Monochromatic 4-term arithmetic progressions in 2-colorings of Z n

Transcription

1 Monochromatic 4-term arithmetic progressions in 2-colorings of Z n Linyuan Lu Xing Peng University of South Carolina Integers Conference, Carrollton, GA, October 26-29, 2011.

2 Three modules We will consider monochromatic k-term Arithmetic Progressions (k = 3, 4, 5) in the following: [n] Z p Z n 2 / 32

3 Monochromatic progressions We consider the monochromatic 3-term arithmetic progression (3-APs) in a 2-coloring of first 9 integers: / 32

4 Monochromatic progressions We consider the monochromatic 3-term arithmetic progression (3-APs) in a 2-coloring of first 9 integers: Red 3-APs: 2, 5, 8 3 / 32

5 Monochromatic progressions We consider the monochromatic 3-term arithmetic progression (3-APs) in a 2-coloring of first 9 integers: Red 3-APs: 2, 5, 8 Blue 3-APs: 3, 6, 9 3 / 32

6 Monochromatic progressions We consider the monochromatic 3-term arithmetic progression (3-APs) in a 2-coloring of first 9 integers: Red 3-APs: 2, 5, 8 Blue 3-APs: 3, 6, 9 Here is an alternative representation using 0-1 string / 32

7 Van der Waerden s theorem Theorem [Van der Waerden, 1927] For any given positive integers r and k, there is some number N such that if the integers in [N] = {1, 2,..., N} are colored, each with one of r different colors, then there are at least k integers in arithmetic progression all of the same color / 32

8 Van der Waerden s theorem Theorem [Van der Waerden, 1927] For any given positive integers r and k, there is some number N such that if the integers in [N] = {1, 2,..., N} are colored, each with one of r different colors, then there are at least k integers in arithmetic progression all of the same color The least such N is the Van der Waerden number W(r, k). W(2, 3) = 9. 4 / 32

9 Van der Waerden numbers Some values of W(r, k) r\ k ,132 > 3, > 292 > 2, 173 > 11, 191 > 48, > 1, 048 > 17, 705 > 91, 331 > 420, > 170 > 2, 254 > 98, 740 > 540, > 223 > 9, 778 > 98, 748 > 816, / 32

10 Van der Waerden numbers Some values of W(r, k) r\ k ,132 > 3, > 292 > 2, 173 > 11, 191 > 48, > 1, 048 > 17, 705 > 91, 331 > 420, > 170 > 2, 254 > 98, 740 > 540, > 223 > 9, 778 > 98, 748 > 816, 981 Gowers [2001] 2 k+9 W(r, k) 2 2r2 5 / 32

11 Monochromatic k-aps For any fixed r and k, let c r,k be the greatest number such that any r-coloring of [n] (for n sufficiently large) contains at least (c r,k + o(1))n 2 monochromatic k-aps. Question: What can we say about c r,k? 6 / 32

12 An upper bound on c r,k The set [n] has ( 1 2(k 1) + o(1))n2 k-aps. If we colors [n] randomly using r colors, then each k-ap being monochromatic has the probability 1 r k 1. 7 / 32

13 An upper bound on c r,k The set [n] has ( 1 2(k 1) + o(1))n2 k-aps. If we colors [n] randomly using r colors, then each k-ap being monochromatic has the probability 1 r k 1. The expected number of monochromatic k-aps is ( 1 r k 1 + o(1)) 1 2(k 1) n2. 7 / 32

14 An upper bound on c r,k The set [n] has ( 1 2(k 1) + o(1))n2 k-aps. If we colors [n] randomly using r colors, then each k-ap being monochromatic has the probability 1 r k 1. The expected number of monochromatic k-aps is ( 1 r k 1 + o(1)) 1 2(k 1) n2. We have c r,k 1 + o(1). 2(k 1)rk 1 7 / 32

15 A lower bound on c r,k Let K := W(r, k). There are about 1 2(K 1) n2 of K-APs in [n]. 8 / 32

16 A lower bound on c r,k Let K := W(r, k). There are about 1 2(K 1) n2 of K-APs in [n]. Every K-AP contains at least one monochromatic k-ap. 8 / 32

17 A lower bound on c r,k Let K := W(r, k). There are about 1 2(K 1) n2 of K-APs in [n]. Every K-AP contains at least one monochromatic k-ap. Each monochromatic k-ap is in at most K 2 of K-APs. 8 / 32

18 A lower bound on c r,k Let K := W(r, k). There are about 1 2(K 1) n2 of K-APs in [n]. Every K-AP contains at least one monochromatic k-ap. Each monochromatic k-ap is in at most K 2 of K-APs. Putting together, we have c r,k 1 2W(r, k) 3 > 0. 8 / 32

