Diffy六邊形之探討 - 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. 政治大. 立六邊形之探討 Diffy. ‧ 國. 學 ‧. A Study about Diffy Hexagons. n. er. io. sit. y. Nat. al. Ch. engchi. i n U. v. 碩士班學生：王偉名撰指導教授：李陽明博士中華民國 103 年 6 月 25 日.

(2) 謝水井的外面，是怎麼樣的世界? 那是為了明白，值得費盡心思去相見的東西嗎？水井的外面，是怎麼樣的世界？那是數度墜落，依然充滿魅力值得挑戰的嗎？. 治政我們一起為了明白而努力，一起品嘗掉下時的痛苦吧。大立在最後來臨的世界，那裡一定是很棒的世界。水井的外面，是怎麼樣的世界？. ‧ 國. 學. 即使那裡只是水井的底端。. 踏出水井之外的決心，是通往新世界的鑰匙。不論踏得出去或踏不出去，. ‧. 新的世界一定會來臨…. Frederica Bernkastel. y. Nat. sit. n. al. er. io. 首先，我要感謝生我育我的父母。如果沒有他們的支持與鼓勵，就沒有今天的我。接下來，我要感謝的是政大應數系對我的栽培，把我從一個完全不懂得如何寫好一個證明的毛頭小子，栽培到足以寫出一份碩士論文的程度。在這其中，我要特別感謝我的指導老師李陽明老師：在我剛進入碩士班時，老師給了我人生中第一個擔任 TA 的機會，讓我成為老師在大學部所開設的離散數學課程的 TA，並從此開啟了我碩士班的 TA 生涯。在擔任 TA 的過程中，我學習到了不少東西，也更加堅定了我將來想繼續從事教學與研究工作的決心。此外，當我陷入低潮時，老師總是對我不離不棄，並且適時的給予我鼓勵與幫助，甚至不吝惜撥出暑假休息的時間陪我釐清問題的所在並將問題一一解決，這也是我在日後的種種考試中能夠平安順利通過的主要原因，真的非常謝謝老師。. Ch. engchi. i n U. v. 謹以此論文獻給愛我的父母與我愛的母校王偉名謹誌于國立政治大學應用數學系中華民國一百零三年六月 i.

(3) 中文. 要. 在這篇論文裡，我們研究 Diffy 六邊形。本文一開始將 Diffy 六邊形視為 Ducci 序列，然後我們討論關於 Ducci 序列的一些性質。然而，Diffy 六邊形事實上是可以旋轉與翻轉的，但是，我們所考慮的 Ducci 序列並不具備這樣的性質。所以，在本文的最後，我們討論在考慮旋轉與翻轉情況下的 Ducci 序列。關鍵字：循環、Diffy 六邊形、Ducci 過程、Ducci 序列、週期、相似循環. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. ii. i n U. v.

(4) Abstract In this thesis, we study the Diffy Hexagons: Initially, we regard a Ducci sequence as a Diffy Hexagon game and discuss some properties about Ducci sequences. However, a Ducci sequence isn’t actually a Diffy Hexagon game due to. 政治大. the fact that regular hexagons has some symmetries under rotations and reflections,. 立. but the Ducci sequences don’t. So, we apply an identification in the end.. ‧ 國. 學. Keywords: Cycles, Diffy Hexagons, Ducci processes, Ducci sequences, Periods, Similar cycles. ‧. n. er. io. sit. y. Nat. al. Ch. engchi. iii. i n U. v.

(5) Contents 謝要. Abstract. 政治大. ‧ 國. Similar Cycles. 4. Diffy Hexagons. al. n. Bibliography. 3. y. 3. sit. Ducci Sequences. ‧. 2. iv 1. io. Introduction. Nat. 1. iii. 學. Contents. 立. ii. 14. er. 中文. i. Ch. engchi. iv. i n U. v. 31 42.

(6) Chapter 1 Introduction. 政治大. The Diffy Hexagons which are generalized Diffy Boxes (see [3]) are games with the follow-. 立. ing procedures:. ‧ 國. 學. Step 1. Arrange six nonnegative integers around a regular hexagon. Step 2. Produce another regular hexagon of six nonnegative integers from the one obtained. ‧. in Step 1: For each adjacent pair of numbers, compute the absolute value of their difference and. Nat. sit. Finally, form the new regular hexagon with the remaining numbers.. y. place it between them. Then, remove the original numbers and the original regular hexagon.. al. n. Step 2. over and over.. er. io. Step 3. To obtain a sequence of regular hexagons of six nonnegative integers by performing. Ch. engchi. i n U. v. Figure 1.1: Diffy Hexagon. Without loss of generality, we denote a regular hexagon of six nonnegative integers as Figure 1.1. 1.

(7) From now on, let N be a positive integer with N ≥ 2. We denote the set of all N-tuples of nonnegative integers by A N . Define D : A N → A N by D ( a 1 , a 2 , · · · , a N ) = ( | a 1 − a 2 | , · · · , | a N −1 − a N | , | a N − a 1 | ) for all ( a1 , a2 , · · · , a N ) ∈ A N . Then, D is a well-defined function. Definition 1.1. The function D : A N → A N defined by D ( a 1 , a 2 , · · · , a N ) = ( | a 1 − a 2 | , · · · , | a N −1 − a N | , | a N − a 1 | ). 政治大. for all ( a1 , a2 , · · · , a N ) ∈ A N is called a Ducci process.. 立. Definition 1.2. Let⃗a = ( a1 , a2 , · · · , a N ) ∈ A N . A sequence of the form that⃗a, D (⃗a), D2 (⃗a), · · ·. ‧ 國. 學. is called the Ducci sequence of ⃗a. On the other hand, we denote ⃗a by D0 (⃗a).. ‧. Remark 1.3. Note that a 6-tuple of nonnegative integers is regarded as written in a regular hexagon, and hence a Ducci sequence of 6-tuples in A6 is regarded as a sequence of regular. y. Nat. n. er. io. al. sit. hexagons, that is, a Diffy Hexagon game.. Ch. engchi. 2. i n U. v.

(8) Chapter 2 Ducci Sequences. 政治大. Lemma 2.1. Let ⃗a ∈ A N . Then, there are nonnegative integers n, k with n > k such that. 立. D n (⃗a) = D k (⃗a).. ‧ 國. 學. Proof.. Write ⃗a = ( a1 , a2 , · · · , a N ). ‧. Let M = max{ a1 , a2 , · · · , a N }. sit. y. Nat. =⇒ There are at most ( M + 1) N different N-tuples which are obtained by perform-. io. D n (⃗a) = D k (⃗a). er. ing Ducci processes on ⃗a, and hence there are nonnegative integers n, k with n > k such that. al. n. v i n C h n is the positive integer Definition 2.2. Let⃗a ∈ A N . Suppose that e n g c h i U such that⃗a, D(⃗a), D2 (⃗a), · · · , D n−1 (⃗a) are all distinct and D n (⃗a) = D k (⃗a), where 0 ≤ k ≤ n − 1. We define the period of⃗a to. be n − k and the (n − k)-cycle of⃗a (or simply the cycle of⃗a) to be D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a). Definition 2.3. Let ⃗a ∈ A N . The largest component of ⃗a is denoted by max⃗a. Remark 2.4. If 0 ≤ x, y ≤ M, then | x − y| ≤ M. Proof. Note that − M ≤ x − y ≤ M, then we obtain | x − y| ≤ M Lemma 2.5. Let ⃗a ∈ A N . For all nonnegative integers r, s with r ≥ s, then we have max Dr (⃗a) ≤ max D s (⃗a). 3.

(9) Proof. Given nonnegative integers r, s with r ≥ s If r = s, there is nothing to prove Now, we may assume that r > s It suffices to show that max D s+1 (⃗a) ≤ max D s (⃗a): Write D s (⃗a) = ( x1 , x2 , · · · , x N ) and D s+1 (⃗a) = (y1 , y2 , · · · , y N ), where y 1 = | x 1 − x 2 | , y 2 = | x 2 − x 3 | , · · · , y N −1 = | x N −1 − x N | , y N = | x N − x 1 |. ∵ 0 ≤ x1 , x2 , · · · , x N ≤ max D s (⃗a). 政治大 (⃗a) ≤ max立 D (⃗a). ∴ By Remark 2.4, yi ≤ max D s (⃗a) for all i = 1, 2, · · · , N =⇒ max D s+1. s. ‧ 國. Then, max Dr (⃗a) = max D s (⃗a) for k ≤ r, s ≤ n − 1.. ‧. Proof.. 學. Lemma 2.6. Let ⃗a ∈ A N . Suppose D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) is the (n − k )-cycle of ⃗a.. Nat. sit. y. Given k ≤ r, s ≤ n − 1. er. io. We may assume that r ≤ s. al. n. v i n Suppose r < s, then max D s (⃗a) C ≤h max Dr (⃗a) by Lemma e n g c h i U 2.5. If r = s, then it is trivial. Now, look at the Ducci sequence of D s (⃗a):. D s (⃗a), D s+1 (⃗a), · · · , D n−1 (⃗a), D n (⃗a) = D k (⃗a), D k+1 (⃗a), · · · , Dr (⃗a). By Lemma 2.5, we know that max Dr (⃗a) ≤ max D k (⃗a) = max D n (⃗a) ≤ max D n−1 (⃗a) ≤ max D s (⃗a) Therefore, max Dr (⃗a) ≤ max D s (⃗a) So, we conclude that max Dr (⃗a) = max D s (⃗a) 4.

