Davenport-Schinzel Sequences and their Geometric Applications

Transcription

1 Advaced Computatioal Geometry Sprig 2004 Daveport-Schizel Sequeces ad their Geometric Applicatios Prof. Joseph Mitchell Scribe: Mohit Gupta 1 Overview I this lecture, we itroduce the cocept of Daveport-Schizel sequeces. We start by defiig them ad givig some results for the bouds o the legth of these sequeces, followed by explorig a few of their geometric applicatios. 2 Defiitio Let, s be positive itegers. A sequece U =< u 1,, u m > of symbols(derived from a alphabet Λ of size ) is a (, s) Daveport-Schizel sequece (a DS(, s)-sequece for short), if it satisfies the followig coditios : 1. For each i < m, we have u i u i There do ot exist s + 2 idices 1 i 1 < i 2 < < i s+2 m such that u i1 = u i3 = u i5 = = a, u i2 = u i4 = u i6 = = b, ad a b. I other words, there is o sub-sequece of two alteratig symbols of legth greater tha or equal to s + 2. We refer to s as the order of the sequece. 2.1 Example < 1, 2, 1, 3, 2, 1, 2, 3, 1, 4, 2, 3, 2, 1 > would be a DS(4, 8) sequece. It does t have ay two cosecutive symbols which are same ad it has o sub-sequece of two alteratig symbols of legth greater tha or equal to 10. 1

2 3 Maximum Legth of a DS(,s) Sequece We will write U = m for the legth of the sequece U. Defie λ s () = max{m U is a DS(,s)-sequece }. Followig are some results for bouds o λ s () for varyig s: 3.1 DS(,1)-Sequeces I such sequeces, each of the symbols ca appear atmost oce. If ay symbols is repeated at idex i ad idex j, the there should be aother symbol betwee u i ad u j ( due to property 1). Let i < k < j. The, sice u i = u j, < u i... u k... u j > form a alteratig sequece of legth 3, prevetig U from beig a order 1 sequece. So, the maximum legth of a order 1 DS-sequece is. λ 1 () =. 3.2 DS(,2)-Sequeces I this subsectio, we prove that λ 2 () = 2 1. We first start with givig a proof by iductio for the claim that λ 2 () 2 1. We the follow it up with a costructio, which establishes that λ 2 () 2 1, thereby provig that λ 2 () = λ 2 () 2 1 Base Case λ 2 (1) = 1. So, the statemet is true for = 1. Iductio Hypothesis Assume the claim to be true for = k 1. So, λ 2 (k 1) 2k 3. Iductio Step Prove the claim for = k : Let U be a DS(2,k)-sequece of maximum possible legth. So, U = λ 2 (k). Let µ a = First idex i U =< u 1, u 2,, u m > where a occurs (u µa = a). Let b be the symbol which maximises µ b, i.e. µ b µ a aɛλ (Λ is the alphabet from which the symbols 2

3 are derived). Lemma: b occurs oly oce i U. Proof: Proof by cotradictio. Let b were to appear twice i U. The two occureces of b ca t be cosecutive, so let there be aother symbol c betwee the two occureces of symbol b. Now, the first occurece of c should be before that of b, as µ b > µ c from the way we have chose b. This gives us a alteratig sub-sequece of legth 4 < c ldotsb...c...b, prevetig U from beig a order 2 DS sequece. Thus, if µ b µ a aɛλ, the b ca appear oly oce i U. Now, if we delete b from U, ad U looks like <...xbx... >, the we have to remove oe of the occureces of x for the remaiig list to be still a DS sequece. If b is ot flaked o both sides by the same symbol, the we ca safely remove it ad the remaiig sequece, which would be a sequece o k 1 symbols, would still be a DS sequece. So λ 2 (k) λ 2 (k 1) + 2 (2k 3) + 2 = 2k 1. Hece, λ 2 () λ 2 () 2 1 Give symbols, a 1, a 2,..., a, we ca costruct a DS(,2)-sequece of legth 2 1. Cosider the sequece U = < a 1, a 2,..., a, a 1, a 2,..., a 1 >.. Now, U = 2 1, ad U satisfies the two properties for beig a DS(,2)-sequece. So, we have costructed a DS(,2)-sequece of legth 2 1. Thus, λ 2 () 2 1. The above two subsectios imply that λ 2 () = DS(,3)-Sequeces Daveport-Schizel showed i 1965 that λ 3 () 2(l() + O(1)). Proof: Let U be the logest possible DS(,3)-sequece. So, U = λ 3 (). So, o a average, each symbol occurs U λ 3() times. Let a be the symbol with fewest occureces i the give U. 3

