Estimating the Size of the Largest Families not Containing Tree-like - - PowerPoint PPT Presentation
Estimating the Size of the Largest Families not Containing Tree-like Posets Wei-Tian Li Jerrold R. Griggs & Linyuan Lu Department of Mathematics University of South Carolina 24th Cumberland Conference May 14th 2011 Theorem (Sperner, 1928)
Wei-Tian Li
Jerrold R. Griggs & Linyuan Lu
Department of Mathematics University of South Carolina
24th Cumberland Conference May 14th 2011
Theorem (Sperner, 1928)
Let A be an inclusion-free family of subsets of [n]. Then |A| ≤ n ⌊ n
2⌋
Theorem (Sperner, 1928)
Let A be an inclusion-free family of subsets of [n]. Then |A| ≤ n ⌊ n
2⌋
Poset P = (P, ≤) A poset P1 = (P1, ≤1) contains another poset P2 = (P2, ≤ 2) as a subposet if there exists an injection f from P2 to P1 such that f (a) ≤1 f (b) whenever a ≤2 b. Example:
t t t
a b c P2
❅
a − → A f : b − → B c − → C
t t t
A B C P1
The Boolean lattice Bn = (2[n], ⊆) is the poset consisting of the power set of [n] and the inclusion relation as the partial order. A full chain C in Bn is a collection of n + 1 subsets as follows: ∅ ⊂ {a1} ⊂ {a1, a2} · · · ⊂ {a1, . . . , an}. A P-free family F is a collection of subsets of [n] such that it does not contain P as a subsposet. The largest size of a P-free family of subsets of [n] is denoted by La(n, P). The sum of middle k binomial coefficients is denoted by Σ(n, k) and B(n, k) is the family of subsets of middle k sizes.
D: De Bonis E: Erd˝
G: Griggs K: Katona L: Lu S: Sperner Sw: Swanepoel T: Tarj´ an Th: Thanh
La(n, P) for various posets
r r r r r r
. . .
k{
r r r ✓ ✓ ❙ ❙ r ✓ ✓ ❙ ❙ r ❙ ❙ ✓ ✓ r r r r r r r .... r r
✑ ◗ ◗
r
. . .
k{
r r r r r .... ✁ ✁ ❆ ❆ ✓ ✓ ❙ ❙
r
r r r r ❅ ❅ ❅
r r r ❅ ❅ ❅ r r r r r .... r r r r r .... r r ✟ ❍✏ ✏ P P ❍ ✟P P ✏ ✏
t s
. . .
k − 2{ k ≥ 3
r r r r ❇ ❇ ✧✧ ✧ ✡ ✡ ❜ ❜ ❜ ✂ ✂ ❏ ❏
... ...
r s r, s ≥ 2
r r r r r r ✄ ✄ ✄ ✄ ✄ ✄ ✄ ✄ ◗ ◗ ◗
... ...
k ≥ 2 2k 2k S, 1928 = Σ(n, 1) E, 1945 = Σ(n, k − 1) KT, 1983 ∼
⌊ n 2 ⌋
≤ 2 3
11
⌊ n 2 ⌋
DK, 2007 ∼ k
⌊ n 2 ⌋
∼
⌊ n 2 ⌋
= Σ(n, 2) GK, 2008 ∼
⌊ n 2 ⌋
∼ (k − 1)
⌊ n 2 ⌋
∼ 2
⌊ n 2 ⌋
∼
⌊ n 2 ⌋
⌊ n 2 ⌋
Let F ⊂ 2[n]. For each full chain C in Bn, if a subset E ∈ F ∩ C, then we associate E and C together to get a pair (E, C). Count the number of pairs in two different ways: (1) There are |E|!(n − |E|)! full chains containing a set E ∈ F. (2) For each full chain C, it contains |F ∩ C| subsets in F.
Let F ⊂ 2[n]. For each full chain C in Bn, if a subset E ∈ F ∩ C, then we associate E and C together to get a pair (E, C). Count the number of pairs in two different ways: (1) There are |E|!(n − |E|)! full chains containing a set E ∈ F. (2) For each full chain C, it contains |F ∩ C| subsets in F. The Lubell function ¯ h(F)(or ¯ hn(F)) of F is defined to be the number of pairs (E, C) over n!. ¯ h(F) =
|E|!(n − |E|!) n! =
1 n
|E|
= ave
C |F ∩ C|
(ave. no. of times C meet F)
Lemma
Let F be a collection of subsets of [n]. If ¯ h(F) ≤ m, for some real number m > 0, then |F| ≤ m n ⌊ n
2⌋
Moreover, if m is an integer, then |F| ≤ Σ(n, m), and equality holds if and only if (1) F = B(n, m) (when n + m is odd), or (2) F = B(n, m − 1) together with any
(n+m)/2
(n ± m)/2 when n + m is even.