(10) Remark 2.7. If 0 ≤ x, y ≤ M with | x − y| = M, then x, y ∈ {0, M } and at least one of them is M. Proof. Since | x − y| = M, we obtain x − y = ± M Case 1: x − y = M. =⇒ M + y = x ≤ M =⇒ y ≤ 0 By assumption, y ≥ 0. ∴y=0. 治政 Therefore, x, y ∈ {0, M} and at least one of them is M 大立 Case 2: x − y = − M =⇒ x = M. ‧ 國. 學. =⇒ x + M = y ≤ M. y. sit. io. =⇒ y = M. Nat. ∴x=0. ‧. Note that x ≥ 0. Hence, x, y ∈ {0, M} and at least one of them is M. n. al. Ch. engchi. er. =⇒ x ≤ 0. i n U. v. Lemma 2.8. Let ⃗a = ( a1 , a2 , · · · , a N ), ⃗b = (b1 , b2 , · · · , b N ) ∈ A N such that D (⃗b) = ⃗a and max⃗a = max⃗b = M. If ai ∈ {0, M}, ∀ i = 1, 2, · · · , t and at least one of them is M, then bi ∈ {0, M }, ∀ i = 1, 2, · · · , t, t + 1 and at least one of them is M. Proof. We prove it by induction on t: t=1: Note that M = a1 = |b1 − b2 | and 0 ≤ b1 , b2 ≤ M By Remark 2.7, b1 , b2 ∈ {0, M } and at least one of them is M, holds Suppose t = 1, 2, · · · , K holds Then, t = K + 1: 5.

(11) By assumption, we have the following four cases: Case 1: a1 = M and a2 = a3 = · · · = aK = aK +1 = 0. =⇒ b2 = b3 = · · · = bK = bK+1 = bK+2 , since D (⃗b) = ⃗a ∵ M = a1 = |b1 − b2 | and 0 ≤ b1 , b2 ≤ M ∴ By Remark 2.7, b1 , b2 ∈ {0, M} and at least one of them is M Therefore, we obtain b1 , b2 , · · · , bK , bK +1 , bK +2 ∈ {0, M}. and at least one of them is M, holds. 政治大 , since D (⃗b) = ⃗a. Case 2: aK +1 = M and a1 = a2 = a3 = · · · = aK = 0. 立. =⇒ b1 = b2 = · · · = bK = bK+1. ‧ 國. 學. ∵ M = aK+1 = |bK+1 − bK+2 | and 0 ≤ bK+1 , bK+2 ≤ M. ∴ By Remark 2.7, bK+1 , bK+2 ∈ {0, M} and at least one of them is M. ‧. =⇒ b1 , b2 , · · · , bK , bK+1 , bK+2 ∈ {0, M} and at least one of them is M, holds Case 3: a1 = aK +1 = M and a2 = a3 = · · · = aK = 0. y. Nat. io. ∵ M = a1 = |b1 − b2 | and 0 ≤ b1 , b2 ≤ M. n. al. er. sit. =⇒ b2 = b3 = · · · = bK = bK+1 , since D (⃗b) = ⃗a. v. ∴ By Remark 2.7, b1 , b2 ∈ {0, M} and at least one of them is M. Ch. engchi. i n U. Note that M = aK +1 = |bK +1 − bK +2 | and 0 ≤ bK +1 , bK +2 ≤ M By Remark 2.7, we have bK +1 , bK +2 ∈ {0, M } and at least one of them is M So, we conclude that b1 , b2 , · · · , bK , bK +1 , bK +2 ∈ {0, M}. and at least one of them is M, holds Case 4: ∃ 2 ≤ i ≤ K such that ai = M. ∵ a1 , a2 , · · · , ai ∈ {0, M} and ai = M with 2 ≤ i ≤ K. 6.

(12) ∴ By induction hypothesis, b1 , b2 , · · · , bi , bi+1 ∈ {0, M}. and at least one of them is M Note that ai = M, ai+1 , · · · , aK +1 ∈ {0, M } and 2 = ( K + 1) − ( K − 1). ≤ ( K + 1) − ( i − 1) ≤ ( K + 1) − (2 − 1). 政治大. =K. 立. bi , bi+1 , · · · , bK , bK +1 , bK +2 ∈ {0, M}. Nat. sit. y. and at least one of them is M, by induction hypothesis. ‧. ‧ 國. 學. Since Ducci processes are cyclic, we obtain. n. al. er. io. Hence, we conclude that. Ch. i n U. v. b1 , b2 , · · · , bi , bi+1 , · · · , bK , bK +1 , bK +2 ∈ {0, M }. engchi. and at least one of them is M, holds Remark 2.9. In Lemma 2.8, we know that 1 ≤ t ≤ N − 1, since A N is a collection of N-tuples of nonnegative integers. Lemma 2.10. Let ⃗a ∈ A N . Suppose D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) is the (n − k )-cycle of ⃗a. Then, there are at least i + 1 cyclic consecutive components of D (n−1)−i (⃗a) taken from 0 or M such that at least one of them is M, where M = max D k (⃗a). Proof. We prove it by induction on i: 7.

(13) i = 0: By Lemma 2.6, max D n−1 (⃗a) = max D k (⃗a) = M. =⇒ there is a component of D n−1 (⃗a) is M =⇒ there is one cyclic consecutive component of D n−1 (⃗a) which is taken from 0 or M such that at least one of them is M, holds Suppose i = K holds Then i = K + 1: We must prove that there are at least (K + 1) + 1 cyclic consecutive components of D (n−1)−(K +1) (⃗a) taken from 0 or M such that at least one of them is M: Write D (n−1)−(K +1) (⃗a) = ( x1 , · · · , x N ) and D (n−1)−K (⃗a) = (y1 , · · · , y N ). 政治大. =⇒ D ( x1 , x2 , · · · , x N ) = (y1 , y2 , · · · , y N ). 立. By induction hypothesis, we know that there are at least the K + 1 cyclic consecutive com-. ‧ 國. 學. ponents of D (n−1)−K (⃗a) are taken from 0 or M such that at least one of them is M Since Ducci processes are cyclic, we may assume y1 , y2 , · · · , yK , yK +1 are K + 1 cyclic. ‧. consecutive components of D (n−1)−K (⃗a) which are taken from 0 or M such that at least one of. y. Nat. them is M without loss of generality. n. al. er. io. is M. sit. By Lemma 2.8, we know that x1 , x2 , · · · , xK , xK +1 , xK +2 ∈ {0, M } and at least one of them. i n U. v. Hence, we conclude that x1 , x2 , · · · , xK , xK +1 , xK +2 are (K + 1) + 1 cyclic consecutive components of. D (n−1)−(K +1) (⃗a). Ch. engchi. which are taken from 0 or M such that at least one of them is. M, holds Remark 2.11. In Lemma 2.10, we observe that: (a) 0 ≤ i ≤ N − 1. (b) If i ≤ min{n − k − 1, N − 1}, then D (n−1)−i (⃗a) is in the (n − k )-cycle of ⃗a. Proof. The first statement follows from the fact that A N is a collection of N-tuples of nonnegative integers 8.

(14) Now, we prove the last statement:. ∵ i ≤ min{n − k − 1, N − 1} ∴i ≤ n−k−1 By (a), we have 0 ≤ i ≤ n − k − 1. =⇒ k = ( n − 1) − ( n − k − 1). ≤ ( n − 1) − i ≤ ( n − 1) − 0 = n−1. 立. 政治大. Therefore, D (n−1)−i (⃗a) is in the (n − k)-cycle. ‧ 國. 學. Theorem 2.12. Let ⃗a ∈ A N . Suppose D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) is the (n − k )-cycle of ⃗a.. ‧. Then, the components of Di (⃗a) are all equal to either 0 or M for each i = k, k + 1, · · · , n − 1, where M = max D k (⃗a).. n. al. er. io. By Lemma 2.6, we obtain. sit. y. Nat. Proof.. Ch. engchi. i n U. v. max D j (⃗a) = max D k (⃗a) = M for all j = k, k + 1, · · · , n − 1 In particular, D n−1 (⃗a) = M. By Lemma 2.10, we know that there are at least N cyclic consecutive components of D (n−1)−( N −1) (⃗a) taken from 0 or M. =⇒ the components of D n− N (⃗a) are all equal to either 0 or M which follows from D n− N (⃗a) ∈ A N. 9.

(15) Now, look at the Ducci sequence of D n− N (⃗a): D n− N (⃗a), D n− N +1 (⃗a), · · · , D n−1 (⃗a), D n (⃗a) = D k (⃗a), D k+1 (⃗a), D k+2 (⃗a), · · · , D n− N −1 (⃗a), · · ·. =⇒ the components of D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) are all equal to either 0 or M, since the components of D n− N (⃗a) are all equal to 0 or M Hence, we complete this proof Remark 2.13. If N ̸= 2, then there are ⃗a,⃗b ∈ A N with D (⃗a) = ⃗b such that. 立. 政治大. max⃗a = max⃗b = M. ‧. ‧ 國. Proof.. 學. and the components of ⃗a, ⃗b aren’t all equal to either 0 or M.. Let M > 1 be an integer and ⃗a = ( M, 0, 1, · · · , 1, 1) ∈ A N. sit. y. Nat. Choose ⃗b = D (⃗a) ∈ A N. er. io. =⇒ ⃗b = ( M, 1, 0, · · · , 0, M − 1) Note that max⃗a = max⃗b = M. n. al. Ch. i n U. v. Therefore, the components of ⃗a, ⃗b aren’t all equal to either 0 or M. engchi. Definition 2.14. Let ⃗a ∈ A N with ⃗a ̸= ⃗0. If ⃗a = ( a1 , a2 , · · · , a N ), then gcd⃗a is the number gcd( a1 , a2 , · · · , a N ). Lemma 2.15. Let ⃗a ∈ A N with ⃗a ̸= ⃗0 and n be a nonnegative integer. Then, we obtain that gcd⃗a | max D n (⃗a). Proof. Let gcd⃗a = d Write ⃗a = d⃗b with gcd⃗b = 1 Note that d > 0 and D (⃗a) = D (d⃗b) = dD (⃗b) 10.

