Properties of Posets in Non-crossing Pairings on Bitstrings

44 Pages·2009·0.355 MB·English

by  Wei Quan Julius Poh

CORNELL UNIVERSITY MATHEMATICS DEPARTMENT SENIOR THESIS
Properties of Posets in Non-crossing
Pairings on Bitstrings
A THESISPRESENTEDINPARTIALFULFILLMENT
OFCRITERIAFOR HONORSIN MATHEMATICS
Wei Quan Julius Poh
May 2009
BACHELOROF ARTS, CORNELL UNIVERSITY
THESIS ADVISOR(S)
Edward Swartz
Department of Mathematics
1
Acknowledgements
I would like to thank Professor Swartz for his priceless advice in
writing this thesis and for his continual guidance, as well as Profes-
sorKempandtheotherREUparticipantsforinspiringthisthesis.
Iwouldalsoliketothankmyparentsforworkinghardandsaving
every penny so that I could afford college, as well as for their moral
support.
ABSTRACT. Thereisastandardbijectionbetweenthenon-crossing
partitionlatticeNC(n)andthesetofnon-crossingpairingsNC (2n)
2
inthelargerlatticeNC(2n).Westudytheimageofthisbijectionon
subsets of NC (2n) corresponding to bit strings of length 2n. We
2
show that the resulting posets are connected, convex sub-posets
ofNC(n). Wegivesimplealgorithmstocalculatetheextremalele-
mentsintheseposets.Weshowthateachsetofpartitionsmapped
fromsubsetsof NC (2n)correspondingtobitstringsoflength2n
2
isuniqueuptoinversionof1sand0s. Finally, weshowthatthe
resultingposetisconstructible.
1. INTRODUCTION
1.1. PosetsofNon-CrossingPartitions. Wefollowtheintroduction
writtenin[1]byKempasbackgroundtothisthesis.
Let [n] denote the set {1,2,...,n}, and let P(n) denote the set of
all partitions of [n]; that is, π ∈ P(n) means that π is a collection
{B ,...,B }ofnon-emptydisjointsubsets B ⊆ [n]suchthat B ∪···∪
1 r j 1
B = [n]. (The subsets B are called the blocks of π.) There is a natural
r j
poset ordering on P(n): namely, we say π ≤ σ if π is a refinement of
σ (that is, each block of σ is contained in a block of π). It is a clas-
sical result (see, for instance [7], page 127) that (P(n),≤) is actually
a graded lattice, graded by the number of blocks in each partition;
the maximal and minimal elements (usually denoted 0 and 1 ) are
n n
picturedbelow.
1 0
6 6
FIGURE 1. The maximal and minimal elements, 1
6
and 0 , in the lattice P(6), along with another (cross-
6
ing)partitionof[6]. Theyarerepresentedhereaslinear
partition diagrams, which we will use through much
ofthisthesis.
2
3
There is a natural linear order on [n], and from it we can define
crossings: given a partition π of [n], two blocks B ,B ∈ π are said
1 2
to cross if there exist i , j ∈ B and i , j ∈ B such that i < i <
1 1 1 2 2 2 1 2
j < j . An example of a partition with a crossing is also given in
1 2
Figure 1. A partition is called non-crossing if no two of its blocks
cross. The set of non-crossing partitions of [n] is denoted NC(n), and
is a graded lattice in its own right under the refinement ordering.
It is, in fact, more symmetric than P(n), as it is self-dual (we will
discusstheassociatedcomplementationmapbelow).
Both P(n) and NC(n) were explicitly enumerated long ago; |P(n)|
is the Bell number B , while |NC(n)| is counted by the Catalan num-
(cid:16) (cid:17) n
ber C = 1 2n . This last result may be proved recursively quite
n n+1 n
easily, but for our purposes there is a somewhat simpler proof that
goesthroughthenon-crossingpairings. Apairingisapartitioneachof
whose blocks has two elements. Evidently, there are no pairings of
[n] if n is odd; P (2n) denotes the set of pairings of [2n] and NC (2n)
2 2
denotes the non-crossing pairings of [2n]. It is easy to check that
|P (2n)| = (2n−1)!!. It turns out that |NC(2n)| is also counted by the
2
Catalan number C , as Figure 2 demonstrates. This result affords a
n
slightly different proof that |NC(n)| = C , which motivates our main
n
resultsthatfollow. WerecordthisproofinFigure2.
