Classification of the Second Minimal Odd Periodic Orbits in the Sharkovskii Ordering

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by  Ugur G. Abdulla

CLASSIFICATION OF THE SECOND MINIMAL ODD PERIODIC
ORBITS IN THE SHARKOVSKII ORDERING
UGURG.ABDULLA,RASHADU.ABDULLA,MUHAMMADU.ABDULLA,
7 ANDNAVEEDH.IQBAL
1
0
2 Abstract. Thispaperpresentsfullclassificationofsecondminimaloddperiodic
orbitsofacontinuousendomorphismsontherealline. A(2k+1)-periodicorbit
n
a (k≥3)iscalledsecondminimalforthemap f,if2k−1isaminimalperiodof
J f in the Sharkovskii ordering. We prove that there are 4k−3 types of second
0 minimal(2k+1)-orbits,eachcharacterizedwithuniquecyclicpermutationand
1 directedgraphoftransitionswithaccuracyuptoinverses.

]
S
D
h. 1. IntroductionandMainResult
t
a Let f :I→I beacontinuousendomorphism,andI beanon-degenerateinterval
m
on the real line. Let fn :I →I be an n-th iteration of f. A point c∈I is called a
[ periodic point of f with period m if fm(c)=c, fk(c)(cid:44)c for 1≤k<m. The set of

1 mdistinctpoints
v
c,f(c),···,fm−1(c)
5
9
is called the orbit of c, or briefly m-orbit or periodic m-cycle. In his celebrated
6
2 paper [7], Sharkovskii discovered a law on the coexistence of periodic orbits of
0 continuousendomorphismsontherealline.
.
1
0 Theorem 1.1 (Sharkovski). [7] Let the positive integers be totally ordered in the
7 followingway:
1
:
v
i (1) 1(cid:47)2(cid:47)22(cid:47)23(cid:47)···(cid:47)22·5(cid:47)22·3(cid:47)···(cid:47)2·5(cid:47)2·3(cid:47)···(cid:47)9(cid:47)7(cid:47)5(cid:47)3.
X
r Ifacontinuousendomorphism, f :I →I,hasacycleofperiodnandm(cid:47)n,then f
a
alsohasaperiodicorbitofperiodm.
This result played a fundamental role in the development of the theory of dis-
cretedynamicalsystems. Aconceptuallynovelproofwasgivenin[6]. Following
thestandardapproach,wecharacterizeeachperiodicorbitwithcyclicpermutations
anddirectedgraphsoftransitionsordigraphs. Considerthem-orbit:
B={β <β <···<β }
1 2 m
DepartmentofMathematics,FloridaInstituteofTechnology,Melbourne,FL32901.
1
2 U.G.ABDULLA,R.U.ABDULLA,M.U.ABDULLA,ANDN.H.IQBAL
Definition1.2. If f(β)=β for1≤s ≤m,withi=1,2,...,m,thenBisassociated
i si i
withcyclicpermutation
(cid:32) (cid:33)
1 2 ... m
π=
s s ... s
1 2 m
Definition1.3. Letωbetheorderreversingpermutation
(cid:32) (cid:33)
1 2 ... m−1 m
ω=
m m−1 ... 2 1
Then,givenacyclicpermutationπ,it’sinverseisobtainedasπ−1=ω◦π◦ω.
Inthesequel<a,b>meanseither[a,b]or[b,a].
Definition 1.4. Let Ji = [βi,βi+1]. The digraph of m-orbit is a directed graph
of transitions with vertices J ,J ,···,J and oriented edges J → J if J ⊂ <
1 2 m−1 i s s
f(βi),f(βi+1)>.
Definition1.5. Theinversedigraphofm-orbitisadigraphassociatedwithinverse
cyclicpermutationπ−1. Equivalently,inversedigraphisobtainedfromthedigraph
ofm-orbitbyreplacingeach J with J .
i m−i
(cid:2) (cid:3) (cid:2) (cid:3)
Definition 1.6. A continuous function P : β ,β → β ,β is called the P-
f 1 m 1 m
linearizationof f ifP (β)= f(β)andPisalinearfunctionineachinterval J
f i i i
Definition 1.7. The arrangement of the minimums and maximums of the map P
f
intheopeninterval(β ,β )willbecalledthetopologicalstructureoftheperiodic
1 m
orbit.
