A relationship between generalized Davenport-Schinzel sequences and interval chains

Transcription

1 A relationhip between generalized Davenport-Schinzel equence and interval chain The MIT Faculty ha made thi article openly available. Pleae hare how thi acce benefit you. Your tory matter. Citation A Publihed Publiher Geneon, Jee. "A relationhip between generalized Davenport-Schinzel equence and interval chain." Electronic Journal of Combinatoric (3) (Augut 015). p19 European Mathematical Information Service (EMIS) Verion Final publihed verion Acceed Mon Dec 03 14:05:5 EST 018 Citable Link Term of Ue Detailed Term Article i made available in accordance with the publiher' policy and may be ubject to US copyright law. Pleae refer to the publiher' ite for term of ue.

2 A relationhip between generalized Davenport-Schinzel equence and interval chain Jee Geneon Department of Mathematic Maachuett Intitute of Technology Maachuett, U.S.A. Submitted: Oct 0, 014; Accepted: Jul 4, 015; Publihed: Aug 14, 015 Mathematic Subject Claification: 05D99 Abtract Let an (r, )-formation be a concatenation of permutation of r ditinct letter, and let a block of a equence be a ubequence of conecutive ditinct letter. A k-chain on [1, m] i a equence of k conecutive, dijoint, nonempty interval of the form [a 0, a 1 ][a 1 + 1, a ]... [a k 1 + 1, a k ] for integer 1 a 0 a 1 <... < a k m, and an -tuple i a et of ditinct integer. An -tuple tab an interval chain if each element of the -tuple i in a different interval of the chain. Alon et al. (008) oberved imilaritie between bound for interval chain and Davenport-Schinzel equence, but did not identify the caue. We how for all r 1 and 1 k m that the maximum number of ditinct letter in any equence S on m + 1 block avoiding every (r, + 1)-formation uch that every letter in S occur at leat k + 1 time i the ame a the maximum ize of a collection X of (not necearily ditinct) k-chain on [1, m] o that there do not exit r element of X all tabbed by the ame -tuple. Let D,k (m) be the maximum number of ditinct letter in any equence which can be partitioned into m block, ha at leat k occurrence of every letter, and ha no ubequence forming an alternation of length. Nivach (010) proved that D 5,d+1 (m) = Θ(mα d (m)) for all fixed d. We how that D +1, (m) = ( m ) for all. We alo prove new lower bound which imply that D5,6 (m) = Θ(m log log m) and D 5,d+ (m) = Θ(mα d (m)) for all fixed d 3. Keyword: alternation, formation, generalized Davenport-Schinzel equence, interval chain, invere Ackermann function, permutation Supported by the NSF Graduate Reearch Fellowhip under Grant No the electronic journal of combinatoric (3) (015), #P3.19 1

3 1 Introduction A equence contain a equence u if ome ubequence of can be changed into u by a one-to-one renaming of it letter. If doe not contain u, then avoid u. A equence i called r-pare if any r conecutive letter in are ditinct. Collection of contiguou ditinct letter in are called block. A generalized Davenport-Schinzel equence i an r-pare equence avoiding a fixed forbidden equence (or et of equence) with r ditinct letter. Bound on the length of generalized Davenport-Schinzel equence were ued to bound the complexity of lower envelope of olution et of linear homogeneou differential equation [5], the complexity of face in arrangement of arc [1], and the maximum number of edge in k-quaiplanar graph with no pair of edge interecting in more than O(1) point [6, 14]. Let λ (n) be the maximum length of any -pare equence with n ditinct letter which avoid alternation of length +. For = 1 and =, λ (n) = Θ(n). However, Hart and Sharir [7] howed that λ 3 (n) = Θ(nα(n)), uch that α(n) denote the invere λ Ackermann function. Nivach [10] and Klazar [9] proved that lim 3 (n) n =. nα(n) Agarwal, Sharir, Shor [] and Nivach [10] proved that λ (n) = n 1 t! α(n)t ±O(α(n) t 1 ) for