一個卡特蘭等式的組合證明 - 政大學術集成

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(1)國⽴政治⼤學應⽤數學系碩⼠學位論⽂. 立. 政治大. ⼀個卡特蘭等式的組合證明 ‧ 國. 學 ‧. A Combinatorial Proof of a Catalan Identity n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 碩⼠班學⽣：劉映君撰指導教授：李陽明博⼠中華民國 106 年 6 ⽉ 16 ⽇.

(2) 國⽴政治⼤學應⽤數學系劉映君君所撰之碩⼠學位論⽂⼀個卡特蘭等式的組合證明 A Combinatorial Proof of a Catalan Identity. 政治大業經本委員會審議通過立論⽂考試委員會委員： ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 指導教授：系主任：. 中華民國 106 年 6 ⽉ 16 ⽇.

(3) 謝感謝這三年來中所有指導過我的⽼師，還有系上幫助過我的學⾧姐及同學們，幫助我順利完成學業。⾸先感謝我的指導教授李陽明⽼師，謝謝⽼師對於我所投⼊的時間及⼼⾎，無論是平常上課、準備論⽂以及最後的論⽂修改，因為有⽼師的耐⼼指導，才能有這篇論⽂的誕⽣。接著要感謝的是我最愛的家⼈，讀研究所的這三年來總是讓他們操煩，⽽且上研究所後與家⼈相處的時間減少許多，但他們仍給於我精神上的關懷，並給我很多⽀持與協助。最後，感謝研究所的同學們，在我⾯對困難的時候伸出援⼿，也陪伴我度過這三年的研究所⽣活。. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. ii. i n U. v.

(4) 中⽂. 要. 在這篇論⽂裡，我們探討卡塔蘭等式 (n + 2)Cn+1 = (4n + 2)C2 的證明⽅法。以往都是⽤計算的⽅式來呈現卡塔蘭等式的由來，但我們選擇⽤組合的⽅法來證明卡塔蘭等式。這篇論⽂主要是應⽤ Cn+1 壞路徑對應到打點 Cn 好路徑以及 Cn+1 好路徑對應到打點 Cn 壞路徑的⽅式來證明卡特蘭等式。. 立. 政治大. 關鍵字：卡塔蘭等式. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. iii. i n U. v.

(5) Abstract In this thesis, we give another approach to prove Catalan identity,. (n + 2)Cn+1 = (4n + 2)C2 . In the past we use the method of computation to show. 政治大. Catalan Identity, in this thesis we choose a combinatorial proof of the Catalan iden-. 立. tity.. ‧ 國. 學. This thesis is primary using the functions from Cn+1 totally bad path to Cn dotted good path, and from Cn+1 good path to Cn dotted totally bad path.. ‧. n. al. er. io. sit. y. Nat. Keywords: Catalan Identity. Ch. engchi. iv. i n U. v.

(6) Contents 試委員會審. i. 謝. 立. 要. sit. al. n. Introduction. er. io. 1. iv v. y. Nat. List of Figures. ‧. Contents. 學. Abstract. ii iii. ‧ 國. 中⽂. 政治大. Ch. engchi U. v ni. vi 1. 2. Paths Start with North. 3. 3. Paths Start with East. 14. 4. Summary. 25. A Some examples of Catalan identity (n + 2)Cn+1 = (4n + 2)Cn. 27. Bibliography. 30. v.

(7) List of Figures 2.1. GT⃗eQ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.2. Q⃗eGT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4. 2.4. 政治 .................. Q⃗eGT . . . . . . . . . . . . . . . . . . . . .大立 Q ⃗eG T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.5. Q⃗eGT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.6. Q ⃗eG T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.7. Q⃗eGT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.8. Q ⃗eG T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 2.9. Q ⃗eG T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ′. ′. ′. ′. ‧. ′. Nat. ′. io. al. n. Ch. engchi U. 6 6 7 7 8. 2.10 Q⃗eGT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . preimage under f 1. 5. y. ′. 5. sit. ′. ′. 學. ′. ′. er. ′. ‧ 國. 2.3. v ni. 8. ˆ −−−−−−−−−→ SeR ˆ 2.11 ReS . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 9. 2.12 Lemma2.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 10. 2.13 n = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 11. 3.1. TG⃗nQ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15. 3.2. Q⃗nTG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 15. 3.3. Q⃗nTG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 16. 3.4. Q ⃗nT G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 3.5. Q ⃗nT G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 17. 3.6. Q⃗nTG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 17. 3.7. Q⃗nTG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 18. 3.8. Q ⃗nT G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ′. ′. ′. ′. ′. ′. ′. ′. ′. vi. 16. 18.

