1. 1 Tighter bounds and simpler constructions for Davenport-Schinzel sequences Haim Kaplan, Gabriel Nivasch, Micha Sharir Tel-Aviv University

2. 2 What are Davenport-Schinzel sequences? A Davenport-Schinzel sequence of order s on n symbols does not contain adjacent equal symbols (aa) does not contain any alternating subsequence a … b … of length s + 2. Examples:a c d b c c aNot a DS sequence a c a b a d a c d c bNot a DS seq. of order 3

3. 3 Motivation: Lower envelopes Suppose we have n x-monotone curves in the plane. Each pair of curves intersects at most s times. Then the lower envelope corresponds to a DS sequence of order s. Why: An alternation of length s + 2 would imply s + 1 intersections. a b c d dcabd abab s = 2

4. 4 Motivation: Lower envelopes What if we have n segments of x-monotone curves? (Each pair of curves intersects at most s times.) Then we get a DS sequence of order s + 2. Example: Straight-line segments. Alternation ababa is impossible: a b a b a Straight-line segments give DS sequences of order 3.

5. 5 Bounds for DS sequences Let λ s (n) be the maximum length of a DS sequence of order s on n symbols. (We take s to be a constant.) Max sequence: 1 2 3 … n Max sequence: 1 2 3 … n … 3 2 1 or: 1 2 1 3 1 … 1 n 1 s = 1 (no aba): s = 2 (no abab): What about s ≥ 3?Not so simple anymore! Involves inverse Ackermann function.

6. 6 The inverse Ackermann function Inverse Ackermann hierarchy: # times you have to apply α k–1, starting from x, until you get a result ≤ 1. We have:etc. Each function in the hierarchy grows much slower than the preceding one.

7. 7 The inverse Ackermann function For every fixed x ≥ 6, the sequence decreases strictly, until it settles at 3. Example: x = 9876!: Inverse Ackermann function: We havefor every fixed k.

8. 8 For s ≥ 4: [Agarwal, Sharir, Shor ’89] Bounds for s ≥ 3 [Hart, Sharir ’86; Wiernik, Sharir ’88]

9. 9 Stabbing interval chains (a completely unrelated problem) [Alon, Kaplan, N., Sharir, Smorodinsky ’08]: Given parameters j, k, and n, An interval chain of length k (or a k-chain) is a sequence of k consecutive, disjoint, nonempty intervals in [1,n]. 10-chain stabbed by a 5-tuple A j-tuple stabs a k-chain if its j elements fall on j different intervals. 1 n Problem: Build a family of j-tuples, as small as possible, that stab all k-chains in [1,n]. 10-chain

10. 10 Stabbing interval chains (a completely unrelated problem) Stabbing with 2j-tuples: If we let the number of intervals be then stabbing all k-chains in [1,n] requires 2j-tuples (roughly speaking). Stabbing with (2j+1)-tuples: If we let then stabbing all k-chains in [1,n] requires (2j+1)-tuples (roughly speaking). Solution:

11. 11 Back to DS sequences Current bounds for DS sequences: Conjecture: By analogy from interval chains, the DS bounds should be: Need 2 things: 1/(s–1)! coefficient in the upper bounds. Lower bounds with log for odd s. We will show

12. 12 Back to DS sequences Theorem: We’ve got two different proofs!

13. 13 Bounding DS sequences Let S be a DS sequence of order s on n symbols, of maximum length. Partition S greedily, from left to right, into maximal blocks of order s – 2. Consider a block. s – 1 s-th ababbaab We can’t have a b before the block and an a after the block. Every block is the first or last for at least one symbol. Number of blocks is at most 2n. order s – 2 (Also works for s odd.) (s even)

14. 14 Bounding DS sequences Let be the maximum length of a DS sequence of order s on n symbols that can be partitioned into m or fewer blocks of order r. We just showed: (This is all [ASS89].) Lemma 1: If r = 1 then we write

