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Background Four letters Five letters Conclusion Circular repetition thresholds for small alphabets: Last cases of Gorbunovas Conjecture Lucas Mol Joint work with James D. Currie and Narad Rampersad Prairie Discrete Math Workshop Brandon

  1. Background Four letters Five letters Conclusion Progress on Dejean’s Conjecture k = 3 Dejean 1972 k = 4 Pansiot 1984 5 ≤ k ≤ 11 Moulin-Ollagnier 1992 12 ≤ k ≤ 14 Currie, Mohammad-Noori 2004 27 ≤ k ≤ 32 Currie, Rampersad 2008 k ≥ 33 Carpi 2007

  2. Background Four letters Five letters Conclusion Progress on Dejean’s Conjecture k = 3 Dejean 1972 k = 4 Pansiot 1984 5 ≤ k ≤ 11 Moulin-Ollagnier 1992 12 ≤ k ≤ 14 Currie, Mohammad-Noori 2004 15 ≤ k ≤ 26 Rao and Currie, Rampersad 2009 27 ≤ k ≤ 32 Currie, Rampersad 2008 k ≥ 33 Carpi 2007

  3. Background Four letters Five letters Conclusion Progress on Dejean’s Conjecture k = 3 Dejean 1972 k = 4 Pansiot 1984 5 ≤ k ≤ 11 Moulin-Ollagnier 1992 12 ≤ k ≤ 14 Currie, Mohammad-Noori 2004 15 ≤ k ≤ 26 Rao and Currie, Rampersad 2009 27 ≤ k ≤ 32 Currie, Rampersad 2008 k ≥ 33 Carpi 2007

  4. Background Four letters Five letters Conclusion Circular words

  5. Background Four letters Five letters Conclusion Circular words • Intuitively, a circular word is obtained from a linear word by linking the ends, giving a cyclic sequence of letters.

  6. Background Four letters Five letters Conclusion Circular words • Intuitively, a circular word is obtained from a linear word by linking the ends, giving a cyclic sequence of letters. • Factors don’t “wrap around” more than once.

  7. Background Four letters Five letters Conclusion Circular words • Intuitively, a circular word is obtained from a linear word by linking the ends, giving a cyclic sequence of letters. • Factors don’t “wrap around” more than once. • i.e. The longest factors of a circular word of length n have length n .

  8. Background Four letters Five letters Conclusion Circular words • Intuitively, a circular word is obtained from a linear word by linking the ends, giving a cyclic sequence of letters. • Factors don’t “wrap around” more than once. • i.e. The longest factors of a circular word of length n have length n . • As a linear word, onion is 2 -free.

  9. Background Four letters Five letters Conclusion Circular words • Intuitively, a circular word is obtained from a linear word by linking the ends, giving a cyclic sequence of letters. • Factors don’t “wrap around” more than once. • i.e. The longest factors of a circular word of length n have length n . • As a linear word, onion is 2 -free. • However, the circular word ( onion ) has factor onon , so it is not 2 -free.

  10. Background Four letters Five letters Conclusion Circular Repetition Threshold Definition Let k ≥ 2 . The circular repetition threshold for k letters, denoted CRT( k ) , is the infimum of the set of all β such that there are β -free circular words of every length on k letters.

  11. Background Four letters Five letters Conclusion Known values of the circular repetition threshold

  12. Background Four letters Five letters Conclusion Known values of the circular repetition threshold • CRT(2) = 5 2 (Aberkane, Currie, 2004)

  13. Background Four letters Five letters Conclusion Known values of the circular repetition threshold • CRT(2) = 5 2 (Aberkane, Currie, 2004) • CRT(3) = 2 (Currie, 2002)

  14. Background Four letters Five letters Conclusion Known values of the circular repetition threshold • CRT(2) = 5 2 (Aberkane, Currie, 2004) • CRT(3) = 2 (Currie, 2002) Conjecture (Gorbunova, 2012) For all k ≥ 4 , CRT( k ) = ⌈ k/ 2 ⌉ + 1 ⌈ k/ 2 ⌉

  15. Background Four letters Five letters Conclusion Known values of the circular repetition threshold • CRT(2) = 5 2 (Aberkane, Currie, 2004) • CRT(3) = 2 (Currie, 2002) Conjecture (Gorbunova, 2012) For all k ≥ 4 , CRT( k ) = ⌈ k/ 2 ⌉ + 1 ⌈ k/ 2 ⌉ • Gorbunova confirmed her conjecture for all k ≥ 6 .

  16. Background Four letters Five letters Conclusion Known values of the circular repetition threshold • CRT(2) = 5 2 (Aberkane, Currie, 2004) • CRT(3) = 2 (Currie, 2002) Conjecture (Gorbunova, 2012) For all k ≥ 4 , CRT( k ) = ⌈ k/ 2 ⌉ + 1 ⌈ k/ 2 ⌉ • Gorbunova confirmed her conjecture for all k ≥ 6 . • Last remaining cases: CRT(4) and CRT(5) .