(16) =⇒ D n (⃗a) = D n (d⃗b) = dD n (⃗b) by induction on n =⇒ d | D n (⃗a) Therefore, we know that gcd⃗a | D n (⃗a) Corollary 2.16. Let ⃗a ∈ A N with ⃗a ̸= ⃗0. Suppose that D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) is the. (n − k)-cycle of ⃗a. Then, the components of Di (⃗a) are all equal to either 0 or M for each i = k, k + 1, · · · , n − 1, where M is a multiple of gcd⃗a. Proof. It follows from Theorem 2.12 and Lemma 2.15 Example 2.17. Let d, K ≥ 1 be integers and ⃗a = (d, d, d, d, d, Kd) ∈ A6 . Then, we have gcd⃗a = d.. 立. 政治大. Now, apply Ducci processes on ⃗a:. ‧ 國. 學. D (⃗a) = (0, 0, 0, 0, (K − 1)d, (K − 1)d). ‧. D2 (⃗a) = (0, 0, 0, (K − 1)d, 0, (K − 1)d). sit. y. Nat. D3 (⃗a) = (0, 0, (K − 1)d, (K − 1)d, (K − 1)d, (K − 1)d). io. er. D4 (⃗a) = (0, (K − 1)d, 0, 0, 0, (K − 1)d). D5 (⃗a) = ((K − 1)d, (K − 1)d, 0, 0, (K − 1)d, (K − 1)d). n. al. i n 6 C D (⃗a) = (0, (K − 1)h d, 0, (K − 1)d, 0, 0U engchi ). v. D7 (⃗a) = ((K − 1)d, (K − 1)d, (K − 1)d, (K − 1)d, 0, 0) D8 (⃗a) = (0, 0, 0, (K − 1)d, 0, (K − 1)d). = D2 (⃗a) So, the period of ⃗a is 6 and the 6-cycle of ⃗a is D2 (⃗a), D3 (⃗a), · · · , D7 (⃗a). Note that the components of Di (⃗a) are taken from 0 or (K − 1)d, ∀ i = 2, 3, · · · , 7 and. (K − 1)d is a multiple of d = gcd⃗a, where the multiple is (K − 1). Lemma 2.18. Let ⃗a ∈ A N with ⃗a ̸= ⃗0. For all nonnegative integers r, s with r ≤ s, then gcd Dr (⃗a) | gcd D s (⃗a). In particular, we have gcd Dr (⃗a) ≤ gcd D s (⃗a). 11.

(17) Proof. It suffices to show that gcd Dr (⃗a) | gcd D s (⃗a) Given nonnegative integers r, s with r ≤ s If r = s, there is nothing to prove Now, we may assume that r < s It is reduced to prove that gcd Dr (⃗a) | gcd Dr+1 (⃗a): Write Dr (⃗a) = ( x1 d, x2 d, · · · , x N d) such that gcd( x1 , x2 , · · · , x N ) = 1, where gcd Dr (⃗a) = d. =⇒ d > 0, since Dr (⃗a) ∈ A N =⇒ Dr+1 (⃗a) = ( | x1 − x2 |d, · · · , | x N −1 − x N |d, | x N − x1 |d ). 政治大. Let gcd( | x1 − x2 |, · · · , | x N −1 − x N |, | x N − x1 | ) = d∗. 立. =⇒ gcd Dr+1 (⃗a) = d∗ · d = d∗ · gcd Dr (⃗a). ‧ 國. 學. ∴ gcd Dr (⃗a) | gcd Dr+1 (⃗a). Lemma 2.19. Let ⃗a ∈ A N with ⃗a ̸= ⃗0. Suppose that D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) is the. ‧. (n − k)-cycle of ⃗a. Then, we have gcd Dr (⃗a) = gcd D s (⃗a) for all k ≤ r, s ≤ n − 1.. sit. y. Nat. Proof.. n. al. er. io. Given k ≤ r, s ≤ n − 1. We may assume that r ≤ s If r = s, then it is trivial. Ch. engchi. i n U. v. Suppose r < s, then gcd Dr (⃗a) ≤ gcd D s (⃗a) by Lemma 2.18 Now, look at the Ducci sequence of D s (⃗a): D s (⃗a), D s+1 (⃗a), · · · , D n−1 (⃗a), D n (⃗a) = D k (⃗a), D k+1 (⃗a), · · · , Dr (⃗a). By Lemma 2.18, we know that gcd D s (⃗a) ≤ gcd D k (⃗a) = gcd D n (⃗a) ≤ gcd D k+1 (⃗a) ≤ gcd Dr (⃗a) Therefore, gcd D s (⃗a) ≤ gcd Dr (⃗a) 12.

(18) So, we conclude that gcd Dr (⃗a) = gcd D s (⃗a) Example 2.20. Let ⃗e1 = (1, 0, 0, 0, 0, 0) ∈ A6. =⇒. D (⃗e1 ) = (1, 0, 0, 0, 0, 1) D2 (⃗e1 ) = (1, 0, 0, 0, 1, 0) D3 (⃗e1 ) = (1, 0, 0, 1, 1, 1) D4 (⃗e1 ) = (1, 0, 1, 0, 0, 0). 政治大. D5 (⃗e1 ) = (1, 1, 1, 0, 0, 1). 立D (⃗e ) = (0, 0, 1, 0, 1, 0) 6. 1. ‧ 國. 學. D7 (⃗e1 ) = (0, 1, 1, 1, 1, 0) D8 (⃗e1 ) = (1, 0, 0, 0, 1, 0). ‧. = D2 (⃗e1 ). y. Nat. n. al. er. io. D6 (⃗e1 ), D7 (⃗e1 ). sit. =⇒ the period of ⃗e1 is (8 − 2) = 6, and the 6-cycle of ⃗e1 is D2 (⃗e1 ), D3 (⃗e1 ), D4 (⃗e1 ), D5 (⃗e1 ),. i n U. v. Note that gcd Di (⃗e1 ) = 1 = max Di (⃗e1 ) for all i = 0, 1, · · · , 7 However,. D0 (⃗e. 1). Ch. engchi. = ⃗e1 , D (⃗e1 ) are not in the cycle of ⃗e1. 13.

(19) Chapter 3 Similar Cycles. 政治大 ⃗a is said to be similar to the cycle of ⃗b, if ∃ m ∈ N such that (n − k)-cycle of ⃗b. The cycle of立. Definition 3.1. Let ⃗a ∈ A N and ⃗b ∈ (Z2 ) N . Suppose D k (⃗b), D k+1 (⃗b), · · · , D n−1 (⃗b) is the. ‧ 國. 學. Dr (⃗a) = mD s (⃗b), where r, s are nonnegative integers with k ≤ s ≤ n − 1. Theorem 3.2. Let ⃗a ∈ A N . Then, the cycle of ⃗a is similar to the cycle of ⃗b, where ⃗b ∈ (Z2 ) N. sit. y. Nat. Proof.. ‧. and the period of ⃗b is equal to the period of ⃗a.. n. a kl. er. io. By Lemma 2.1, we may assume that the period of ⃗a is n − k and. i n U. D (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a). Ch. engchi. v. is the (n − k )-cycle of ⃗a Let D k (⃗a) = ( x1 , x2 , · · · , x N ) By Theorem 2.12, x1 , x2 , · · · , x N ∈ {0, M}, where M = max D k (⃗a) Since D k (⃗a) ∈ A N , we know that M is a nonnegative integer Case 1: M = 0. =⇒ D k (⃗a) = (0, 0, · · · , 0) ∈ (Z2 ) N =⇒ D k+1 (⃗a) = D ( D k (⃗a)) = D (0, 0, · · · , 0) = (0, 0, · · · , 0) = D k (⃗a) =⇒ n − 1 ≤ k, since ⃗a, D (⃗a), · · · , D k (⃗a), · · · , D n−1 (⃗a) are all distinct ∵ the period of ⃗a is n − k 14.

