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Stacking Blocks and Counting Permutations Hana Mizuno Occidental College mizuno@oxy.edu December 3rd, 2015 Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 1 / 55 Overview Background 1 Original Question and Solution 2

  1. Every p ∈ A n can be obtained from some p ′ ∈ A n − 1 Let p ∈ A n , and let p ′ be obtained from p by removing the two copies of n contained in p . Notice, p ′ ∈ A n − 1 , because if p ′ / ∈ A n − 1 ,then some subsequence s i of p ′ reduces to a sequence in { 132, 231, 213 } . But s i is also a subsequence of p . This contradicts the fact that p ∈ A n . Thus, every p n can be obtained by adding 2 copies of n to some p ′ ∈ A n − 1 . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 15 / 55

  2. Ways to insert two n ’s into p ′ . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 16 / 55

  3. Ways to insert two n ’s into p ′ . 1 “between” the digits of p ′ Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 16 / 55

  4. Ways to insert two n ’s into p ′ . 1 “between” the digits of p ′ 2 two n ’s in the beginning of p ′ Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 16 / 55

  5. Ways to insert two n ’s into p ′ . 1 “between” the digits of p ′ 2 two n ’s in the beginning of p ′ 3 two n ’s at the end of p ′ Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 16 / 55

  6. Ways to insert two n ’s into p ′ . 1 “between” the digits of p ′ 2 two n ’s in the beginning of p ′ 3 two n ’s at the end of p ′ 4 one n in the beginning and the other at the end of p ′ Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 16 / 55

  7. 1) “between” the digits of p ′ In the case where at least one n has a digit of p ′ to its left and a digit of p ′ to its right. Let a , b and c each represent a digit in p ′ . There are six different cases to consider. Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 17 / 55

  8. 1) “between” the digits of p ′ Forbidden Patterns Case 1: a < b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 18 / 55

  9. 1) “between” the digits of p ′ Forbidden Patterns Case 1: a < b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 18 / 55

  10. 1) “between” the digits of p ′ Forbidden Patterns Case 1: a < b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 18 / 55

  11. 1) “between” the digits of p ′ Forbidden Patterns Case 1: a < b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 18 / 55

  12. 1) “between” the digits of p ′ Forbidden Patterns Case 1: a < b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 18 / 55

  13. 1) “between” the digits of p ′ Forbidden Patterns Case 2: a > b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 19 / 55

  14. 1) “between” the digits of p ′ Forbidden Patterns Case 2: a > b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 19 / 55

  15. 1) “between” the digits of p ′ Forbidden Patterns Case 2: a > b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 19 / 55

  16. 1) “between” the digits of p ′ Forbidden Patterns Case 2: a > b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 19 / 55

  17. 1) “between” the digits of p ′ Forbidden Patterns Case 2: a > b Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 19 / 55

  18. 1) “between” the digits of p ′ Forbidden Patterns Case 5: a < b = c Case 3: a = b > c Case 4: a = b < c Case 6: a > b = c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 20 / 55

  19. 1) “between” the digits of p ′ Forbidden Patterns Case 3: a = b > c Case 5: a < b = c Case 4: a = b < c Case 6: a > b = c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 20 / 55

  20. 1) “between” the digits of p ′ Forbidden Patterns Case 3: a = b > c Case 5: a < b = c Case 4: a = b < c Case 6: a > b = c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 20 / 55

  21. 1) “between” the digits of p ′ Forbidden Patterns Case 3: a = b > c Case 5: a < b = c Case 4: a = b < c Case 6: a > b = c → Never generates a member of A n . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 20 / 55

  22. 2) two n’s in the beginning of p ′ Let p ′ ∈ A n − 1 and let p be obtained from p ′ by placing two n ’s at the beginning of p ′ . We show that p ∈ A n . There are two parts to look at: 1) Subsequences which contain n ’s 2) Subsequences without n ’s Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 21 / 55

  23. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  24. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  25. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  26. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  27. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  28. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  29. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  30. 2) two n’s in the beginning of p ′ Subsequences which contain n : Any subsequence contains exactly one n will typically be reduced to a sequence r which begins with a 3, thus, r / ∈ { 132 , 231 , 213 } . (A subsequence which contains both n ’s will be reduced to 221.) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 22 / 55

  31. 2) two n’s in the beginning of p ′ No subsequence which contains an n will reduce to { 132,231,213 } Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 23 / 55

  32. 2) two n’s in the beginning of p ′ Subsequences without n ’s: Since p ′ ∈ A n − 1 , no subsequence of p ′ reduces to a sequence from { 132,231,213 } . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 24 / 55

