Ordered Partitions and Drawings of Rooted Plane Trees

0.15 MB·English

by  Qingchun Ren

ORDERED PARTITIONS AND
DRAWINGS OF ROOTED PLANE TREES
3
1 QINGCHUNREN
0
2
Abstract. Westudytheboundedregionsinagenericsliceofthehyperplane

n arrangementinRnconsistingofthehyperplanesdefinedbyxiandxi+xj. The
a boundedregionsareinbijectionwithseveralclassesofcombinatorialobjects,
J includingthe orderedpartitions of [n]all ofwhoseleft-to-rightminimaoccur

7 at odd locations and the drawings of rooted plane trees with n+1 vertices.
2 These are sequences of rooted plane trees such that each tree in a sequence
canbeobtainedfromthenextonebyremovingaleaf.

]
O
C
. 1. Introduction
h
t We define the combinatorialobjects to be studied in this paper. The firstone is
a
m the following hyperplane arrangementon Rn:
[ = x ,1 i n x +x ,1 i<j n .
Hn { i ≤ ≤ }∪{ i j ≤ ≤ }
1 Let P be the hyperplane in Rn defined by
v
P = l x +l x + +l x =1 ,
7 1 1 2 2 n n
{ ··· }
2 where l l l >0. We are interested in the set of bounded regions of
1 2 n
3 ≫ ≫ ··· ≫
the hyperplane arrangement P = H P :H in the affine space P.
6 Hn∩ { ∩ ∈Hn}
. Definition 1. An ordered partition of [n]= 1,2,...,n (also calleda preferential
1
{ }
0 arrangementbyGross[2])isanorderedsequence(A1,...,Ak)ofdisjointnon-empty
3 subsets whose union is [n]. Each A is called a block. A left-to-right minimum of
i
1 (A ,...,A )ism =min(A A ),where1 i k. Wesaythataleft-to-right
1 k i 1 i
: ∪···∪ ≤ ≤
v minimum mi occurs at an odd location if mi Aj for some odd j.
∈
i
X Definition 2. A signed permutation ((a ,a ,...,a ),σ) of [n] is a permutation
1 2 n
r (a1,a2,...,an) of [n] together with a map σ: [n] 1 . σ(i) is called the sign
a → {± }
of i. It has decreasing blocks if a > a for any two a ,a with the same
i i+1 i i+1
sign. Aleft-to-right minimumof((a ,a ,...,a ),σ)ism =min(a ,...,a ),where
1 2 n i 1 i
1 i n. For simplicity, we indicate the sign of a by writing a+ or a−.
≤ ≤ i i i
Definition 3. Abuild-tree codeisasequencec c c ofpairsc =(a ,σ )where
1 2 n i i i
···
0 a i 1 and σ 1 such that (a ,σ )=(0, 1). For simplicity, we write
i i i i
a+≤or a≤− in−stead of (a∈,σ{±).} 6 −
i i i i
Definition 4. Anincreasing labelingofarootedplanetreeT ofn+1vertices,also
called a simple drawing or a heap order, is a bijection λ: T 0,1,...,n such
→ { }
that if u,v T and u is a child of v, then λ(u) > λ(v). λ(v) is called the label of
∈
v. An increasingly labeled tree is a rooted plane tree together with an increasing
labeling. For simplicity, we identify a vertex with its label if there is no confusion.
Date:January29,2013.
1
2 QINGCHUNREN
Definition 5. Let (T,λ) be an increasingly labeled tree. The right associate of
a vertex v is the sibling u to the right of v with the smallest label such that
λ(u) > λ(v), and λ(u) is smaller than the labels of all siblings between u and v,
if such a u exists. (T,λ) is a Klazar tree if it satisfies the following property: for
any vertex v with a right associate u, v is not a leaf, and λ(u) is larger than the
minimum of all labels of children of v.
Figure 1 shows two different linear extensions of the same tree. The tree in the
right of Figure 1 is a Klazar tree. The tree in the left is not a Klazar tree, because
the vertex 3 has a right associate 4, but 3 is a leaf.