h(F) ≤ m, then |F| =
1 ≤
n
⌊ n
2 ⌋
|E|
≤ m n ⌊ n
2⌋
If m is an integer, view ¯ h(F) as the weighted sum of the sets in F.
q q s s s... s s... s s s
(n + m)/2 (n − m)/2
s s s s s s s s s s s s s s s ⌊n/2⌋ s s s s s s s s s s s s s s s s s s s s s s s s s s s s
Given a poset P, let e(P) be the maximum m such that for all n, the union of any m consecutive levels m
i=1
[n]
s+i
contain P as a subposet. Example: For all n, the union of two levels [n]
k
[n]
k+1
0 ≤ k ≤ n − 1, contains no butterfly B = q q
q q
both contain two same (k − 1)-subsets, so e(B) ≥ 2. However, the union of three consecutive levels must contain B when n ≥ 3. So e(B) = 2.
The limit lim
n→∞
La(n, P) n
⌊ n
2 ⌋
Conjecture (Griggs and Lu, 2009)
For any finite poset, π(P) exists and is an integer.
The limit lim
n→∞
La(n, P) n
⌊ n
2 ⌋
Conjecture (Griggs and Lu, 2009)
For any finite poset, π(P) exists and is an integer. fake line The family B(n, e(P)) contains no P, so when it exists, π(P) must be at least e(P). Observation (Saks and Winkler) All posets with π(P) determined satisfied e(P) = π(P).
Let λn(P) be max ¯ h(F) over all P-free families F ⊂ 2[n]. Suppose F is P-free and |F| = La(n, P). Then La(n, P) n
⌊ n
2 ⌋
1 n
|E|
= ¯ h(F) ≤ λn(P). Define λ(P) = lim
n→∞ λn(P). We have
e(P) ≤ π(P) ≤ λ(P) if both limits exist.
Let λn(P) be max ¯ h(F) over all P-free families F ⊂ 2[n]. Suppose F is P-free and |F| = La(n, P). Then La(n, P) n
⌊ n
2 ⌋
1 n
|E|
= ¯ h(F) ≤ λn(P). Define λ(P) = lim
n→∞ λn(P). We have
e(P) ≤ π(P) ≤ λ(P) if both limits exist. There are posets that have π(P) < λ(P). Example: V2 =
q q q ❆✁
1 = π(V2) < λ(V2) = 2.
There are posets that have λn(P) ≤ e(P) for all n. For such posets e(P) = π(P) = λ(P). Example: The chain poset of size k, Pk is a poset that has such property.
There are posets that have λn(P) ≤ e(P) for all n. For such posets e(P) = π(P) = λ(P). Example: The chain poset of size k, Pk is a poset that has such property. We call such a poset a uniformly L-bounded poset.
There are posets that have λn(P) ≤ e(P) for all n. For such posets e(P) = π(P) = λ(P). Example: The chain poset of size k, Pk is a poset that has such property. We call such a poset a uniformly L-bounded poset. Various uniformly L-bounded posets:
s s s s s s
❅
❍ ❍ s s s s s s
s s s s s
s s s s s s ❅
✡ ❏ ❏ s s s s s s
❅
s s s s s ✂ ✂ ❇ ❇
❅ ❇ ❇ ✂ ✂ ❅ ❅
Proposition (Griggs, Li, and Lu, 2011)
For a uniformly L-bounded poset P with e(P) = m, La(n, P) = Σ(n, m) for all n. If F is a P-free family of the largest size, then F = B(n, m).
Theorem (Griggs, Li, and Lu, 2011)
The k-diamond poset Dk is a uniformly L-bounded poset if k is an integer in [2m−1 − 1, 2m − m
⌊ m
2 ⌋
s s s s s s
❅ ❅ ❅ ❅ ❅
❩ ❩ ❩ ❩ ❩❩❩❩ ✚ ✚ ✚ ✚
.............
Dk Example: k = 3, 4, 7, 8, 9, 15, 16, ....
