Post on 02-Jan-2016
description
Combinatorial Expansions for Paths, Chung-Feller Theorem
and Hankel Matrix
Speaker: Yeong-Nan Yeh
Institute of mathemetics, Academia sinica
2
Online
Part I. Functions of uniform-partition type Part II. Combinatorial interpretations for a class
of function equations Part III. Lattice paths and Fluctuation theory Part IV Paths with some avoiding sets shift
equivalence Part V. Addition formulas of polynomials and
Hankel determinants
3
Part I.
Functions of
uniform-partition type
4
Catalan paths
An n-Catalan path is a lattice path in the first quadrant starting at (0,0) and ending at (2n,0) with only two kinds of steps---up-step: U=(1,1) and down- step: D=(1,-1).
5
Catanlan number
1, 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, … ,
6
Catanlan number
The Catalan sequence was first described in the 18th century by Leonhard Euler, who was interested in the number of different ways of dividing a polygon into triangles.
7
Eugène Charles Catalan (May 30, 1814 – February 14, 1894) was a French and Belgian mathematician.
The sequence is named after Eugène Charles Catalan, who discovered the connection to parenthesized expressions during his exploration of the Towers of Hanoi puzzle.
E.C. Catalan, Note surune equation aux di erences finies, J. Math.Pures Appl. ff3(1838), 508-515.
((ab)c)d (ab)(cd) (a(bc))d a((bc)d) (ab)(cd)
8
Catanlan number
The counting trick for Catalan words was found by D. André in 1887
D. André, Solution directe du problème résolu par M. Bertrand, Comptes Rendus de l’Académie des Sciences, Paris 105 (1887) 436–437.
9
Chung-Feller Theorem (The number of Dyck path of semi-length n with m nonpositive up-steps is the n-th Catalan number and independent on m.)
K.L. Chung, W. Feller, On fluctuations in-coin tossing, Proc. Natl. Acad. Sci. USA 35 (1949) 605-608
We say Chung-Feller theorem is an uniform
partition of up-down type.
10
The classical Chung-Feller theorem was proved by Macmahon.
MacMahon, P. A. Memoir on the theory of the partitions of numbers, Philos. Trans. Roy. Soc. London, Ser. A, 209 (1909), 153-175.
Chung and Feller reproved the theorem by analytic method.
Chung, K. L. and Feller, W. On fluctuations in-coin tossing, Proc. Natl. Acad. Sci. USA 35 (1949) 605-608.
A combinatorial proof.
Narayana, T. V. Cyclic permutation of lattice paths and the Chung-Feller theorem, Skand. Aktuarietidskr. (1967) 23-30
Eu, Liu and Yeh proved the Chung-Feller theorem by using the Taylor expansions of generating functions.
Eu, S. P. Liu, S. C. and Yeh, Y. N. Taylor expansions for Catalan and Motzkin numbers, Adv. Appl. Math. 29 (2002) 345-357
Eu, Fu and Yeh gave a strengthening of the Chung-Feller Theorem and a weighted version for schroder paths.
Eu, S. P. Fu, T. S. and Yeh, Y. N. Refined Chung-Feller theorems for lattice paths, J. Combin. Theory Ser. A 112 (2005) 143-162
11
Bijection proofs.
D. Callan, Pair them up! A visual approach to the Chung-Feller theorem, Coll. Math. J. 26(1995)196-198.R.I. Jewett, K. A. Ross, Random walk on Z, Coll. Math. J. 26(1995)196-198.
Mohanty’s book devotes an entire section to exploring the Chung-Feller theorem.
Mohanty, S. G. Lattice path counting and applications, NewYork : Academic Press, 1979.
Narayana's book introduced a refinement of this theorem.
T.V. Narayana, Lattice path combinatorics, with statistical applications,Toronto;Buffalo : University of Toronto Press, c1979.
Callan reviewed and compared combinatorial interpretations of three different expressions for the Catalan number by cycle method.
D. Callan, Why are these equal? http://www.stat.wisc.edu/~callan/notes/ Huq developed generalized versions of this theorem for lattice paths.