Figure 1. Two increasingly labeled trees
0 0
1 2 2 1
3 4 4 3
The last object is the set of drawings of rooted plane trees with n+1 vertices:
Definition 6. A drawing of a rooted plane tree T is a sequence of rooted plane
trees T = root ,T ,...,T =T such that for each 0 i n 1, the tree T can
0 1 n i
{ } ≤ ≤ −
be obtained from T by removing a leaf together with its pendant edge.
i+1
The main result of this paper is
Theorem 7. The following seven sets are in bijection:
(1) The set of bounded regions in the affine hyperplane arrangement P.
n
H ∩
(2) The set of ordered partitions of [n] all of whose left-to-right minima occur
at odd locations.
(3) The set of signed permutations of [n] with decreasing blocks all of whose
left-to-right-minima have positive signs.
(4) The set of build-tree codes of length n such that there is a v+ after (but
not necessarily adjacent to) each v−.
(5) The set of build-tree codes of length n such that there is a v+ before (but
not necessarily adjacent to) each v−.
(6) The set of Klazar trees with n+1 vertices.
(7) The set of drawings of rooted plane trees with n+1 vertices.
Let b be the common cardinality of these sets. Set b = 1. Then, the sequence
n 0
b has exponential generating function
n
{ }
∞ xn ex
b = .
nn! 2 ex
n=0 r −
X
The above generating function is due to Klazar [3] in a paper that discusses
various counting problems of rooted plane trees. The bijections between (5), (6)
and(7)arestudiedbyCallan[1]. OurnotationsaredifferentfromCallan’sbecause
we use the top-to-bottom convention for trees in this paper. Callan shows that b
n
ORDERED PARTITIONS AND DRAWINGS OF ROOTED PLANE TREES 3
alsoequalsthenumberofperfectmatchingsontheset[2n]inwhichnoevennumber
is matched to a larger odd number. The sequence b begins with
n
{ }
1,1,2,7,35,226,1787,16717,....
This sequence can be found in the On-Line Encyclopedia of Integer Sequences [4,
A014307].
Ourinvestigationoriginatesfromalatent allocation modelingenomics[5]. Max-
imum likelihoodestimation for this statisticalmodel involves finding localmaxima
of the the function
n
xi ui xi+xj uij
| | | |
i=1 1≤i<j≤n
Y Y
onthe hyperplane P′ = x +x + +x =1 ,where u ,u are genericpositive
1 2 n i ij
{ ··· }
integers. ByatheoremofVarchenko[8],theMLdegreeofthestatisticalmodel,i.e.
the number of local maxima of the above function equals the number of bounded
regions in P′. Our hyperplane P can be considered as a deformation of P′:
n
H ∩
P = l x +l x + +l x =1 ,
1 1 2 2 n n
{ ··· }
wherel ,...,l aregenericparameters. The numberofbounded regionsin P
1 n n
H ∩
givesanupperboundonthenumberofboundedregionsin P′,andthusgives
n
H ∩
an upper bound on the maximum likelihood degree of the latent allocation model.
Without loss of generality, we will assume that l l l >0.
1 2 n
≫ ≫ ··· ≫
Our hyperplane arrangement is refined by the well-studied hyperplane ar-
n
H
rangement of type B :
n
= x ,1 i n x x ,1 i<j n .
n i i j
B { ≤ ≤ }∪{ ± ≤ ≤ }
Section2discussestheanalogousproblemfor ,anditshowsthatthebounded
n
B
regionsin P areinbijectionwithincreasinglylabeledtreeswithn+1vertices.
n
B ∩
Based on this, Section 3 proves our main result, Theorem 7.
2. Bounded regions in a slice of
n
B
First, we consider the regions of the central hyperplane arrangement . The
n
hyperplanes x in divide Rn into 2n orthants. In each orthant, thBe hyper-
i n
B
planes x x divides the orthant into n! regions, one for each total ordering of
i j
±
x ,..., x . Thus, for each signed permutation ((a ,a ,...,a ),σ), we can asso-
1 n 1 2 n
| | | |
ciate it with a region R of :
n
B
R= (x ,...,x ) Rn : x > x > > x ,sgn(x )=σ(i) .
{ 1 n ∈ | a1| | a2| ··· | an| i }
Clearly, this is a bijection between regions of and signed permutations of [n].
n
B
Lemma 8. Let R be a regionof . Let ((a ,a ,...,a ),σ) be the corresponding
n 1 2 n
B
signed permutation. Then
(a) The polyhedronR P is nonemptyandbounded ifallleft-to-rightminima
∩
of (a ,a ,...,a ) have positive signs.