(8) 3.9. Q⃗nTG . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ′. ′. ′. 3.10 Q ⃗nT G . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . preimage under f 2. 19 19. ˆ −−−−−−−−−→ SnR ˆ . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.11 RnS. 20. 3.12 Lemma3.7 2nd part . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 21. 3.13 n = 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 22. g1. h. g2. h. 4.1. 1 A1 − → B1 − → C1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 4.2. 2 → B2 − → C2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A2 −. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. vii. i n U. v. 25 26.

(9) Chapter 1 Introduction 政治大 Definition 1.1. A segment is either an east(e) or a north(n). 立. ‧ 國. 學. segments.. A path consists of consecutive. Definition 1.2. An (n, n) path is a path with n e’s and n n’s segments.. ‧. Definition 1.3. A good path means that all segments are below diagonal y = x. A bad path is. y. Nat. sit. a path that is not a good path.. n. al. er. io. Note : A bad path has at least one segment above diagonal y = x.. Ch. i n U. v. Definition 1.4. A totally bad path means that all segments are above diagonal y = x.. engchi. Catalan numbers are the number of good paths from the origin to the point (n, n), and we define Catalan number, Cn , by Cn =. 1 2n n+1 Cn ,. for n ≥ 0. In this thesis, we focus on a combinatorial. proof of a Catalan identity,. (n + 2)Cn+1 = (4n + 2)Cn . [6]. 1.

(10) In general, we obtain this formula by. (n + 2)Cn+1 = = = = =. 立. (n + 2)Cn2n++1 2 n+2 (2n + 2)! ( n + 1) ! ( n + 1) ! (2n + 1)(2n)!(2n + 2) (n + 1)n!n!(n + 1) 2(2n + 1)(2n)!(2n + 2) (n + 1)n!n!(2n + 2) (4n + 2)(2n)! (n + 1)n!n! (4n + 2)Cn2n = (4n + 2)Cn n+1. = 治政大. ‧ 國. 學. It is well-known that the number of paths with n + 1 flaws, which has n + 1 east and n + 1 north. ‧. segments above the diagonal y = x, is equal to the number of such paths with n flaws, which is. y. Nat. equal to the number of such paths with n − 1 flaws, and so on. In other words, we have split up. al. n. By Pascal Identity, Cn2n++1 2 =. So the left side (n + 2)Cn+1 = Cn2n++1 2 .. er. 1 C2n+2 . ( n +2) n +1. io. desired formula Cn+1 =. sit. the set of all paths into n + 2 equally sized classes. Since there are Cn2n++1 2 paths, we obtain the Cn2n+1 | {z }. Ch. +. i vn2n++11 C n U | {z }. e n g c h i Paths starting with east. Paths starting with north. On the right-hand side, (4n + 2)Cn = 2(2n + 1)Cn = (2n + 1)Cn + | {z } Dotted good paths. (2n + 1)Cn | {z }. .. Dotted totally bad paths. In Chapter 2, we give a bijective proof between ”Paths start with north” and ”Dotted good paths”. In Chapter 3, we give a bijective proof between ”Paths start with east” and ”Dotted totally bad paths”. Therefore, we complete the proof of (n + 2)Cn+1 = (4n + 2)Cn combinatorially. For more details, we refer to [1–5, 7–10]. 2.

(11) Chapter 2 Paths Start with North Definition 2.1. The set X1. 政治大 consists of all (n + 1, n + 1) paths which have at least one flaw. 立 ⇀. ‧ 國. 學. Each path in X1 can be factorized into GT e Q. The set Y1 consists of all (n + 1, n + 1) paths which has at most n flaws.. ‧. Define a function f 1 from X1 into Y1 by the following:. y. Nat. n. al. er. io. y = x.. sit. 1. Starting from the bottom left, (0, 0), follow the path until it first travels above the diagonal. Ch. i n U. v. ⇀. 2. Continue to follow the path until it touches the diagonal y = x again. Denote by e , the. engchi. ⇀. first such segment that touches the diagonal y = x, in fact, e must be an east segment. ⇀. ⇀. 3. Swap the portion of the path before e with portion after e . ⇀. ⇀. i.e. f 1 ( GT e Q) = Q e GT, where G is an (i, i ) good path, 0 ≤ i ≤ n + 1, T is a ( j, j + 1) totally bad path, 0 ≤ j ≤ ⇀. n − i, the east segment e is the first east which touches the diagonal y = x, and Q is an. (n − i − j, n − i − j) path. NOTE: After using f 1 , the flaws of path P decrease one. i.e. If P has k flaws, k ≥ 1, then f 1 ( P) has k − 1 flaws.. 3.

(12) ⇀. ⇀. To show f 1 ( GT e Q) = Q e GT by graph, we have:. 學. ⇀. ‧ 國. 立. 政治大 Figure 2.1: GT⃗eQ. ‧. Fix e and switch GT with Q, we have:. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. Figure 2.2: Q⃗eGT. 4. v.

