Nested Chain Movement of length 1 of Beta Number in James Abacus Diagram

18 Pages·2016·0.22 MB·English

GlobalJournalofPureandAppliedMathematics.
ISSN0973-1768Volume12,Number 4(2016),pp. 2953-2969
©ResearchIndiaPublications
http://www.ripublication.com/gjpam.htm
Nested Chain Movement of length 1 of
Beta Number in James Abacus Diagram
EmanF.Mohomme
SchoolofQuantitativeSciences,
CollegeofArtsandSciences,
UniversitiUtaraMalaysia06010Sintok,
Kedah,Malaysia.
NazihahAhmad
SchoolofQuantitativeSciences,
CollegeofArtsandSciences,
UniversitiUtaraMalaysia06010Sintok,
Kedah,Malaysia.
HaslindaIbrahim
SchoolofQuantitativeSciences,
CollegeofArtsandSciences,
UniversitiUtaraMalaysia06010Sintok,
Kedah,Malaysia.
AmmarSeddiqMahmood
DepartmentofMathematicsCollegeofEducation,
UniversityofMusul.
Abstract
Jamesabacusdiagramisagraphicalrepresentationforanypartitionµofapositive
integer t. One way of producing the diagram is by using beta number with the
special number of even columns e, where e (cid:1) 2. This paper constructs a new
methodforpartitionµwithasinglemotionofnestedchainmovementoflength1
inJamesabacusdiagram. First,theestablishmentofanarithmeticsequenceamong
the nested chains of the diagram position is considered. Then, for the movement,
we select several beta numbers as the initial points in every chain. The location
of the rest of the beta numbers in the James abacus diagram would be changed
anticlockwise by length 1 accordingly when the positions of initial beta numbers
are changed. Using these new nested chains, a new diagram Atc1 that displays
a new partition, is constructed. Furthermore, guides, which are finite number of
partitions that are obtained from the original partition after adding zeros, are de-
veloped. Thenumberofcommonbetanumbersamongthesedevelopedguidesare
then determined. We have established rules to obtain new diagram using a single
motion of nested chain movement of length 1. The new diagram can be used in
areas of number theory and design. Finally, the proposed method is employed as
a special type of James abacus diagram where the number of columns is an even
integersmallerthanthenumberofrowsinthediagram.
AMSsubjectclassification:05A17.
Keywords:Jamesabacusdiagram,Betanumber,ArithmeticsequenceandPartition.
1. Introduction
James abacus diagram is a graphical representation of a special type of non-increasing
(cid:1)b
sequenceµ = (µ1,µ2,...,µb)calledpartitionoft if|µ| = µi = t withµ(b+1) = 0
i=1
where b is the number of partition parts and t is any positive integer [6]. James abacus
diagramcamefromtheideaoftheabacusdiagram. Foranynumberofcolumn(runner)
e, James abacus diagram is seen as an important component in modern algebra which
plays a key role in Iwahori−Hecke andq-Schur algebras [11], [1], [2], [9]. The James
abacus diagram configuration for beta numbers, β ,β ,β ,...,β can be created by
1 2 3 b
rearranging them on the runners, where β = µ + b − i for 1 (cid:2) i (cid:2) b [8]. There
i i
are infinitely many James abacus diagrams since one or more zeros can be added to the
partition. The diagram of b is obtained after adding s − 1 zeros to µ where James
s
abacus diagram of b have b + s − 1 beta numbers [7]. Furthermore, several types
s
of movement of beta numbers have been constructed by previous researchers such as
movingallofthemashighaspossibleintherunner[3],addingemptyrunner[4],adding
full runner [2], removing runner [1], reflecting their positions in leading diagonal [12],
scanningmovement[9],usingjustifiedpositionsmovement[10]andmovingsingle-step
bead[7]. Guidesormaindiagramsarefinitenumberofpartitionsthatareobtainedafter
addingzeros. Astudyhasproventhattherearecommonbetanumbersamongguides[7].
Some studies have applied the properties of James abacus diagram in countless fields
and constructed several diagrams by transforming the diagram position that preserves
the original structure of the diagram [14], [15], [16], [18]. Among the transformation
constructed are reflection in line x and y, rotation by 90o,180o and 270o as well as the
composition of both geometric transformations of reflection and rotation by 90o,180o
and270o. Asinglechainmovementwasappliedincasee = 2 ofanylength[19].
In this paper, a new diagram will be constructed by applying a single motion of
nested chain movement of length 1 and is called diagram Atc1. The idea of nested
chain movement is adopted from graph theory [8]. A nested chain comprises one outer