1 2 3 4 5 6 ··· 2k 2k+1 ··· 2n
FIGURE 2. Inanynon-crossingpairingof[2n],1must
pair to an even number (else there would be an odd
number of indices enclosed by the pairing, forcing a
crossing). Hence, 1 can pair to any of 2,4,...,2n. If it
pairs to 2k then, as the figure demonstrates, the non-
crossing condition factors the remaining possible pair-
ings into two independent factors: the pairings of the
2(k − 1) indices enclosed by the block {1,2k}, and the
remaining 2(n−k) indices to the right. So, conditioned
on1pairingto2k,thereare|NC (2(k−1))·NC (2(n−k))
2 (cid:80) 2
pairings. In total, then, NC (2n) = n |NC (2(k −1))|·
2 k=1 2
|NC (2(n − k))|; this is the recurrence relation for the
2
Catalannumbers.
4
Proposition1.1. Thereisabijection θ : NC (2n) → NC(n).
n 2
Proof. Theideaistomap[2n]onto[n]ina2−1fashion,suchthatthe
induced map on partitions preserves the non-crossing conditions.
There are many ways to do this; we will use the map h(j) = (cid:100)j/2(cid:101).
For notational clarity we will use boldface when referring to the el-
ements of [n]. Explicitly, given any partition π ∈ P(2n), we form the
partition θ (π) ∈ P(n) as follows: i,j ∈ [n] are in the same block of
n
θ (π)iffthesetsh−1{i} = {2i−1,2i},h−1{j} = {2j−1,2j}shareacommon
n
block. Figure3belowdemonstrates.
FIGURE 3. Three partitions in P(8), and their images
underθ . Theactionofhistocollapseeachconsecutive
4
pair {2j − 1,2j} to the single index j, hence we can vi-
sualizetheactionofθ byconnectingtheseneighbours
n
(aswehavedonewithdottedlinesabove).
Now, let π ∈ NC(2n). Suppose (for a contradiction) that θ (π) has a
n
crossing – say B ,B ∈ θ (π) cross: i < i < j < j , where i ,j ∈ B
1 2 n 1 2 1 2 1 1 1
and i ,j ∈ B . By definition, i ,j ∈ B ∈ θ (π) means that one of the
2 2 2 1 1 1 n
two numbers h−1{i } = {2i − 1,2i } shares a common block V ∈ π
1 1 1 1
with one of the two numbers h−1{j } = {2j − 1,2j }. Similarly, one
1 1 1
of the two numbers h−1{i } = {2i − 1,2i } shares a common block
2 2 2
V ∈ π with one of the two numbers h−1{j } = {2j − 1,2j }. Notice
2 2 2 2
that 2i −1 < 2i < 2i −1 < 2i < 2j −1 < 2j < 2j −1 < 2j . It is
1 1 2 2 1 1 2 2
easy to see, therefore, that the blocks V ,V ∈ π must cross, which is
1 2
acontradiction. Hence, θ maps NC(2n)into NC(n).
n
In particular, the image θ (NC (2n)) of θ on non-crossing pairings
n 2 n
is contained in NC(n). We will now show that this mapping is a bi-
jection, by explicitly producing its inverse. From Figure 3 above,
we see that θ (π) is produced from π by “collapsing” each consec-
n
utive pair of indices in [2n] into a single index in [n]. To reverse
5
this, we must “double” each line in the partition diagram of the tar-
get image. Let σ ∈ NC(n), and consider any block B = {i ,...,i } in
1 k
σ, where i < ··· < i . Doubling these lines simply means that we
1 k
must include the following 2-blocks in ϕ (σ): {2i − 1,2i },{2i ,2i −
n 1 k 1 2
1},{2i ,2i − 1},...,{2i ,2i − 1}. An argument much like the one
2 3 k−1 k
aboveshowsthatthepairingϕ (σ)isnon-crossing,anditisasimple
n
matter to check that ϕ : NC(n) → NC (2n) and θ : NC (2n) → NC(n)
n 2 n 2
areinverses. (cid:3)
FIGURE 4. Theactionofϕ onapartitionπinNC(12).
12
Topologically, we take a tubular neighbourhood of the
1-skeleton that represents the partition diagram of π,
and then take the boundary of this minus the top
boundary segments. The resulting 1-skeleton is the
partitiondiagramof ϕ (π)in NC (24).