(13) ⇀. Theorem 2.2. Let P = GT e Q is an (n + 1, n + 1) path. Define f 1 : X1 −→ Y1 ⇀. ⇀. by f 1 ( GT e Q) = Q e GT, where G is an (i, i ) good path, 0 ≤ i ≤ n + 1, T is a ( j, j + 1) ⇀. totally bad path, 0 ≤ j ≤ n − i, the east segment e is the first east which touches the diagonal y = x, and Q is an (n − i − j, n − i − j) path. The function f 1 is one-to-one and onto. Proof. Claim: f 1 is one-to-one. ⇀. ′. ′. ′⇀. ′. ′. ′. Let P = GT e Q, P = G T e Q , where T is a an (l, l + 1) totally bad path, G is a (k, k ) good ′. path, and Q is an (n − k − l, n − k − l ) path. ⇀. ′. ′⇀. ⇀. ′. ′⇀. ′. ′. f 1 ( GT e Q) = f 1 ( G T e Q ) ⇒ Q e GT = Q e G T . Claim: T = T. ′. Case1: l > j. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. ′. Figure 2.3: Q⃗eGT. ′. Figure 2.4: Q ⃗eG T. ′. When we start on (n + 1, n + 1), trace back the path. We let two path both trace back to the point (n − j + 1, n − j + 1), in the next step, the path in Figure 2.3 is below the diagonal y = x, but the path in Figure 2.4 is still above the diagonal y = x. This is a contradiction, as two paths are the same. 5.

(14) Case2: l < j. 立. 政治大. ‧ 國. 學 ′. ′. Figure 2.6: Q ⃗eG T. ′. ‧. Figure 2.5: Q⃗eGT. Nat. sit. y. When we start on (n + 1, n + 1), trace back the path.. n. al. er. io. We let two paths both trace back to the point (n − l + 1, n − l + 1), in the next step, the path. i n U. v. in Figure 2.6 is below the diagonal y = x, but the path in Figure 2.5 is still above diagonal y = x.. Ch. engchi. This is a contradiction, as two paths are the same. Thus, we have proved that l = j. ′. ∴T=T. ′. Claim: G = G . Case1: i > k. 6.

(15) 政治大. 立. ′. ′. Figure 2.8: Q ⃗eG T. ′. ‧ 國. 學. Figure 2.7: Q⃗eGT We both start on (n − j + 1, n − j).. ‧. Since i > k, when we let the path in Figure 2.8 trace back to (n − j − k + 1, n − j − k), the ⇀. Nat. ⇀. sit. y. next segment is e , but the path in Figure 2.7 is not e .. al. n. Case2: i < k. er. io. This is a contradiction, as two paths are the same.. Ch. engchi. 7. i n U. v.

(16) ′. 立 ′. ′. Figure 2.10: Q⃗eGT. ‧ 國. 學. Figure 2.9: Q ⃗eG T. 政治大. We both start on (n − j + 1, n − j).. ‧. Since i < k, when we let the path in Figure 2.10 trace back to (n − j − i + 1, n − j − i ), the ⇀. Nat. ⇀. sit. y. next segment is not e , but the path in Figure 2.9 is e .. al. n. Thus, we have proved that i = k.. er. io. This is a contradiction, as two paths are the same.. ′. ∴G=G. ′. Ch. ′. Since T = T and G = G . ⇀. ′⇀. ′. ⇀. ′. engchi. i n U. v. ′⇀. ∵ Q e GT = Q e G T ⇒ Q e = Q e ′. ⇀. ′. ′⇀. ∴ Q = Q ⇒ GT e Q = G T e Q. ′. ∴ f 1 is one-to one. Claim: f 1 is onto. i.e. For any path in Y1 , which has at most n flaws, we choose the last east leaving the diagonal ˆ then we switch the portions before eˆ and after e. ˆ We can get a new path y = x, denoted by e, with at least one flaw, and the path is in X1 . To show by graph:. 8.

(17) 立. 政治大. preimage under f 1. ˆ −−−−−−−−−→ SeR ˆ Figure 2.11: ReS. ‧ 國. 學. To show by formula:. preimage under f 1. ‧. ˆ = P, ˆ −−−−−−−−−→ SeR Q = ReS where Q has at most n flaws, P has at least one flaw.. y. Nat. io. sit. In fact, if Q has k flaws, P has k + 1 flaws.. n. al. er. So, for every path Q in Y1 , we can find a path P in X1 such that f 1 ( P) = Q. Therefore, f 1 is one-to-one and onto.. Ch. engchi. i n U. v. NOTE: Let f 1−1 be the inverse function of f 1 .. ⇀. Lemma 2.3. The first east e touching the diagonal y = x in X1 is below the diagonal y = x in Y1 after using f 1 . ⇀. ⇀. Proof. Let P = S e R, where S is a ( j, j + 1) path, e is a (1, 0) east path, and R is an (n − j, n − j) path. ⇀. After using f 1 , we swap R and S, since R is an (n − j, n − j) path, the next segment e is below the diagonal y = x. ⇀. Thus, we have proved that the first east e touching the diagonal y = x in X1 and it is below the 9.