15. 15 dfcabbcfdbafcab Bounding DS sequences Let S be a max-length sequence of order s on n symbols that is partitionable into m blocks of order r. rrrrr s n 1 distinct symbols n 2 distinct symbols n 3 distinct symbols n 4 distinct symbols n 5 distinct symbols Leave in each block just the first occurrence of each symbol: dgbagefbec Remove adjacent repetitions. (Remove at most m symbols.) n1n1 n2n2 n3n3 n4n4 n5n5 S' dfcab S'' S dbafe gc agebcf

16. 16 Bounding DS sequences rrrrr s n 1 distinct symbols n 2 distinct symbols n 3 distinct symbols n 4 distinct symbols n 5 distinct symbols Let be a nondecreasing function of n such that S Lemma 2:

17. 17 Bounding DS sequences Lemma 1 + Lemma 2: Let be a nondecreasing function of n such that Then, So all we need to do is bound and then substitute m = 2n. (This is all [ASS89].)

18. 18 Recurrence for ψ s (m,n) Let S be a max-length sequence of order s on n symbols that is partitionable into m blocks of order 1. Group the blocks into b layers of m/b blocks each. (b = free parameter) block 1 layer Classify each symbol into: Local – appears in only one layer Global – appears in more than one layer Classify occurrences of global symbols into: Starting (first layer) Middle (intermediate layers) Ending (last layer) aaabbbbbbb startingendingmiddle n i – # symbols local to i-th layer. n* – # global symbols. block m

19. 19 Recurrence for ψ s (m,n) Decompose S into 4 subsequences: T local – local symbols T start – starting global symbols T middle – middle global symbols T end – ending global symbols Remove adjacent repetitions from each seq. (at interface bet. blocks). Removed at most 4m symbols. Get sequences T' local, T' start, T' middle, T' end.

20. 20 Bounding |T' local | The i-th layer in T' local is a DS sequence of order s on n i symbols, partitionable into m/b blocks of order 1. i-th layer

21. 21 Bounding |T' start | (and |T' end |) Each layer in T' start is a DS sequence of order s – 1 on n* symbols. layer Why: Each symbol of T' start appears in S in a subsequent layer. So no layer in T' start can have an alternation of length s + 1. aaabbba s + 1 (s + 2)-nd in S Moreover, all of T' start is a DS sequence of order s – 1 on n* symbols. Why: Each symbol appears in only one layer. Similarly:

22. 22 Bounding |T' middle | Each layer in T' middle is a DS sequence of order s – 2 on n* symbols. layer Why: If a layer in T' middle had an alternation of length s, then S would have an alternation of length s + 2. s (Warning: We can’t say all of T' middle is of order s – 2. A symbol can appear in many layers.) [Agarwal, Sharir, Shor ’89] said: aaabbbaba We can do better!

23. 23 Bounding |T' middle | Let n* i be the number of global symbols that appear in the i-th layer. (So n* i ≤ n* for all i.) We will get a bound on their sum Better bound for |T' middle |: We need to say something about the n* i ’s.

24. 24 Bounding n * 1 + … + n * b Given original sequence S, form U by taking, from each layer, just the first occurrence of each global symbol. Form U' by removing adjacent repetitions from U. We remove at most b symbols. Each layer becomes a block of order 1. i-th layerS n* i global symbols U n* i U'

25. 25 The recurrence Putting it all together, we have just proven: Theorem: Given integers m, s, and n, and given an integer b, there exists a partition of n into and there exist integerssatisfying such that

26. 26 Using the recurrence Theorem: For every s and k there exist integers P s,k and Q s,k, both of the form such that The recurrence yields the following: (Increasing k diminishes α k (m), but makes the coefficients P s,k and Q s,k larger.)