A. Huq, Generalized Chung-Feller Theorems for Lattice Paths(Thesis), http://arxiv.org/abs/0907.3254
12
W.J. Woan, Uniform partitions of lattice paths and Chung-Feller Generalizations, Amer. Math. Monthly 108(2001) 556-559.
Another uniform partition for Dyck paths
The number of up-steps at the left of the rightmost lowest point of a dyck path
We say this uniform partition
is of left-right type.
13
Motzkin paths
An n-Motizkin path is a lattice path in the first quadrant starting at (0,0) and ending at (n,0) with only two kinds of steps---level-step: (1,0), up-step: U=(1,1) and down- step: D=(1,-1).
14
An uniform partition for Motzkin paths
Shapiro found an uniform partition for Motzkin path.
L. Shapiro, Some open questions about random walks, involutions, limiting distributions, and generating functions, Advances in Applied Math. 27 (2001), 585-596.
The number of steps at the left of the rightmost lowest point of a lattice path
This uniform partition is of left-right type.
Eu, Liu and Yeh proved this proposition.
Eu, S. P. Liu, S. C. and Yeh, Y. N. Taylor expansions for Catalan and Motzkin numbers, Adv. Appl. Math. 29 (2002) 345-357
15
Another uniform partition of up-down type for Motzkin paths.
The number of steps touching x-axis and under x-axis
第 16页
Our main results1. Eu, Liu and Yeh proved the Chung-Feller theorem by using the Taylor
expansions of generating functions.
Eu, S. P. Liu, S. C. and Yeh, Y. N. Taylor expansions for Catalan and Motzkin numbers, Adv. Appl. Math. 29 (2002) 345-357
2. Eu, Fu and Yeh gave a strengthening of the Chung-Feller Theorem and a weighted version for schroder paths.
Eu, S. P. Fu, T. S. and Yeh, Y. N. Refined Chung-Feller theorems for lattice paths, J. Combin. Theory Ser. A 112 (2005) 143-162
3. Ma and Yeh gave a generalizations of Chung-Feller theorems
J. Ma, Y.N. Yeh, Generalizations of Chung-Feller theorems, Bull. Inst. Math., Acad. Sin.(N.S.)4(2009) 299-332.
第 17页
Our main results
4. Ma and Yeh gave a characterization for uniform partitions of cyclic permutations of a sequence of real number
J. Ma, Y.N. Yeh, Cyclic permutations ofsequences and uniform partitions, The electronic journal ofcombinatorics 17 (2010), #R117.
5. Liu, Wang, Yeh gave the concepts of functions of Chung-Feller type
S.C. Liu, Y. Wang, Y.N. Yeh, Chung-Feller Property in View of Generating Functions, Electron. J. Comb. 18(2011), #P104.
6. Ma and Yeh gave a refinement of Chung-Feller theorems
J. Ma, Y.N. Yeh, Refinements of (n,m)-Dyck paths, European. J. Combin. 32(2011) 92-99.
第 18页
Our main results7. Ma and Yeh generalized the cycle lemma.
J. Ma, Y.N. Yeh, Generalizations of the cycle lemma, (Accepted 2014).8. Ma and Yeh gave a characterization for uniform partitions of cyclic
permutations of a sequence of real number
J. Ma, Y.N. Yeh, Rooted cyclic permutations of a lattice paths and uniform partitions, submitted.
9. Ma and Yeh studied a class of generating functions and their functions of Chung-Feller typeJ.Ma, Y.N.Yeh, Combinatorial interpretations for a class of functions of Chung-Feller theorem. submitted
19
Part II.
Combinatorial interpretations
for a class of
function equations
20
Uniform-partition Extension
),(
21
Liu, S. C. Wang, Y. and Yeh, Y. N. The function of uniform-partition type, submitted
the function of uniform-partition type for :
.)()(
1
)()(),( Then
.1
)(
1
1
),(CS
function generating heconsider t We
.0any for Suppose .)(Let
0
001
0 0 000 0,
,0
zyz
zzSyzyzS
y
zSyzySzyCS
y
yzsyzsz
y
ys
zyszyszyfzy
nksfzszS
n
n
nn
n
nn
nn
n
n n
n
k
nkn
n
k
nkn
n
n
k
nkkn
nknn
nn
22
An example for catalan sequence (up-down type)
23
An example for Motzkin sequence (left-right type)
221 MzzMM 22 ][zMzMzzzM 21 AA
Az
2**
*
][1 AA
Aw
rightmost lowest point
第 24页
• In general, given a combinatorial structure, let f(z) be a generating function correspoding with this combinatorial structure. We can obtain a functional equation which f(z) satisfies.