even 4 with t =. Pettie [1] proved that λ 5(n) = Θ(nα(n) α(n) ) and that λ (n) = n 1 t! α(n)t ±O(α(n) t 1) for odd 7 with t = 3. In [10], imilar bound were alo derived for a different equence extremal function. Let an (r, )-formation be a concatenation of permutation of r ditinct letter. For example abcddcbaadbc i a (4, 3)-formation. Define F r, (n) to be the maximum length of any r-pare equence with n ditinct letter which avoid all (r, )-formation. Klazar [8] proved that F r, (n) = Θ(n) and F r,3 (n) = Θ(n) for every r > 0. Nivach proved that F r,4 (n) = Θ(nα(n)) for r. Nivach [10] howed that F r, (n) = n 1 t! α(n)t ±O(α(n) t 1) for all r and odd 5 with t = 3. Pettie [13] proved for r = that F r,6 (n) = Θ(nα(n) α(n) ) and that F r, (n) = n 1 t! α(n)t ±O(α(n) t 1) for even 8 with t = 4. When r 3, Pettie proved that F r,(n) = n 1 t! α(n)t log α(n)±o(α(n) t) for even 6 with t = 4. To define the Ackermann hierarchy let A 1 (n) = n and for k, A k (0) = 1 and A k (n) = A k 1 (A k (n 1)) for n 1. To define the invere function let α k (x) = min {n : A k (n) x} for all k 1. We define the Ackermann function A(n) to be A n (3) a in [10]. The invere Ackermann function α(n) i defined to be min {x : A(x) n}. Nivach bound on λ (n) were derived uing an extremal function which maximize number of ditinct letter intead of length. Let D,k (m) be the maximum number of ditinct letter in any equence on m block avoiding alternation of length uch that every letter occur at leat k time. Clearly D,k (m) = 0 if m < k and D,k (m) = if k < 1 and k m. Nivach proved that Ψ (m, n) k(d,k (m)+n) [10], where Ψ (m, n) denote the maximum poible length of any equence on m block with n ditinct letter that avoid alternation of length. Nivach alo proved that D 5,d+1 (m) = Θ(mα d (m)) for each fixed d (but noted that the bound on D 5,k (m) were not tight for even k). Sundar [15] derived imilar bound the electronic journal of combinatoric (3) (015), #P3.19

4 in term of m on function related to the Deque conjecture. Let F r,,k (m) be the maximum number of ditinct letter in any equence on m block avoiding every (r, )-formation uch that every letter occur at leat k time. Clearly F r,,k (m) = 0 if m < k and F r,,k (m) = if k < and k m. Every (r, )-formation contain an alternation of length + 1 for every r, o D +1,k (m) F r,,k (m) for every r. Nivach proved that Ψ r,(m, n) k(f r,,k (m) + n) [10], where Ψ r,(m, n) denote the maximum poible length of any equence on m block with n ditinct letter that avoid every (r, )-formation. Nivach alo proved for r that F r,4,d+1 (m) = Θ(mα d (m)) for each fixed d. The recurive inequalitie for the upper bound on F r,4,d+1 (m) in [10] alo imply that F r,4,6 (m) = O(m log log m) and that F r,4,d+ (m) = O(mα d (m)). Similar bound were alo derived on an extremal function related to interval chain. A k-chain on [1, m] i a equence of k conecutive, dijoint, nonempty interval of the form [a 0, a 1 ][a 1 + 1, a ]... [a k 1 + 1, a k ] for integer 1 a 0 a 1 <... < a k m. An -tuple i a et of ditinct integer. An -tuple tab an interval chain if each element of the -tuple i in a different interval of the chain. Let ζ,k (m) denote the minimum ize of a collection of -tuple uch that every k-chain on [1, m] i tabbed by an -tuple in the collection. Clearly ζ,k (m) = 0 if m < k and ζ,k (m) i undefined if k < and k m. Alon et al. [3] howed that ζ, (m) = ( m ) for 1, ζ3,4 (m) = Θ(m log m), ζ 3,5 (m) = Θ(m log log m), and ζ 3,k (m) = Θ(mα k (m)) for k 6. Alon et al. [3] oberved imilaritie between bound for interval chain and Davenport-Schinzel equence, but did not identify the caue. Let η r,,k (m) denote the maximum ize of a collection X of not necearily ditinct k-chain on [1, m] o that there do not exit r element of X all tabbed by the ame -tuple. Clearly η r,,k (m) = 0 if m < k and η r,,k (m) = if k < and k m. In Section we how