12 2
Remark 1.2. Aside from giving a concrete bijection between NC (2n)
2
andNC(n)forenumerationpurposes,themapθ allowsustoimport
n
the lattice structure of NC(n) into NC (2n). Note that every pairing
2
in NC (2n) has n blocks, and so all pairings are incomparable in the
2
refinement order on NC(2n). But we can pull back the ordering of
NC(n)viaθ ,asfollows.
n
Definition 1.3. Given π,σ ∈ NC (2n), say that π ≤ σ iff θ (π) ≤ θ (σ)
2 θ n n
in NC(n). Since θ is a bijection, this means that (NC (2n),≤ ) is a lattice
n 2 θ
isomorphicto NC(n). Referto ≤ asthethickenedordering on NC (2n).
θ 2
Hence,anysubsetT ⊆ NC (2n)hasanaturalposetstructureinduced
2
by the thickened ordering on NC (2n). This brings us to the special
2
subsets of non-crossing pairings we are motivated to study in this
thesis.
1.2. Knights and Ladies of the Round Table. King Arthur is host-
ing a dinner for his n Knights, to which he has invited n Ladies. The
2nguestsseatthemselvesrandomlyaroundtheRoundTable,before
Arthurcanassignseatstothem. Giventheirseatingarrangement,in
how many distinct ways can the Knights and Ladies pair off to chat
withoutanyconversationscrossing?
6
Put in slightly more precise terms, the pairings in question are de-
finedasfollows.
Definition1.4. Lets = s ···s beabitstringthatisbalanced: |{j; s =
1 2n j
0}| = |{j; s = 1}| = n. The set of non-crossing pairings of s, denoted
j
NC (s), is the set of non-crossing pairings π ∈ NC (2n) with the property
2 2
that, for any block B = {i, j} ∈ π, s (cid:44) s . That is, NC (s) consists of those
i j 2
non-crossingpairingsthatmatch 1sto0sins.
1 1
0 1 0 1
0 0 0 0
0 0 0 0
1 0 1 0
1 1 1 1
1 1
FIGURE 5. Twonon-crossingpairingsofthebitstring
s = 110001111000. They are represented here in circu-
larform.
Remark 1.5. Cyclic permutations of [n] induce lattice isomorphisms
of NC(n) (and these, together with the reflection [n] (cid:55)→ [n]∗ = (n,n −
1,...,2,1), generate all lattice isomorphisms, see [6]). Hence, it is
possibletodrawpartitiondiagramseitheronalineaswehavebeen
doinguptonow,oronacircleasinFigure5above. However,aswe
will discuss in Section 3, the thickened ordering ≤ on NC (2n) is not
θ 2
preserved under all cyclic permutations of [2n], and so we typically
preferthelinearrepresentationwhenconvenient.
The problem of the Knights and Ladies of the Round Table is the
question of enumerating NC (s) for any given balanced bit string s.
2
This is a surprisingly difficult problem, which has been addressed
in different ways in the recent papers [2, 3, 4, 5]. This set of pair-
ingshasimportantconnectionswithrandommatrixtheoryandfree
probability theory, which provided the motivation for its study. In
this thesis, we are primarily interested in studying the poset struc-
ture(NC (s),≤ ). Tothatend,weintroducesomeusefulnotation.
2 θ
7
Notation 1.6. For a balanced bit string s ∈ {0,1}2n, let T = T(s) denote
s
thesub-posetθ (NC (s))ofNC(n). Alternatively,T denotesthesetNC (s)
n 2 s 2
equippedwiththethickenedordering ≤ .
θ
Example1.7. Inanynon-crossingpairingof[2n],eachblockmustpair
twoindicesofoppositeparity,orelsetherewouldbeanoddnumber
of indices enclosed by that block resulting in a crossing. In other
words,anynon-crossingpairingisautomaticallyin NC (1010···10).
2
ThusT((10)n)isequal(asaposet)to NC(n).
Example 1.8. The reader may readily verify that the bit string 1n0n
has only one pairing, pictured in Figure 6 below. Hence, T(1n0n) is
thetrivialsingletonposet.
(cid:36) =
FIGURE 6. Thesinglepairing (cid:36)of1n0n.
Example 1.9. Consider the bit string s = 1011010010. The set NC (s)
2
contains 10 pairings. These pairings, their image under θ in NC(5),
5
andtheposet T ,aredisplayinginFigure7.
s
2. THE KREWERAS COMPLEMENT
Followingtheabovebackground,wewillnowproveafewtheorems
classifyingthestructureofalltheposetsT . Inordertodoso,wefirst
s
define the Kreweras complement. Let {1,2,...,n} and {1(cid:48),2(cid:48),...,n(cid:48)} be
twodisjointsetsofnelements,andordertheunionofthesetwosets
{1 < 1(cid:48) < 2 < 2(cid:48) < ... < n < n(cid:48)}.