1 2 n
(b) ThepolyhedronR P isemptyifallleft-to-rightminimaof(a ,a ,...,a )
1 2 n
∩
have negative signs.
(c) The polyhedron R P is nonempty and unbounded if neither of the above
∩
holds.
4 QINGCHUNREN
Proof. Lete ,e ,...,e betheunitvectors(1,0,...,0),(0,1,...,0),...,(0,0,...,1)
1 2 n
inRn. Then,theregionRistheconespannedoverR bythefollowingnvectors:
≥0
v =σ(a )e ,
1 1 a1
v =σ(a )e +σ(a )e ,
2 1 a1 2 a2
,
···
v =σ(a )e + +σ(a )e .
n 1 a1 ··· n an
Let L be the ray R (v ). Fix the linear form f(x) = l x + +l x on Rn.
i ≥0 i 1 1 n n
···
Then,
f(v )=σ(a )l + +σ(a )l .
i 1 a1 ··· i ai
Since l l >0, f(v ) has the same sign as σ(min(a ,...,a )).
1 n i 1 i
≫ ··· ≫
(a) By assumption, each left-to-right minima min(a ,...,a ) has positive sign.
1 i
Therefore, each f(v ) is positive. Hence, P = f(x) = 1 intersects the ray L at
i i
{ }
v /f(v ). Then, R P is the simplex with vertices v /f(v ),...,v /f(v ), which
i i 1 1 n n
∩
is nonempty and bounded.
(b) Similarly, each f(v ) is negative. Then, f is negative on R. Thus, P =
i
f(x)=1 does not intersect R.
{ }
(c) In this case, some f(v ) are positive and the others are negative. Say
i
f(v ) > 0 and f(v ) < 0. Then, P = f(x) = 1 intersects L at v /f(v ). More-
i j i i i
{ }
over, f(f(v )v f(v )v )) = 0. Hence, R P contains the affine ray v /f(v )+
i j j i i i
R+(f(v )v f(−v )v )). Thus, R P is non∩empty and unbounded. (cid:3)
i j j i
− ∩
Theorem 9. The following four sets are in bijection:
(1) The set of bounded regions in P.
n
B ∩
(2) The setofsignedpermutationsof[n]allofwhoseleft-to-rightminima have
positive signs.
(3) The set of build-tree codes of length n.
(4) The set of increasingly labeled trees with n+1 vertices.
Moreover,thecommoncardinalityofthesesetsequals(2n 1)!!=1 3 5 (2n 1).
− · · ··· −
Proof. (1) (2). The regions of P are exactly R P for regionR of such
n n
↔ B ∩ ∩ B
that R P is nonempty. Therefore, it follows from Lemma 8 that the cardinalities
∩
of the sets (1) and (2) are equal.
(2) (3). We constructabijectionbetweenthesetofbuild-treecodesoflength
↔
nandthesetofsignedpermutations[n]allofwhoseleft-to-rightminimahavepos-
itive signs. Given a build-tree code c c c , we construct a signed permutation.
1 2 n
···
We start from the empty signed permutation. In each step, we look at c and add
i
one element to the signed permutation:
(i) If c =0+, then add i to the beginning with positive sign.
i
(ii) If c =j+ for j >0, then add i immediately after j with the opposite sign
i
from j.
(iii) Ifc =j− forj >0,thenaddiimmediatelyafterj withthesamesignasj.
i
We obtaina signedpermutation of[n] in this way. In eachstep, if (ii) or (iii) hold,
the left-to-right minima stay the same. If (i) holds, then i becomes a new left-to-
right minimum, and we construct it to have the positive sign. Thus, the signed
permutation we constructed has the property that all of its left-to-right minima
have positive signs. On the other hand, given a signed permutation of [n] all of
ORDERED PARTITIONS AND DRAWINGS OF ROOTED PLANE TREES 5
whoseleft-to-rightminima havepositivesigns,wecanreversethe constructionand
obtain a build-tree code. It is straightforwardto verify that this gives a bijection.
(3) (4). See Callan [1]. We construct a bijection between the set of build-
↔
tree codes of length n and the set of increasingly labeled trees with n+1 vertices.