19 Questions What is the best constant c r,k? 9 / 32

20 Questions What is the best constant c r,k? Does random coloring have asymptotically the least number of monochromatic APs? 9 / 32

21 Questions What is the best constant c r,k? Does random coloring have asymptotically the least number of monochromatic APs? In this talk, we we only consider the case using 2 colors. Write c k for c 2,k. 9 / 32

22 History Parillo, Robertson, and Saracino [2007] proved c / 32

23 History Parillo, Robertson, and Saracino [2007] proved c There is a 2-coloring on [n], which has fewer monochromatic 3-APs than a random 2-coloring. 10 / 32

24 History Parillo, Robertson, and Saracino [2007] proved c There is a 2-coloring on [n], which has fewer monochromatic 3-APs than a random 2-coloring. Butler, Constello, and Graham [2010] proved c , c / 32

25 Our results on [n] Theorem 1 [Lu-Peng 2011]: c 4 1 = , 72 c 5 1 = / 32

26 Monochromatic APs on Z n Z n : the cyclic group of order n. 12 / 32

27 Monochromatic APs on Z n Z n : the cyclic group of order n. k-ap : a, a + d,...,a + (k 1)d for some a, d Z n. 12 / 32

28 Monochromatic APs on Z n Z n : the cyclic group of order n. k-ap : for some a, d Z n. a, a + d,...,a + (k 1)d Here we allow degenerated k-aps, i.e., d = n i for some i {1,...,k 1}. 12 / 32

29 Monochromatic APs on Z n Z n : the cyclic group of order n. k-ap : for some a, d Z n. a, a + d,...,a + (k 1)d Here we allow degenerated k-aps, i.e., d = n i for some i {1,...,k 1}. The total number of k-aps in Z n is n / 32

30 Definition of m k Let m k (n) be the largest number such that any 2-coloring of Z n has at least m k (n)n 2 monochromatic k-aps. 13 / 32

31 Definition of m k Let m k (n) be the largest number such that any 2-coloring of Z n has at least m k (n)n 2 monochromatic k-aps. A random 2-coloring of Z n has ( 1 2 k 1 + o(1))n 2 monochromatic k-aps. 13 / 32

32 Definition of m k Let m k (n) be the largest number such that any 2-coloring of Z n has at least m k (n)n 2 monochromatic k-aps. A random 2-coloring of Z n has ( 1 2 k 1 + o(1))n 2 monochromatic k-aps. We have m k (n) 1 2 k 1 + o n(1). 13 / 32

33 3-APs in Z n Fact: For n sufficiently large, we have m 3 (n) = o(1). 14 / 32

34 3-APs in Z n Fact: For n sufficiently large, we have m 3 (n) = o(1). I.e., random colorings have asymptotically fewer number of APs. 14 / 32

35 The value m 4 (Z p ) for prime p Wolf [2010] proved for sufficient large prime p o(1) m 4(Z p ) 1 8 ( ) + o(1). 15 / 32

36 The value m 4 (Z p ) for prime p Wolf [2010] proved for sufficient large prime p o(1) m 4(Z p ) 1 8 ( ) + o(1). This result shows that random colorings are not optimal. It uses quadratic Fourier analysis and probabilistic method. 15 / 32

37 Our result on Z p Theorem 2 [Lu-Peng 2011]: For sufficient large prime p, we have o(1) m 4(Z p ) o(1). 16 / 32

38 Our result on Z p Theorem 2 [Lu-Peng 2011]: For sufficient large prime p, we have o(1) m 4(Z p ) o(1). We increase the lower bound by a factor of / 32

39 Our result on Z p Theorem 2 [Lu-Peng 2011]: For sufficient large prime p, we have o(1) m 4(Z p ) o(1). We increase the lower bound by a factor of 1 6. The gap 1 8 m 4(Z p ) is increased from Wolf s bound to / 32

40 Our result on Z p Theorem 2 [Lu-Peng 2011]: For sufficient large prime p, we have o(1) m 4(Z p ) o(1). We increase the lower bound by a factor of 1 6. The gap 1 8 m 4(Z p ) is increased from Wolf s bound to Our upper bound uses an explicit construction. 16 / 32

41 Related results Theorem [Cameron-Cilleruelo-Serra (2003)] Any 2-coloring of a finite abelian graph with order n relatively prime to 6 has at least monochromatic APs. ( o(1))n2 17 / 32