that η r,,k (m) = F r,+1,k+1 (m+1) for all r 1 and 1 k m. Lemma 1.1. ζ,k (m) η,,k (m) for all 1 k m. Proof. Let X be a collection of k-chain on [1, m] o that there do not exit element of X that are tabbed by the ame -tuple. Then by the pigeonhole principle, X i the minimum poible ize of a collection Y of -tuple uch that every k-chain in X i tabbed by an -tuple in Y. The lat lemma and the equality in Section imply that D +,k+1 (m + 1) ζ,k (m) for all 1 k m. Thi implie that D +1, (m) ( m ) for all, D5,6 (m) = O(m log log m), and D 5,d+ (m) = O(mα d (m)) for d 3. In Section 3 we contruct alternation-avoiding equence to prove lower bound on D,k (m). We prove that D +1, (m) = ( m ) for all. Furthermore we how that D 5,6 (m) = Ω(m log log m) and D 5,d+ (m) = Ω( 1 d mα d(m)) for d 3. Thu the bound on D 5,d (m) have a multiplicative gap of O(d) for all d. the electronic journal of combinatoric (3) (015), #P3.19 3

5 Proof that F r,+1,k+1 (m + 1) = η r,,k (m) We how that F r,+1,k+1 (m + 1) = η r,,k (m) for all r 1 and 1 k m uing map like thoe between matrice and equence in [4] and [11]. Lemma.1. F r,+1,k+1 (m + 1) η r,,k (m) for all r 1 and 1 k m. Proof. Let P be a equence with F r,+1,k+1 (m + 1) ditinct letter and m + 1 block 1,..., m + 1 uch that no ubequence i a concatenation of + 1 permutation of r different letter and every letter in P occur at leat k + 1 time. Contruct a collection of k-chain on [1, m] by converting each letter in P to a k-chain: if the firt k +1 occurrence of letter a are in block a 0,..., a k, then let a be the k-chain with i th interval [a i 1, a i 1]. Suppoe for contradiction that there exit r ditinct letter q 1,..., q r in P uch that q 1,..., q r are tabbed by the ame -tuple 1 j 1 <... < j m. Let j 0 = 0 and j +1 = m + 1. Then for each 1 i + 1, q n occur in ome block b n,i uch that b n,i [j i 1 + 1, j i ] for every 1 n r. Hence the letter q 1,..., q r make an (r, + 1)- formation in P, a contradiction. Corollary.. D +,k+1 (m + 1) ζk (m) for all 1 k m. The bound on ζ,k (m) in [3] imply the next corollary. Corollary.3. D +1, (m) ( m ) for, D5,5 (m) = O(m log m), D 5,6 (m) = O(m log log m), and D 5,k (m) = O(mα k 1 (m)) for k 7. To prove that F r,+1,k+1 (m + 1) η r,,k (m) for all r 1 and 1 k m, we convert collection of k-chain into equence with a letter correponding to each k-chain. Lemma.4. F r,+1,k+1 (m + 1) η r,,k (m) for all r 1 and 1 k m. Proof. Let X be a maximal collection of k-chain on [1, m] o that there do not exit r element of X all tabbed by the ame -tuple. To change X into a equence P create a letter a for every k-chain a in X, and put a in every block i uch that either a ha an interval with leat element i or a ha an interval with greatet element i 1. Order the letter in block tarting with the firt block and moving to the lat. Let L i be the letter in block i which alo occur in ome block j < i and let R i be the letter which have firt occurrence in block i. All of the letter in L i occur before all of the letter in R i. If a and b are in L i, then a appear before b in block i if the lat occurrence of a before block i i after the lat occurrence of b before block i. The letter in R i may appear in any order. P i a equence on m + 1 block in which each letter occur k + 1 time. Suppoe for contradiction that there exit r letter q 1,..., q r which form an (r, + 1)-formation in P. Lit all (r, + 1)-formation on the letter q 1,..., q r in P lexicographically, o that formation f appear before formation g if there exit ome i 1 uch that the firt i 1 element of f and g are the ame, but the i th element of f appear before the i th element of g in P. the electronic journal of combinatoric (3) (015), #P3.19 4