27. 27 Using the recurrence Proof outline: Want to show: Apply induction on s, k, and m.Choose bound by (ind. on s) (ind. on k) (ind. on m)

28. 28 Using the recurrence We get:if (d s and d' s – constants). Initial conditions:Then: In general:

29. 29 Wrapping up If we take k = constant (say k = 10), then ψ s (m,n) is slightly superlinear in m and linear in n. The constants P s,k and Q s,k don’t matter. If we let k grow very slowly with m, namely k = α(m), then the term α k (m) becomes 3. But now P s,k and Q s,k are functions of α(m). where: We get:

30. 30 Wrapping up We have finally bounded ψ s (m,n): Back to λ s (n): where contributes lower- order terms QED

31. 31

32. 32 We promised a second proof! An ADS sequence of order s on m blocks has no alternation of length s + 2, is composed of at most m blocks, where each block contains distinct symbols. We define Almost Davenport-Schinzel (ADS) sequences: We don’t disallow adjacent repetitions at the interface between blocks. But we limit the number of blocks. … c e gg b a …

33. 33 ADS sequences Suppose we have an ADS sequence of order s on m blocks, in which each symbol appears at least k times (in k different blocks). We turn the problem around: Example 1: We what is the maximium value of n, the number of distinct symbols? If k = s (each symbol must appear s times), then we can have arbitrarily many symbols. … 1 2 3 …… 3 2 11 2 3 …… 3 2 1 block 1block 2block 3block s Maximum alternation is of length s + 1.

34. 34 ADS sequences Example 2: If k = s + 1 (each symbol must appear s + 1 times), then there can be at most symbols. Proof: Suppose there aresymbols. There are only m blocks. There must be two symbols a and b with their s – 1 inner occurrences in the same s – 1 blocks. In the best case: a bb aa bb a a b … Alternation of length s + 2. Contradiction. block 1block 2block 3block s – 1 s

35. 35 ADS sequences Theorem: Let k 1, k 2, k 3 be integers, and let k = k 2 k 3 + 2k 1 – 3k 2 – k 3 + 2. Then, Let n k s (m) be the maximum number of distinct symbols in an ADS sequence of order s on m blocks in which each symbol appears at least k times. where t is a free parameter. For fixed s, n k s (m) is increasing in m and decreasing in k. This is very similar to interval chains. We already know what to do!

36. 36 ADS sequences Proof: Let S be an ADS sequence of order s on m blocks, in which each symbol appears at least k times, with maximum number of distinct symbols. Group the m blocks into m/t layers, with t blocks per layer. # distinct symbols: n k s (m). block 1 layer aaabbbbbbb block m Local symbol – appears in only one layer Global symbol – appears in more than one layer

37. 37 ADS sequences Classify each global symbol into: Left-concentrated: The first k 1 occurrences fall in one layer. Middle-concentrated: Some inner k 2 occurrences fall in one layer, and the symbol also appears in previous and subsequent layers. Right-concentrated: The last k 1 occurrences fall in one layer. Scattered: The symbol appears in at least k 3 different layers. Each global symbol must satisfy at least one of these properties. Why: If a global symbol violates all properties, it occurs at most (k 1 – 1) + (k 3 – 3)(k 2 – 1) + (k 1 – 1) times.But : k = k 2 k 3 + 2k 1 – 3k 2 – k 3 + 2. = k – 1

38. 38 Bounding # local symbols For each layer, take just the symbols local to that layer. # local symbols: At most We get an ADS sequence of order s on t blocks. Each symbol appears k times. Layer contains t blocks. per layer. At mosttotal.

39. 39 Bounding # left-concentrated (and right-concentrated) global symbols For each layer, take just the left-concentrated global symbols that start at that layer. # left-concentrated global symbols: At most We get an ADS sequence of order s – 1 on t blocks. Each symbol appears k 1 times. # right-concentrated global symbols: Same. usual argument

40. 40 Bounding # middle-concentrated global symbols For each layer, take just the middle-concentrated global symbols that appear at least k 2 times in that layer, and aslo appear in previous and subsequent layers. # middle-concentrated global symbols: At most We get an ADS sequence of order s – 2 on t blocks. Each symbol appears k 2 times. usual argument

41. 41 Bounding # scattered global symbols For each scattered global symbol, leave just one occurrence per layer. # scattered global symbols: At most We get an ADS sequence of order s on m/t blocks. Each symbol appears k 3 times. Each layer becomes a block (has distinct symbols). Adding up: QED