NestedChainMovementoflength1ofBetaNumberinJamesAbacusDiagram 3
chainandseveralinnerchains. Thispaperseekstoaddressseveralquestions. CanJames
abacusdiagrampositionsbedividedintonestedchains? Howmanypositionsaretherein
everychain? Whataretherelationshipsamongthediagrampositions? Isthemovement
ofthebetanumberpositionsinthenewmaindiagramsregularornot? Ifitisregular,can
the new main diagrams of b that depend on the new main diagram for b be designed?
2 1
In general, is the intersection of the new main diagrams equal to the intersection of the
originalmaindiagrams?
2. Preliminaries
Thissectionbrieflydiscussessomebasicdefinitionsandtheorems. Thepartitionofeach
James abacus diagram can be connected by e runners which are labelled from left to
right as 0 to e − 1, where e = 2x, x ∈ Z+. Beta numbers will be referred to as bead
positions. ThebeadpositionsontheJamesabacusdiagramwhicharelabelledfromleft
torightandcontinuesfromtoptobottomstartingwith0arelocatedacrosstherunners.
Thebeadpositionsme,me+1,...,(m+1)e−1arelocatedinrowmofthediagram,
ascanbeseeninFigure1aforcasee = 4.
Definition 2.1. Let µ = (µ ,µ ,...,µ ) represents the partition of a positive integer
1 2 b
(cid:1)b
t such that |µ| = µi = t with µ(b+1) = 0 where b is the number of partition parts.
i=1
µ µ µ
Theβ-sequenceofpartitionµ,β ,β ,...,β ,isdefinedby
1 2 b
β(µ,b) = {µ +b−1,µ +b−2,...,µ +b−b}
1 2 b
wheretheconstructionisdiscussedin[8].
The abacus configuration for µ is called James abacus diagram of b where each b
s s
hasb+s −1beadsands isapositiveintegergreaterthanorequaltoone.
Example2.2. Letµ = (8,8,6,3,2,1,1,1,1,0,0)beapartitionof31whereb = 9. If
s = 3,thenµ = 0,µ = 0andthesetofβ-numbersis{18,17,14,10,8,6,5,4,3,1,0}.
10 11
Every beta number will be represented by (o) for a bead position in the diagram
while the empty bead position will be represented by (-). Figure 1 shows a general
diagram and final diagram of e = 4 displaying the arrangement of beads representing
µ = (8,8,6,3,2,1,1,1,1).
Diagramsofb wheres = 1,2,...,earedefinedasmaindiagrams. Theintersection
s
ofthemaindiagramsisalsoexamined.
Theorem2.3. [13] Let
(µ ,µ ,...,µ ) = (µτ1,µτ2,...,µτk).
1 2 b 1 2 k
The numerical value of the resulting intersection of main James abacus diagrams is
4 EmanF.Mohomme,etal.
0 1 2 3 o o – o
4 5 6 7 o o o –
8 9 10 11 −→ o – o –
12 13 14 15 – – o –
16 17 18 19 – o o –
a b
Figure1: Thebeadpositionsontheabacuswith4runners(a)Generaldiagram(b)Final
diagram.
(cid:2)e
denoted by # m·d· and it is equal to φ in the case of no existence of common bead
s=1
positions.
(cid:2)e
1. # m·d· = φ ifτ = 1,k ∈ N.
k
s=1
2. Let(cid:4)bethenumberofpartsofλwhichsatisfiestheconditionτ ≥ eforsomek,
k
(cid:2)e (cid:1)(cid:4)
then# m·d· = τ −(cid:4)(e−1).
k
s=1 k=1
3. Method
TheconstructionofthenewdiagramisdevelopedbyfirstconvertingtheoriginalJames
abacusdiagramtomatrixform. Jamesabacusdiagrampositionsaredividedintoseveral
nested chains. Then, a new method is constructed for partitioning µ with a single
motion of nested chain movement of length 1. Next, main diagrams are developed and
theintersectionpointsofthemaindiagramsareobtained.
3.1. ConvertJamesabacusdiagramtomatrixform
In order to formulate this work easily, the James abacus diagram is converted to matrix
form. If (β ,β ,...,β ) is a set of beta numbers of a partition µ of a positive integer t
1 2 b
andr isthenumberofrowsinJamesabacusdiagramofb then
1
(cid:3) (cid:4)
β
r = 1 .
e
ThenumberofrowsinaJamesabacusdiagramofb isr +s −1wheres ∈ N.
s
Lemma 3.1. Suppose that {β ,β ,··· ,β } a set of beta numbers of a partition µ =
1 2 b
(µ ,µ ,··· ,µ )ofapositiveintegert whereb ∈ Z,theneveryJamesabacusdiagram
1 2 b
NestedChainMovementoflength1ofBetaNumberinJamesAbacusDiagram 5
positioncanbeconvertedtoamatrixAr×e by
me+n ⇒ a(m+1)(n+1)
Proof. InJamesabacusdiagram,thebeadpositionsincolumnnandrowmarenumbered
as (me +n) for e (cid:1) 2. The row numbers are from 0 to r −1 and column numbers are
from0toe−1whileeverymatrix(m×n)consistsofmrowsfrom1tor andncolumns
from 1 to e where m and n are positive integers; so any position me + n in the James
abacusdiagramisanelementa(m+1)(n+1) inthematrix(r ×e). Then
βb = me+n ⇒ a(m+1)(n+1)
wherer referstothenumberofrowsinthediagram. (cid:3)
3.2. Nestedchains
Nested chains are formed from James abacus diagram position. These chains are num-
beredfrom1toi wherei isapositiveintegerandchain1istheouterchain. Thereisno