6 Let f 0 be the firt (r, + 1)-formation on the lit and let π i (repectively ρ i ) be the number of the block which contain the lat (repectively firt) element of the i th permutation in f 0 for 1 i +1. We claim that π i < ρ i+1 for every 1 i. Suppoe for contradiction that for ome 1 i, π i = ρ i+1. Let a be the lat letter of the i th permutation and let b be the firt letter of the (i + 1) t permutation. Then a occur before b in block π i and the b in π i i not the firt occurrence of b in P, o the a in π i i not the firt occurrence of a in P. Otherwie a would appear after b in π i. Since the a and b in π i are not the firt occurrence of a and b in P, then the lat occurrence of a before π i mut be after the lat occurrence of b before π i. Let f 1 be the ubequence obtained by deleting the a in π i from f 0 and inerting the lat occurrence of a before π i. Since the lat occurrence of a before π i occur after the occurrence of b in the i th permutation of f 0, f 1 i alo an (r, + 1)-formation. Moreover, f 1 occur before f 0 on the lit. Thi contradict the definition of f 0, o for every 1 i, π i < ρ i+1. For every 1 j r and 1 i + 1, the letter q j appear in ome block between ρ i and π i incluive. Since π i < ρ i+1 for every 1 i, the -tuple (π 1,..., π ) tab each of the interval chain q 1,..., q r, a contradiction. Hence P contain no (r, + 1)- formation. The idea for the next lemma i imilar to a proof about doubled formation free matrice in [4]. Lemma.5. F r,, (m) (r 1) ( m ) for every r 1 and 1 m. Proof. Let P be a equence with m block uch that no ubequence of P i a concatenation of permutation of r ditinct letter and every letter of P occur at leat time. An occurrence of letter a in P i called even if there are an odd number of occurrence of a to the left of it. Otherwie the occurrence of a i called odd. Suppoe for contradiction that P ha at leat 1 + (r 1) ( m ) ditinct letter. The number of ditinct tuple (m 1,..., m ) for which a letter could have even occurrence in block m 1,..., m i equal to the number of poitive integer olution to the equation (1+x 1 )+...+(1+x )+x 1+ = m+1 if i even and (1+x 1 )+...+(1+x )+x 1+ = m if i odd. Since for each the equation ha ( m ) poitive integer olution, by the pigeonhole principle there are at leat r ditinct letter q 1,..., q r with even occurrence in the ame block of P. Then P contain a concatenation of permutation of the letter q 1,..., q r, a contradiction. The lat lemma i an alternative proof that D +1, (m) ( m ) ince D+1, (m) F r,, (m) for all, m 1 and r. 3 Lower bound In the lat ection we howed that D +1, (m) ( m ) for. The next lemma provide a matching lower bound. the electronic journal of combinatoric (3) (015), #P3.19 5

7 Lemma 3.1. D +1, (m) ( m ) for all and m + 1. Proof. For every 1 and m + 1 we build a equence X (m) with ( m ) ditinct letter. Firt conider the cae of even. The equence X (m) i the concatenation of m 1 fan, o that each fan i a palindrome coniting of two block of equal length. Firt aign letter to each fan without ordering them. Create a letter for every -tuple of non-adjacent fan, and put each letter in every fan in it -tuple. Then order the letter in each fan tarting with the firt fan and moving to the lat. Let L i be the letter in fan i which occur in ome fan j < i and let R i be the letter which have firt occurrence in fan i. In the firt block of fan i all of the letter in L i occur before all of the letter in R i. If a and b are in L i, then a occur before b in the firt block of fan i if the lat occurrence of a before fan i i after the lat occurrence of b before fan i. If a and b are in R i, then a occur before b in the firt block of fan i if the firt fan which contain a without b i before the firt fan which contain b without a. Conider for any ditinct letter x and y the maximum alternation contained in the ubequence of X (m) retricted to x and y. The maximum alternation i contructed tarting with a fan that contain only x, and ucceively adding adjacent fan. Any other fan which contain x without y or y without x add