42. 42 ADS sequences where:k = k 2 k 3 + 2k 1 – 3k 2 – k 3 + 2. Compare to recurrence for interval chains: where:k = k 2 k 3 + 2k 1 – 2k 2. Same thing basically! Recurrence:

43. 43 ADS sequences For fixed s, as k increases, n k s (m) becomes closer to linear in m. Namely: If we letthen we get If we letthen we get Question: Very nice, but what does this have to do with bounding λ s (n)? The recurrence yields the following:

44. 44 ADS sequences Answer: Let S be a max-length Davenport-Schinzel sequence of order s on n symbols partitionable into m blocks. Regularize the sequence, so that each symbol appears exactly k times. Get sequence S'. (k – free parameter) aaaaaaaaaaaaaaaaaaa'a'a' a'a' a'a' a'a' a '' kkk< k (This doesn’t introduce any forbidden alternations.) We removed at most kn symbols.S' is an ADS sequence on m blocks.

45. 45 ADS sequences Applying our bounds on n k s (m), we get: (d – free parameter) Let d = α(m). Same bounds as before, obtained differently.

46. 46 END OF UPPER BOUNDS

47. 47 Lower bounds for DS sequences [Agarwal, Sharir, Shor ’89]: Construction that gives the lower bound We will show a simpler version of this construction.

48. 48 The Ackermann hierarchy Each function is repeated application of the previous one. We haveetc. Put differently, for k ≥ 2: The Ackermann function:

49. 49 The Ackermann hierarchy The Ackermann hierarchy is robust to perturbations. If we use a modified recurrence relation like The Ackermann recurrence relation: then we will have for some c.

50. 50 The DS construction S k s (n) has a recursive definition similar to the Ackermann recurrence relation. S k s (n) will be a DS sequence of order s, for s even. S k s (n) satisfies the following properties: It consists of blocks of length n. Each block – distinct symbols. Each symbol occurs exactly μ s (k) times (function independent of n).

51. 51 The DS construction (We omit the base cases of the construction.) We want to construct 1.Let S' n – 1 f General case: S' has blocks of length n – 1. Let f be the number of blocks in S'.

52. 52 The DS construction 2.Let S f g 12345678 S is of order s – 2. Let g be the number of distinct symbols in S. 12345678 S has blocks of length f.

53. 53 The DS construction 3.Let S* g S* is of order s. S* has blocks of length g.

54. 54 The DS construction S* has blocks of length g. S has g distinct symbols. S* 12345678 Replace each block in S* by a copy of S using the same symbols, making their first appearances in the same order. S 12345678 S* We get S*. It has blocks of length f. f Let h be the number of blocks in S*. g g distinct symbols

55. 55 The DS construction S* has h blocks of length f. 1234567 S' has f blocks of length n – 1. Make h copies of S', one for each block in S*. Use fresh symbols for each copy of S'. For each block in S*, insert the symbols at the end of each block of the copy of S'. copy of S' S* This is the desired sequence S. It has blocks of length n. n S

56. 56 The DS construction Summary of the construction: S* g S g distinct symbols f S* n – 1  n f f (We still need to prove correctness.) S copy of S' S …… ……

57. 57 Analysis: Multiplicity of the symbols Recall: In S k s (n) each symbol occurs μ s (k) times. μ s–2 (k – 1) μ s (k – 1) μ s–2 (k – 1)μ s (k – 1) μs(k)μs(k) We have: μ s (k) = μ s–2 (k – 1)μ s (k – 1) Base case: μ 2 (k) = 2 (we didn’t show) copy of S*

58. 58 Analysis: Sequence length Meaning, take sequences For a fixed value of 2s, fix n = 2 and let k = 1, 2, 3, … We have N(k) ≤ A(k + c) for some c. Multiplicity of symbols in k-th sequence: We want: Sequence length as a function of # distinct symbols. N k - # distinct symbols in k-th sequence. As claimed

59. 59 Proof of correctness We want to show that S k s (n): does not contain adjacent repetitions does not contain an alternation of length s + 2. Recall: S k s (n) is composed of one copy of S*, and many copies of S'. Each copy has different symbols. In each block of length n, the first n – 1 symbols come from S', and the last symbol comes from S*. S' S* n

60. 60 Proof of correctness We’ll first prove a few things by induction: Corollary: If n ≥ 2, there are no adjacent repetitions. 1.Each symbol always occurs in the same position within the blocks. We call it the depth of the symbol. aa Proof: If the symbol comes from a copy of S', by induction. If it comes from S*, it always has depth n.