Given a build-tree code c c c , we construct an increasingly labeled tree. We
1 2 n
···
start from the rooted plane tree with no non-root vertices. In each step, we look
at c and add one leaf to the tree with label i:
i
(i) If c =0+, then add i as the leftmost child of the root.
i
(ii) If c =j+ for j >0, then add i as the leftmost child of vertex j.
i
(iii) If c =j− for j >0, then add i as the immediate right neighbor of j.
i
On the other hand, given an increasingly labeled tree with n+1 vertices, we can
reversetheconstructionandobtainabuild-treecode. Itisstraightforwardtoverify
that this gives a bijection.
It is easy to see that there are (2n 1)!! build-tree codes of length n, because
each c has exactly 2i 1 independent−choices. (cid:3)
i
−
Example10. Table1illustrateshowweobtainasignedpermutationof 1,2,3,4,5,6
{ }
from the build-tree code 0+1−1+1+0+3+ with the construction above.
Table 1. Constructing a signed permutation from a build-tree code
Step Build-tree code Rule applied Signed permutation
0
1 0+ (i) beginning, + sign 1+
2 0+1− (iii) after 1, same sign 1+2+
3 0+1−1+ (ii) after 1, opposite sign 1+3−2+
4 0+1−1+1+ (ii) after 1, opposite sign 1+4−3−2+
5 0+1−1+1+0+ (i) beginning, + sign 5+1+4−3−2+
6 0+1−1+1+0+3+ (ii) after 3, opposite sign 5+1+4−3−6+2+
Example 11. Table 2 illustrates how we obtain an increasingly labeled tree with
7 vertices from the build-tree code 0+1+1+1−0+3+ with the construction above.
Remark 12. Stanley [7, Section 5.1] computes the characteristic polynomial for
thehyperplanearrangement . Thenumber(2n 1)!!isthesignedconstantterm
n
B −
of the characteristic polynomial.
3. Bounded regions in a slice of
n
H
First, we consider the regions of the central hyperplane arrangement . Let
n
(A ,A ,...,A ) be an ordered partition of [n]. We define a cone in Rn byH
1 2 k
R+ = (x ,...,x ) Rn: x >0 for i A for odd j,
1 n i j
{ ∈ ∈
x <0 for i A for even j,
i j
∈
x > x for i A ,i A ,1 j k 1 .
| i1| | i2| 1 ∈ j 2 ∈ j+1 ≤ ≤ − }
Equivalently,the3rdconditionabovecanbereplacedbytheconditionthatx +x
i1 i2
has the same sign as ( 1)j+1. Let R− = R+.
− −
6 QINGCHUNREN
Table 2. Constructing an increasingly labeled tree from a build-
tree code
Step Build-tree code Rule applied Tree
0
0
0
(i) leftmost
1 0+
child of 0
1
0
(ii) leftmost
2 0+1+ 1
child of 1
2
0
(ii) leftmost
3 0+1+1+ 1
child of 1
3 2
0
(iii) right
4 0+1+1+1− 1 4
neighbor of 1
3 2
0
(i) leftmost
5 0+1+1+1−0+ 5 1 4
child of 0
3 2
0
5 1 4
(ii) leftmost
6 0+1+1+1−0+3+
child of 3
3 2
6
ORDERED PARTITIONS AND DRAWINGS OF ROOTED PLANE TREES 7
Lemma 13. There is a 2 to 1 map from the set of regions of to the set of
n
H
ordered partitions of [n].
Proof. First, we notice that R+ and R− are defined by inequalities involving only
the linear forms in . Also, allsigns of x areimplied by the defining inequalities
n i
H
of R+ and R−. These inequalities also imply the order of the x except those i
i
| |
in the same block. Since the x with i in the same block have the same sign, all
i
signsof x +x areimplied by the defining inequalities. Therefore, R+ andR− are
i j
indeed regions of .
n
Onthe otherhaHnd, givena genericpoint (x ,...,x ) Rn, we claimthatit lies
1 n
∈
in a region of the form R+ or R−. Indeed, we order the x by their absolute value:
i
x > x > > x . So, we get a permutation (p ,...,p ) of [n]. Then, we
| p1| | p2| ··· | pn| 1 n
grouptogether consecutive segments of the p such that the x has the same sign.
i pi
In this way, we get an ordered partition of [n]. It follows that the point lies in R+
or R−, depending on the sign of x . Thus, we get a 2 to 1 correspondence from
p1
the set of regions of to the set of ordered partitions of [n]. (cid:3)
n
H
Lemma 14. Let (A ,A ,...,A ) be an ordered partition of [n]. Let R+, R− be
1 2 k
the two corresponding regions of . Then
n
H
(a) If(A ,A ,...,A )hasallleft-to-rightminimaatoddlocations,thenR+ P
1 2 k
∩
is nonempty and bounded, and R− P is empty.