第 25页
Combinatorial structure Generating
function
f(z)
Functional equation which
f(z) satisfies
Catalan path:(1,1),(1,-1) in the first quadrant
C(z)
Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant
M(z)
Schroder path:(1,1),(1,-1),(2,0) in the first quadrant
S(z)
第 26页
Combinatorial structure Generating
function
f(z)
Functional equation which
f(z) satisfies
Catalan path:(1,1),(1,-1) in the first quadrant
C(z) C(z)=1+z[C(z)]2
Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant
M(z) M(z)=1+zM(z)+z2[M(z)]2
Schroder path:(1,1),(1,-1),(2,0) in the first quadrant
S(z) S(z)=1+zS(z)+z[S(z)]2
第 27页
Combinatorial structure Generating
function
f(z)
Functional equation which
f(z) satisfies
Catalan path:(1,1),(1,-1) in the first quadrant
C(z) C(z)=1+z[C(z)]2
Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant
M(z) M(z)=1+zM(z)+z2[M(z)]2
Schroder path:(1,1),(1,-1),(2,0) in the first quadrant
S(z) S(z)=1+zS(z)+z[S(z)]2
Given a functional equation , how to find a combinatorial structure and its corresponding generating function f(z) such that f(z) satisfies this functional equation?
第 28页
Combinatorial structure Generating
function
f(z)
Functional equation which
f(z) satisfies
Catalan path:(1,1),(1,-1) in the first quadrant
C(z) C(z)=1+z[C(z)]2
Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant
M(z) M(z)=1+zM(z)+z2[M(z)]2
Schroder path:(1,1),(1,-1),(2,0) in the first quadrant
S(z) S(z)=1+zS(z)+z[S(z)]2
Given a functional equation , how to find a combinatorial structure and its corresponding generating function f(z) such that f(z) satisfies this functional equation?
第 29页
Combinatorial structure Generating
function
f(z)
Functional equation which
f(z) satisfies
Catalan path:(1,1),(1,-1) in the first quadrant
C(z) C(z)=1+z[C(z)]2
Motzkin path:(1,1),(1,-1),(1,0) in the first quadrant
M(z) M(z)=1+zM(z)+z2[M(z)]2
Schroder path:(1,1),(1,-1),(2,0) in the first quadrant
S(z) S(z)=1+zS(z)+z[S(z)]2
??? ??f(z)
Given a functional equation , how to find a combinatorial structure and its corresponding generating function f(z) such that f(z) satisfies this functional equation?
第 30页
Let . The recurrence relation which the sequence satisfies is independent on a0(z) . Hence, let a0(z) =1..
We focus on the following functional equation.
Let S be a set of vector in the plane Z×Z. We also call the set S step set and vectors in S steps.
Let L be a function from S to N, where N is the set of nonnegative integers . We call L a step-length function of the set S and L(s) the step length of the step s in the set S repectively.
Let W be a function from S to R, where R is the set of real numbers. We call W a weight function of the set S and W(s) the weight of the setp s in the set S respectively,
第 31页
Let P be a sequence of vectors (x1,y1)…(xn,yn) in the set S such that y1+…yn=0, y1+…yi≥0 for all i. We call P an S-path. Let Ω(S) be the set of all S-paths.
Define the L-length of a S-path P= (x1,y1)…(xn,yn) , denoted by L(P), as L(P)=L(x1,y1)+…L(xn,yn).
Define the W-length of a S-path P= (x1,y1)…(xn,yn) , denoted by W(P), as W(P)=W(x1,y1)…W(xn,yn).