The proof of Sharkovskii’s theorem significantly uses the concept of minimal
orbit.
Definition 1.8. n-orbit of f is called minimal if n is the minimal period of f in
Sharkovski’sordering.
Definition1.9. Digraphofthem-orbitcontainstherededgeJi→JsifJs=< f(βi),f(βi+1)>.
The structure of the minimal orbits is well understood [8, 3, 5, 4, 1]. Minimal
oddorbitsarecalledStefanorbits,duetothefollowingcharacterization:
Theorem1.10(Stefan). [8,4]Thedigraphofam=2k+1minimaloddorbithas
theuniquestructuregiveninfig.1andcyclicpermutation2uptoaninverse.
(cid:32) (cid:33)
1 2 3 ··· k k+1 k+2 k+3 ··· 2k 2k+1
(2)
k+1 2k+1 2k ··· k+3 k+2 k k−1 ··· 2 1
Themaingoalofthispaperisthecharacterizationofsecondminimaloddorbits.
Definition 1.11. A (2k+1)-orbit, k ≥3 is called second minimal if 2k−1 is the
minimalperiodof f intheSharkovskiiordering.
SECONDMINIMALODDPERIODICORBITS 3
Jk+2 Jk+3 ... J4 J2k
Jk+1 J1
Jk Jk−1 ... J3 J2
Figure1. DigraphofMinimalOddOrbit
TopologicalStructure Count
max 1
min-max 1
min-max-min 1
max-min 2
max-min-max 2k−3
max-min-max-min-max 2k−5
Table1. TopologicalStructureofall4k−3secondminimal(2k+
1)-orbits
Let B(i,j),1≤i≤ j≤2k+1besubsetsofa(2k+1)-orbitdefinedas
(3) B(i,j)={β ∈B:i≤k≤ j}
k
Definition1.12. A(2k+1)-orbitiscalledsimpleifeither
(1) B(k+2,2k+1)ismappedto B(1,k+1);and B(1,k+1)ismappedto B(k+
2,2k+1)exceptonepoint;or
(2) B(1,k)ismappedtoB(k+1,2k+1);andB(k+1,2k+1)ismappedtoB(1,k)
exceptonepoint.
We say a simple (2k+1)-orbit is of type + (resp. type −) if (1) (resp. (2)) is
satisfied.
Ourmainresultreads:
Theorem1.13. Secondminimal(2k+1)-orbits,k≥3,aresimple. Thereare4k−3
possibletypesofsecondminimal(2k+1)-orbitsofsimplepositivetype, eachwith
unique digraph and cyclic permutation. Their inverses represent all second mini-
mal (2k+1)-orbits of simple negative type. The topological structure of all 4k−3
types of second minimal (2k+1)-orbits is presented in table 1. The topological
structure of their inverses is obtained by replacing “max” and “min” with each
other respectively. The P-linearization of the 4k−3 types (and their inverses)
presentsanexampleofacontinuousmapwithasecondminimal(2k+1)-orbit.
Theorem 1.13 in the particular case k =3 was proved in [2]. Second minimal
orbits play important role in the problem on the distribution of periodic windows
within the chaotic regime of the bifurcation diagram of the one-parameter family
4 U.G.ABDULLA,R.U.ABDULLA,M.U.ABDULLA,ANDN.H.IQBAL
of unimodal maps. It is revealed in [2] that the first appearance of all the orbits is
alwaysaminimalorbit,whilethesecondappearanceisasecondminimalorbit.
Thestructureoftheremainderofthepaperisasfollows: InSection2,werecall
somepreliminaryfacts. Theorem1.13isprovedinSection3.
2. PreliminaryResults
Lemma2.1. Thedigraphofanm-orbit,B={β <β <···<β },m>2,possesses
1 2 m
thefollowingproperties[4]:
(1) Thedigraphcontainsaloop: ∃r suchthat J →J .
∗ r∗ r∗
(2) ∀r, ∃r(cid:48) and r(cid:48)(cid:48) such that Jr(cid:48) → Jr → Jr(cid:48)(cid:48); moreover, it is always possible
tochooser(cid:48) (cid:44)r unlessmisevenandr=m/2,anditisalwayspossibleto
chooser(cid:48)(cid:48)(cid:44)runlessm=2.