  17. Background Four letters Five letters Conclusion Last Cases of Gorbunova’s Conjecture k RT( k ) CRT( k ) 2 2 5 / 2 3 7 / 4 2 4 7 / 5 3 / 2 5 5 / 4 4 / 3 6 6 / 5 4 / 3 7 7 / 6 5 / 4 8 8 / 7 5 / 4 9 9 / 8 6 / 5 10 10 / 9 6 / 5

  18. Background Four letters Five letters Conclusion The lower bound

  19. Background Four letters Five letters Conclusion The lower bound Proposition (Gorbunova, 2012) For any k ≥ 4 , there are no circular ⌈ k/ 2 ⌉ +1 ⌈ k/ 2 ⌉ -free words of length k + 1 over a k letter alphabet.

  20. Background Four letters Five letters Conclusion The lower bound Proposition (Gorbunova, 2012) For any k ≥ 4 , there are no circular ⌈ k/ 2 ⌉ +1 ⌈ k/ 2 ⌉ -free words of length k + 1 over a k letter alphabet. Sketch of Proof.

  21. Background Four letters Five letters Conclusion The lower bound Proposition (Gorbunova, 2012) For any k ≥ 4 , there are no circular ⌈ k/ 2 ⌉ +1 ⌈ k/ 2 ⌉ -free words of length k + 1 over a k letter alphabet. Sketch of Proof. • Pigeonhole principle.

  22. Background Four letters Five letters Conclusion The lower bound Proposition (Gorbunova, 2012) For any k ≥ 4 , there are no circular ⌈ k/ 2 ⌉ +1 ⌈ k/ 2 ⌉ -free words of length k + 1 over a k letter alphabet. Sketch of Proof. • Pigeonhole principle. So to prove the last two cases of Gorbunova’s Conjecture, it suffices to find

  23. Background Four letters Five letters Conclusion The lower bound Proposition (Gorbunova, 2012) For any k ≥ 4 , there are no circular ⌈ k/ 2 ⌉ +1 ⌈ k/ 2 ⌉ -free words of length k + 1 over a k letter alphabet. Sketch of Proof. • Pigeonhole principle. So to prove the last two cases of Gorbunova’s Conjecture, it suffices to find + -free circular words of every length on 4 letters, and • 3 2

  24. Background Four letters Five letters Conclusion The lower bound Proposition (Gorbunova, 2012) For any k ≥ 4 , there are no circular ⌈ k/ 2 ⌉ +1 ⌈ k/ 2 ⌉ -free words of length k + 1 over a k letter alphabet. Sketch of Proof. • Pigeonhole principle. So to prove the last two cases of Gorbunova’s Conjecture, it suffices to find + -free circular words of every length on 4 letters, and • 3 2 + -free circular words of every length on 5 letters. • 4 3

  25. Background Four letters Five letters Conclusion Plan Background Four letters Five letters Conclusion

  26. Background Four letters Five letters Conclusion Morphisms

  27. Background Four letters Five letters Conclusion Morphisms • An r -uniform morphism takes a word as input and replaces every letter by a word of length r .

  28. Background Four letters Five letters Conclusion Morphisms • An r -uniform morphism takes a word as input and replaces every letter by a word of length r . • A morphism f preserves β -freeness if f ( w ) is β -free whenever w is β -free.

  29. Background Four letters Five letters Conclusion Morphisms • An r -uniform morphism takes a word as input and replaces every letter by a word of length r . • A morphism f preserves β -freeness if f ( w ) is β -free whenever w is β -free. • Iterating gives β -free words of arbitrarily long length.

  30. Background Four letters Five letters Conclusion Four letters Idea:

  31. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words.

  32. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem:

  33. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem: • If a linear word is β -free, then so are all of its factors.

  34. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem: • If a linear word is β -free, then so are all of its factors. • This is not the case for circular words.

  35. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem: • If a linear word is β -free, then so are all of its factors. • This is not the case for circular words. • e.g. ( discrete ) is 2 -free, but ( ete ) is not.

  36. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem: • If a linear word is β -free, then so are all of its factors. • This is not the case for circular words. • e.g. ( discrete ) is 2 -free, but ( ete ) is not. • Starting with a single letter, and iteratively applying an r -uniform morphism only gives words of length r n .

  37. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem: • If a linear word is β -free, then so are all of its factors. • This is not the case for circular words. • e.g. ( discrete ) is 2 -free, but ( ete ) is not. • Starting with a single letter, and iteratively applying an r -uniform morphism only gives words of length r n . Solution:

  38. Background Four letters Five letters Conclusion Four letters Idea: + -freeness for • Find a uniform morphism that preserves 3 2 circular words. Problem: • If a linear word is β -free, then so are all of its factors. • This is not the case for circular words. • e.g. ( discrete ) is 2 -free, but ( ete ) is not. • Starting with a single letter, and iteratively applying an r -uniform morphism only gives words of length r n . Solution: • Use two different morphisms: an r -uniform morphism and an s -uniform morphism (where r and s are relatively prime).