intersection between any two of the chains as shown in Figure 2 where there are three
nestedchainsforµ = (35,16,8,8,6,3,2,1,1,1,1)ande = 6.
b
1
- o o o o -
o - o - - -
o - - o o -
- - - - - -
- o - - - -
- - - - - -
- - - - - -
- - - o - -
Figure2: Nestedchaindiagram.
FromFigures1band2,wehavethefollowingcases.
i. Ife = 2,thediagramhasonlyonechainwhichconsistsofalldiagrampositions.
ii. Ife = 4 andr (cid:1) 4,thisimpliesthatthediagramhastwochains.
(cid:127) Chain1={a ,a ,a ,a : 1 (cid:2) m (cid:2) 10,1 < n < 4}.
m1 m4 1n rn
(cid:127) Chain2={a ,a : 2 (cid:2) m (cid:2) 9}
m2 m3
1. Ife = 6andr ≥ 6,thisimpliesthatthediagramhasthreechains.
(cid:127) Chain1={a ,a ,a ,a : 1 (cid:2) m (cid:2) 7,1 < n < 6}.
m1 m6 1n rn
6 EmanF.Mohomme,etal.
(cid:127) Chain2={am2,am5,a2n,a((r−1)n) : 2 (cid:2) m (cid:2) 6,2 < n < 5}.
(cid:127) Chain3={a ,a : 3 (cid:2) m (cid:2) 5}
m3 m4
Finally,wecanmakeageneralizationforanyevennumbere.
Definition 3.2. Let e be any even column number and r the number of rows in James
abacusdiagramwheree < r. Foranypartitionµofapositiveintegert,
chaini = {ami,am(e−i+1),ain,a(r−i+1)n : i (cid:2) m (cid:2) (r −i +1),i < n < e−i +1}
where
1. Everychainisderivedfromtwocolumns,i ande−i +1.
2. Thelastchainisderivedfromtwoconsecutivecolumns.
Thegeneralizationofourmainresultisstatedinthefollowingtheorem.
Theorem3.3. LetebeanyevencolumnnumberandithenumberofchainsintheJames
abacus diagram where e2 (cid:2) β . Then, for any partition µ of a positive integer t, then
1
e
thenumberofchainsinthediagramis .
2
Proof. By Definition 3.2, every chain is derived from two columns, which are i and
e−i +1. Since the last chain is derived from two consecutive columns, the difference
betweenthesetwocolumnnumbersis
e−i +1−i = 1.
Thus,
e
i = .
2
(cid:3)
Hence, for Example 2.2, if e = 4 we have two chains as shown in Figure 1.b. The
nexttheoremshowsthatthenumberofpositionsineverychainis2r +2e−4(2i −1).
Theorem3.4. Letebeanyevencolumnnumber,r thenumberofrowsandi thenumber
of chains in the James abacus diagram where e2 (cid:2) β . Then, for any partition µ of a
1
positiveintegert,thenumberofpositionsineachchainis
2r +2e−4(2i −1).
Proof. Sincethechainsformarectanglethenthelengthofchaini is
(r −i +1)i = r −2i +1.
NestedChainMovementoflength1ofBetaNumberinJamesAbacusDiagram 7
Thewidthofchaini is
(e−i +1)−i = e−2i +1.
Thentheperimeterofchaini isgivenby
2[(r −2i +1)+(e−2i +1)] = 2r +2e−4(2i −1).
(cid:3)
Consider Example 2.2 and the partition µ = (8,8,6,3,2,1,1,1,1). If e = 4, the
diagramhastwochainsandchain1has14positionswhilechain2has6positions.
An arithmetic sequence among the nested chains can be obtained as given in the
followingTheorem3.5.
Theorem3.5. Letebeanyevencolumnnumber,r thenumberofrowsandi thenumber
of chains in the James abacus diagram where e2 (cid:2) β . Then, for any partition µ of a
1
positiveintegert
(cid:5) Vi (cid:6)=(cid:5) V1,V2,V3,...,Ve (cid:6)
2
is the arithmetic sequence for the number of positions in the nested chains with −8 as
thecommondifferenceofsuccessivetermswhereV isthenumberofpositionsinchain
i
i.
Proof. LetVi+1,Vi representthenumberofpositionsinchaini +1andchaini respec-
e−2
tivelywherei = 1,2,..., byTheorem3.3. Thus
2
[2r +2e−4(2(i −1)−1)][2r +2e−4(2i −1)] = −8.
(cid:3)
Inthenextsection,theapplicationofasinglemotionofthenestedchainmovement
onJamesabacusdiagramoflength[1,1,...,1]willbeexamined.
3.3. Nestedchainmovementoflength[1,1,...,1]
Thissectiondescribesthenewmethodforpartitioningµwithasinglemotionofnested
chainmovementoflength1inJamesabacusdiagramwhereeisanevencolumnnumber
and e2 (cid:2) β . First, a beta number on every chain in the James diagram is selected
1
randomly. Thebeadthatrepresentstheselectedbetanumberisdenotedastheinitialpoint
ofeverychain. Asalltheinitialbeadpositionsaresimultaneouslymovedanticlockwise
by length 1 in a single motion to the next positions in the chains that they belong to,
the location of the remaining beads in the diagram will be changed accordingly. A new
diagram Atc1 that displays a new partition is constructed. Applying such chain nested
movement of length 1 yields the following results where the notation → as in j → k
meansthatj representstheoriginallocationandk representsthenewlocation.
8 EmanF.Mohomme,etal.
Rule3.6. Lete beanyevencolumnnumber,r thenumberofrowsandi thenumberof
chainsintheJamesabacusdiagramwheree2 (cid:2) β . Then
1