(3) If(cid:2)β(cid:48),β(cid:48)(cid:48)(cid:3)(cid:44)(cid:2)β1,βm(cid:3),β(cid:48),β(cid:48)(cid:48)∈B,then∃Jr(cid:48) ⊂(cid:2)β(cid:48),β(cid:48)(cid:48)(cid:3)and∃Jr(cid:48) (cid:42)(cid:2)β(cid:48),β(cid:48)(cid:48)(cid:3)such
that Jr(cid:48) →Jr(cid:48)(cid:48).
(4) Thedigraphofacyclewithperiodm>2containsasubgraph J →···J
r∗ r
forany1≤r≤m−1.
Definition 2.2. A cycle in a digraph is said to be primitive if it does not consist
entirelyofacycleofsmallerlengthdescribedseveraltimes.
Lemma 2.3 (Straffin). [9, 4] If f has a periodic point of period n > 1 and its
associateddigraphcontainsaprimitivecycleJ →J →···→J →J oflength
0 1 m−1 0
m,thenfhasaperiodicpointyofperiodmsuchthat fk(y)∈J ,(0≤k<m).
k
Lemma 2.4 (Converse Straffin). [4] Let f have a periodic point of period n>1
with digraph D. Suppose f is strictly monotonic on each subinterval Ji =[βiβi+1]
for 1≤i≤n−1. If f has an orbit of period m in the open interval (β ,β ) then
1 n
either D contains a primitive cycle of length m, or m is even and D contains a
primitivecycleoflengthm/2.
3. ProofsofMainResults
3.1. DigraphsofSecondMinimalOddOrbits.
Proof. Let f :I →I beacontinuousendomorphismthathasa2k+1-orbit(k≥4)
which is second minimal. Let B={β1<β2<···<β2k+1} be the ordered elements
of this orbit; Let r =max{i| f(β)>β}. Such an r exists since f(β )>β and
∗ i i ∗ 1 1
f(β2k(cid:12)+1)(cid:12)<β(cid:12)2k+(cid:12)1. So,Jr∗→Jr∗;Let(cid:12)B−=(cid:12) (cid:8)β(cid:12) ∈B(cid:12) |β≤βr∗(cid:9),B+=(cid:8)β∈B|β>βr∗(cid:9);We
have(cid:12)(cid:12)B−(cid:12)(cid:12)+(cid:12)(cid:12)B+(cid:12)(cid:12)=2k+1andhence(cid:12)(cid:12)B−(cid:12)(cid:12)(cid:44)(cid:12)(cid:12)B+(cid:12)(cid:12). Assume,withoutlossofgenerality,
(cid:12)(cid:12)(cid:12)B−(cid:12)(cid:12)(cid:12)>(cid:12)(cid:12)(cid:12)B+(cid:12)(cid:12)(cid:12). Let r =max(cid:8)i<r∗| f(βi)≤βr∗(cid:9). We have f(βr)≤βr∗, f(βr+1)>βr∗,
andhence J →J . FromLemma2.1itfollowstheexistenceofthesubgraph
r r∗
(4) (cid:8)J →···→J →J
r∗ r r∗
Assumethat(4)presentstheshortestpath. Sincethereare2kintervals,itslengthis
atmost2k+1andatleast2k−1. Indeed,ifitslengthis2k−2orless,thenLemma
2.3 implies the existence of an odd periodic orbit of period 2k−3 or less. Let us
SECONDMINIMALODDPERIODICORBITS 5
change the indices of intervals in (4) successfully as r =r ,···,r =r and write
∗ 1 m
path(4)as
(5) (cid:8)J →···→J →J
r1 rm r1
where m=2k−2, 2k−1, or 2k; For simplicity we are going to use the notation
a
i for β. In the sequel the notation in the second row of the cyclic permutation
i b
means that either of the entries a or b are valid choices for the image of the node
inthesamecolumnofthefirstrow; J →[a,b]means f(r)=aand f(r +1)=b,
ri i i
thenotation(cid:104)J ,J (cid:105)meanstheunionof J , J , andalltheintervalsbetweenthem.
r s r s(cid:68) (cid:69)
Notethatif J →J and J →J then J → J ,J .