at mot 1 to the alternation length. Any fan which contain both x and y add to the alternation length. If x and y occur together in i fan, then the length of their alternation i at mot ( i) + ( i) + i =. Every pair of adjacent fan have no letter in common, o every pair of adjacent block in different fan can be joined a one block when the m 1 fan are concatenated to form X (m). Thu X (m) ha m block and ( m ) letter, and each letter occur time. For odd 3, contruct X (m) by adding a block r after X 1 (m 1) containing all of the letter in X 1 (m 1) uch that a occur before b in r if the lat occurrence of a in X 1 (m 1) i after the lat occurrence of b in X 1 (m 1). Then X (m) contain no alternation of length +1 ince X 1 (m 1) contain no alternation of length. Moreover X (m) ha m block and ( m +1 1 ) letter, and each letter occur time. The next lemma how how to extend the lower bound on D +1, (m) to F r,, (m). Lemma 3.. F r,,k (m) (r 1)F,,k (m) for all r 1 and 1 k m. Proof. Let P be a equence with F,,k (m) ditinct letter and m block uch that no ubequence i a concatenation of permutation of two ditinct letter and every letter occur at leat k time. P i the equence obtained from P by creating r 1 new letter a 1,..., a r 1 for each letter a and replacing every occurrence of a with the equence a 1... a r 1. Suppoe for contradiction that P contain an (r, )-formation on the letter q 1,..., q r. Then there exit indice i, j, k, l and ditinct letter a, b uch that q i = a j and q k = b l. P contain a (, )-formation on the letter q i and q k, o P contain a (, ) formation on the letter a and b, a contradiction. Then P i a equence with (r 1)F,,k (m) ditinct the electronic journal of combinatoric (3) (015), #P3.19 6

8 letter and m block uch that no ubequence i a concatenation of permutation of r ditinct letter and every letter occur at leat k time. Corollary 3.3. F r,, (m) = (r 1) ( m ) for all r 1 and m 1. The proof of the next theorem i much like the proof that D 5,d+1 (m) = Ω( 1 d mα d(m)) for d in [10]. The only ubtantial difference i the bae cae. Theorem 3.4. D 5,6 (m) = Ω(m log log m) and D 5,d+ (m) = Ω( 1 d mα d(m)) for d 3. We prove thi theorem by contructing a family of equence that avoid ababa. For all d, m 1, we inductively contruct equence G d (m) in which each letter appear d + time and no two ditinct letter make an alternation of length 5. Thi proof ue a different definition of fan: fan will be the concatenation of two palindrome with no letter in common. Each palindrome conit of two block of equal length. For d, m 1 the block in G d (m) containing only firt and lat occurrence of letter are called pecial block. Let S d (m) be the number of pecial block in G d (m). Every letter ha it firt and lat occurrence in a pecial block, and each pecial block in G d (m) ha m letter. Block that are not pecial are called regular. No regular block in G d (m) ha pecial block on both ide, but every pecial block ha regular block on both ide. The equence G 1 (m) are the concatenation of m + 1 fan. In each fan the econd block of the firt palindrome and the firt block of the econd palindrome make one block together ince they are adjacent and have no letter in common. The firt palindrome in the firt fan and the econd palindrome in the lat fan are empty. There i a letter for every pair of fan and the letter i in both of thoe fan. The letter with lat appearance in fan i are in the firt palindrome of fan i. They appear in fan i firt palindrome firt block in revere order of the fan in which they firt appear. The letter with firt appearance in fan i are in the econd palindrome of fan i. They appear in fan i econd palindrome firt block in order of the fan in which they lat appear. For example, the equence G 1 (4) i iomorphic to the following equence, uch that () denote pecial block boundarie, [] denote regular block boundarie, and