(18) diagonal y = x in Y1 after using f 1 .. ⇀. Lemma 2.4. The last east eˆ leaving from diagonal y = x in Y1 is the first east e touching the diagonal y = x in X1 . ⇀. ⇀. i.e. If e is the first east touching the diagonal y = x in X1 , then e is the last east leaving the diagonal y = x in Y1 .. 立. 政治大. ‧. ‧ 國. 學. n. al. er. io. sit. y. Nat ⇀. Figure 2.12: Lemma2.4. i n U. v. Proof. Since e is the first east touching the diagonal y = x, we can observe that there is a empty. Ch. engchi. ⇀. area enclosed by the first north that leaves the diagonal y = x, denoted by n , the diagonal y = x, ⇀. e , and the diagonal y = x + 1. ⇀. ⇀. After swapping two portions, another empty is enclosed by e , the diagonal y = x, n , and the ⇀. diagonal y = x − 1, so that there is no east segment can touch the diagonal y = x between e ⇀. and n . ⇀. And the remain segments which are behind n are at most touching the diagonal y = x but not be below the diagonal y = x. Therefore, the last east eˆ leaving from diagonal y = x in Y1 is the ⇀. first east e touching the diagonal y = x in X1 . Definition 2.5. The set A1 consists of all paths which first segment is north, and the first touching the diagonal y = x east is marked. The set B1 consists of all paths which are good path. 10.

(19) (k). (k ). Define g1 from A1 into B1 by g1 ( P) = f 1 ( P), where P has k flaws, and f 1. = f 1 ◦ f 1 ◦ ... ◦ f 1 . | {z } k. Example 2.6. The following example is one of g1 ( P):. 政治大. 學. ‧ 國. 立. Figure 2.13: n = 3. ‧. ⇀. Lemma 2.7. In g1 ( P), after using the first f 1 , the first east which is denoted by e , connects. io. Proof. First, we prove the first part. ⇀. n. al. er. sit. y. Nat. with the first segment of P in A1 . And this part will not be separated afterward.. i n U. v. Since using f 1 will swap the portion berfore and after e , and the first segment of path is north, ⇀. Ch. engchi. after using f 1 , e connects with the north segment. Thus, we have proved.. Next, we prove the second part. ⇀. Notice that after using the first f 1 , the part of e and the north segment is below the diagonal y = x. There is another first touching the diagonal y = x east, and the part is in a ( j, j) path after that east, in the next step, we use f 1 again, so this part will be swapped to before that east, since it is. ( j, j) path, the part is still below the diagonal y = x. ⇀. Therefore, no matter how many times we use f 1 , e connects with the first segment of P in A1 and they are not be separated afterward.. 11.

(20) Theorem 2.8. g1 is one-to-one and onto. Proof. Claim: g1 is one-to-one. (k ). (h). g1 ( P) = g1 ( Q) ⇒ f 1 ( P) = f 1 ( Q), where P has k flaws and Q has h flaws. Case1: k < h (k ). ( k −1). (h). f 1 ( P) = f 1 ( Q) ⇒ f 1 ( f 1. ( h −1). ( P)) = f 1 ( f 1. ( Q)). ∵ f 1 is one-to-one. ( k −1). ∴ f1. ( h −1). ( P) = f 1. ( k −2). f1 ( f1. ( Q ). ( h −2). ( P)) = f 1 ( f 1. ( k −2). ( Q)) ⇒ f 1. ( h −2). ( P) = f 1. ( Q), since f 1 is one-to-one.. (h−(k−1)). (h−k). ⇒ P = f1. ( Q). 立. ⇀. (h−k). 政治大. Use this way for k − 1 times, we have f 1 ( P) = f 1. ( Q) = f 1 ( f 1 (h−k). ⇀. The first e of P is above the diagonal y = x, but the first e of f 1. ( Q)). ( Q) is below the diagonal. ‧ 國. 學. y = x by Lemma 2.7.. This is a contradiction.. ( k −1). ( h −1). io. ∵ f 1 is one-to-one.. ( Q)). y. ( h −1). ( P)) = f 1 ( f 1. sit. ( k −1). f 1 ( P) = f 1 ( Q) ⇒ f 1 ( f 1. er. (h). Nat. (k ). ‧. Case2: k > h1. al. ( P) = f 1. n. v i n C ( k −2) ( k −2) ( h −2) ( h −2) f1 ( f1 ( P)) = f 1 ( f 1 ( Q)) ⇒hf 1e ( P) =hfi1 U ( Q), since f 1 is one-to-one. ngc. ∴ f1. ( Q ).. (k−(h−1)). Use this way for h − 1 times, we have f 1 (k−h). ⇒ f1. (k −h). ( P) = f 1 ( f 1. ( P) = Q. ⇀. ⇀. (k−h). The first e of Q is above the diagonal y = x, but the first e of f 1 y = x by Lemma 2.7. This is a contradiction. Case3: k = h (k ). ( k −1). (h). f 1 ( P) = f 1 ( Q) ⇒ f 1 ( f 1 ( k −1). ⇒ f1. ( P)) = f 1 ( Q). ( h −1). ( P) = f 1. ( h −1). ( P)) = f 1 ( f 1. ( Q)). ( Q) (∵ f 1 is one-to-one.). Use this way for k − 1 times, we have f 1 ( P) = f 1 ( Q) ⇒ P = Q Therefore, g1 is one-to-one. 12. ( P) is below the diagonal.

