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Lower Envelopes (Cont.)

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Lower Envelopes (Cont.)

Yuval Suede

- Lower Envelope is the graph of the pointwise minimum of the (partially defined) functions.
- Letbe the maximum number of pieces in the lower envelope.

- Each Davenport-Schinzel sequence of order s over n symbols corresponds to the lower envelope of a suitable set of n curves with at most s intersections between each pair.
- DS sequence important property:
- There is no subsequence of the form

- Let be the maximum possible length of Davenport-Schinzel sequence of order s over n symbols.
- Upper bound:

- Let W = a1a2 .. al be a sequence
- A non-repetitive chain in W is contiguous subsequence U = aiai+1 .. ai+k consisting of k distinct symbols.
- A sequence W is m-decomposable if it can be partitioned to at most mnon-repetitive chains.

- Let denote the maximum possible length of m-decomposable DS(3,n).
- Lemma (7.4.1): Every DS(3,n) is 2n-decomposable and so

- Proof:
- Let w be a sequence. We define a linear orderingon the symbols of w: we set a b if the first occurrence of a in w precedes the first occurrence of b in w.
- We partition w into maximal strictly decreasing chains to the ordering
- For example: 123242156543 -> 1|2|32|421|5|6543

- Proof (Cont.)
- Each strictly decreasing chain is non-repetitive.
- It is sufficient to show that the number of non-repetitive chains is at most 2n.

- Proof (Cont.)
- Let Uj and Uj+1be two consecutive chains: U1 .. UjUj+1 ..
- Let a be the last symbol of Uj and and(i) its indexand let b be the first symbol of Uj+1 and (i+1) its index :a b U1 .. UjUj+1 ..

- Claim:
- The i-th position is the last of aor the first of b
- if not, there should be b before a (b .. ab)
- And there should be a after the b (b .. ab .. a)
- And because of there should be a before the first b (otherwise the (i+1)-th position could be appended to Uj).
- So we get the forbidden sequence ababa !!

- Proof (Cont.)
- We have at most 2n Uj chains, because each sybol is at most once first, and at most once last.

- Proof (Cont.)
- We have at most 2n Uj chains, because each sybol is at most once first, and at most once last.

- Proposition (7.4.2) : Let m,n ≥ 1 and p ≤ m be integers, and let m = m1 + m2 + .. mpbe a partition of m into p addends, then there is partition n = n1 + n2 + .. + np + n* such that:

- Proof:
- Let w = DS(3,n) attaining
- Let u1u2 .. umbe a partition of w into non-repetitive chains where :
w1 = u1u2 .. Um1

w2 = um1+1um1+2 .. Um2

…

wp

- We divide the symbols of w into 2 classes:
- A symbol a is local if it occurs in at most one of the parts wk
- A symbol a is non-local if it appears in at least two distinct parts.
- Let n* be the number of distinct non-local symbols
- Letnk be the number of local symbols in wk

- By deleting all non-local symbols from wk we get mk-decomposable sequence over nk symbols (no ababa)
- This can contains consecutive repetitions, but at most mk-1 (only at the boundaries of uj)
- We remain with DS sequence with length at most (the contribution of local-symbols):

- A non-local symbol is middle symbol in a part of WK if it appears before and after Wk
- Otherwise it is non-middlesymbol in Wk

- For each Wk:
- Delete all local symbols.
- Deleteall non-middlesymbols.
- Delete all symbols (but one) of each contiguous repetition (we delete at most m middle symbols)
- The resulting sequence is DS(3,n*)

- Claim: The resulting sequence is p-decomposable.

- Each sequence Wk cannot contain b .. a .. b there is a before and a after
- Remaining sequence of Wk is non-repetitive chain.
- Total contribution of middle symbols in W is at most m +

- We divide non-middle symbols of Wk to starting and ending symbols.
- Let be the number of distinct starting symbols in Wk. A symbol is starting in at most one part, so we have
- We remove from Wk all but starting symbols and all contiguous repetitions in each Wk.

- The remaining starting symbols contain no abab because there is a following Wk
- What is left of Wk is DS(2, ) that has length at most 2-1
- Total number of starting symbols in all W is at most

- Summing all together:

- The recurrence can be used to prove better and better bound.
- 1st try: we assume m is a power of 2.
- We choose p=2, m1=m2= and we get :
Using we estimate the last expression by

- 2nd try: we assume (the tower function) for an integer
- We choose and
- Estimate using the previous bound.
- This gives:

- If then
- We chose so
- Recall that
- And since
- We get that

- It is possible to show that :
- But not today …