Theorem (Griggs, Li, and Lu, 2011)
The harp poset H(ℓ1, ..., ℓk) is a uniformly L-bounded poset if ℓ1 > · · · > ℓk ≥ 3.
s ✘✘✘✘ ❳❳❳❳ s s s s s ✏✏✏ s s s PPP s ✑✑ s s ◗◗ s s ✁ ✁ s ❆ ❆s s ❇ ❇ ❇ ❇ ✂ ✂ ✂ ✂
H(7, 6, 5, 4, 3)
Let Q be a subset of poset P. Then Q+ := {p ∈ P | q ≤ p for some q ∈ Q}. Q− := {p ∈ P | p ≤ q for some q ∈ Q}.
Lemma
For a poset P, if there exists an element p ∈ P such that P = P1 ∪ P2, where P1 = {p}− and P2 = {p}+, then e(P) ≥ e(P1) + e(P2) and λn(P) ≤ λn(P1) + λn(P2).
s s s s s s
P1
❅
❍ ❍ sp s s s s s P2 ❅
✡ ❏ ❏
If a poset has the unique maximal(minimal) element, then the element is denoted by ˆ 1(ˆ 0).
If a poset has the unique maximal(minimal) element, then the element is denoted by ˆ 1(ˆ 0).
Theorem
Let P1, . . . , Pk be uniformly L-bounded posets such that each of which has ˆ 0 and ˆ
1 of Pi to the ˆ 0 of Pi+1, we
s s s sP1 s
❅
s s sP2 s
❅
. .
s s s sPk s
❅
Proof(Sketch). Induction on the number k.
s s s s s s
❅
❍ ❍ s s s s s s ❅
✡ ❏ ❏
←as p in the lemma
✞ ✝ ☎ ✆
. . .
s s s s s s ✂ ✂ ❇ ❇
❅ ❇ ❇ ✂ ✂ ❅ ❅
Theorem
Given uniform L-bounded posets P1,...,Pk such that each Pi has ˆ 0, let P be the union of them with the ˆ 0’s identified. W.l.o.g, assume that e(P1) ≥ e(P2) · · · ≥ e(Pk). Then e(P) = e(P1) and La(n, P) = (e(P) + o(1)) n ⌊ n
2⌋
and hence, e(P) = π(P).
PP P ❅ ❅
P1 ❅
❅ ❇ ❇ ❇
P2
✂ ✂ ✂
✏ ✏
Pk
rˆ
· · ·
Theorem
Given uniform L-bounded posets P1,...,Pk such that each Pi has ˆ 0, let P be the union of them with the ˆ 0’s identified. W.l.o.g, assume that e(P1) ≥ e(P2) · · · ≥ e(Pk). Then e(P) = e(P1) and La(n, P) = (e(P) + o(1)) n ⌊ n
2⌋
and hence, e(P) = π(P).
PP P ❅ ❅
P1 ❅
❅ ❇ ❇ ❇
P2
✂ ✂ ✂
✏ ✏
Pk
rˆ
· · ·
Theorem
Let P be the collection of uniformly L-bounded posets such that each poset P contains ˆ 0 and ˆ
0. Let PT be a poset obtained by replacing each “edge” of T by some poset in P. Then La(n, PT) = (e(PT) + o(1)) n ⌊ n
2⌋
and hence, e(PT) = π(PT).
t t t t t t t t t ✟✟✟✟ ❍ ❍ ❍ ❍ ❅ ❅
❆ ✁ ✁
Example:
t ❢ t t ❢ t t t t t t t t t t t t ❢ ❢ ❢ t t t t t t ❢ t t t t t ❢ t t t t ❢ t t t t t t t t t ❢
✏✏✏✏✏✏ ✘✘✘✘✘✘✘✘ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❵ ❅ ❅ ❍ ❍ ❍ ❍ P P P P P P ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❳ ❅ ❅
❆ ✁ ✁ ✑ ✑ ✑ ◗◗◗ ❅ ❅
✁ ❆ ❆ ❆ ❆ ✁ ✁
❅ ◗◗◗ ✁ ✁ ❆ ❆ ❆ ❆ ✁ ✁ ✑ ✑ ✑ ❆ ❆ ✁ ✁ ✁ ✁ ❆ ❆ ✁ ✁ ❆ ❆ ❆ ❆ ✁ ✁
❅
✟✟✟✟ ✦✦✦✦✦ ❈ ❈ ❙ ❙ ✄ ✄ ✓ ✓
e(PT) = π(PT) = 7, La(n, PT) ∼ 7 n ⌊ n
2⌋
Question Can we generalize the result to any tree?
Question Can we generalize the result to any tree? Question How to characterize the uniformly bounded posets?