(21) Claim: g1 is onto. Given Q ∈ B1 . (−k). Define f 1. = f 1−1 ◦ f 1−1 ◦ ... ◦ f 1−1 . | {z } k. f 1−1 ( Q) is a preimage of Q under f 1 and f 1−1 ( Q) has 1 flaw. ⇀. We can use this way for n + 1 times, until the segment e is above the diagonal y = x by Lemma 2.7. −(n+1). So we have f 1. ( Q) = P, where P has n + 1 flaws, P ∈ A1 .. Thus, g1 is onto. Therefore, g1 is one-to-one and onto.. 政治大 the next north segment in B 立 with a dot, and all paths in B are (n + 1, n + 1) path.. Definition 2.9. The set C1 consists of all (n, n) paths which are replaced the marked east and 1. 1. ‧ 國. ⇀⇀. 學. Let h1 be the function from B1 into C1 .. i.e. P = R e n S is an (n + 1, n + 1) path, where R and S are all good paths. ⇀⇀. ‧. h1 ( P ) = h1 ( R e n S ) = R • S. al. n. Given Q = R • S ∈ C1 .. ⇀⇀. We can change • into e n .. Ch. engchi. ⇀⇀. Thus, R • S ⇒ R e n S ∈ B1 . Therefore, h1 is one-to-one and onto.. 13. er. io. Proof. It is clearly obvious that h1 is one-to-one and onto.. sit. y. Nat. Theorem 2.10. h1 is one-to-one and onto.. i n U. v.

(22) Chapter 3 Paths Start with East Definition 3.1. The set X2. 政治大 consists of all (n + 1, n + 1) paths which have at most n flaw. 立 ⇀. ‧ 國. 學. Each path in X2 can be factorized into TG n Q. The set Y2 consists of all (n + 1, n + 1) paths which has at least one flaws.. ‧. Define a function f 2 from X2 into Y2 by the following:. y. Nat. n. al. er. io. y = x.. sit. 1. Starting from the bottom left, (0, 0), follow the path until it first travels below the diagonal. Ch. i n U. v. ⇀. 2. Continue to follow the path until it touches the diagonal y = x again. Denote by n , the. engchi. ⇀. first such segment that touches the diagonal y = x, in fact, n must be an north segment. ⇀. ⇀. 3. Swap the portion of the path before n with portion after n . ⇀. ⇀. i.e. f 2 ( TG n Q) = Q n TG, where T is an (i, i ) totally bad path, 0 ≤ i ≤ n + 1, G is a ( j + 1, j) good path, 0 ≤ j ≤ n − i, ⇀. the north segment n is the first north touching diagonal y = x, and Q is an (n − i − j, n − i − j) path. NOTE: After using f 2 , the flaws of path P increase one. i.e. if P has k flaws, k ≤ n, then f 2 ( P) has k + 1 flaws.. 14.

(23) ⇀. ⇀. To show f 2 ( TG n Q) = Q n TG by graph, we have:. 立. ‧ 國. 學 Figure 3.1: TG⃗nQ. ‧. ⇀. 政治大. n. al. er. io. sit. y. Nat. Fix n and switch TG with Q, we have:. Ch. engchi. i n U. Figure 3.2: Q⃗nTG. 15. v.