denote fan boundarie: [](abcd)[dcba] [a](aefg)[gfe] [eb](behi)[ih] [hfc](cfhj)[j] [jigd](dgij)[] Lemma 3.5. G 1 (m) contain no alternation of length 5. Proof. Let x and y be any two letter in G 1 (m). If x and y do not occur in any fan together, then clearly the ubequence of G 1 (m) retricted to x and y contain no alternation of length 5. If x and y occur together in a fan, then their common fan contain an alternation on x and y of length or 3. By the contruction of G 1 (m), thi alternation i extended to length at mot 4 by the occurrence of x and y outide the common fan. the electronic journal of combinatoric (3) (015), #P3.19 7

9 For all d 1 the equence G d (1) i defined to conit of d + copie of the letter 1. The firt and lat copie of 1 are both pecial block, and there are empty regular block before the firt 1 and after the lat 1. The equence G d (m) for d, m i contructed inductively from G d (m 1) and G d 1 (S d (m 1)). Let f = S d (m 1) and g = S d 1 (f). Make g copie X 1,..., X g of G d (m 1) and one copy Y of G d 1 (f), o that no copie of G d (m 1) have any letter in common with Y or each other. Let Q i be the i th pecial block of Y. If the l th element of Q i i the firt occurrence of the letter a, then inert aa right after the l th pecial block of X i. If the l th element of Q i i the lat occurrence of a, then inert aa right before the l th pecial block of X i. Replace Q i in Y by the modified X i for every i. The reulting equence i G d (m). Lemma 3.6. For all d and m, G d (m) avoid ababa. Proof. Given that the alternation in G 1 (m) have length at mot 4 for all m 1, then the ret of the proof i the ame a the proof in [10] that Z d (m) avoid ababa. Let L d (m) be the length of G d (m). Oberve that L d (m) = (d + 1)mS d (m) ince each letter in G d (m) occur d + time, twice in pecial block, and each pecial block ha m letter. Define N d (m) a the number of ditinct letter in G d (m) and M d (m) a the number of block in G d (m). Alo let X d (m) = M d(m) S d (m) V d (m) a in [10]. Lemma 3.7. For all m, d 1, X d (m) d + and V d (m) m. and V d (m) = L d(m) M d (m). We bound X d(m) and Proof. By contruction S 1 (m) = m + 1 for m 1, S d (1) = for d, and S d (m) = S d (m 1)S d 1 (S d (m 1)) for d, m. Furthermore M 1 (m) = 3m + 3, M d (1) = d + 4 for d, and M d (m) = M d (m 1)S d 1 (S d (m 1)) + M d 1 (S d (m 1)) S d 1 (S d (m 1)) for d, m. Thu S (m) = S (m 1)(S (m 1) + 1) (S (m 1)) and S (1) =. Since m 1 atifie S (m) = the recurrence S (1) = and S (m + 1) = (S (m)), then m 1 S (m) m 1. For d, S d () = S d 1 () and S 1 () = 3. So S d () = 3 d 1. For d, M d () = (d + 3)(3 d ) + M d 1 () and M 1 () = 9. Hence M d () = (6d + 3) d 1. Then X 1 (m) = 3 for all m, X d (1) = d + for all d, and X d () = d + 1 for all d. For d, m, X d (m) = X d (m 1) + X d 1(S d (m 1)) 1 S d. (m 1) We prove by induction on d that X d (m) d + for all m, d 1. Oberve that the inequality hold for X 1 (m), X d (1), and X d () for all m, d. Fix d and uppoe X d 1 (m) d for all m. Then X d (m) X d (m 1) + Hence X d (m) X d () + (d 1) n= S d(n) 1 = d (d 1) n= S d(n) 1. Since S d (m) S d (m 1) for all d, m, then n= S d(n) 1 S d () 1 = 1 1 for all d, o X d 1 d(m) d +. Hence V d (m) = L d(m) = (d+1)ms d(m) M d (m) M d m. (m) d 1. S d (m 1) 3 d The lower bound on D 5,d+ (m) for each d are proved in the following ubection. the electronic journal of combinatoric (3) (015), #P3.19 8