(20) ∴k ≤ n−1 Therefore, we have n − 1 = k So, the period of ⃗a is n − k = 1 Choose ⃗b = ⃗0 ∈ (Z2 ) N. =⇒ D (⃗b) = D (0, 0, · · · , 0) = (0, 0, · · · , 0) = ⃗b = D0 (⃗b) =⇒ the period of ⃗b is (1 − 0) = 1 and the 1-cycle of ⃗b is D0 (⃗b) = ⃗0 Therefore, the period of ⃗a is equal to the period of ⃗b Note that D k+1 (⃗a) = ⃗0 = D0 (⃗b). =⇒ the cycle of ⃗a is similar to the cycle of ⃗b Hence, we are done Case 2: M > 0. 立. =⇒ M ∈ N. 政治大. ‧ 國. 學. Write D k (⃗a) = ( x1 , x2 , · · · , x N ) = M(y1 , y2 , · · · , y N ), where y1 , y2 , · · · , y N are taken from 0 or 1. ‧. Choose ⃗b = (y1 , y2 , · · · , y N ) ∈ (Z2 ) N. y. Nat. =⇒ D k (⃗a) = M⃗b. =⇒ D k+i (⃗a) = MDi (⃗b), ∀ i = 1, 2, · · · , n − k. n. al. In particular, M⃗b = D k (⃗a) = D n (⃗a) = MD n−k (⃗b). =⇒. D0 (⃗b). = ⃗b = D n−k (⃗b). Ch. engchi. er. io. sit. =⇒ D k+1 (⃗a) = D ( D k (⃗a)) = D ( M⃗b) = MD (⃗b). i n U. v. By assumption, D k (⃗a) = M⃗b, D k+1 (⃗a) = MD (⃗b), · · · , D n−1 (⃗a) = D n−k−1 (⃗b) are all distinct. =⇒ D0 (⃗b) = ⃗b, D (⃗b), · · · , D n−k−1 (⃗b) are all distinct Therefore, the period of ⃗b is (n − k ) − 0 = n − k and the (n − k)-cycle of ⃗b is D0 (⃗b) = ⃗b, D (⃗b), · · · , D n−k−1 (⃗b) So, the period of ⃗a is equal to the period of ⃗b. ∵ D k (⃗a) = M⃗b = MD0 (⃗b) 15.

(21) ∴ the cycle of ⃗a is similar to the cycle of ⃗b Hence, we complete this proof Remark 3.3. When we discuss cycles of N-tuples in A N , it is enough to cope with N-tuples in. (Z2 ) N according to Theorem 3.2. Lemma 3.4. Let ⃗a = ( a1 , a2 , · · · , a N ) ∈ A N and max⃗a = M. Suppose that the cycle of ⃗a is similar to the cycle of ⃗b, where ⃗b ∈ (Z2 ) N and the period of ⃗b is equal to the period of ⃗a. If. ⃗ac = ( M − a1 , M − a2 , · · · , M − a N ), then the cycle of ⃗ac is similar to the cycle of ⃗b. Proof. Note that D (⃗ ac ) = (| a1 − a2 |, | a2 − a3 |, · · · , | a N − a1 |) = D (⃗a). 治政 ⃗ Since the cycle of ⃗a is similar to the cycle of b, ∃ m ∈ N大 such that 立 ‧ 國. 學. Dr (⃗a) = mD s (⃗b),. n. al. er. io. D r +1 ( ⃗ ac ) = Dr ( D (⃗ ac )) = Dr ( D (⃗a)) = Dr+1 (⃗a). sit. y. Nat. Then, we have:. ‧. where r, s are nonnegative integers with k ≤ s ≤ n − 1. i n U. v. = D ( Dr (⃗a)) = D (mD s (⃗b)) = mD ( D s (⃗b)) = mD s+1 (⃗b). Ch. engchi. which completes this proof Remark 3.5. In the proof of Lemma 3.4, D s+1 (⃗b) is in the cycle of ⃗b. Proof. If k ≤ s < n − 1, then it is trivial Now, we may assume that s = n − 1:. =⇒ s + 1 = n =⇒ D s+1 (⃗b) = D n (⃗b) = D k (⃗b) is in the cycle of ⃗b. 16.

(22) Lemma 3.6. Let ⃗a = ( a1 , a2 , · · · , a N ) ∈ A N and max⃗a=M. Suppose that D k (⃗a), D k+1 (⃗a), · · · , D n−1 (⃗a) is the (n − k )-cycle of ⃗a. Then, ⃗ ac = ( M − a1 , M − a2 , · · · , M − a N ) is in the (n − k )-cycle of ⃗a if and only if ⃗a = ⃗0. Proof. Note that D (⃗ ac ) = (| a1 − a2 |, | a2 − a3 |, · · · , | a N − a1 |) = D (⃗a) “⇒” Suppose the condition holds. 政治大. =⇒ ∃ k ≤ r ≤ n − 1 such that ⃗ac = Dr (⃗a). 立. =⇒ D (⃗a) = D (⃗ac ) = D ( Dr (⃗a)) = Dr+1 (⃗a). ‧. ‧ 國. distinct. 學. =⇒ n − 1 ≤ r, since D0 (⃗a) = ⃗a, D (⃗a), · · · , D k (⃗a), · · · , Dr (⃗a), · · · , D n−1 (⃗a) are all. Therefore, r = n − 1. n. al. Claim: k = 0. Ch. engchi. er. io. =⇒ D k (⃗a) = D n (⃗a) = Dr+1 (⃗a) = D (⃗ac ) = D (⃗a). sit. y. Nat. =⇒ n = r + 1. i n U. (∗). v. Proof. If not, suppose k ≥ 1 By assumption, D0 (⃗a) = ⃗a, D (⃗a), · · · , D k−1 (⃗a), D k (⃗a), · · · , D n−1 (⃗a) are all distinct. =⇒ n − 1 ≤ k − 1, by (∗) =⇒ n ≤ k which is a contradiction to n > k By Claim and (∗), we obtain ⃗a = D0 (⃗a) = D (⃗ ac ). =⇒ ∀ 1 ≤ i ≤ N, ai = M − ai 17.

(23) M 2. =⇒ ∀ 1 ≤ i ≤ N, ai = =⇒ M = max⃗a =. M 2. =⇒ M = 0 ∵ 0 ≤ a1 , a2 , · · · , a N ≤ M = 0 ∴ a1 = a2 = · · · = a N = 0 =⇒ ⃗a = ⃗0 “⇐” Suppose ⃗a = ⃗0. =⇒ a1 = a2 = · · · = a N = 0 =⇒ M = 0 =⇒ ⃗ac = (0, 0, · · · , 0). 立. 政治大. ‧ 國. 學. Note that D (⃗a) = D (0, 0, · · · , 0) = (0, 0, · · · , 0) = ⃗a = D0 (⃗a). n. Hence, we complete this proof. Ch. y. er. io. al. sit. Nat. D0 (⃗a) = (0, 0, · · · , 0) = ⃗ ac. ‧. =⇒ the period of ⃗a is (1 − 0) = 1 and the 1-cycle of ⃗a is. engchi. i n U. v. Now, we define T : A N → A N by. T ( x 1 , x 2 , · · · , x N −1 , x N ) = ( x 2 , x 3 , · · · , x N , x 1 ) for all ( x1 , x2 , · · · , x N −1 , x N ) ∈ A N . Clearly, T is well-defined. On the other hand, we fix the following notations: D = D |(Z2 ) N , T = T |(Z2 ) N , and D 0 = T 0 = I , where I is the identity on (Z2 ) N . Lemma 3.7. Let ⃗x, ⃗y ∈ A N and c be a nonnegative integer, then. 18.

(24) (a) T (c⃗x + ⃗y) = cT (⃗x ) + T (⃗y). (b) D ◦ T = T ◦ D. Proof. Write ⃗x = ( x1 , x2 , · · · , x N ), ⃗y = (y1 , y2 , · · · , y N ) ∈ A N (a) T (c⃗x + ⃗y) = T (cx1 + y1 , cx2 + y2 , · · · , cx N + y N ). = (cx2 + y2 , · · · , cx N + y N , cx1 + y1 ) = c ( x2 , · · · , x N , x1 ) + ( y2 , · · · , y N , y1 ). 治政 = cT (⃗x ) + T (⃗y) 大立 ‧ 國. 學 ‧. Given ( a1 , a2 , · · · , a N ) ∈ A N. sit. y. Nat. io. D ◦ T ( a1 , a2 , · · · , a N ) = D ( T ( a1 , a2 , · · · , a N )). n. al. er. (b). Ch. = D ( a2 , · · · , a N , a1 ). i n U. v. =n (| ag e 2− cah3 |,i· · · , |aN − a1 |, |a1 − a2 |). T ◦ D ( a1 , a2 , · · · , a N ) = T ( D ( a1 , a2 , · · · , a N )). = T (| a1 − a2 |, · · · , | a N −1 − a N |, | a N − a1 |) = (| a2 − a3 |, · · · , | a N − a1 |, | a1 − a2 |) =⇒ D ◦ T ( a1 , a2 , · · · , a N ) = T ◦ D ( a1 , a2 , · · · , a N ) ∴D◦T = T◦D. 19.

(25) Remark 3.8. Let x, y ∈ Z2 . Then, | x − y| = x + y. Proof. Given x, y ∈ Z2. =⇒ 2y = 0 =⇒ x − y = x − y + 2y = x + y =⇒ | x − y| = | x + y| =⇒ | x − y| = x + y, since x, y ∈ Z2 Remark 3.9. Let L : (Z2 ) N → (Z2 ) N be a function. Then, we know that L is a linear transformation if and only if L (⃗x + ⃗y) = L (⃗x ) + L (⃗y) for all ⃗x, ⃗y ∈ (Z2 ) N . Proof.. 立. ‧ 國. 學. “⇒” It is trivial. 政治大. “⇐” Suppose the condition holds. ‧. Given ⃗x, ⃗y ∈ (Z2 ) N , and c ∈ Z2. sit. n. er. io. If c = 1, then there is nothing to prove. al. Now, we may assume that c = 0:. =⇒. Ch. y. Nat. We must show that L (c⃗x + ⃗y) = cL (⃗x ) + L (⃗y):. engchi. i n U. v. L (c⃗x + ⃗y) = L (⃗0 + ⃗y) = L (⃗y) and cL (⃗x ) + L (⃗y) = ⃗0 + L (⃗y) = L (⃗y). ∴ L (c⃗x + ⃗y) = cL (⃗x ) + L (⃗y). Lemma 3.10. T i is a linear transformation for each i = 0, 1, 2, · · · .. 20.