Define a generating function f(z) as
第 32页
.1
equation functional thesatisfies
function generating Then the
. (n,0) to(0,0) from
paths-S of weightsof sum thebe and 1Let
.1110
)},,2,1,,2,1)1,{()}1,0{(Let
:Theorem
1 1
0
0
ijr
i
m
jij
n
nn
n
ij
[f(z)]zaf(z)
zff(z)
ff
a),w(j,-i),w(
mjriijS
第 33页
• A decomposition of a S-path.
P=(0,1)P1(0,1)P2(0,1)P3…Pi-1(j,-i+1)Pi
W(1,1)=1,W(j,-i+1)=ai,j
第 34页
Part III. Lattice paths
and Fluctuation theory
35
• Fluctuation theory is the name given to that part of probability theory which deals with the fluctuations of the partial sums sn=x1+...+xn of a sequence of random variables x1,…,xn.
36
• Consider x=(r1,…rn). Let s0=0,si=r1+…+ri
• Let p(x) be the number of positive sums si
• Let m(x) be the index where the maximum is attained for the first time.
37
x partial sum p(x) m(x)
(1,2,3) (3,1,-2) (3,4,2) 3 2
(1,3,2) (3,-2,1) (3,1,2) 3 1
(2,1,3) (1,3,-2) (1,4,2) 3 2
(3,1,2) (-2,3,1) (-2,1,2) 2 3
(2,3,1) (1,-2,3) (1,-1,2) 2 3
(3,2,1) (-2,1,3) (-2,-1,2) 1 3
r1=3,r2=1,r3=-2
38
x partial sum p(x) m(x)
(1,2,3) (1,2,-2) (1,3,1) 3 2
(1,3,2) (1,-2,2) (1,-1,1) 2 1
(2,1,3) (2,1,-2) (2,3,1) 3 2
(3,1,2) (-2,1,2) (-2,-1,1) 1 3
(2,3,1) (2,-2,1) (2,0,1) 2 1
(3,2,1) (-2,2,1) (-2,0,1) 1 3
r1=1,r2=2,r3=-2
39
• Fix X=(r1,…rn).
• Let Xi=(ri,…rn,r1,…,ri-1) (cyclic permutations.)
• Let P(X)={p(Xi)| i=1,2,…,n}
M(X)={m(Xi)| i=1,2,…,n}
40
• F. Spitzer, (1956)
• Let X be a sequence of real numbers of length n such that sn=0 and no other partial sum of distinct elements vanishes. Then P(X)=M(X)=[0,n-1].
41
Remark
• Fix X=(r1,…rn). Suppose r1+…+rn=m.• Let m=0.
The conditions in the results of Spitzer are necessary and sufficient conditions for P(X)=[0,n-1]
The conditions in the results of Spitzer are not necessary for M(X)=[0,n-1].
42
• T.V. Narayana, (1967)
• Let n be a positive integer and X be a sequence of integers with -n<ri< 2 for all i=1,2,…,n such that sn=1. Then P(X)=[n].
43
• J. Ma, Y.N. Yeh, Generalizations of The Chung-Feller Theorem II, submitted.
• Let n be a positive integer and X be a sequence of integers with -n<ri< 2 for all i=1,2,…,n such that sn=1. Then M(X)=[n].
44
Two natural problems
• What are necessary and sufficient conditions for M(X)=[n] and P(X)=[n] if m>0?
• What are necessary and sufficient conditions for M(X)=[0,n-1] and P(X)=[0,n-1] if m<=0?
45
• Fix X=(r1,…rn). Given an index j=1,…,n, define
LP(X;j)={i|sj>si,i=1,…,j-1} and
RP(X;j)={i|sj>=si i=j+1,…,n}
46
• Let m>0.The necessary and sufficient conditions for M(X)=[n] are sm(X)-si>=m for all i in LP(X;m(X))The necessary and sufficient conditions for P(X)=[n] are sj-si>=m for any j in [n] and any all i in [0,j-1]\LP(X;j)
47
• Let m<=0.The necessary and sufficient conditions for M(X)=[0,n-1] are si -sm(X)<m for all i in RP(X;m(X))
The necessary and sufficient conditions for P(X)=[0,n-1] are sj-si<m for any j in [n] and any all i in [0,j-1]\LP(X;j)
48
Part IV. Paths with some avoiding
sets shift equivalence
49
Let M be a Motzkin path.