ri rj ri rk ri rj rk
Since(5)istheshortestpathwehave
(6a) J (cid:57)J for j>i+1,1≤i≤m−2;
ri rj
(6b) J (cid:57)J for2≤i≤m−1;
ri r1
wealsohave
(7a) J (cid:57)J for1< j<i,4≤i≤m,i− jeven;
ri rj
unless i=m=2k, j=2. Indeed, otherwise according to Lemma 2.3 an odd orbit
of length less than 2k−1 must exist. From (6) and (7) we can infer the relative
positionoftheintervalstobeeither
··· Jr5 Jr3 Jr1 Jr2 Jr4 Jr6 ···
Figure2. Relativepositionsofintervalsinthesub-graphoflength
mwhen J totherightof J
r2 r1
or
··· Jr6 Jr4 Jr2 Jr1 Jr3 Jr5 ···
Figure3. Relativepositionsofintervalsinthesub-graphoflength
mwhen J totheleftof J
r2 r1
If m=2k then the path (5) contains all 2k intervals. From the proof of Theo-
rem 1.10 (for example, see Proposition 8 in [4]) it follows that the corresponding
periodic orbit is minimal. Thus, we need only to focus in cases m=2k−1 and
m=2k−2.
6 U.G.ABDULLA,R.U.ABDULLA,M.U.ABDULLA,ANDN.H.IQBAL
Lemma 3.1. The case when m=2k−1 produces exactly 2 second minimal (2k+
1)-orbits. Cyclic permutations are given in (8) and (9), and the corresponding
digraphsarepresentedinfig.4andfig.5respectively.
(cid:32) (cid:33)
1 2 ··· k−2 k−1 k k+1 k+2 ···
(8)
k+1 2k+1 ··· k+5 k+2 k+4 k+3 k ···
(cid:32) (cid:33)
1 2 ··· k−2 k−1 k k+1 k+2 k+3 ···
(9)
k 2k+1 ··· k+5 k+4 k+2 k+3 k+1 k−1 ···
Jk+2 Jk+3 Jk+4 Jk+5 ... J2k−2 J2k−1 J2k
Jk+1 J1
Jk Jk−1 Jk−2 Jk−3 ... J4 J3 J2
Figure4. Digraphoffirstcyclicpermutation(8)whenm=2k−1.
Jk+2 Jk+3 Jk+4 Jk+5 ... J2k−2 J2k−1 J2k
Jk+1 J1
Jk Jk−1 Jk−2 Jk−3 ... J4 J3 J2
Figure5. Digraphofsecondcyclicpermutation(9)whenm=2k−1
Proof. Whenthelengthofthepath(5)is2k−1wehaveexactlyoneinterval, call
it J˜, missing. Since J → J , J (cid:57) J for i>1 odd, one of the endpoints
r2k−1 r1 r2k−1 ri
of J mustbemappedtosomeelementoftheorbitwhichseparates J and J .
r2k−1 r1 r3
Since(cid:8)J →J ,but J (cid:57)J itfollowsthattheendpointof J whichseparates
r1 r2 r1 r3 r1
J and J , must be mapped to the element of the orbit which separates J and
r1 r2 r1
J . Therefore, unless there is an interval between J and J , the element of the
r3 r1 r3
orbit which separates J and J will be an image of two distinct elements of the
r1 r3
orbit. Hence, J˜is between J and J , and the distribution of the intervals is as
r1 r3
in fig. 6 or fig. 6 reflected about the center point k+1. Note that the distribution
(cid:12) (cid:12) (cid:12) (cid:12)
describedinfig.6isrelevantduetoourassumption(cid:12)(cid:12)B−(cid:12)(cid:12)>(cid:12)(cid:12)B+(cid:12)(cid:12). Theothercasewill
(cid:12) (cid:12) (cid:12) (cid:12)
provide the associated inverse digraph with (cid:12)(cid:12)B+(cid:12)(cid:12)>(cid:12)(cid:12)B−(cid:12)(cid:12). Hence, the structure is as
itisdescribedinfig.6.
This implies the following cyclic permutation. Note the possible alternation of
imagesofelements1,k+2,k−1,andk.