  39. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2

  40. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Find a 9 -uniform morphism f 9 and an 11 -uniform morphism + -freeness. f 11 that preserve 3 2

  41. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Find a 9 -uniform morphism f 9 and an 11 -uniform morphism + -freeness. f 11 that preserve 3 2 • Define f 9 by: 0 �→ 012132310 1 �→ 123203021 2 �→ 230310132 3 �→ 301021203

  42. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Find a 9 -uniform morphism f 9 and an 11 -uniform morphism + -freeness. f 11 that preserve 3 2 • Define f 9 by: 0 �→ 012132310 1 �→ 123203021 2 �→ 230310132 3 �→ 301021203 • Define f 11 by: 0 �→ 01213231210 1 �→ 12320302321 2 �→ 23031013032 3 �→ 30102120103

  43. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2

  44. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Use a strong inductive argument.

  45. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Use a strong inductive argument. • Find some short words by computer search to get things started.

  46. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Use a strong inductive argument. • Find some short words by computer search to get things started. + -free circular word of every length • Assume we have found a 3 2 less than n .

  47. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Use a strong inductive argument. • Find some short words by computer search to get things started. + -free circular word of every length • Assume we have found a 3 2 less than n . • For n sufficiently large, using the Postage Stamp Lemma, we can write n = 9 k + 11 ℓ, for k ≥ 8 and 2 ≤ ℓ ≤ 10 .

  48. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Use a strong inductive argument. • Find some short words by computer search to get things started. + -free circular word of every length • Assume we have found a 3 2 less than n . • For n sufficiently large, using the Postage Stamp Lemma, we can write n = 9 k + 11 ℓ, for k ≥ 8 and 2 ≤ ℓ ≤ 10 . + -free circular word ( w ) of length k + ℓ , and write it • Take a 3 2 as w = uv , where | u | = k and | v | = ℓ .

  49. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 • Use a strong inductive argument. • Find some short words by computer search to get things started. + -free circular word of every length • Assume we have found a 3 2 less than n . • For n sufficiently large, using the Postage Stamp Lemma, we can write n = 9 k + 11 ℓ, for k ≥ 8 and 2 ≤ ℓ ≤ 10 . + -free circular word ( w ) of length k + ℓ , and write it • Take a 3 2 as w = uv , where | u | = k and | v | = ℓ . + -free. • Claim: ( f 9 ( u ) f 11 ( v )) is 3 2

  50. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 .

  51. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | .

  52. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) .

  53. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases:

  54. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 9 ( u ) y x x

  55. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 9 ( u ) y x x

  56. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 11 ( v ) y x x

  57. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 11 ( v ) y x x

  58. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 9 ( u ) f 11 ( v ) y x x

  59. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 9 ( u ) f 11 ( v ) y x x

  60. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 11 ( v ) f 9 ( u ) y x x

  61. Background Four letters Five letters Conclusion + -free words on four letters Constructing circular 3 2 Sketch of Proof. Suppose otherwise that ( f 9 ( u ) f 11 ( v )) contains some factor with exponent greater than 3 2 . • Then ( f 9 ( u ) f 11 ( v )) has some factor of the form xyx , where | x | > | y | . • Argue that if | x | is sufficiently large, it appears in only one of f 9 ( u ) or f 11 ( v ) . • Then have several cases: f 11 ( v ) f 9 ( u ) y x x

  62. Background Four letters Five letters Conclusion Therefore,

  63. Background Four letters Five letters Conclusion Therefore, CRT(4) = 3 2 .

  64. Background Four letters Five letters Conclusion Plan Background Four letters Five letters Conclusion

  65. Background Four letters Five letters Conclusion Gorbunova’s technique for larger alphabets • e.g. seven letter alphabet { 0 , 1 , 2 , 3 , 4 , 5 , 6 } . k RT( k ) CRT( k ) 2 2 5 / 2 3 7 / 4 2 4 7 / 5 3 / 2 5 5 / 4 4 / 3 6 6 / 5 4 / 3 7 7 / 6 5 / 4 8 8 / 7 5 / 4 9 9 / 8 6 / 5 10 10 / 9 6 / 5

  66. Background Four letters Five letters Conclusion Gorbunova’s technique for larger alphabets • e.g. seven letter alphabet { 0 , 1 , 2 , 3 , 4 , 5 , 6 } . k RT( k ) CRT( k ) 2 2 5 / 2 3 7 / 4 2 4 7 / 5 3 / 2 5 5 / 4 4 / 3 6 6 / 5 4 / 3 7 7 / 6 5 / 4 8 8 / 7 5 / 4 9 9 / 8 6 / 5 10 10 / 9 6 / 5

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