LM: the set of the height of the level stepsLM={0,3}
PM: the set of the height of the peaks PM={2,1}
VM: the set of the height of the valleys VM={0,1}
50
Motzkin paths from (0,0) to (2(n-1),0) without level of height larger than 0
51
Peaks-, Valleys- and Level-avoiding Sets CPBA ,,
CVBPAL MMM ,,
we consider the Motzkin path such that Given the sets
A: level-avoiding set B: peak-avoiding sest C: valley-avoiding set
CVBPAL MMM ,,
A: level-restricting set B: peak-restricting setC: valley-restricting set
(1):
(2):
52
Generating FunctionsCBAslkn
m,,;,,,
the number of the Motzkin path of length n withk levles, l peaks and s valleys
slknCBAslknCBA
qyxzmqzyxM,,;,,,,,
),,,(
53
)}0(])0(1[
)0(1}{)1(1{
)0(1
,,,,
,,1,1,12
,,
,,
CqxzMAM
xzMAyBMz
xzMA
M
CBACBA
CBACBA
CBA
CBA
54
Some results
,}1{,;Pnm
,}{,; hPnm
E. Deutsch, Dyck path enumeration, Discrete Math. 204 (1999), 167--202.
P. Peart and W-J. Woan, Dyck paths with no peaks at height k,J. Integer Seq. 4 (2001), Article 01.1.3.
,}2{,;Pnm
the n-th Fine number
the (n-1)-th Catalan number
55
Shu-Chung Liu, Jun Ma, Yeong-Nan Yeh, Dyck Paths with Peak- and Valley-Avoiding Sets, Stud. Appl. Math. 121:263-289
S.-P. Eu, S.-C. Liu, and Y.-N. Yeh, Dyck Paths with Peaks Avoiding or Restricted to a Given Set, Stud. Appl. Math. 111 Iss 4 (2003), 453--465.
CBPknm
,,;,
,,;, BPknm
• Continued fractions• Close forms• Shift equivalence
56
0
)(n
nn zazH
0
)(n
nn zbzG
nsn ba
)()( zGzzH m
Suppose and
Then
if and only if there is a positive integer m such that
is a polynomial.
nsn ba
nqnp ba
If there exist non-negative integers p and q such that
57
Some interesting shift equivalence
numberCatalanmm
mmm
ssNnsNns
sNnsNnsNn
...
...
,}3,2{,;,}2{,;
}1,0{,,;}0{,,;,,;
numberCatalandGeneralizemm ssENnsNn ...
,O,;O,E,;
numberCatalandGeneralizem
mmm
mmm
sNns
sNnsNnsNns
sNnsNnsNn
...
...
...
33
333333
333
{1},}1{,;
{0},}0{,;{2},}2{,;{1},}1\{}1{,;
{0,2},,;{0,1},,;{1,2},,;
58
Some interesting shift equivalence
SloaneinAmm ssNnsNn 025265...
3333 }0{,}1{,;}2{,}0{,;
SloaneinAmm ssNnsNn 127389...
3333 }0{,}2{,;}2{,}1{,;
SloaneinAmmm ssENnsENnsNn 126120...
1,}2{)1E(,;,E,;O,O,;
numberMotzkinmm
mm
mmm
ssNnsNns
sNnsNns
sNnsNnsNn
...
...
...