(24) ⇀. Theorem 3.2. Let P = TG n Q is an (n + 1, n + 1) path. Define f 2 : X2 −→ Y2 ⇀. ⇀. by f 2 ( TG n Q) = Q n TG, where T is an (i, i ) totally bad path, 0 ≤ i ≤ n + 1, G is a ( j + 1, j) ⇀. good path, 0 ≤ j ≤ n − i, the north segment n is the first north which touches the diagonal y = x, and Q is an (n − i − j, n − i − j) path. The function f 2 is one-to-one and onto. Proof. Claim: f 2 is one-to-one. ⇀. ′. ′⇀. ′. ′. ′. ′. Let P = TG n Q, P = T G n Q , where G is a an (l + 1, l ) good path, T is a (k, k ) totally ′. bad path, and Q is an (n − k − l, n − k − l ) path. ⇀. ′⇀. ′. ′. ′⇀. ⇀. ′. ′. If f 2 ( TG n Q) = f 2 ( T G n Q ) ⇒ Q n TG = Q n T G . ′. Claim: G = G . Case1: l > j. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. ′. Figure 3.3: Q⃗nTG. ′. Figure 3.4: Q ⃗nT G. ′. When we start on (n + 1, n + 1), trace back the path. We let two paths both trace back to the point (n − j, n − j + 1), in the next step, the path in Figure 3.3 is above the diagonal y = x, but the path in Figure 3.4 is still below diagonal y = x. This is a contradiction, as two paths are the same. 16.

(25) Case2: l < j. 政治大. 學. ′. ′. Figure 3.5: Q ⃗nT G. ′. Figure 3.6: Q⃗nTG. ‧. ‧ 國. 立. Nat. sit. y. When we start on (n + 1, n + 1), trace back the path.. n. al. er. io. Ww let two paths both trace back to the point (n − l, n − l + 1), in the next step, the path in. i n U. v. Figure 3.5 is above the diagonal y = x, but the path in Figure 3.6 is still below diagonal y = x.. Ch. engchi. This is a contradiction,as two paths are the same. Thus, we have proved that l = j. ′. ∴G=G. ′. Claim: T = T . Case1: i > k. 17.

(26) 政治大. 立. ′. ′. Figure 3.8: Q ⃗nT G. ′. ‧ 國. 學. Figure 3.7: Q⃗nTG We both start on (n − j, n − j + 1).. ‧. Since i > k, when we let the path in Figure 3.8 trace to (n − j − k, n − j − k + 1), the next ⇀. Nat. ⇀. sit. y. segment is n , but the path in Figure 3.7 is not n .. al. n. Case2: i < k. er. io. This is a contradiction,as two paths are the same.. Ch. engchi. 18. i n U. v.

(27) 政治大. 立. ′. ′. Figure 3.10: Q ⃗nT G. ′. ‧ 國. 學. Figure 3.9: Q⃗nTG We both start on (n − j + 1, n − j).. ‧. Since i < k, when we let the path in Figure 3.10 trace back to (n − j − i, n − j − i + 1), the ⇀. Nat. ⇀. sit. y. next segment is not n , but the path in Figure 3.9 is n ,.. al. n. Thus, we have proved that i = k.. er. io. This is a contradiction, as two paths are the same.. ′. ∴T=T. ′. Ch. ′. Since G = G and T = T . ⇀. ′⇀. ′. ⇀. ′. engchi. i n U. v. ′⇀. ∵ Q n TG = Q n T G ⇒ Q n = Q n ′. ⇀. ′. ′⇀. ∴ Q = Q ⇒ TG n Q = T G n Q. ′. ∴ f 2 is one-to one. Claim: f 2 is onto. i.e. For any path in Y2 , which has at least one flaw, we choose the last north leaving the diagonal ˆ then we switch the portions before nˆ and after n. ˆ We can get a new path y = x, denoted by n, with at most n flaws which is in X2 . To show by graph:. 19.

(28) 立. 政治大. preimage under f 2. ˆ −−−−−−−−−→ SnR ˆ Figure 3.11: RnS. ‧ 國. 學. To show by formula:. preimage under f 2. ‧. ˆ −−−−−−−−−→ SnR ˆ = P, Q = RnS where Q has at least one flaw, P has at most n flaws.. y. Nat. sit. In fact, if Q has k flaws, P has k − 1 flaws.. n. al. er. io. So, for every path Q in Y2 , we can find a path P in X2 such that f 2 ( P) = Q. Therefore, f 2 is one-to-one and onto.. Ch. engchi. i n U. v. ⇀. Lemma 3.3. The first north n touching the diagonal y = x in X2 is above the diagonal y = x in Y2 after using f 2 . ⇀. ⇀. Proof. Let P = S n R, where S is a ( j + 1, j) path, n is a (0, 1) north path, and R is an (n − j, n − j) path. ⇀. After using f 2 , we swap R and S, since R is an (n − j, n − j) path, the next segment n is above the diagonal y = x. ⇀. Thus, we have proved that the first north n touching the diagonal y = x in X2 and it is above the diagonal y = x in Y2 after using f 2 .. 20.

