Tighter bounds and simpler constructions for Davenport-Schinzel sequences

1. Tighter bounds andsimpler constructions forDavenport-Schinzel sequences Haim Kaplan, Gabriel Nivasch, Micha Sharir Tel-Aviv University

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 … a … b … • of length s + 2. Examples: a c d b c c a Not a DS sequence a c a b a d a c d c b Not a DS seq. of order 3

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

4. Motivation: Lower envelopes What if we have nsegments 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. Straight-line segments give DS sequences of order 3. Alternation ababa is impossible: a b a b a

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

6. The inverse Ackermann function Inverse Ackermann hierarchy: # times you have to apply αk–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. The inverse Ackermann function For every fixed x ≥ 6, the sequence Example: x = 9876!: decreases strictly, until it settles at 3. Inverse Ackermann function: We have for every fixed k.

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

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 10-chain n 1 A j-tuple stabs a k-chain if its j elements fall on j different intervals. Problem: Build a family of j-tuples, as small as possible, that stab allk-chains in [1,n].

10. Stabbing interval chains(a completely unrelated problem) Solution: 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).

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

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

13. order s – 2 order s – 2 order s – 2 order s – 2 order s – 2 b a b a b a b a s – 1 s-th Bounding DS sequences Let S be a DS sequence of order s on n symbols, of maximum length. (s even) Partition Sgreedily, from left to right, into maximalblocks of order s – 2. Consider a block. 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. (Also works for s odd.)

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. If r = 1 then we write We just showed: Lemma 1: (This is all [ASS89].)

15. n1distinct symbols n2distinct symbols n3distinct symbols n4distinct symbols n5distinct symbols n1 n2 n3 n4 n5 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. s r r r r r d a e b f b b a a c c f f a b d d f c b d a g g e g b e c c f b S Leave in each block just the first occurrence of each symbol: S' Remove adjacent repetitions. (Remove at most m symbols.) S'' b a c f d a g e b c f

16. n1distinct symbols n2distinct symbols n3distinct symbols n4distinct symbols n5distinct symbols Bounding DS sequences s r r r r r S Let be a nondecreasing function of n such that Lemma 2:

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

19. Recurrence for ψs(m,n) Decompose S into 4 subsequences: Tlocal – local symbols Tstart – starting global symbols Tmiddle – middle global symbols Tend – 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. Bounding |T'local| The i-th layer in T'local is a DS sequence of order s on ni symbols, partitionable into m/b blocks of order 1. i-th layer

21. layer a b a b a a b s + 1 (s + 2)-nd in S Bounding |T'start| (and |T'end|) Each layer in T'start is a DS sequence of order s – 1 on n* symbols. 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. Moreover, all ofT'start is a DS sequence of order s – 1 on n* symbols. Why: Each symbol appears in only one layer. Similarly:

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

23. Bounding |T'middle| Better bound for |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 need to say something about the n*i’s. We will get a bound on their sum

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. S i-th layer n*i global symbols U n*i U' Each layer becomes a block of order 1. Form U' by removing adjacent repetitions from U. We remove at most b symbols.

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 integers satisfying such that

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

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

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

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

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

32. … c e g g b a … We promised a second proof! We define Almost Davenport-Schinzel (ADS) sequences: • 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 don’t disallow adjacent repetitions at the interface between blocks. But we limit the number of blocks.

33. 1 2 3 … 1 2 3 … … 3 2 1 … 3 2 1 ADS sequences We turn the problem around: 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 what is the maximium value of n, the number of distinct symbols? Example 1: If k = s (each symbol must appear s times), then we can have arbitrarily many symbols. … block 1 block 2 block 3 block s Maximum alternation is of length s + 1.

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

35. ADS sequences Let nks(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. For fixed s, nks(m) is increasing in m and decreasing in k. This is very similar to interval chains. We already know what to do! Theorem: Let k1, k2, k3 be integers, and let k = k2k3 + 2k1 – 3k2 – k3 + 2. Then, where t is a free parameter.

36. b b b a a b b a b b 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. block 1 block m layer # distinct symbols: nks(m). Group the m blocks into m/tlayers, with t blocks per layer. • Local symbol – appears in only one layer • Global symbol – appears in more than one layer

37. ADS sequences • Classify each global symbol into: • Left-concentrated: The firstk1 occurrences fall in one layer. • Middle-concentrated: Some innerk2 occurrences fall in one layer, and the symbol also appears in previous and subsequent layers. • Right-concentrated: The lastk1 occurrences fall in one layer. • Scattered: The symbol appears in at least k3 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 (k1 – 1) + (k3 – 3)(k2 – 1) + (k1 – 1) = k – 1 times. But : k = k2k3 + 2k1 – 3k2 – k3 + 2.

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

39. Bounding # left-concentrated (and right-concentrated) global symbols For each layer, take just the left-concentrated global symbols that start at that layer. We get an ADS sequence of order s – 1 on t blocks. Each symbol appears k1 times. usual argument # left-concentrated global symbols: At most # right-concentrated global symbols: Same.

40. Bounding # middle-concentrated global symbols For each layer, take just the middle-concentrated global symbols that appear at least k2 times in that layer, and aslo appear in previous and subsequent layers. We get an ADS sequence of order s – 2 on t blocks. Each symbol appears k2 times. usual argument # middle-concentrated global symbols: At most

41. Bounding # scattered global symbols For each scattered global symbol, leave just one occurrence per layer. Each layer becomes a block (has distinct symbols). We get an ADS sequence of order s on m/t blocks. Each symbol appears k3 times. # scattered global symbols: At most Adding up: QED

42. ADS sequences Recurrence: where: k = k2k3 + 2k1 – 3k2 – k3 + 2. Compare to recurrence for interval chains: where: k = k2k3 + 2k1 – 2k2. Same thing basically!

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

44. a a a a a a a a a' a a' a a' a a a' a' a a a a a a k k k < k a'' a'' a'' a'' a'' 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 exactlyk times. Get sequence S'. (k – free parameter) (This doesn’t introduce any forbidden alternations.) We removed at most kn symbols. S' is an ADS sequence on m blocks.

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

46. END OF UPPER BOUNDS

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

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

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