42 Related results Theorem [Cameron-Cilleruelo-Serra (2003)] Any 2-coloring of a finite abelian graph with order n relatively prime to 6 has at least monochromatic APs. ( o(1))n2 It implies for if gcd(n, 6) = 1 then m 4 (Z n ) o(1). 17 / 32

43 Lower bound on m 4 (Z n ) Theorem 2 [Lu-Peng 2011]: If n is not divisible by 4 and large enough, then we have m 4 (n) o(1). 18 / 32

44 Lower bound on m 4 (Z n ) Theorem 2 [Lu-Peng 2011]: If n is not divisible by 4 and large enough, then we have m 4 (n) o(1). If n is divisible by 4 and large enough, then we have m 4 (Z n ) o(1). 18 / 32

45 Upper bound on m 4 (Z n ) Theorem 3 [Lu-Peng 2011]: If 20 n, then m 4 (n) = / 32

46 Upper bound on m 4 (Z n ) Theorem 3 [Lu-Peng 2011]: If 20 n, then m 4 (n) = If 22 n, then m 4 (Z n ) < / 32

47 Upper bound on m 4 (Z n ) Theorem 3 [Lu-Peng 2011]: If 20 n, then m 4 (n) = If 22 n, then m 4 (Z n ) < If n is odd, then m 4 (Z n ) o(1) < o(1). 19 / 32

48 Upper bound on m 4 (Z n ) Theorem 3 [Lu-Peng 2011]: If 20 n, then m 4 (n) = If 22 n, then m 4 (Z n ) < If n is odd, then m 4 (Z n ) o(1) < o(1). If n is even, then m 4 (Z n ) o(1) < o(1). 19 / 32

49 Upper bound on m 4 (Z n ) Theorem 3 [Lu-Peng 2011]: If 20 n, then m 4 (n) = If 22 n, then m 4 (Z n ) < If n is odd, then m 4 (Z n ) o(1) < o(1). If n is even, then m 4 (Z n ) o(1) < o(1). For sufficiently large n, there is a 2-coloring of Z n with substantially fewer monochromatic APs than random 2-colorings have. 19 / 32

50 Block construction method Let n = bt and B be a good 2-coloring/bit-string in Z b with x monochromatic 4-APs. Consider a 2-coloring of Z b defined as follows. BB B }{{} t 20 / 32

51 Block construction method Let n = bt and B be a good 2-coloring/bit-string in Z b with x monochromatic 4-APs. Consider a 2-coloring of Z b defined as follows. BB B }{{} t The number of monochromatic 4-APs in this coloring is exactly xt / 32

52 Block construction method Let n = bt and B be a good 2-coloring/bit-string in Z b with x monochromatic 4-APs. Consider a 2-coloring of Z b defined as follows. BB B }{{} t The number of monochromatic 4-APs in this coloring is exactly In particular, if b n, we have xt 2. m 4 (Z n ) m 4 (Z b ). 20 / 32

53 Block construction with extra bits Let n = bt + r (0 < r < b) and B be a good 2-coloring/bit-string in Z b. How many monochromatic 4-APs in the following construction? BB }{{ B} R t Here R is any bit string of length r. 21 / 32

54 Block construction with extra bits Let n = bt + r (0 < r < b) and B be a good 2-coloring/bit-string in Z b. How many monochromatic 4-APs in the following construction? BB }{{ B} R t Here R is any bit string of length r. The number of all 4-APs which pass through some bit(s) in R is O(n). The major term in the number of all monochromatic 4-APs only depends on B and r. 21 / 32

55 Classes of 4-APs We divide the set of all 4-APs in Z n into 8 classes. type meaning 0 a < a + d < a + 2d < a + 3d < n 1 a < a + d < a + 2d < n < a + 3d < 2n 2 a < a + d < n < a + 2d < a + 3d < 2n 3 a < a + d < n < a + 2d < 2n < a + 3d < 3n 4 a < n < a + d < a + 2d < a + 3d < 2n 5 a < n < a + d < a + 2d < 2n < a + 3d < 3n 6 a < n < a + d < 2n < a + 2d < a + 3d < 3n 7 a < n < a + d < 2n < a + 2d < 3n < a + 3d < 4n 22 / 32

56 A graphical view of 8 classes Every 4-AP a, a + d,...,a + (k 1)d is determined by a pair (a, d). The 8 classes can be viewed as 8 regions shown below. d n 7 2n/ n/3 n/2 2 5 n/2 n/3 1 n/ n a 23 / 32

57 The number of 4-APs The number of 4-APs in each class is proportional to the area a i of the corresponding i-th region as shown below. d/n 1 1/6 2/3 1/12 1/6 1/12 2/3 1/2 1/2 1/6 1/3 1/12 1/12 1/3 1/6 0 1 a/n 24 / 32