(29) ⇀. Lemma 3.4. The last north nˆ leaving from diagonal y = x in Y2 is the first north n touching the diagonal y = x in X2 . ⇀. ⇀. i.e. If n is the first north touching the diagonal y = x in X2 , then n is the last north leaving the diagonal y = x in Y2 .. 政治大. 立. y. ‧. ‧ 國. 學. Nat. ⇀. Figure 3.12: Lemma3.7 2nd part. io. sit. Proof. Since n is the first north touching the diagonal y = x, we can observe that there is a ⇀. n. al. er. empty area enclosed by the first east that leaves the diagonal y = x, denoted by e , the diagonal ⇀. y = x, n , and the diagonal y = x − 1.. Ch. engchi. i n U. v. ⇀. ⇀. After swapping two portions, another empty area is enclosed by n , the diagonal y = x, e , and the diagonal y = x + 1, so that there is no north segment can touch the diagonal y = x between ⇀. ⇀. n and e . ⇀. And the remain segments which are behind e are at most touching the diagonal y = x but not be above the diagonal y = x. Therefore, the last north nˆ leaving from diagonal y = x in Y ; is ⇀. the first north n touching the diagonal y = x in Y2 . Definition 3.5. The set A2 consists of all paths which first segment is east, and the first touching the diagonal y = x north is marked. The set B2 consists of all paths which are totally bad path. Define g2 from A2 into B2 by g2 ( P) = f (k) ( P), where P has k flaws, and f (k) = f 2 ◦ f 2 ◦ ... ◦ f 2 . | {z } k. 21.

(30) Example 3.6. The following example is one of g2 ( P):. 治n = 3 Figure 3.13: 政大. 立. ⇀. Lemma 3.7. In g2 ( P), after using the first f 2 , the first north which is denoted by n , connects. ‧ 國. 學. with the first segment of P in A2 . And this part will not be separated afterward.. ‧. Proof. First, we prove the first part. ⇀. Since using f 2 will swap the portion berfore and after n , and the first segment of path is east,. y. Nat. ⇀. n. al. er. io. Thus, we have proved.. sit. after using f 2 , n connects with the east segment.. Then we prove the second part.. Ch. engchi. i n U. v. ⇀. Notice that after using the first f 2 , the part of n and the east segment is above the diagonal y = x. There is another first touching the diagonal y = x north, and the part is in a ( j, j) path after that north, in the next step, we use f 2 again, so this part will be swapped to before that north and since it is ( j, j) path, the part is still above the diagonal y = x. ⇀. Therefore, no matter how many times you use f 2 , n connects with the first segment of P in A2 and they are not be separated afterward. Theorem 3.8. g2 is one-to-one and onto. Proof. Claim: g2 is one-to-one. (k ). (h). g2 ( P) = g2 ( Q) ⇒ f 2 ( P) = f 2 ( Q), where P has k flaws and Q has h flaws. 22.

(31) Case1: k < h (k ). ( k −1). (h). f 2 ( P) = f 2 ( Q) ⇒ f 2 ( f 2. ( h −1). ( P)) = f 2 ( f 2. ( Q)). ∵ f 2 is one-to-one. ( k −1). ∴ f2. ( h −1). ( P) = f 2. ( k −2). f2 ( f2. ( Q ). ( h −2). ( P)) = f 2 ( f 2. 2( k −2). ( Q)) ⇒ f 2. ( h −2). ( P) = f 2. ( Q). Since f 2 is one-to-one.. (h−(k−1)). Use this way for k − 1 times, we have f 2 ( P) = f 2 (h−k). ⇒ P = f2. (h−k). ( Q) = f 2 ( f 2. ( Q). ⇀. (h−k). ⇀. The first n of P is below the diagonal y = x, but the first n of f 2 y = x by Lemma 3.7.. 立. Case2: k > h. ( k −1). ‧ 國. f 2 ( P) = f 2 ( Q) ⇒ f 2 ( f 2. ( h −1). ( P)) = f 2 ( f 2. ( Q)). 學. (h). ∵ f 2 is one-to-one.. ( h −2). Nat. ( P)) = f 2 ( f 2. ( k −2). ( Q)) ⇒ f 2. ( h −2). ( P) = f 2. (k−(h−1)). al. n. ( P) = Q. io. (k−h). ⇒ f2. (k −h). ( P) = f 2 ( f 2. ⇀. Ch. ⇀i n U. v. (k−h). The first n of Q is below the diagonal y = x, but the first n of f 2 y = x by Lemma 3.7.. engchi. This is a contradiction. Case3: k = h (k ). ( k −1). (h). f 2 ( P) = f 2 ( Q) ⇒ f 2 ( f 2 ( k −1). ⇒ f2. ( h −1). ( P) = f 2. ( h −1). ( P)) = f 2 ( f 2. ( Q)). ( Q) (∵ f 2 is one-to-one.). Use this way for k − 1 times, we have f 2 ( P) = f 2 ( Q) ⇒ P = Q Therefore, g2 is one-to-one. Claim: g2 is onto. Given Q ∈ B2 . (−k). Define f 2. ( P)) = f 2 ( Q). er. Use this way for h − 1 times, we have f 2. ( Q). Since f 2 is one-to-one.. y. ( k −2). f2 ( f2. ( Q ).. sit. ( h −1). ( P) = f 2. ‧. ( k −1). ∴ f2. ( Q) is above the diagonal. 政治大. This is a contradiction.. (k ). ( Q)). = f 2−1 ◦ f 2−1 ◦ ... ◦ f 2−1 . | {z } k. 23. ( P) is above the diagonal.