4 umber of occureces of a λ 3(). Now, remove a from U. This step will iduce some adjacet pairs of same symbols i the remaider of the list. For example, if U =< 1, 2, 3, 2, 4, 5, 3, 5, 6, 3, 6 >, the removal of all the occureces of 3 from the list will give a adjacet pair each of 2, 6 ad 5 i the remaiig list. Lemma: If all the occureces of a symbol a are removed from a DS(,3)-sequece, the remaiig list cosists of at most two adjacet pairs of same symbols, amely those iduced by the removal of first ad last occureces of a from the origial list. Proof: Proof by cotradictio. Assume there is a adjacet pair of same symbols i the remaider of the list, which is ot iduced by either the first, or the last occurece of a. So the, the origial list U would look like <... a...xax...a... >. This sequece has a alteratig sub-sequece of legth 5, cotradictig the fact that U is a order-3 DS sequece. Thus, removig all the occureces of a ad at most 2 other symbols from U gives a valid DS(-1,3)-sequece. So λ 3 () λ 3 ( 1) + λ 3() + 2 λ 3 ()(1 1 ) λ 3( 1) + 2 λ 3() λ 3( 1) λ 3() λ 3( 2) λ 3() λ 3(1) λ 3() O(1) + 2l() λ 3 () 2(l() + O(1)) Note: The tight boud for λ 3 () has bee achieved sice, ad is as follows: λ 3 () = Θ[α()] 4

5 3.4 Higher Order DS-Sequeces Followig table gives the bouds for the maximum legths of higher order DS-sequeces: Table 1: Bouds o λ s () for varyig values of s Order of the sequece(s) Boud o Legth(λ s ()) 3 Θ(α()) 4 Θ(2 α()).. 2s O(2 O(α()s 1) ) Ω(2 Ω(α()s 1) ) 2s + 1 O(α() O(α()s 1) ) 4 Motivatio: Lower Evelopes Let F = {f 1, f 2,..., f } be a collectio of real-valued cotious fuctios defied o a commo iterval I. Suppose that for each i j, the fuctios f i ad f j itersect i at most s poits (as a example, this would be the case with polyomials of degree s). The lower evelope of F is defied as: L(x) = mi(f i (x)), 1 i, x ɛi Let m be the smallest umber of subitervals I 1, I 2,..., I m of I such that for each I k, there exists a idex u k with L(x) = f uk (x) for all x ɛ I k. I other words, m is the umber of (maximal) coected portios of the graphs of the f i s which costitute the graph of L(x). The edpoits of the itervals I k are called the breakpoits or the trasitio poits of the evelope L(x). Assumig that I 1, I 2,..., I m are arraged i this order from left to right, put U(f 1, f 2,..., f ) =< u 1,..., u m > See the followig figure for a illustratio of this lower evelope sequece. 5

6 4.1 Lower Evelopes ad DS Sequeces Lemma U(f 1,..., f ) is a DS(, s)-sequece. Proof: For the first property, ote that, by defiitio, U = U(f 1,..., f ) does ot cotai a pair of adjacet idetical elemets (as the sub-itervals are take to be maximal coected portios). For secod property, suppose that there exist two distict idices a b so that U cotais a alteratig sub-sequece of a ad b of legth s + 2. By defiitio of the lower evelop, this would imply existece of s + 2 sub-itervals of I (I 1, I 2,..., I s+2 ), such that f a (x) < f b (x) for x ɛ (I 1 I 3...) ad f b (x) < f a (x) for x ɛ (I 2 I 4...). Sice f a ad f b are cotious ad their lower evelop forms s + 2 distict maximal coected itervals, there must exist s + 1 distict poits where f a crosses f b. This cotradicts the fact that f a ad f b itersect i at most s poits. So U(f 1,..., f ) is a DS(, s)-sequece. 6

7 5 Geometric Applicatios for DS Sequeces DS sequeces, as illustrated above, fid their applicatios i fidig the complexity of lower evelopes of a collectio of curves. Followig are some of the examples: Evelop of lies: Lies are degree 1 curves, so the lower evelop of lies would cosist of at most λ 1 () = distict maximal itervals, such that each of these itervals is a part of a sigle lie. Hece the complexity of the lower evelop of lies is. This boud may also be applied to fid the complexity of a face i a arragemet of lies. Lower Evelop of parabolas: Cosider parabolas, all of them opeig towards the positive y directio. Sice, a parabola is a degree-2 curve, so the lower evelop of this arragemet of parabolas would cosist of at most λ 2 () = 2 1 distict maximal itervals, such that each of these itervals is a part of a sigle parabola. This kid of arragemet arises i the Fortue Sweep algorithm to fid the vorooi diagram of poits. This boud helps i provig the ruig time of the aforesaid algorithm. Lower Evelop of curve segmets: As is illustrated i the figure, lower evelop of two lie segmets ca have 4 separate(maximal) sub-itervals such that each belogs to oe of the segmets. So, 2 lie segmets ca be said to itersect i 3 differet poits. So, the lower evelop of a arragemet of lie segmets would cosist of at most λ 3 () = Θ[α()] distict maximal itervals, such that each of these itervals is a part of a sigle lie segmet. 7

8 So, although the lower evelop of degree-s curves will correspod to a DS(,s)- sequece, the lower evelop of degree-s curved segmets will correspod to a DS(,s+2)-sequece due to the iductio of two extra trasitio poits (at the eds of the curved segmets). 8