(26) Proof. We prove it by induction on i: i = 0: T 0 = I is a linear transformation, holds Suppose i = K holds Then, i = K + 1: By Remark 3.9, it suffices to show that T. K +1. (⃗x + ⃗y) = T K+1 (⃗x ) + T K+1 (⃗y), ∀ ⃗x, ⃗y ∈ (Z2 ) N. 政治大. Given ⃗x = ( x1 , x2 , · · · , x N ), ⃗y = (y1 , y2 , · · · , y N ) ∈ (Z2 ) N K +1. 立. (⃗x + ⃗y) = T K ( T (⃗x + ⃗y)). 學. ‧ 國. T. = T K (T ( x1 + y1 , x2 + y2 , · · · , x N + y N )). ‧. = T K ( x2 + y2 , · · · , x N + y N , x1 + y1 ). sit. y. Nat. = T K ( x2 , · · · , x N , x1 ) + T K (y2 , · · · , y N ), by induction hypothesis. io. er. = T K (T ( x1 , x2 , · · · , x N )) + T K (T (y1 , y2 , · · · , y N )). al. n. v i n =T (C x1 , x2 , · · · , x N ) + T h e n g c h i U ( y1 , y2 , · · · , y N ) K +1 K +1 K +1. =T. K +1. (⃗x ) + T. (⃗y). By induction, we complete this proof Lemma 3.11. D i is a linear transformation for each i = 0, 1, 2, · · · . Proof. We prove it by induction on i: i = 0: D 0 = I is a linear transformation, holds Suppose i = K holds 21.

(27) Then, i = K + 1: By Remark 3.9, it suffices to show that: D K +1 (⃗x + ⃗y) = D K +1 (⃗x ) + D K +1 (⃗y), ∀ ⃗x, ⃗y ∈ (Z2 ) N Given ⃗x = ( x1 , x2 , · · · , x N ), ⃗y = (y1 , y2 , · · · , y N ) ∈ (Z2 ) N D K +1 (⃗x + ⃗y) = D K (D (⃗x + ⃗y)). = D K (D ( x1 + y1 , x2 + y2 , · · · , x N + y N )) = D K ( |( x1 + y1 ) − ( x2 + y2 )|, · · · , |( x N −1 + y N −1 ) − ( x N + y N )|,. 政治大. |( x N + y N ) − ( x1 + y1 )| ). 立. = D K ( ( x 1 + y 1 ) + ( x 2 + y 2 ) , · · · , ( x N −1 + y N −1 ) + ( x N + y N ) ,. ‧ 國. 學. ( x N + y N ) + ( x1 + y1 ) ), by Remark 3.8. = D K (( x1 + x2 ) + (y1 + y2 ), · · · , ( x N −1 + x N ) + (y N −1 + y N ),. ‧. ( x N + x1 ) + (y N + y1 )). Nat. sit. y. = D K ( x 1 + x2 , · · · , x N −1 + x N , x N + x1 ) + D K ( y1 + y2 , · · · ,. n. al. er. io. y N −1 + y N , y N + y1 ), by induction hypothesis. i n U. v. = D K ( | x1 − x2 |, · · · , | x N −1 − x N |, | x N − x1 | )+. Ch. engchi. D K ( |y1 − y2 |, · · · , |y N −1 − y N |, |y N − y1 |), by Remark 3.8. = D K (D ( x1 , x2 , · · · , x N )) + D K (D (y1 , y2 , · · · , y N )) = D K +1 ( x 1 , x 2 , · · · , x N ) + D K +1 D ( y 1 , y 2 , · · · , y N ) = D K+1 (⃗x ) + D K+1 (⃗y). By induction, we complete this proof Lemma 3.12. Let ⃗a ∈ A N . Suppose that r, s, t are nonnegative integers such that s ≤ r and. 22.

(28) s ≤ t. If Dr (⃗a) = D s (⃗a), then D (r−s)i ( D t (⃗a)) = D t (⃗a) for each i = 0, 1, 2, · · · . Proof. We prove it by induction on i: i = 0: It is trivial Suppose i = K holds Then, i = K + 1:. 立. 政治大. D (r−s)(K +1) ( D t (⃗a)) = Dr−s ( D (r−s)K ( D t (⃗a))). = Dr−s+t (⃗a). ‧. ‧ 國. 學. = Dr−s ( D t (⃗a)), by induction hypothesis. = D t−s ( Dr (⃗a)), since s ≤ t. n. al. Ch. By induction, we complete this proof. engchi. y. sit. io. = D t (⃗a), holds. er. Nat. = D t−s ( D s (⃗a)). i n U. v. Theorem 3.13. Let ⃗ei = (δi1 , δi2 , · · · , δiN ) ∈ A N , where    1, if j = i δij =   0, if j ̸= i for all i, j ∈ {1, 2, · · · , N }. Then, we have: (a) If Dr (⃗e1 ) = D s (⃗e1 ) for some nonnegative integers r, s, then we have: Dr (⃗b) = D s (⃗b), ∀⃗b ∈ (Z2 ) N .. 23.

(29) (b) The period of ⃗e1 , ⃗e2 , · · · , ⃗e N are all identical. (c) If ⃗a ∈ A N , then the period of ⃗a divides the period of ⃗e1 . In particular, the maximal period of N-tuples in A N is equal to the period of ⃗e1 . Proof. Note that {⃗e1 , ⃗e2 , · · · , ⃗e N } is a basis of (Z2 ) N over Z2 (a) Claim: Dr (⃗ei ) = D s (⃗ei ) for each i = 1, 2, · · · , N Proof. Given i ∈ N with 1 ≤ i ≤ N. 立. 政治大. ‧ 國. 學. If i = 1, then there is nothing to prove Now, we may assume that 2 ≤ i ≤ N. ‧. Note that ⃗ei = T N −i+1 (⃗e1 ). Dr (⃗ei ) = Dr ( T N −i+1 (⃗e1 )). n. al. er. io. sit. y. Nat. =⇒. v i n =CT h ( D (⃗e )), by Lemma 3.7(b) e n g c1 h i U N − i +1. r. = T N −i+1 ( D s (⃗e1 )). = D s ( T N −i+1 (⃗e1 )), by Lemma 3.7(b) = D s (⃗ei ) ∴ Dr (⃗ei ) = D s (⃗ei ) for each i = 1, 2, · · · , N Given ⃗b ∈ (Z2 ) N Finally, we must show that Dr (⃗b) = D s (⃗b):. ∵ {⃗e1 , ⃗e2 , · · · , ⃗e N } is a basis of (Z2 ) N over Z2 24.

(30) ∴ ∃ c1 , c2 , · · · , c N ∈ Z2 such that ⃗b = c1⃗e1 + c2⃗e2 + · · · + c N⃗e N =⇒ Dr (⃗b) = D r (⃗b). = D r (c1⃗e1 + c2⃗e2 + · · · + c N⃗e N ) = c1 D r (⃗e1 ) + c2 D r (⃗e2 ) + · · · + c N D r (⃗e N ), by Lemma 3.11 = c1 Dr (⃗e1 ) + c2 Dr (⃗e2 ) + · · · + c N Dr (⃗e N ) = c1 D s (⃗e1 ) + c2 D s (⃗e2 ) + · · · + c N D s (⃗e N ), by Claim = c1 D s (⃗e1 ) + c2 D s (⃗e2 ) + · · · + c N D s (⃗e N ). 政治大. = D s (c1⃗e1 + c2⃗e2 + · · · + c N⃗e N ), by Lemma 3.11. 立. = D s (⃗b). (b). ‧. ‧ 國. 學. = D s (⃗b). sit. y. Nat. Given i ∈ N with 2 ≤ i ≤ N. n. al. er. io. It suffices to show that the period of ⃗ei is equal to the period of ⃗e1 :. v. By Lemma 2.1, we may assume that the period of ⃗e1 = n − k. Ch. engchi. i n U. =⇒ D0 (⃗e1 ) = ⃗e1 , D (⃗e1 ), D2 (⃗e1 ), · · · , D n−1 (⃗e1 ) are all distinct and D n (⃗e1 ) = D k (⃗e1 ). (∗). By (a), we know that D n (⃗ei ) = D k (⃗ei ) Claim: D0 (⃗ei ) = ⃗ei , D (⃗ei ), D2 (⃗ei ), · · · , D N −1 (⃗ei ) are all distinct Proof. If not, suppose ∃ a, b ∈ Z with 0 ≤ a < b ≤ n − 1 such that D a (⃗ei ) = D b (⃗ei ). =⇒ T i−1 ( D a (⃗ei )) = T i−1 ( D b (⃗ei )) 25.