Definition 2.1. The Kreweras complement of π ∈ NC(1,2,...,n) is the
largestelementσ ∈ NC(1(cid:48),2(cid:48),...,n(cid:48))suchthatπ∪σ ∈ NC(1,1(cid:48),2,2(cid:48),...,n,n(cid:48))
[6].
Rotation of π ∈ NC (2n) by an even number of places corresponds to
2
a rotation of θ (π), since we are essentially rotating the n underlying
n
elements of the partition θ (π). Rotation by one element to the left,
n
showninFigure8,givestheKrewerascomplement.
Lemma 2.2. Let π(cid:48) be π rotated one element to the left. Then, θ (π(cid:48)) is the
n
Krewerascomplementof θ (π).
n
8
1 0 1 1 0 1 0 0 1 0
π
0
π
0
T(1011010010)
π ∈ NC (1011010010) θ (π)
2 5
FIGURE 7. The ten pairings in NC (s) with s =
2
1011010010, and their images under the thickening θ .
5
Also shown is the poset T in NC(5). The astute ob-
s
server will notice that T is actually equal to the whole
s
interval[π ,1 ]in NC(5).
0 5
Proof. Thefunctionθ actingonπ(cid:48)joinsthe(2i)thand(2i+1)thelements
n
oftheoriginalπfori ≥ 1,andbringsthefirstelementtotherightand
joinsitwiththe(2n)th element,withblocksasdescribedearlier. Now,
if we place these n elements between our original n elements in the
way described in the definition of the Kreweras complement, θ (π(cid:48))
n
isapartitioninNC(n)whoseunionwiththeoriginalpartitionθ (π)is
n
in NC(2n). Toseethis,notethatinorderforacrossingtooccur,there
9
FIGURE 8.
must be some elements i < j of θ (π) and i(cid:48) < j(cid:48) of θ (π(cid:48)) such that i
n n
and j are in a block a, and i(cid:48) and j(cid:48) are in a block b with i < i(cid:48) < j < j(cid:48).
Furthermore,theseelementscanbechosensothat
• Therearenoelementsofeither aorbbetweeni(cid:48) andj.
• iisthegreatestelementof awhichislessthan i(cid:48).
• j(cid:48) istheleastelementof bwhichisgreaterthan j.
Butthiswouldimplythatthereisapairingbetweenelement2i(cid:48) and
2j(cid:48)−1inπ(cid:48),andanotherbetween2j−1and2iinπ,whichcorresponds
tocrossingpairingsin π.
Now we show that θ (π(cid:48)) is the largest partition whose union with
n
θ (π) is in NC(2n), which tells us that θ (π(cid:48)) is the Kreweras comple-
n n
ment of θ (π). We prove this by contradiction. If such a partition σ
n
is strictly larger than θ (π(cid:48)), then at least one of its blocks contains
n
more than one block of θ (π(cid:48)). Since the union of σ with θ (π) is non-
n n
crossing, any refinement in the interval [θ (π(cid:48)),σ] has a union with
n
θ (π) that is non-crossing, so we can take two blocks of θ (π(cid:48)) that lie
n n
inthesameblockof σandjointhemwithoutintroducinganycross-
ings. Letaandbbetwosuchblocks. Findapairofelements, i(cid:48) from
aandj(cid:48) fromtheb,suchthatnoelementsfromeitherblockareinbe-
tweenthose. Withoutlossofgenerality,assumethati(cid:48) < j(cid:48). Consider
theunderlyingpairingofπ(cid:48) wherethebits2i(cid:48)−1and2i(cid:48) correspond
to element i(cid:48) in θ (π(cid:48)). Note that these are really the bits 2i and 2i+1
n
in π. We see that 2i + 1 is paired to some bit 2k. Then in θ (π), the
n
element i+1 is in the same block as the element k. Since the union
ofθ (π(cid:48))andθ (π)isnon-crossing,kisneithertotheleftofinortothe
n n
rightofj. Thus,inθ (π),i < k < j. Butthenahasanelementbetween
n
i and j, a contradiction. Hence, rotation by an odd number of places
givesusarotationoftheKrewerascomplementofthepartition. (cid:3)