SECONDMINIMALODDPERIODICORBITS 7
Jr2k−1 Jr2k−3 ··· Jr3 J˜ Jr1 Jr2 ··· Jr2k−4 Jr2k−2
1 2 3 k−1 k k+1 k+2 2k−1 2k 2k+1
(cid:12) (cid:12) (cid:12) (cid:12)
Figure6. Completeintervalforcasewhenm=2k−1with(cid:12)(cid:12)B−(cid:12)(cid:12)>(cid:12)(cid:12)B+(cid:12)(cid:12)
(10)  k+1k1 ······ (cid:104)kk−+12, k+k4(cid:105) kk++13 kk++k21 kk+−31 kk+−42 ······ 
(1) Case(1): f(k−1)=k+2⇒ f(k)=k+4
(2) Case(1) : f(1)=k+1⇒ f(k+2)=k;Thisproducesasimplepositivetype(2k+
1
1)-orbit given in (8) and fig. 4 with topological structure max-min-max. Next
weanalyzethedigraphtoshowthattherearenoprimitivecyclesofevenlength
≤2k−2,whichwouldimplybyStraffin’slemmaanexistenceofoddperiodicorbit
oflength≤2k−3. FromLemms2.4itthenfollowsthatthetheP-linearizationof
theorbit(8)presentsanexampleofcontinuousmapwithsecondminimal(2k+1)-
orbit. Wesplittheanalysisintotwocases:
(a) Considerprimitivecyclesthatcontain J . Withoutlossofgeneralitychoose
1
J as the starting vertex. First assume that cycle doesn’t start with chain
1
J1→Jk+1. Since J2k+1−i→Ji, i=1,...,k−1,anysuchcyclecanbeformed
only by adding on to the starting vertex J1 pairs (J2k+1−i,Ji),i=1,...,k−1.
Therefore, thelengthofthecycle(bycounting J twice)willbealwaysan
1
odd number. On the contrary, if cycle starts with chain J1 → Jk+1, then to
closeitatJ thesmallestrequiredevenlengthis2k.
1
(b) Consider primitive cycles that don’t contain J . Obviously, such a cycle
1
doesn’tcontainJ2,J3,...,Jk−3orJk+4,Jk+5,...,J2k−1,J2ksincethesevertices
have red edges connecting them all the way to J . Additionally, this cycle
1
cannotcontainJk+1orJksinceJ1istheonlyvertex(besidesJk+1itself)with
adirectededgeto Jk+1,and Jk+1 istheonlyvertexwithdirectededgeto Jk.
This leaves 4 vertices: Jk−2,Jk−1,Jk+2,Jk+3. Since Jk+2→Jk−1, Jk+3→Jk−2
and Jk+2(cid:11)Jk−1, Jk+3(cid:11)Jk−2,anycycleformedbythesefourverticeswill
consist of a starting vertex followed (or ending vertex preceded) by pairs
(Jk+2,Jk−1), (Jk+3,Jk−2) added arbitrarily many times, and hence no cycles
ofevenlengthcanbeproduced.
(3) Case(1) : f(1)=k⇒ f(k+2)=k+1,thenwehavetheperiod4-suborbit
2
{k−1,k+1,k+2,k+3},acontradiction.
(4) Case(2): f(k−1)=k+4⇒ f(k)=k+2
(5) Case(2) : f(1)=k+1⇒ f(k+2)=kthenwehavetheperiod2-suborbit{k,k+2},
1
acontradiction.
(6) Case(2) : f(1)=k⇒ f(k+2)=k+1;Thisproducesasimplepositivetype(2k+
2
1)-orbitgivenin(9)andfig.5withtopologicalstructuremax-min-max.Werepeat
theargumentfromCase(1) . Firstweanalyzethedigraphtoshowthatthereare
1
noprimitivecyclesofevenlength≤2k−2. Wesplittheanalysisintotwocases:
(a) Considerprimitivecyclesthatcontain J . Withoutlossofgeneralitychoose
1
J as a starting vertex. First assume that cycle starts with the edge J →
1 1
Jj, with j taking any value between k+3 and 2k. Since J2k+1−i→Ji, i=
8 U.G.ABDULLA,R.U.ABDULLA,M.U.ABDULLA,ANDN.H.IQBAL
1,...,k−2, any such cycle can be formed only by adding to starting vertex
J1 pairs (J2k+1−i,Ji),i=1,...,k−2. Therefore, the length of the cycle (by
counting J twice) will be always an odd number. If the cycle starts with
1
theedge J1→Jk+2,thentheonlydifferencefromthepreviouscasewillbe
theadditionofthepairs(Jk+2,Jk−1)and/or(Jk+2,Jk)arbitrarilymanytimes.