∩
(b) Otherwise, both R+ P and R− P are nonempty and unbounded.
∩ ∩
Proof. Since refines ,aregionof P isnonempty(resp. unbounded)ifand
n n n
B H H ∩
onlyifitcontainsa nonempty(resp. unbounded) regionof P. Fromthe proof
n
B ∩
of Lemma 13, R+ (resp. R−) contains exactly the regions in corresponding to
n
B
signed permutations ((a ,...,a ),σ) such that σ(a )=1, (resp. σ(a )= 1) and
1 n 1 1
−
(A ,A ,...,A )canbeobtainedfrom(a ,...,a )bygroupingtogetherconsecutive
1 2 k 1 n
elements with the same signs.
(a) Let (x ,...,x ) R+. Since the x for i A have sign ( 1)j−1, an odd
1 n i j
∈ ∈ −
location in (A ,A ,...,A ) corresponds to elements in ((a ,...,a ),σ) with posi-
1 2 k 1 n
tive signs. Therefore,(A ,A ,...,A )has allleft-to-rightminima atoddlocations
1 2 k
ifandonlyifallleft-to-rightminima of((a ,...,a ),σ)havepositivesigns. There-
1 n
fore, R+ contains only regions of type (a) in Lemma 8, which are nonempty and
bounded. Thus, R+ P is nonempty and bounded. Similarly, R− contains only
∩
regions of type (b) in Lemma 8. Thus, R− P is empty.
∩
(b) Similarly, both R+ and R− contains only regions of type (c) in Lemma 8.
Thus, both R+ P and R− P are nonempty and unbounded. (cid:3)
∩ ∩
Now we prove our main result.
Proof of Theorem 7. (1) (2). Theregionsof P areexactlyR P forregions
n
↔ H ∩ ∩
R of such that R P is nonempty. Therefore, it follows from Lemma 14 that
n
H ∩
the cardinalities of the sets (1) and (2) are equal.
(2) (3). For each ordered partition (A ,A ,...,A ), we construct a signed
1 2 k
↔
permutation with decreasing blocks by writing elements of each A in decreas-
i
ing order and concatenating them to form a permutation. The signs of the ele-
ments of A is ( 1)i−1. For example, the ordered partition (15,246,3) is sent to
i
−
5+1+6−4−2−3+. It is clear that ordered partitions of [n] all of whose left-to-right
minimaoccuratoddlocationsareinbijectionwithsignedpermutationsof[n]with
decreasing blocks all of whose left-to-right minima have positive signs.
8 QINGCHUNREN
(3) (4). A signed permutation fails to have decreasing blocks if and only
↔
if there are two adjacent elements u,v with the same sign such that u < v. In
other words, v is added after u with the same sign, and no more element is added
after u afterwards. Under the bijection described in the proof of Theorem 9, this
translates exactly to the condition that there is no u+ after some u−. Thus, the
bijectionsendssignedpermutationswithdecreasingblocksallofwhoseleft-to-right
minima have positive signs to build-tree codes such that there is a v+ after each
v−, and vice versa.
(4) (5). Given a build-tree code, we keep the numerals in the build-tree
↔
code, and reverse the order of the signs over each fixed numeral. For example,
0+1+1−2+2+2− goes to 0+1−1+2−2+2+. In this way, the build-tree codes such
that there is a v+ before eachv− aresent exactly to the build-tree codes suchthat
there is a v+ after each v−, and vice versa.