a(Amtc−1b11)n if i +1 (cid:2) m (cid:2) (r −i +1),n = e−i +1




aAtc1b1 if i (cid:2) m (cid:2) (r −i),n = i
(m+1)n
ab1 →
mn 
aAtc1b1 if m = i,i +1 < n (cid:2) e−i +1
 m(n−1)




aAtc1b1 if m = (r −i +1),i (cid:2) n < e−i
m(n+1)
Fig.3illustratestheaboveruleforµ = (35,16,8,8,6,3,2,1,1,1,1)ande = 6.
- o o o o - o o o o o -
o - o - - - - o - - o -
o - - o o - o - o - - -
- - - - - - o - - - - -
- o - - - - - - - - - -
- - - - - - - o - - - -
- - - - - - - - - - - -
- o o o o - - - o o o -
a b
Figure3: (a)Jamesabacusdiagram(b)DiagramAtc1.
3.4. Developmentofnewmaindiagrams
WehaveobtaineddiagramAtc1 ofb fromJamesabacusdiagramusingasinglemotion
1
of nested chain movement of length 1. Next, we can find the diagram of b from the
2
diagramofb afterapplyingthesimilarchainmovementoflength1again. Thefollowing
1
tworulesapplyforthecaseofe = 2 ande = 4x.
Rule3.7. Lete = 2beanyevencolumnnumber,r thenumberofrowsandi thenumber
NestedChainMovementoflength1ofBetaNumberinJamesAbacusDiagram 9
ofchainsintheJamesabacusdiagramwheree2 (cid:2) β then
1