(32) Since f 2−1 ( Q) is a preimage of Q under f 2 and f 2−1 ( Q) has n − 1 flaw. ⇀. We can use this way for n + 1 times, until the segment n is below the diagonal y = x by Lemma 3.7. −(n+1). So we have f 2. ( Q) = P, where P has no flaw, P ∈ A2 .. Thus, g2 is onto. Therefore, g2 is one-to-one and onto. Definition 3.9. The set C2 consists of all (n, n) paths which are replaced the marked north and the next east segment in B2 with a dot, and all paths in B2 are (n + 1, n + 1) path. Let h2 be the function from B2 into C2 . ⇀⇀. 政治大. i.e. P = R n e S is an (n + 1, n + 1) path, where R and S are all totally bad paths. ⇀⇀. h2 ( P ) = h2 ( R n e S ) = R • S. 立. ‧ 國. 學. Theorem 3.10. h2 is one-to-one and onto. Proof. It is clearly obvious that h2 is one-to-one and onto.. ‧. Given Q = R • S ∈ C2 .. ⇀⇀. sit. y. Nat. We can change • into n e . ⇀⇀. al. er. io. Thus, R • S ⇒ R n e S ∈ B2 .. n. Therefore, h2 is one-to-one and onto.. Ch. engchi. 24. i n U. v.

(33) Chapter 4 Summary 政治大 In this thesis, we prove the Catalan identity in a combinatorial way. We split the paths into 立 g1. h. ‧ 國. 學. 1 → B1 − → C1 two portions according to the first segment. Then we construct the functions in A1 −. which the first segment is north.. ‧ er. io. sit. y. Nat. n. al. Ch. engchi. g1. i n U. v. h. 1 → B1 − → C1 Figure 4.1: A1 −. g2. h. 2 → C2 which the first segment is east. → B2 − And the other functions in A2 −. 25.

(34) g2. h. 2 Figure 4.2: A2 − → B2 − → C2. 政治大. In chapter 3, we can also use reflection along the diagonal y = x to prove the paths with. 立. east segment, since the paths in chapter 3 is reflection along the diagonal y = x to the paths. ‧. ‧ 國. io. sit. y. Nat. n. al. er. understand.. 學. in chapter 2. But it will be less clear. In this thesis, we can obverse more details and easier to. Ch. engchi. 26. i n U. v.

(35) Appendix A Some examples of Catalan identity. (n + 2)Cn+1 = (政 4n +治2)C大n 立. ‧. ‧ 國. 學. io. sit. y. Nat. n. al. er. n=1. Ch. engchi. 27. i n U. v.

(36) n = 2, the first segment is north.. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 28. i n U. v.

(37) n = 2, the first segment is east.. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 29. i n U. v.

(38) Bibliography [1] Ronald Alter. Some remarks and results on Catalan numbers. pages 109–132, 1971. [2] Ronald Alter and K. K. Kubota. Prime and prime power divisibility of Catalan numbers.. 政治大. J. Combinatorial Theory Ser. A, 15:243–256, 1973.. 立. [3] Federico Ardila. Catalan numbers. Math. Intelligencer, 38(2):4–5, 2016.. ‧ 國. 學. [4] Young-Ming Chen. The Chung-Feller theorem revisited. Discrete Math., 308(7):1328–. ‧. 1329, 2008.. Nat. sit. y. [5] Ömer E ̆gecioğlu. A Catalan-Hankel determinant evaluation. In Proceedings of the Fortieth. n. al. er. io. Southeastern International Conference on Combinatorics, Graph Theory and Computing, volume 195, pages 49–63, 2009.. Ch. engchi. i n U. v. [6] R. Johnsonbaugh. Discrete Mathematics. Pearson/Prentice Hall, 2009. [7] Thomas Koshy. Catalan numbers with applications. Oxford University Press, Oxford, 2009. [8] Tamás Lengyel. On divisibility properties of some differences of the central binomial coefficients and Catalan numbers. Integers, 13:Paper No. A10, 20, 2013. [9] Youngja Park and Sangwook Kim. Chung-Feller property of Schröder objects. Electron. J. Combin., 23(2):Paper 2.34, 14, 2016. [10] Matej ̌Crepin ̌sek and Luka Mernik. An efficient representation for solving Catalan number related problems. Int. J. Pure Appl. Math., 56(4):589–604, 2009.. 30.

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