Hence, only cycles of odd length will be produced. On the contrary, if the
cycle starts with the chain J1 → Jk+1 or J1 → Jk, then to close it at J1 the
smallestrequiredevenlengthis2k.
(b) Consider primitive cycle that doesn’t contain J . Obviously, such a cycle
1
doesn’t contain J2,J3,...,Jk−2 or Jk+3,Jk+4,...,J2k−1,J2k since these ver-
tices have red edges connecting them all the way to J . Additionally, this
1
cycle cannot contain Jk+1 since J1 is the only vertex (besides Jk+1 itself)
withdirectededgetoJk+1. Thisleaves3vertices: Jk,Jk−1,Jk+2connectedas
Jk(cid:11)Jk+2(cid:11)Jk−1. Therefore,thistriplecanonlyproducecycleswhenpairs
(Jk+2,Jk−1)and(Jk+2,Jk)areaddedtoastartingvertex. Therefore,nocycle
ofevenlengthcanbeproduced.
(cid:3)
Now,observethatwhenthelength,m,ofpath(5)is2k−2itiscomprisedof2k−
2 distinct intervals and thus there are 2 additional intervals required to complete
the periodic orbit of period 2k+1. From path (5) and the rules (6) it follows that
the relative distribution of the 2k−2 intervals is in one of the following 2 forms
illustrated in fig. 7 or fig. 8. Call the two missing intervals J˜ and Jˆ. There are
2k−1 slots in fig. 7 or fig. 8 where we can place each of these extra intervals for
a total of (2k−1)2 pairs. However, since swapping the locations of J˜and Jˆdoes
not affect the analysis let us consider the distribution given in fig. 8 in the upper
triangularmatrix(11)where(i,j)indicatesplacing J˜inpositioniand Jˆinposition
j.
Jr2k−3 ··· Jr5 Jr3 Jr1 Jr2 Jr4 Jr6 ··· Jr2k−2
Figure7. Relativepositionsofintervalsinthesub-graphoflength
m=2k−2when J totherightof J
r2 r1
or
Jr2k−2 ··· Jr6 Jr4 Jr2 Jr1 Jr3 Jr5 ··· Jr2k−3
Figure8. Relativepositionsofintervalsinthesub-graphoflength
m=2k−2when J totheleftof J
r2 r1
SECONDMINIMALODDPERIODICORBITS 9
(1,1) (1,2) (1,3) ··· (1,2k−3) (1,2k−2) (1,2k−1)
(2,2) (2,3) ··· (2,2k−3) (2,2k−2) (2,2k−1)
. .
. .
(3,3) ··· . .
(11) ... (i,i) ··· (i,2k−i+2)
... ... ...
... (2k−1,2k−1)
Thenextlemmaspecifiesalltheentries(i,j)inmatrix(11),suchthatinsertion
of(J˜,Jˆ)in(i,j)canproducesecondminimaloddorbits.
Lemma 3.2. Fix the entry point, i, for J˜, then to produce second minimal 2k+1
orbit, Jˆcanonlybeplacedin
(1) position2k−1wheni=1,
(2) positions2k−ior2k−i+1for1<i<k,
(3) andpositionk+1wheni=k.
··· Jrw+2 Jrw J˜ Jrw−2 Jrw−4 ··· Jrw−5 Jrw−3 Jrw−1 Jrw+1 ···
i−2 i−1 i i+1 i+2 2k−i−1 2k−i+2
Figure9. Relativeorderingwhenm=2k−2and J˜isinpositioni
Proof. Let1<i<kfork>3. Lettheintervalimmediatelytotheleftof J˜becalled
J , or, the w-th distinct interval in path (5). From the relative positions of the
rw
intervalsinfig.3itisclearthatw=2(k−i+1). Asdepictedinfig.9,theintervals
totherightof J˜,haveasetpath. J mapsonlyto J , J mapsonlyto J ,
rw−4 rw−3 rw−2 rw−1
etc. Notethatwhiletheexactpointsthattheseintervalsmaptomightchangeupon
the insertion of Jˆ, the overall structure must remain the same in order to preserve
path(5).