aAtc1b2 if ≤ m ≤ r,n = 1
 (m−2)2





a(Amtc+1b32)1 if m = 1,2,··· ,r −2,n = 2





aAtc1b2 if m = r −1,n = 2
r2
aAtc1b1 →
mn 

aAtc1b2 if m = r,n = 2
 (r−1)2




aAtc1b2 if m = 2,n = 1
 11




aAtc1b2 if m = 1,n = 1 .
31
where ab1 is any position in the diagram Atc1 of b in column n, rows m and m =
mn 1
1,2,...,r −1.
Rule 3.8. Let e = 4x where x is any positive integer, r the number of rows and i the
numberofchainsintheJamesabacusdiagramwheree2 (cid:2) β . Applyingasinglemotion
1
ofnestedchainmovementoflength1willgivethefollowingresults:
e
1. Ifm = i,i (cid:2) n (cid:2) e−i +1andi (cid:8)= then
2

 amA(tcn1+b21) if m = i,n+1 < e−i




aAtc1b2 if m = i,n = e−i,i (cid:8)= 1
 (m−1)(n+2)


amAntc1b1 →  amA(tcn1+b21) if m = i,n = e−i +1,i (cid:8)= 1





 aAtc1b2 if m = 1,n = e−i
 31




aAtc1b2 if m = 1,n = e
41
e
2. Ifn = e−i +1,i < m < r −i +1andi (cid:8)= then
2

amA(tcn1+b21) if i < m (cid:2) r −i +1,n = e−i +1,i (cid:8)= 1




aAtc1b1 → a(Amtc+1b32)1 if 1 < m (cid:2) r −2,n = e
mn 



a(Art+c11b)22 if m = r −1,n = e
10 EmanF.Mohomme,etal.
e
3. Ifm = r −i +1,i (cid:2) n (cid:2) e−i +1andi (cid:8)= then
2

aAtc1b2 if m = r −i +1,i (cid:2) n < e−i +2
m(n+1)
aAtc1b1 →
mn 

aAtc1b2 if m = r −i +1,e−i (cid:2) n (cid:2) e−i +1
(m−1)n
e
4. Ifn = i,i (cid:2) m (cid:2) e−i +1andi (cid:8)= then
2

amA(tcn1+b21) if n = i,r −i +1 (cid:2) m−1 (cid:2) i


aAtc1b1 → aAtc1b2 if n = i,m = i +1
mn  m(n+1)




aAtc1b2 if m = r −i +1,n = i
(m−1)n
e
5. i (cid:2) n (cid:2) i +1,i (cid:2) m (cid:2) r −i +1andi =
2

a(Amtc−1b11)(n+2) if n = 2e,m = 2e




 a(Amtc−1b11)n if n = 2e,m = e+2 2




 e e+4 2r −e+2
aAtc1b1 if n = , ≤ m ≤
aAtc1b1 → (m−2)(n+1) 2 2 2
mn 



 e+2 e 2r −e+2
 amA(tcn1+b11) if n = 2 , 2 ≤ m < 2




 e+2 2r −e+2
 a(Amtc−1b11)n if n = 2 ,m = 2
(cid:9) (cid:10)
Atc1 Atc1
6. a b2 ,a b2 |m = r +1,2 < n (cid:2) e arebeadpositions.
mn re
Figure 4(a) depicts the original James abacus diagram for b ,b ,b and b while
1 2 3 4
Figure4(b)showsallresultsaccordingtoules3.6and3.8forµ = (8,8,6,3,2,1,1,1,1)
ande = 4.
In general, Rules 3.7 and 3.8 can be further applied to find the guides or the main
diagramsAtc1 ofb ,b ,....
3 4