10 3.1 Bound for d = If d =, let m i = M (i) and n i = N (i). Then m i = X (i)s (i) 6S (i) 6( i 1 ) i + for i 1. Then i = Ω(log log m i ), o n i = L (i) 6 = V (i)m (i) 6 im (i) 1 = Ω(m i log log m i ). We ue interpolation to extend the bound from m i to m. Let i and t atify m i m < m i+1 and t = m m i. Concatenate t copie of G (i) with no letter in common for a total of at leat m m i n i = Ω(m log log m) letter. Hence D 5,6 (m) = Ω(m log log m). 3. Bound for d = 3 We prove that S 3 (m) A 3 (m) following the method of [10]. Since S 3 (m) = S 3 (m 1)S (S 3 (m 1)) S (S 3 (m 1)) S 3 (m 1)+1, then let F (m) = m+1 and G(m) = m. Then F (m) = m+1 1 m = G(m) for every m 0. Thu S 3 (m) F (m 1) (S 3 (1)) < F (m 1) (S 3 (1)) G (m 1) (S 3 (1)) = A 3 (m). Let m i = M 3 (i) and n i = N 3 (i). Therefore m i = X 3 (i)s 3 (i) 8S 3 (i) A 3 (i + ) for i 1. So i = Ω(α 3 (m i )) and n i = N 3 (i) = L 3(i) = V 3(i)M 3 (i) im 3(i) = Ω(m i α 3 (m i )). Then D 5,8 (m) = Ω(mα 3 (m)) by interpolation. 3.3 Bound for d 4 For each d 4 we prove that S d (m) A d (m + ) for m 1 by induction on d. Since S 4 (m) = S 4 (m 1)S 3 (S 4 (m 1)) S 3 (S 4 (m 1)), then let F (m) = S 3 (m). Since 4F (m) A 3 (4m) and A 4 (3) > 4S 4 (1), then S 4 (m) F (m 1) (S 4 (1)) < 4F (m 1) (S 4 (1)) A (m 1) 3 (4S 4 (1)) < A 4 (m + ). Fix d > 4 and uppoe that S d 1 (m) A d 1 (m + ). Define F (m) = S d 1 (m). Since 4F (m) A d 1 (4m) and A d (3) > 4S d (1), then S d (m) F (m 1) (S d (1)) < 4F (m 1) (S d (1)) A (m 1) d 1 (4S d (1)) < A d (m + ). Fix d 4. Let m i = M d (i) and n i = N d (i). Then m i = X d (i)s d (i) (d + )S d (i) (d + )A d (i + ) A d (i + 3). Then i α d (m i ) 3, o n i = L d(i) d+ = V d(i)m d (i) d+ im d(i) 4d+4 = Ω( 1 d m iα d (m i )). By interpolation D 5,d+ (m) = Ω( 1 d mα d(m)) for d 4. Corollary 3.8. If r, then F r,4,6 (m) = η r,3,5 (m) = Ω(m log log m) and F r,4,d+ (m) = η r,3,d+1 (m) = Ω( 1 d mα d(m)) for d 3. Acknowledgment Thi reearch wa upported by the NSF graduate reearch fellowhip under Grant No The author thank Seth Pettie for comment on thi paper and Peter Shor for improving the bound on S (m) in Lemma 3.7. The author alo thank the anonymou referee for their helpful uggetion. the electronic journal of combinatoric (3) (015), #P3.19 9

11 Reference [1] M. Sharir and P.K. Agarwal. Davenport-Schinzel equence and their geometric application. Cambridge Univerity Pre, Cambridge, [] P.K. Agarwal, M. Sharir, and P. Shor. Sharp upper and lower bound for the length of general Davenport-Schinzel equence. Journal of Combinatorial Theory Serie A, 5 (1989) [3] N. Alon, H. Kaplan, G. Nivach, M. Sharir, and S. Smorodinky. Weak epilon-net and interval chain. J. ACM, 55 (008). [4] J. Cibulka and J. Kynčl. Tight bound on the maximum ize of a et of permutation with bounded vc-dimenion. Journal of Combinatorial Theory Serie A, 119 (01) [5] H. Davenport and A. Schinzel. A combinatorial problem connected with differential equation. American Journal of Mathematic, 87 (1965) [6] J. Fox, J. Pach, and A. Suk. The number of edge in k-quaiplanar graph. SIAM Journal on Dicrete Mathematic, 7 (013) [7] S. Hart and M. Sharir. Nonlinearity of Davenport-Schinzel equence and of generalized path compreion cheme. Combinatorica, 6 (1986) [8] M. Klazar. A general upper bound in the extremal theory of equence. Commentatione Mathematicae Univeritati Carolinae, 33 (199) [9] M. Klazar. On the maximum length of Davenport-Schinzel equence. Contemporary trend in dicrete mathematic (DIMACS erie in dicrete mathematic and theoretical computer cience), 49 (1999) [10] G. Nivach. Improved bound and new technique for Davenport-Schinzel equence and their generalization. Journal of the ACM, 57 (010). [11] S. Pettie. Degree of nonlinearity in forbidden 0-1 matrix problem. Dicrete Mathematic 311 (011) [1] S. Pettie. Sharp bound on Davenport-Schinzel equence of every order. Sympoium on Computational Geometry (013): [13] S. Pettie. Sharp bound on formation-free equence. Proceeding of the Twenty Sixth Annual ACM-SIAM Sympoium on Dicrete Algorithm, SODA 15 (015) [14] A. Suk and B. Walczak. New bound on the maximum number of edge in k-quaiplanar graph. Twenty-firt International Sympoium on Graph Drawing (013): [15] R. Sundar. On the Deque conjecture for the play algorithm. Combinatorica 1(1): (199). the electronic journal of combinatoric (3) (015), #P