61. 61 Proof of correctness Proof: If both symbols come from the same copy of S', by induction. 2.No two symbols appear together in more than one block. bbaa This copy of S' receives only one b. S' S* b aaaaaa If they come from different copies of S', or both from S*, they never appear together. If a comes from a copy of S' and b comes from S*:

62. 62 Proof of correctness Proof: If both symbols have depth ≤ n – 1, they come from the same copy of S'. So by induction. Or else they come from different copies of S', so they don’t alternate at all. 3.Symbols at different depths make alternations of length at most 5 (ababa). This copy of S' receives at most one a. S' S* aaaaaaa bbbbbb ababa Suppose a has depth n (comes from S*), and b has depth ≤ n – 1 (comes from a copy of S').

63. 63 Proof of correctness Claim: 1.S k s (n) has no forbidden alternation of length s + 2. Now what we want to prove. We need to prove by induction a stronger property. 2.Even if we replace each block in S k s (n) by a sequence of order s – 2 with the same symbols, making their first appearances in the same order, we still get no alternation of length s + 2. Sks(n)Sks(n) 1 2 3 4 5 order s – 2

64. 64 Proof of correctness Claim: 1.S k s (n) has no forbidden alternation of length s + 2. 2.Even if we replace each block by a sequence of order s – 2 (with 1 st appearances in same order), we still get no forbidden alternation. Proof of property 1: Suppose for a contradiction that a, b make a forbidden alternation. They must have the same depth (since s + 2 ≥ 6). Say they have depth ≤ n – 1. They can’t come from different copies of S'. And if they come from the same copy of S', then contradiction by induction. Say they have depth n (they come from S*). But S* is obtained from S* by block replacements. Contradiction to property 2 on S* by induction.

65. 65 Proof of correctness Proof of property 2: Suppose for a contradiction that after block replacements on S k s (n), we get a forbidden alternation a…b…. a and b must have appeared together in a block. Claim: 1.S k s (n) has no forbidden alternation of length s + 2. 2.Even if we replace each block by a sequence of order s – 2 (with 1 st appearances in same order), we still get no forbidden alternation. 4 options: They appear together in only one block. a b a b a s – 1 babab bab bab bab s + 2 ba ba ba

66. 66 Proof of correctness This can’t happen! If a and b have depth ≤ n – 1, then they belong to the same copy of S'. The same block replacement would have created a forbidden alternation in S'. Contradiction to property 2 for S'. QED correctness. babab bab bab bab ba ba ba Suppose a has depth ≤ n – 1 (belongs to a copy of S') and b has depth n (comes from S*). In each of the 4 cases, this copy of S' received two b’s from S*. Contradiction.

67. 67 Problem with s odd Why doesn’t the construction work for s odd? The last argument breaks: a b a b a b s – 1 baaba bba baa bab s + 2 ba ba ba No contradiction

68. 68 Advantages over previous construction [ASS89] Each block is a sequence of n distinct symbols. In the old construction, each block was of the form 1 2 3 … n … 3 2 1. All symbols have exactly the same multiplicity. This greatly simplifies calculations. No need to remove adjacent repetitions at the end. The old construction had some “tiny duplications” of symbols. But they are not the cause of the asymptotic growth, so they confuse.

69. 69 Open problems Close the gap for s odd. Do the true bounds have the log factor or not? Improving the upper bound would require a parity- sensitive argument. Geometric realizations: All realizations known so far give Ω(nα(n)) (e.g. segments in the plane). Find a realization that achieves higher-order bounds (e.g. with parabolic segments).

70. 70 THANK YOU!