(5) (6). An increasingly labeled tree (T,λ) can be considered as a process of
↔
constructingthe tree T by adding verticesin the orderdeterminedby λ. The right
associate of a vertex v, if it exists, is the first vertex added as the immediate right
neighbor of v. Under the bijection described in the proof of Theorem 9, the label
of the right associate of v corresponds to the location of the fist appearance of v−
in the build-tree code. The condition that there is a v+ before each v− translates
to the condition that the right associate of v, if it exists, is added after at least
one child of v. This is exactly the defining condition for Klazar trees. Thus, the
bijection sends Klazar trees to build-tree codes such that there is a v+ before each
v−, and vice versa.
(6) (7). See Callan [1]. We elaborate the proof for completion. Given a
↔
drawing T ,T ,...,T = T, we can reconstruct an increasing labeling of T as
0 1 n
follows: suppose we have already constructed an increasing labeling of T . By
n−1
definition, T can be obtained from T by removing a leaf. We label this leaf
n−1
n, and label the rest of the tree in the same way as in T . In this way, we get
n−1
an increasing labeling of T. To make the construction unambiguous, if there are
multipleleavesinT thatcanberemovedtogetT ,wealwayschoosetheleftmost
n−1
possibleone. Weclaimthattheresultingincreasinglylabeledtree(T,λ)isaKlazar
tree. If it is not, then there is a vertex v with a rightassociateu suchthat either v
is a leaf or λ(u) is smaller than the labels of all children of v. Since T contains
λ(u)
exactlythe verticesinT withlabel λ(u), the verticesuandv areadjacentleaves
≤
inT . Therefore,removingeitheruorv inT resultsinthesamerootedplane
λ(u) λ(u)
tree. Sincev istotheleftofu,bytheconstructionabove,wewouldchoosev rather
than u in the λ(u)th step. So we get a contradiction. Thus, (T,λ) is a Klazartree.
An increasingly labeled tree (T,λ) naturally gives a drawing T ,T ,...,T =T,
0 1 n
by setting T to contain exactly the vertices with labels i. Clearly this is a left
i
≤
inverse of the construction process above. It suffices to prove that different Klazar
trees give different drawings. Assume that two different Klazar trees (T,λ) and
(T′,λ′) give the same drawing. Let T (resp. T′) be the subtree of T (resp. T′)
i i
spannedbyverticeswithlabels i. Then, T andT′ areisomorphic. Thus,wecan
≤ i i
identify T with T′. Let k be the smallest positive integer such that (T ,λ ) and
k |Tk
(Tk′,λ′|Tk′)arenotisomorphicincreasinglylabeledtrees. Moreover,bothTk andTk′
are Klazar trees. Without loss of generality, we may assume that k =n.
Let u (resp. u′) be the vertex of T labeled n in (T,λ) (resp. (T,λ′)). Note that
T (resp. T′ ) can be obtained from T by removing u (resp. u′). Then, both
n−1 n−1
ORDERED PARTITIONS AND DRAWINGS OF ROOTED PLANE TREES 9
u and u′ are leaves of T, and u = u′ by the minimality of k. Let v be the lowest
6
common ancestor of u and u′. Let v ,v ,...,v be the children of v, ordered from
1 2 s
lefttoright. Supposethatu(resp. u′)isadescendentofvj (resp. vj′). Thenj =j′
6
by the choice of v. If neither u nor u′ is a child of v, then the size of the subtree
of T rooted at v would be 1 smaller than the subtree of T′ rooted at v . If
n−1 j n−1 j
exactly one of u or u′, say u, is a child of v, then v would have one more child in
T′ than in T . Both cases contradict our assumption that T is isomorphic
n−1 n−1 n−1
to T′ . Thus, both u andu′ are childrenofv. Without loss ofgenerality,we may
n−1
assume that u′ is to the right of u. Since λ′(u′) = n > λ′(u), the vertex u has a
rightassociatein(T,λ′). However,uisaleaf. Thus,the conditionfor(T,λ′)being
a Klazar tree is violated. So we get a contradiction.
It is shown by Klazar [3] that the cardinality of the set (7) has the given expo-
nential generating function. (cid:3)
Wepresentanalternativeproofbycountingthecardinalityoftheset(2). Wesay
that the type of an ordered partition (A ,...,A ) is the set A ,...,A , which
1 k 1 k
{ }
is a partition of [n]. An ordered partition of type 1 , 2 ,..., n is just a
{{ } { } { }}
permutation of [n].
