-colored (2s − 1) × (2(t − 1)Cs2s−1 + 1) checkerboard, there is a. Proof. By pigeonhole principle, there are at least ⌈. n. al. Ch. engchi U. er. io. sit. (s,t)monochromatic-rectangle.. 2(t−1)Cs2s−1 +1 ⌉ 2. v ni. = (t − 1)Cs2s−1 + 1. columns with at least s grids are of the same color. Without loss of generality, let the color be black. Then di ≥ t i = 1, 2, . . . , (t − 1)Cs2s−1 + 1 , where di is the number of black grids of the ith column of the checkerboard. Assume 2-colored 2s − 1 × 2(t − 1)Cs2s−1 + 1 checkerboard has a coloring such that there is no (s,t)monochromatic-rectangle, then any s columns don’t d. contain the same black (s,1)-monochromatic-rectangles, each column contains Cs i distinct black (s,1)-monochromatic-rectangles, and the total number of distinct black (s,1)-monochromaticrectangles is not more than (t − 1)Cs2s−1 . So we have (t−1)Cs2s−1 +1. ∑. Cs k ≤ (t − 1)Cs2s−1 d. k =1. 32.

(41) Because di ≥ s Css. + Css. + · · · + Css. ≤. (t−1)Cs2s−1 +1. ∑. d. Cs k. k =1. combining the two results shows Css + Css + · · · + Css ≤(t − 1)Cs2s−1. ⇒ (t − 1)Cs2s−1 + 1 ≤(t − 1)Cs2s−1 1 ≤ 0 we reach a contradiction in the last inequality. So, If n > 2(t − 1)Cs2s−1 + 1, then in every 2-colored (2s − 1) × (2(t − 1)Cs2s−1 + 1) checkerboard. There is a (s,t)-monochromaticrectangle. □. ‧ 國. The Case of 2s × n Checkerboard. 學. 7.2. 立. 政治大. ‧. Theorem 7.2. In every 2-colored (2s) × (2(t − 1)Cs2s−1 + 1) checkerboard, there is a (s,t)monochromatic-. sit. y. Nat. rectangle.. n. al. er. io. Proof. By Theorem7.1, in every 2-colored (2s − 1) × (2(t − 1)Cs2s−1 + 1) checkerboard,. v. there is a (s,t)monochromatic-rectangle. Therefore, in every 2-colored (2s) × (2(t − 1)Cs2s−1 +. Ch. engchi. i n U. 1) checkerboard, there is a (s,t)monochromatic-rectangle. □. 7.3. Summary. We can convert the above theorems to graphic problems. We have the following proposition. • If n > (2(t − 1)Cs2s−1 + 1), every 2-coloring of K(2s−1),n contains a monochromatic Ks,t . • If n > (2(t − 1)Cs2s−1 + 1), every 2-coloring of K2s,n contains a monochromatic Ks,t .. 33.

(42) Bibliography [1] Ronald L. Graham, Bruce L. Rothschild, and Joel H. Spencer. Ramsey theory. Wiley Series in Discrete Mathematics and Optimization. John Wiley & Sons, Inc., Hoboken, NJ, 2013.. 政治大. [2] Douglas B. West. Introduction to graph theory. Prentice Hall, Inc., Upper Saddle River,. 立. [3] 李炯⽣. 棋盤染⾊問題與⼆部 Ramsey 數. 數學. 學. ‧ 國. NJ, 1996.. , 21(3):63–72, 9 ⽉ 1997.. ‧. [4] 林⼦軒. The coloring of a checkerboard and the monochromatic subgraphs of a complete. n. al. er. io. sit. y. Nat. bipartite graph. 國⽴政治⼤學應⽤數學系碩⼠學位論⽂, 2013.. Ch. engchi. 34. i n U. v.

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