(31) =⇒ D a ( T i−1 (⃗ei )) = D b ( T i−1 (⃗ei )), by Lemma 3.7(b) =⇒ D a (⃗e1 ) = D b (⃗e1 ) which is a contradiction to D0 (⃗e1 ) = ⃗e1 , D (⃗e1 ), D2 (⃗e1 ), · · · , D n−1 (⃗e1 ). are all distinct By Claim and (∗), the period of ⃗ei is equal to n − k. =⇒ the period of ⃗ei is equal to the period of ⃗e1. 政治大 It is enough to prove that the period of ⃗a divides the period of ⃗e 立. 1. 學. By Lemma 2.1, we may assume that the periods of ⃗a and ⃗e1 are n − k and n′ − k′ , respec-. ‧ 國. tively. Nat. y. ‧. Write n′ − k′ = (n − k)q + r, where q, r are nonnegative integers with. io. n. al. sit. 0 ≤ r < n−k. er. (c). So, it suffices to show that r = 0:. Ch. engchi. i n U. v. By Theorem 3.2, ∃⃗b ∈ (Z2 ) N with the period of ⃗b which is equal to the period of ⃗a such that the cycle of ⃗a is similar to the cycle of ⃗b. =⇒ D n (⃗b) = D k (⃗b) Moreover, the period of ⃗e1 is n′ − k′ ′. ′. =⇒ D n (⃗e1 ) = D k (⃗e1 ) ′ ′ By (a), we obtain D n (⃗b) = D k (⃗b) ′ Take ⃗b = D k+k (⃗b) ′. ′. By Lemma 3.12, we obtain D n −k (⃗b) = ⃗b. 26.

(32) =⇒ ⃗b = D n′ −k′ (⃗b) = D (n−k)q+r (⃗b) = Dr ( D (n−k)q (⃗b)) = Dr (⃗b), by Lemma 3.12 ′ ′ =⇒ D k+k (⃗b) = D k+k +r (⃗b). Write k′ = (n − k )q0 + r0 , where q0 , r0 are nonnegative integers with 0 ≤ r0 < n − k. 政治大. 立 ′. D k+k (⃗b) = D (k+(n−k)q0 +r0 ) (⃗b). = D k+r0 (⃗b), by Lemma 3.12. y. sit. n. al. er. io D. ‧. = D (n−k)q0 ( D k+r0 (⃗b)). Nat. and. 學. ‧ 國. =⇒. Ch. k + k ′ +r. engchi. i n U. v. (⃗b) = D k+((n−k)q0 +r0 )+r (⃗b) = D (n−k)q0 ( D k+r0 +r (⃗b)) = D k+r0 +r (⃗b), by Lemma 3.12. ′ ′ =⇒ D k+r0 +r (⃗b) = D k+k +r (⃗b) = D k+k (⃗b) = D k+r0 (⃗b). (∗ ′ ). Note that k + r0 ≤ k + r0 + r < k + r0 + (n − k) = n + r0. =⇒ k + r0 ≤ k + r0 + r ≤ (n − 1) + r0 , since k + r0 + r, n + r0 are integers On the other hand,⃗b, D (⃗b), · · · , D k (⃗b), D k+1 (⃗b), · · · , D n−1 (⃗b) are all distinct and D n (⃗b) = D k (⃗b), since the period of ⃗b is n − k 27.

(33) =⇒ D k+r0 (⃗b), D k+r0 +1 (⃗b), · · · , D k+r0 +r (⃗b), · · · , D (n−1)+r0 (⃗b) are all distinct By (∗′ ), we conclude that k + r0 = k + r0 + r. =⇒ r = 0 Hence, we complete this proof. Lemma 3.14. Let r, s be nonnegative integers. Then, we have: (a) D = I + T .. 政治大. r. (b) If 2r ≡ s (mod N ), then D 2 = I + T s .. 立. Proof.. ‧ 國. 學. (a). ‧. Given ⃗x = ( x1 , x2 , · · · , x N ) ∈ (Z2 ) N. sit. y. Nat. I + T (⃗x ) = I (⃗x ) + T (⃗x ). al. er. io. = I ( x1 , x2 , · · · , x N ) + T ( x1 , x2 , · · · , x N ). v. n. = ( x1 , x2 , · · · , x N ) + ( x2 , · · · , x N , x1 ). Ch. engchi. i n U. = ( x 1 + x 2 , · · · , x N −1 + x N , x N + x 1 ) = ( | x1 − x2 |, · · · , | x N −1 − x N |, | x N − x1 | ), by Remark 3.8 = D ( x1 , x2 , · · · , x N ) = D (⃗x ). ∴D = I +T (b) r. By (a), D 2 = (I + T )2 r. r. Claim: (I + T )2 = I + I 2. r. 28.

(34) Proof. We prove it by induction on r: r = 0: It is trivial Suppose r = K holds Then, r = K + 1:. (I + T )2. K +1. K. = ((I + T )2 )2 K. = (I + T 2 )2 , by induction hypothesis K. K. K. = I 2 + I T 2 + T 2 I + (T 2 )2. 政治大 K. K. = I +T 2 +T 2 +T 2. 立. = I +T 2. K +1. K +1. , since T (Z2 ) ⊆ Z2. ‧ 國. 學. So, r = K + 1 holds. io. r. r. s. r. ∴ D 2 = (I + T )2 = I + T 2 = I + T s. n. al. Ch. Hence, we complete this proof. engchi. sit. Nat. r. By assumption, we know that T 2 = T. y. =T. er. N. ‧. Note that T. i n U. v. Theorem 3.15. Let r be a positive integer. Suppose that N = 2r . If ⃗a ∈ A N , then the cycle of. ⃗a is similar to the 1-cycle of ⃗0. Proof. By Lemma 2.1, we may assume that the period of ⃗a is n − k Note that 2r ≡ 0 (mod N ) By Theorem 3.2, ∃⃗b ∈ (Z2 ) N with the period of ⃗b which is equal to the period of ⃗a such that the cycle of ⃗a is similar to the cycle of ⃗b 29.

(35) =⇒ ∃ m ∈ N such that Dr (⃗a) = mD s (⃗b), where r, s are nonnegative integers with k ≤ s ≤ n−1. ∵ D N (⃗b) = D N (⃗b), since ⃗b ∈ (Z2 ) N. = I + T 0 (⃗b), by Lemma 3.14 = I + I (⃗b) = I (⃗b) + I (⃗b) = ⃗0, since I (⃗b) ∈ (Z2 )6 ∴. 立. 政治大. ‧ 國. 學. Dr+ N (⃗a) = D N ( Dr (⃗a)). = D N (mD s (⃗b)). ‧. = mD N +s (⃗b). y. = mD s (⃗0). n. al. Ch. = m⃗0. e=n⃗0g c h i. er. io. sit. Nat. = mD s ( D N (⃗b)). i n U. v. On the other hand, D (⃗0) = D (0, 0, · · · , 0) = (0, 0, · · · , 0) = ⃗0 = D0 (⃗0). =⇒ the period of ⃗0 is 1 − 0 = 1 and the 1-cycle of ⃗0 is ⃗0 =⇒ the cycle of ⃗a is similar to the 1-cycle of ⃗0 For further informations, please see [1] and [2].. 30.

(36) Chapter 4 Diffy Hexagons. 政治大. According to Remark 1.3, we shall concentrate on the cycles of 6-tuples in A6 in this chapter.. 立. Theorem 4.1. The period of 6-tuples in A6 divides 6. In particular, the maximal period of. ‧ 國. 學. 6-tuples in A6 is equal to 6.. Nat. y. Given ⃗a ∈ A6. ‧. Proof.. n. al. er. io. =⇒. sit. Let ⃗e1 = (1, 0, 0, 0, 0, 0). i n C D (⃗eh 1 ) = (1, 0, 0, 0, 0, 1U engchi ). D2 (⃗e1 ) = (1, 0, 0, 0, 1, 0) D3 (⃗e1 ) = (1, 0, 0, 1, 1, 1) D4 (⃗e1 ) = (1, 0, 1, 0, 0, 0) D5 (⃗e1 ) = (1, 1, 1, 0, 0, 1) D6 (⃗e1 ) = (0, 0, 1, 0, 1, 0) D7 (⃗e1 ) = (0, 1, 1, 1, 1, 0) D8 (⃗e1 ) = (1, 0, 0, 0, 1, 0). = D2 (⃗e1 ) 31. v.

(37) =⇒ the period of ⃗e1 is (8 − 2) = 6 By Theorem 3.13(c), the period of ⃗a divides 6 and the maximal period of 6-tuples in A6 is equal to 6 Lemma 4.2. If ⃗b ∈ (Z2 )6 , then the cycle of ⃗b is one of the followings: (i) (1-cycle) (0, 0, 0, 0, 0, 0). (ii) (3-cycle) (0, 1, 1, 0, 1, 1), (1, 0, 1, 1, 0, 1), (1, 1, 0, 1, 1, 0). (iii) (6-cycle) (0, 1, 0, 0, 0, 1), (1, 1, 0, 0, 1, 1), (0, 1, 0, 1, 0, 0), (1, 1, 1, 1, 0, 0), (0, 0, 0, 1, 0, 1), (0, 0, 1, 1, 1, 1). (iv) (6-cycle) (1, 0, 0, 0, 1, 0), (1, 0, 0, 1, 1, 1), (1, 0, 1, 0, 0, 0), (1, 1, 1, 0, 0, 1), (0, 0, 1, 0, 1, 0), (0, 1, 1, 1, 1, 0).. 政治大 We prove it by enumerating立 as shown in the following diagrams:. Proof.. ‧ 國. (0, 1, 0, 1, 0, 1) (1, 0, 1, 0, 1, 0). ‧. (1, 1, 1,.1, 1, 1). →: a Ducci process. n. al. Ch. engchi. (0, 0, 0, 1, 1, 1). er. io. sit. y. (0, 0, 0, 0, 0, 0). Nat (ii) (3-cycle):. 學. (i) (1-cycle):. i n U. v. (1, 1, 1, 0, 0, 0). (0, 0, 1, 0, 0, 1) (0, 1, 1, 0, 1, 1). . 1, 0, 1) (1, 0, 1,. (1, 1, 0, 1, 1, 0) (0, 1, 0, 0, 1, 0) (1, 1, 0, 0, 0, 1). (1, 0, 1, 1, 0, 0). (0, 0, 1, 1, 1, 0). (0, 1, 1, 1, 0, 0) 32. (1, 0, 0, 0, 1, 1).