  33. 2) two n’s in the beginning of p ′ Neither subsequences with n ’s nor subsequences without n ’s reduce to a sequence from { 132,231,213 } . Therefore p ∈ A n . → Always create a member of A n . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 25 / 55

  34. 3) two n’s at the end of p ′ 4) one n in the beginning and the other at the end of p ′ These cases will create members of A n only when all of the digits of p ′ are in nondecreasing order. Both cases have at least one n at the end of p ′ . The only forbidden pattens with the biggst digit at the end is 213. If p ′ is not in nondecreasing order, then there will always be 21 subsequence in p ′ , which results in generating 213 subsequence in p . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 26 / 55

  35. For each p ′ ∈ A n − 1 , placing two n ’s at the beginning of p ′ generates a p ∈ A n . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 27 / 55

  36. For each p ′ ∈ A n − 1 , placing two n ’s at the beginning of p ′ generates a p ∈ A n . There is exactly one permutation p ′ ∈ A n − 1 whose digits are in nondecreasing order. Placing two n ’s at the end of p ′ or “surrounding” p ′ with one n on each side generates two more permutations from A n . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 27 / 55

  37. Therefore, | A n | = | A n − 1 | + 2 We have seen that A 2 = 6. Therefore, | A 3 | = 6 + 2 = 8 and | A 4 | = 8 + 2 = 10. This indicates that | A n | grows linearly, so | A n | = 2 n + 2, for n ≥ 2 Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 28 / 55

  38. Recurrence in Pudwell’s Permutation Let’s look at the B n = S 2 n (132 , 231 , 2134). Similar to A n = S 2 n (132 , 231 , 213), each q ∈ B n +1 can be generated by inserting two copies of (n+1) into some q ′ ∈ B n . 1 “between” the digits of q ′ 2 two ( n + 1)’s in the beginning of q ′ 3 two ( n + 1)’s at the end of q ′ 4 one ( n + 1) in the beginning and the other at the end of q ′ Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 29 / 55

  39. Let A n = S 2 n (132 , 231 , 213) and let B n = S 2 n (132 , 231 , 2134). Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 30 / 55

  40. Let A n = S 2 n (132 , 231 , 213) and let B n = S 2 n (132 , 231 , 2134). Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 30 / 55

  41. Let A n = S 2 n (132 , 231 , 213) and let B n = S 2 n (132 , 231 , 2134). Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 30 / 55

  42. Let A n = S 2 n (132 , 231 , 213) and let B n = S 2 n (132 , 231 , 2134). Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 30 / 55

  43. Let A n = S 2 n (132 , 231 , 213) and let B n = S 2 n (132 , 231 , 2134). | B 4 | = | B 3 | + 2 | A 3 | Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 30 / 55

  44. | B 4 | = | B 3 | + 2 | A 3 | Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 31 / 55

  45. | B 4 | = | B 3 | + 2 | A 3 | | B n +1 | = | B n | + 2 | A n | Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 31 / 55

  46. | B 4 | = | B 3 | + 2 | A 3 | | B n +1 | = | B n | + 2 | A n | | A n | = 2 n + 2 (From Lemma 1) Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 31 / 55

  47. | B 4 | = | B 3 | + 2 | A 3 | | B n +1 | = | B n | + 2 | A n | | A n | = 2 n + 2 (From Lemma 1) | B n +1 | = | B n | + 4 n + 4 , | B 2 | = 6 Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 31 / 55

  48. Recurrence Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 32 / 55

  49. Theorem n +1 (132 , 231 , 2134) | = 2 n 2 + 6 n − 2 = SA ( n ) for n ≥ 1 | S 2 SA ( n ) = SA ( n − 1) + 4 n + 4 for n ≥ 2 and SA (1) = 6. B n = S 2 n (132 , 231 , 2134). | B n +1 | = | B n | + 4 n + 4 , n ≥ 2 and | B 2 | = 6 Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 33 / 55

  50. Conclusion Relationship between middle school geometry and Combinatorics Application of discrete math to geometry question This is just a part of Pudwell’s study on permutation that avoids other permutations, so it would be interesting to read and investigate other Enumeration of Words with Forbidden Patterns studies. Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 34 / 55

  51. References Pudwell, Lara K. , ”Stacking Blocks and Counting Permutations” , Mathematics Magazine ,83.4,(2008),297-302. Burstein, Alexander , ”Enumeration of Words with Forbidden Patterns” , Dissertation, University of Pennsylvania, 1998. Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 35 / 55