10 U.G.ABDULLA,R.U.ABDULLA,M.U.ABDULLA,ANDN.H.IQBAL
However, this pattern can no longer be continued indefinitely for interval J .
rw
J must be mapped to J , by definition. This implies that one end point of
rw rw+1
J is mapped arbitrarily to the left of J , and the other endpoint is mapped
rw rw+1
arbitrarilytotherightof J . Accordingto(6), J can’tmaptoanyoddinterval
rw+1 rw
greaterthanJ . AssumingthatJ isn’ttherightmostinterval,orinotherwords
rw+1 rw+1
w+1<2k−3,thentheendpointof J thatmapstotherightof J ,mustmapto
rw rw+1
a point separating J and J . This could either be a point directly in between
rw+1 rw+3
J and J ,oranewpointbetween J and J ,upontheinsertionof Jˆ.
rw+1 rw+3 rw+1 rw+3
Notethatifw+1=2k−3,orinotherwords J istherightmostinterval,then
rw+1
J nolongerexists. Ratherthanmappingtoapointseparating J and J ,an
rw+3 rw+1 rw+3
endpointof J willjustmaptotherightof J .
rw rw+1
WhileitisclearhowoneendpointofJ willmaptotherightofJ ,itismuch
rw rw+1
less clear how an endpoint of J will map to the left of J . According to (7),
rw rw+1
J cannotmapto J orany J wherek<wandiseven. Thus,thearbitrarypoint
rw r1 rk
totheleftof J thatanendpointof J mustmapto,mustbeseparating J and
rw+1 rw r1
J . Thus, the missing interval Jˆmust be inserted between J and J . Note
rw+1 r1 rw+1
thataccordingtothepreviouslymentionedpositionalnotation,thisincludesallthe
positionsk+1,k+2,...,2k−i,2k−i+1.
SupposethattheintervalJˆisinsertedintoapositiontotheleftof2k−i. Assume,
without loss of generality, that the interval Jˆ is inserted into position 2k−i−1
(betweenintervals J and J ifw≥6,andbetween J and J ifw=4),asthe
rw−3 rw−5 r1 r2
contradictionwillbethesame. Theintervalimmediatelytotherightof Jˆhassome
trouble mapping to the interval indexed one greater than itself. In the situation
where Jˆis placed in position 2k−i−1, the interval immediately to the right of Jˆ
is J , which has trouble mapping to J . One endpoint of J must map to
rw−3 rw−2 rw−3
the left of J , while the other must map to the right of J . Again, according
rw−2 rw−2
to (7), J can’t map to a lesser odd interval, or J . Thus, the only available
rw−3 r1
pointtotheleftof J andtotherightof J ,isthepointindexedk+1,ortheleft
r1 rw−2
endpoint of J . However, it is important to note that the largest indexed interval,
r1
J ,mustalsomapbackto J . Andaccordingto(7), J can’tmaptoalesser
r2k−2 r1 r2k−2
eveninterval,specificallyincluding J . Therefore,anendpointof J mustmap
r2 r2k−2
to the left of J , but can not map to the left of J . The only point that fits this
r1 r2
description is the one indexed k+1. Thus, the point k+1 is already taken, and an
endpoint of J can not map to the point indexed k+1. Furthermore, there are
rw−3
no open points that are both to the right of J and to the left of J . This is an
rw−2 r1
immediatecontradiction,asitisnowimpossibleforanendpointof J tomapto
rw−3
therightof J ,makingitimpossiblefor J tocontain J .
rw−2 rw−3 rw−2
Now suppose instead w=4. If this is the case, then Jˆis inserted between J
r1
and J . Again,theimageof J mustcontainonly J and J . Thus,oneendpoint
r2 r1 r1 r2
of J must map to the right of J , but can not map to the right of J . Therefore,
r1 r1 r3
theleftendpointof J mustmaptothepointseparating J and J . Now, J must
r1 r1 r3 r2
maptoJ ,butcannotmapbacktoJ . Thus,oneoftheendpointsofJ mustmap
r3 r1 r2
to a point separating J and J . Both the left endpoint of of J and an endpoint
r1 r3 r1