,E,;,\{1}O,;
1E,}2{,;E,,;
2O,}3,2{,;1O,}2{,;O,,;
59
Continued fractionsIt is difficult to represent
as a continued fractions
),,,(,,
qzyxMCBA
60
Close form• Matrix methods
We just consider
)1,,,(),,(,,,
zyxMzyxFBABA
61
),,(]})1(1[
)0({1),,(
,1,1,12
,
zyxFyBMz
xzAzyxF
BACBA
BA
62
||and ||
|| and ||
|| and ||
|| and ||
BA
BA
BA
BA
),,(
),,(
),,(
),,(
,
,
,
,
zyxF
zyxF
zyxF
zyxF
P
N
P
),,(,
zyxFBA
63
)()1(1 22,, BiyzAixzziBA
|| and || BA
xz
Tx 2
10
1
01
xSx
}max,max{max BAm )max(max BA
Let
),,(
),,(),,(
,2
,2
, zyxFzdc
zyxFzbazyxF
mm
mm
BA
mBAiBA
STdc
ba m
imm
mm
,,,,
1
1
Then
where
64
)1(2 m
mm
mm zdc
ba
65
(A,B) being Congruence classes
} somefor )(mod|{: IjkjnnIk
),(),( kk JIBA
),,(
),,(),,(
,2
,2
, zyxFdzc
zyxFbzazyxF
kk
kk
kk
JI
JI
JI
kJIiJIST
dc
ba k
i,,,,
1
1
Define the congruence classes
Let
Then
where
66
then we say that F(x) is algebraic
0)(...)()( 01
1 xayxayxa d
dd
d
If F(x) is a solution of an equation
)(FadThe algebraic degree of F(x):
67
kk JBIA
BA
BA
BA
BA
and
||and ||
|| and ||
|| and ||
|| and ||
),,(,
zyxFBA
),,(,
zyxFBA
is algebraic since it is a solution of aquadratic Equations
2)),,((,
zyxFadBA
68
Problem I• Characterize the set
dzyxFadBABA
)),,((|),(,
69
Problem II• Given a sequence a1,a2,…,an,…, find a
pair (A,B) of the sets such that
nBAn azFz )(][
,
70
1,1,,|),,()( yxBABA
zyxFzF
BAnm
,;We consider the coefficients
in
71
|| and || ii BA
)(
)()(
,2
,2
, zFzdc
zFzbazF
ii
ii
BA ii
2211 ,;,; BAnsBAnmm
))1((
))1((22
11121
2
22222
22
2
2
1
zdzdccz
zdzdcczz r
rk
][| 21212
22121
22121 zdbcazzdbcazddcc k
)]1([
)1()1(
121212
212121
211221zdbcbdaz
zdbcbdazdddcdc
k
1)max(max}max,max{max iiiii ABBAr Let
Suppose
Thenif and only if
(1)
(2)
(3)
k Such that
72
Bijection methods
B1
}2{}1{,}0{)1(;2,; BAnBAn
mm
Suppose
Then
In fact, if B1
1)()(,
2}2{)1(,}0{)1(
zFzzFBABA
73
Problem III
),(),( 2211 BABA s
2211 ,;,; BAnsBAnmm If the sequences
then we say that (A1,B1) and (A2,B2) shift equivalent,
denoted by
• Give a characterization of ),(),( 2211 BABA s
74
}1{},{; inm
is shift equivalent to the Fibonacci numbers
The sequence mn+2i;{i},{1} has Chung-Feller property, i.e., mn+2i;{i},{1} =Fn is independent on i, where Fn is the n-th Fibonacci number.
75
]1,2[},{; iinm
is shift equivalent with the Central binomial coefficients
Replace valleys(DU) of height 0 and level into peak DU and U respectively
Remove the first and final steps
(i=1)
Left factor of Dyck path
第 76页
Addition formulas and Hankel matrix
studied.y been widel has
,
sequence, theof
matrices Hankel theof tsdeterminan theevaluate
toproblem the,0 and }{ sequence aFor
1,0
0
njikji
nn
a
ka
第 77页
101,0
1,01
1,0
2det
1det
1det
,2
1
1 numbersCatalan For the
mjinjimji
njiji
njiji
n
ji
jinc
c
c
n
n
nc
第 78页
)6 (mod 4 ,3 if1
)6 (mod 5 ,2 if0
)6 (mod 1 ,0 if1
det
1det
n,length of pathsMotzkin ofnumber
count the numbersMotzkin For the
1,01
1,0
n
n
n
m
m
m
njiji
njiji
n
第 79页
.2det
2det
paths,der oSchr large ofnumber
count the numbersder oSchr large For the
2
1
1,01
2
1,0
0
n
njiji
n
njiji
nn
r
r
r
identity. Jacobi-Desnanot
by det and det
by evaluated becan det that Notice
10110
10
ni,jjini,jji
ni,jmji
aa
a
第 80页
resultsOur
.)(
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第 81页
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第 85页
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第 87页
第 88页
Thank you for your attention!