  52. Acknowledgements Professor Sundberg Professor Buckmire Megan Liu Kristin Oberiano all of my friends Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 36 / 55

  53. Backup Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 37 / 55

  54. Original Solution a pile of height ( n − 1) glued together with a row of (2n-1) cubes Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 38 / 55

  55. Original Solution a pile of height ( n − 1) glued together with a row of (2n-1) cubes Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 38 / 55

  56. Original Solution a pile of height ( n − 1) glued together with a row of (2n-1) cubes 4(2 n − 1) + 2 = 8 n − 2 Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 38 / 55

  57. Original Solution a pile of height ( n − 1) glued together with a row of (2n-1) cubes 4(2 n − 1) + 2 = 8 n − 2 2 n − 3 sides of cubes overlap (8 n − 2) − 2(2 n − 3) = 4 n + 4 Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 38 / 55

  58. Original Solution a pile of height ( n − 1) glued together with a row of (2n-1) cubes 4(2 n − 1) + 2 = 8 n − 2 2 n − 3 sides of cubes overlap (8 n − 2) − 2(2 n − 3) = 4 n + 4 So the surface area increases by 4 n + 4, when we go from a pile of height ( n − 1) to a pile of height n , thus SA ( n ) − SA ( n − 1) = 4 n + 4 for n ≥ 2. Using this recurrence and the initial condition SA (1) = 6, we can prove that SA ( n ) = 2 n 2 + 6 n − 2 for n ≥ 1 . Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 38 / 55

  59. Original Solution (Inductive Proof) Proof by Induction: Prove: SA ( n ) = 2 n 2 + 6 n − 2 for n ≥ 1 Basis: n = 1 2(1) 2 + 6(1) − 2 = 2 + 6 − 2 = 6 = SA (1) � Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 39 / 55

  60. Original Solution (Inductive Proof) Let’s apply our recurrence ( SA ( n ) − SA ( n − 1) = 4 n + 4) to SA ( n + 1) − SA ( n ), i.e., SA ( n + 1) − SA ( n ) = 4( n + 1) + 4, thus, SA ( n + 1) = SA ( n ) + 4( n + 1) + 4. By induction, SA ( n ) = 2 n 2 + 6 n − 2, thus, SA ( n + 1) = 2 n 2 + 6 n − 2 + (4( n + 1) + 4) = 2 n 2 + 6 n − 2 + 4 n + 4 + 4 = 2 n 2 + 4 n + 2 + 6 n + 6 − 2 = 2( n + 1) 2 + 6( n + 1) − 2 therefore, by the Principle of Mathematical Induction, SA ( n ) = 2 n 2 + 6 n − 2 for n ≥ 1. Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 40 / 55

  61. Permutation Patterns For example, S 2 n ( { 132 , 231 , 2134 } ) = S 2 n (132 , 231 , 2134) will be: When n =2 S 2 2 (132 , 231 , 2134) = { 1122 , 1212 , 1221 , 2112 , 2121 , 2211 } When n =3 S 2 3 (132 , 231 , 2134) = { 112233 , 121233 , 122133 , 211233 , 212133 , 221133 , 311223 , 312123 , 312213 , 321123 , 332211 321213 , 322113 , 331122 , 331212 , 331221 , 332112 332121 } Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 41 / 55

  62. 1) “between” the digits of p ′ LEMMA 1: | S 2 n (132 , 231 , 213) | = 2 n + 2 for n ≥ 2 Case 3: a = b > c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 42 / 55

  63. 1) “between” the digits of p ′ LEMMA 1: | S 2 n (132 , 231 , 213) | = 2 n + 2 for n ≥ 2 Case 3: a = b > c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 42 / 55

  64. 1) “between” the digits of p ′ LEMMA 1: | S 2 n (132 , 231 , 213) | = 2 n + 2 for n ≥ 2 Case 3: a = b > c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 42 / 55

  65. 1) “between” the digits of p ′ LEMMA 1: | S 2 n (132 , 231 , 213) | = 2 n + 2 for n ≥ 2 Case 3: a = b > c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 42 / 55

  66. 1) “between” the digits of p ′ LEMMA 1: | S 2 n (132 , 231 , 213) | = 2 n + 2 for n ≥ 2 Case 3: a = b > c Hana Mizuno (Occidental College) Stacking Blocks December 3rd, 2015 42 / 55

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CEAS AIAE 3AF - AIAE - AIDAA- AAAR-DGLR - FTF - HAES - NVvL- RAeS - SVFW - TsAGI - PSAA - CZAeS

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