58 A lemma For 0 i 7, write i as a bit-string x 1 x 2 x 3. Let c i be the number of sequences in B of form a, a + d + x 1 r, a + 2d + x 2 r, a + 3d + x 3 r. Then the number of monochromatic 4-APs in BB BR is 7 i=0 a i c i t 2 + O(t). In particular, we have m 4 (Z n ) 7 i=0 a i c i b / 32

59 A good 2-coloring in Z 20 In Z 20, consider the 2-coloring given by 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, / 32

60 A good 2-coloring in Z 20 In Z 20, consider the 2-coloring given by 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0. For r = 1, we have type d i / 32

61 A good 2-coloring in Z 20 In Z 20, consider the 2-coloring given by 1, 1, 1, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0. For r = 1, we have This implies type d i m 4 (Z 20k+1 ) o(1). In fact, the same bound works for all odd r. 26 / 32

62 A special 2-coloring of Z 11 In Z 11, consider the 2-coloring given by could be either 0 or 1. B 11 := ( ). 27 / 32

63 A special 2-coloring of Z 11 In Z 11, consider the 2-coloring given by could be either 0 or 1. B 11 := ( ). Property: B 11 contains no non-degenerate monochromatic 4-APs of Z / 32

64 A recursive construction Given a 2-coloring B t of Z t, define B 11 B t to be the following 2-coloring of Z 11t B } 11 B 11 {{ B 11 }, t where t s are replaced by B t. 28 / 32

65 A recursive construction Given a 2-coloring B t of Z t, define B 11 B t to be the following 2-coloring of Z 11t B } 11 B 11 {{ B 11 }, t where t s are replaced by B t. Property: m 4 (Z 11t ) 10 + m 4(Z t ) / 32

66 The limit m 4 (Z 11 s) s / 32

67 The limit m 4 (Z 11 s) s 1. lim n m 4 (Z n ) / 32

68 The limit m 4 (Z 11 s) s 1. lim n m 4 (Z n ) Corollary: c lim n m 4 (Z n ) / 32

69 The limit Corollary: m 4 (Z 11 s) s 1. lim n c lim n m 4 (Z n ) m 4 (Z n ) We conjecture that the equality holds. 29 / 32

70 k-aps (for k > 4) A special 2-coloring exists in Z 37 : B 37 = ( ). 30 / 32

71 k-aps (for k > 4) A special 2-coloring exists in Z 37 : B 37 = ( ). Corollary: c lim n m 5 (Z n ) / 32

72 k-aps (for k > 4) A special 2-coloring exists in Z 37 : B 37 = ( ). Corollary: c lim n m 5 (Z n ) Bulter [2011+]: found a coloring in Z 47 for 6-AP and a coloring in Z 77 for 7-AP. Thus, c and c / 32

73 Lower bounds on m 4 (Z n ) Main idea for Lower bounds: If n is not divisible by 4, then we are able to extend Wolf s proof to Z n. At the same time, we capture the patterns which are thrown away in Wolf s paper. Finally, we use heuristic search to show those patterns should appear with positive density. 31 / 32

74 Lower bounds on m 4 (Z n ) Main idea for Lower bounds: If n is not divisible by 4, then we are able to extend Wolf s proof to Z n. At the same time, we capture the patterns which are thrown away in Wolf s paper. Finally, we use heuristic search to show those patterns should appear with positive density. If n is divisible by 4, then Wolf s proof can not be extended to Z n. However, we are able to extend Cameron-Cilleruelo-Serra s argument to get a lower bound. 31 / 32

75 Questions Conjecture: For k 4, c k = 1 2(k 1) lim n m k (Z n ). 32 / 32

76 Questions Conjecture: For k 4, c k = 1 2(k 1) lim n m k (Z n ). Is c 4 = 1 72? 32 / 32

77 Questions Conjecture: For k 4, c k = 1 2(k 1) lim n m k (Z n ). Is c 4 = 1 72? Is c 5 = 1 304? 32 / 32

78 Questions Conjecture: For k 4, c k = 1 2(k 1) lim n m k (Z n ). Is c 4 = 1 72? Is c 5 = 1 304? Determine m 4 (Z p ) for large p. Currently, we have m 4 (Z p ) / 32

79 Questions Conjecture: For k 4, c k = 1 2(k 1) lim n m k (Z n ). Is c 4 = 1 72? Is c 5 = 1 304? Determine m 4 (Z p ) for large p. Currently, we have m 4 (Z p ) Good lower bound for c 4? 32 / 32