(38) →: a Ducci process. (iii) (6-cycle):. (0, 1, 0, 0, 0, 0). (1, 0, 1, 1, 1, 1). (1, 0, 0, 1, 0, 1). (1, 1, 0, 0, 0, 0). (0, 1, 1, 0, 1, 0). (1, 0, 1, 1, 1, 0). (0, 1, 0, 0, 0, 1) . (1, 1, 0, 0, 1, 1) (1, 0, 1, 0, 0, 1). (1, 1, 1, 0, 1, 1). (1, 1, 1, 0, 1, 0). (0, 0, 1, 1, 1, 1). (0, 0, 1, 1, 0, 0). (0, 1, 0, 1, 0, 0). (0, 1, 0, 1, 1, 0). (0, 0, 0, 1, 0, 0) (0, 0, 0, 1, 0, 1). 立. (0, 1, 1, 0, 0, 1). (1, 0, 0, 1, 1, 0). 學. →: a Ducci process. (0, 1, 1, 1, 1, 1). (1, 0, 0, 0, 0, 0). ‧. (iv) (6-cycle):. (0, 0, 0, 0, 0, 1). ‧ 國. (1, 1, 1, 1, 1, 0). 政治 (1,大0, 1, 0, 1, 1). (0, 0, 1, 0, 1, 1). Nat. y. (1, 1, 0, 1, 0, 0). (1, 0, 0, 0, 0, 1). (0, 1, 1, 1, 0, 1). n. (1, 0, 1, 1, 0, 0) (1, 1, 0, 1, 0, 1). al. er. io. (1, 0, 0, 0, 1, 0) . (1, 0, 0, 1, 1, 1). sit. (0, 0, 0, 0, 1, 1). (1, 1, 1, 1, 0, 0). Ch. i n U. v. e n g c(h1, i0, 1, 0, 0, 0). (0, 1, 1, 1, 1, 0). (0, 1, 0, 0, 1, 1). (1, 1, 0, 1, 1, 1) (0, 1, 1, 0, 0, 0) (0, 0, 1, 0, 0, 0). (0, 0, 1, 0, 1, 0). (1, 1, 1, 0, 0, 1). (0, 0, 0, 1, 1, 0) (1, 1, 1, 1, 0, 1). (0, 1, 0, 1, 1, 1). (0, 0, 0, 0, 1, 0). (0, 0, 1, 1, 0, 1). (1, 1, 0, 0, 1, 0). →: a Ducci process. Theorem 4.3. Let ⃗a ∈ A6 . Then, the cycle of ⃗a is similar to one of the following cycles:. 33.

(39) (i) (1-cycle) (0, 0, 0, 0, 0, 0). (ii) (3-cycle) (0, 1, 1, 0, 1, 1), (1, 0, 1, 1, 0, 1), (1, 1, 0, 1, 1, 0). (iii) (6-cycle) (0, 1, 0, 0, 0, 1), (1, 1, 0, 0, 1, 1), (0, 1, 0, 1, 0, 0), (1, 1, 1, 1, 0, 0), (0, 0, 0, 1, 0, 1), (0, 0, 1, 1, 1, 1). (iv) (6-cycle) (1, 0, 0, 0, 1, 0), (1, 0, 0, 1, 1, 1), (1, 0, 1, 0, 0, 0), (1, 1, 1, 0, 0, 1), (0, 0, 1, 0, 1, 0), (0, 1, 1, 1, 1, 0). Proof. It follows from Theorem 3.2 and Lemma 4.2. As in Remark 1.3, a 6-tuple ( a1 , a2 , a3 , a4 , a5 , a6 ) in A6 is regarded as written in a regular hexagon. However, regular hexagons have symmetries under rotations and reflections, but. ( a1 , a2 , a3 , a4 , a5 , a6 ) does not.. 政治大. Write D6 = {(1)(2)(3)(4)(5)(6), (123456), (135)(246), (14)(25)(36), (153)(264), (165432),. 立. (16)(25)(34), (1)(4)(26)(35), (12)(36)(45), (2)(5)(13)(46), (14)(23)(56), (3)(6)(15)(24)}. ‧ 國. 學. which is the permutation group corresponding to all possible rotations and reflections of the regular hexagon.. ‧. Define ∗ : D6 × A6 → A6 by. y. Nat. n. al. er. io. sit. π ∗ ( a 1 , a 2 , · · · , a 6 ) = ( a π (1) , a π (2) , · · · , a π (6) ). i n U. v. for all π ∈ D6 and ( a1 , a2 , · · · , a6 ) ∈ A6 . Clearly, ∗ is well-defined.. Ch. engchi. Lemma 4.4. ∗ is a left group action of D6 on A6 . Proof.. Write e = (1)(2)(3)(4)(5)(6) which is the identity element of D6 Claim 1: e ∗⃗a = ⃗a, ∀⃗a ∈ A6 Proof. Given ⃗a = ( a1 , a2 , · · · , a6 ) ∈ A6. 34.

(40) =⇒ e ∗⃗a = ( ae(1) , ae(2) , · · · , ae(6) ). = ( a1 , a2 , · · · , a6 ) = ⃗a. Claim 2: (π1 ◦ π2 ) ∗⃗a = π1 ∗ (π2 ∗⃗a), ∀ π1 , π2 ∈ D6 and ⃗a ∈ A6 Proof.. 2. 6. 6. 學. Note that. 1. ‧ 國. Given π1 , π2 ∈ D6. 政治大 and ⃗a = ( a , a , · · · , a ) ∈ A 立. (π1 ◦ π2 ) ∗⃗a = ( aπ1 ◦ π2 (1) , aπ1 ◦ π2 (2) , · · · , aπ1 ◦ π2 (6) ). ‧. and. sit. y. Nat. n. al. er. io. π1 ∗ (π2 ∗⃗a) = π1 ∗ ( aπ2 (1) , aπ2 (2) , · · · , aπ2 (6) ). = ( aπ1 (π2 (1)) , aπ1 (π2 (2)) , · · · , aπ1 (π2 (6)) ). iv n U, · · · , a. C= h(aπ ◦ π (1), aπ ◦ π (2) engchi 1. 2. 1. 2. π1 ◦ π2 (6) ). ∴ (π1 ◦ π2 ) ∗⃗a = π1 ∗ (π2 ∗⃗a) By Claim 1 and Claim 2, we complete this proof For all ⃗x, ⃗y ∈ A6 , define ⃗x ≡ ⃗y by ⃗x = π ∗ ⃗y for some π ∈ D6 . Then, ≡ is the equivalence relation on A6 induced by D6 and we denote an equivalence class of A6 by [( a1 , a2 , · · · , a6 )], where ( a1 , a2 , · · · , a6 ) ∈ A6 . From now on, we identify two 6-tuples ⃗x, ⃗y in A6 , written by ⃗x = ⃗y, if and only if ⃗x ≡ ⃗y. Remark 4.5. In our identification, we observe that: 35.

(41) (a) If ⃗a,⃗b ∈ A6 , then ⃗a = ⃗b if and only if [⃗a ] = [⃗b ]. (b) According to Remark 1.3, a sequence of regular hexagons, that is, a Diffy Hexagon game, is actually a Ducci sequence of 6-tuples in A6 . Definition 4.6. Let ⃗a = ( a1 , a2 , · · · , a6 ) ∈ (Z2 )6 . The complement of ⃗a is defined to be. (1 − a1 , 1 − a2 , · · · , 1 − a6 ) and we denote it by ⃗ac . Remark 4.7. If ⃗a = ( a1 , a2 , · · · , a6 ) ∈ (Z2 )6 , then ⃗ a c ∈ ( Z2 ) 6 . Proof. By assumption, we know that a1 , a2 , · · · , a6 ∈ {0, 1}. 政治大 =⇒ ⃗a = (1 − a , 1 − a , · · · , 1 − a ) ∈ (Z ) 立 =⇒ 1 − a1 , 1 − a2 , · · · , 1 − a6 ∈ {0, 1} c. 2. 1. 6. 2. 6. ‧. ‧ 國. Proof.. 學. Lemma 4.8. If ⃗a = ( a1 , a2 , · · · , a6 ) ∈ (Z2 )6 , then π ∗ ⃗ ac = (π ∗⃗a)c for all π ∈ D6 .. Write ⃗ ac = (b1 , b2 , · · · , b6 ). y. sit. al. n. Note that. er. io. Given π ∈ D6. Nat. =⇒ bi = 1 − ai , ∀ i = 1, 2, · · · , 6. Ch. engchi. π∗⃗ a c = ( bπ (1 ) , bπ (2) , · · · , bπ (6) ). i n U. v. = ( 1 − a π (1) , 1 − a π (2) , · · · , 1 − a π (6) ) and. (π ∗⃗a)c = ( aπ (1) , aπ (2) , · · · , aπ (6) )c = ( 1 − a π (1) , 1 − a π (2) , · · · , 1 − a π (6) ) ∴ π ∗ ⃗ac = (π ∗⃗a)c 36.