Lemma 15. Let p be the number of permutations of [n] whose all left-to-right
n
minima occurs at odd locations. Set p =1. Then
0
∞ xn 1+x
p = .
n
n! 1 x
n=0 r −
X
Proof. We foundthis proofinanunpublishedpostofCallan. For anypermutation
(a ,...,a ) of [n] all of whose left-to-right minima occur at odd locations, we can
1 n
construct a permutation of [n 1] by removing a and decrementing all elements
n
−
greater than a by 1. This new permutation has all left-to right minima at odd
n
locations. On the other hand, for any permutation of [n 1] whose allleft-to-right
−
minima occurs at odd locations and any a [n], we can construct a permutation
n
∈
of [n] by incrementing all elements greater than a by 1 and adding a to the
n n
end. This new permutation has all left-to-right minima at odd locations if and
only if n is odd or n is even and a > 1. Therefore, from this correspondence
n
we get p = np for odd n and p = (n 1)p for even n. By induction,
n n−1 n n−1
−
p =((n 1)!!)2 for even n and p =n!!(n 2)!! for odd n.
n n
− −
Then,
∞ xn xn xn
p = ((n 1)!!)2 + n!!(n 2)!!
n
n! − n! − n!
n=0 neven nodd
X X X
∞ x2m ∞ x2m+1
= ((2m 1)!!)2 + (2m+1)!!(2m 1)!!
− (2m)! − (2m+1)!
m=0 m=0
X X
∞ x2m ∞ x2m+1
= ((2m 1)!!)2 + ((2m 1)!!)2
− (2m)! − (2m)!
m=0 m=0
X X
∞ x2m
=(1+x) ((2m 1)!!)2 .
− (2m)!
m=0
X
10 QINGCHUNREN
On the other hand,
∞
1 1/2
= ( 1)mx2m
√1 x2 m −
− mX=0(cid:18) (cid:19)
∞
(2m 1)!!
= − x2m
2m(m!)
m=0
X
∞ x2m
= ((2m 1)!!)2 .
− (2m)!
m=0
X
So
∞ xn 1+x 1+x
p = = .
n
n! √1 x2 1 x
nX=0 − r −
(cid:3)
Lemma 16. The number of ordered partitions of [n] of type A ,...,A all of
1 k
{ }
whose left-to-right minima occurs at odd locations equals p .
k
Proof. We may replace each A by minA without affecting the locations of the
i i
{ }
left-to-right minima. Therefore, we can reduce the problem to the case of ordered
partitions of k distinct numbers. Thus, the number is p . (cid:3)
k
Letb denotethenumber(2). Setb =0. Then,itfollowsfromthecomposition
n 0
formula [6, Theorem 5.1.4] that
∞ xn 1+(ex 1) ex
b = − = .
n=0 nn! s1−(ex−1) r2−ex
X
Example17. Thebuild-treecodes0+1+1+1−0+3+ and0+1−1+1+0+3+ inExam-
ple 10 and Example 11 can be obtained from each other by reversing the sequence
of signs over each fixed numeral in the build-tree code. Therefore, the objects in
Table 1 and Table 2 are in bijection.
Example 18. Figure 2 shows the 7 bounded regions in P.
3
H ∩
These 7 bounded regions are labeled (1),(2),...,(7). They are:
(1) x >0,x >0,x >0
1 2 3
(2) x >0,x >0,x <0, x , x > x
1 2 3 1 2 3
| | | | | |
(3) x >0,x <0,x >0, x , x > x
1 2 3 1 3 2
| | | | | |
(4) x >0,x <0,x <0, x > x , x
1 2 3 1 2 3
| | | | | |
(5) x >0,x <0,x >0, x > x > x
1 2 3 1 2 3
| | | | | |
(6) x >0,x >0,x <0, x > x > x
1 2 3 1 3 2
| | | | | |
(7) x >0,x >0,x <0, x > x > x
1 2 3 2 3 1
| | | | | |
Table 3 shows various objects that are in bijection with the 7 bounded regions.
Acknowledgements. IwouldliketothankBerndSturmfelsforguidingmethrough
the entire project. Also, I would like to thank Lior Pachter for an insightful dis-
cussionthat motivated the project. This projectwas supported by DARPA (grant
DARPA-11-65-Open-BAA-FP-068).