(42) Lemma 4.9. There are 13 equivalence classes of (Z2 )6 . In fact, they are:. (0, 0, 0, 0, 0, 0), (0, 0, 0, 1, 1, 1), (0, 0, 1, 0, 0, 1), (0, 0, 1, 0, 1, 1), (0, 1, 0, 1, 0, 1), (0, 1, 1, 0, 1, 1), (0, 1, 1, 1, 0, 1), (0, 1, 1, 1, 1, 1), (1, 0, 0, 0, 0, 0), (1, 0, 0, 0, 0, 1), (1, 0, 0, 0, 1, 0), (1, 0, 0, 1, 1, 1), and (1, 1, 1, 1, 1, 1). Proof. For each π ∈ D6 , let Zπ = {z ∈ (Z2 )6 | π ∗ z = z} By the Burnside’s Lemma, we obtain: / |(Z2 )6 ≡| =. 1 | Zπ | |D6 | π∑ ∈D 6. 政治大. 1 = (26 + 2 + 22 + 23 + 22 + 2 + 23 + 24 + 23 + 24 + 23 + 24 ) 12 1 = (64 + 2 + 4 + 8 + 4 + 2 + 8 + 16 + 8 + 16 + 8 + 16) 12 1 = · 156 12. 立. ‧ 國. 學. = 13. ‧ sit. y. Nat. Finally, we enumerate 13 equivalence classes of (Z2 )6 :. io. al. er. By Lemma 4.8, it is reduced to write out [(0, 0, 0, 0, 0, 0)], [(1, 0, 0, 0, 0, 0)], [(0, 0, 1, 0, 0, 1)],. n. [(1, 0, 0, 0, 0, 1)], [(1, 0, 0, 0, 1, 0)], [(0, 0, 0, 1, 1, 1)], [(0, 0, 1, 0, 1, 1)], [(0, 1, 0, 1, 0, 1)]:. Ch. [(0, 0, 0, 0, 0, 0)] = {(0, 0, 0, 0, 0, 0)}. engchi. i n U. v. [(1, 0, 0, 0, 0, 0)] = {(1, 0, 0, 0, 0, 0), (0, 1, 0, 0, 0, 0), (0, 0, 1, 0, 0, 0), (0, 0, 0, 1, 0, 0), (0, 0, 0, 0, 1, 0), (0, 0, 0, 0, 0, 1)} [(0, 0, 1, 0, 0, 1)] = {(0, 0, 1, 0, 0, 1), (1, 0, 0, 1, 0, 0), (0, 1, 0, 0, 1, 0)} [(1, 0, 0, 0, 0, 1)] = {(1, 0, 0, 0, 0, 1), (1, 1, 0, 0, 0, 0), (0, 1, 1, 0, 0, 0), (0, 0, 1, 1, 0, 0), (0, 0, 0, 1, 1, 0), (0, 0, 0, 0, 1, 1)} [(1, 0, 0, 0, 1, 0)] = {(1, 0, 0, 0, 1, 0), (0, 1, 0, 0, 0, 1), (1, 0, 1, 0, 0, 0), (0, 1, 0, 1, 0, 0), (0, 0, 1, 0, 1, 0), (0, 0, 0, 1, 0, 1)}. 37.

(43) [(0, 0, 0, 1, 1, 1)] = {(0, 0, 0, 1, 1, 1), (1, 0, 0, 0, 1, 1), (1, 1, 0, 0, 0, 1), (1, 1, 1, 0, 0, 0), (0, 1, 1, 1, 0, 0), (0, 0, 1, 1, 1, 0)} [(0, 0, 1, 0, 1, 1)] = {(0, 0, 1, 0, 1, 1), (1, 0, 0, 1, 0, 1), (1, 1, 0, 0, 1, 0), (0, 1, 1, 0, 0, 1), (1, 0, 1, 1, 0, 0), (0, 1, 0, 1, 1, 0), (1, 1, 0, 1, 0, 0), (0, 1, 1, 0, 1, 0), (0, 0, 1, 1, 0, 1), (1, 0, 0, 1, 1, 0), (0, 1, 0, 0, 1, 1), (1, 0, 1, 0, 0, 1)} [(0, 1, 0, 1, 0, 1)] = {(0, 1, 0, 1, 0, 1), (1, 0, 1, 0, 1, 0)}. 政治大 Theorem 4.10. Let ⃗b ∈ (Z ) .立 Then, the cycle of ⃗b is one of the followings: 6. 學. ‧ 國. 2. (i) (1-cycle) (0, 0, 0, 0, 0, 0).. ‧. (ii) (1-cycle) (0, 1, 1, 0, 1, 1).. y. sit. al. er. io. Proof.. Nat. (iii) (2-cycle) (0, 0, 1, 0, 1, 0), (0, 1, 1, 1, 1, 0).. v. n. By Lemma 4.9 and Remark 4.5(b), we prove it by enumerating as shown in the following diagrams: (i) (1-cycle):. Ch. engchi. (0, 1, 0, 1, 0, 1) (1, 1, 1,.1, 1, 1) (0, 0, 0, 0, 0, 0). →: a Ducci process (ii) (1-cycle):. 38. i n U.

(44) (0, 0, 0, 1, 1, 1) (0, 0, 1,.0, 0, 1) (0, 1, 1, 0, 1, 1). →: a Ducci process (iii) (2-cycle): (0, 1, 1, 1, 1, 1). (1, 0, 0, 0, 0, 0). (0, 1, 1, 1, 0, 1). (0, 1, 1, 1, 0, 1). (0, 1, 1, 1, 0, 1). 學. ‧ 國. 政治大 (1, 0, 0, 0, 1, 0) . (1, 0, 0, 1, 1, 1) 立 →: a Ducci process. ‧. sit. y. Nat. Corollary 4.11. Let ⃗a ∈ A6 . Then, the cycle of ⃗a is similar to one of the following cycles:. io. n. al. er. (i) (1-cycle) (0, 0, 0, 0, 0, 0).. (ii) (1-cycle) (0, 1, 1, 0, 1, 1).. Ch. engchi. (iii) (2-cycle) (0, 0, 1, 0, 1, 0), (0, 1, 1, 1, 1, 0).. i n U. v. Proof. It follows from Theorem 3.2 and Theorem 4.10 Theorem 4.12. Let ⃗e1 = (1, 0, 0, 0, 0, 0) ∈ A6 . If r is a positive integer, then r is the period of. ⃗a for some ⃗a ∈ A6 if and only if r divides the period of ⃗e1 . Proof. “⇒” Suppose that r is the period of ⃗a for some ⃗a ∈ A6 By Theorem 3.13(c), the period of ⃗a divides the period of ⃗e1 39.

(45) Therefore, r divides the period of ⃗e1 “⇐” Suppose the condition holds Note that. D (⃗e1 ) = (1, 0, 0, 0, 0, 1) D2 (⃗e1 ) = (1, 0, 0, 0, 1, 0) D3 (⃗e1 ) = (1, 0, 0, 1, 1, 1) D4 (⃗e1 ) = (1, 0, 1, 0, 0, 0). 政治大. = (1, 0, 0, 0, 1, 0), by the proof in Lemma 4.9. 立 = D (⃗e ) 2. 1. ‧ 國. 學. By Remark 4.5(a) and Lemma 4.9, ⃗e1 , D (⃗e1 ), D2 (⃗e1 ), D3 (⃗e1 ) are all distinct. ‧. ∴ the period of ⃗e1 is (4 − 2) = 2. y. Nat. io. sit. By assumption, we know that r | 2. n. al. er. So, we have the following two cases: Case 1: r = 1. Ch. engchi. i n U. v. Choose ⃗a = (0, 0, 0, 0, 0, 0) ∈ (Z2 )6 ⊂ A6. By Theorem 4.10, the cycle of ⃗a is (0, 0, 0, 0, 0, 0) and the period of ⃗a is 1. =⇒ the period of ⃗a is r Case 2: r = 2 Choose ⃗a = (0, 0, 1, 0, 1, 0) ∈ (Z2 )6 ⊂ A6 By Theorem 4.10, the cycle of ⃗a is (0, 0, 1, 0, 1, 0), (0, 1, 1, 1, 1, 0) and the period of. ⃗a is 2 =⇒ the period of ⃗a is r. 40.

(46) By Case 1 and 2, we complete this proof. Let r, s be positive integers. Suppose that N = 2s and ⃗e1 = (1, 0, · · · , 0) ∈ A N . By Theorem 3.15, all similar cycles of N-tuples in A N are 1-cycle of ⃗0 which implies the period of every N-tuples in A N is 1, and hence the conclusion in Theorem 4.12, that is r is the period of. ⃗a for some ⃗a ∈ A N if and only if r divides the period of ⃗e1 , is true without identification we use in this chapter. However, above conclusion does not hold in A6 due to Lemma 4.2. If we had enough time, we would like to have further discussions about that at what positive integer N above conclusion holds (even if the identification we use in this chapter is necessary).. 立. 政治大. ‧. ‧ 國. 學. n. er. io. sit. y. Nat. al. Ch. engchi. 41. i n U. v.

(47) Bibliography [1] M. Burmester, R. Forcade, and E. Jacobs. Circles of numbers. Glasgow Mathematical Journal, 19:115–119, July 1978.. 政治大. [2] Amos Ehrlich. Periods in ducci’s n-number game of differences. Fibonacci Quarterly, 28(4):302–305, November 1990.. 立. ‧. ‧ 國. 學. [3] 蔡秀芬. Diffy box. 2008.. n. er. io. sit. y. Nat. al. Ch. engchi. 42. i n U. v.

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