Multipermutation Ulam Sphere Analysis Toward Characterizing Maximal Code Size

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by  Justin Kong

Multipermutation Ulam Sphere Analysis Toward
Characterizing Maximal Code Size
Justin Kong and Manabu Hagiwara
Department of Mathematics and Informatics,
Graduate School of Science, Chiba University
Email: [email protected],[email protected]
7 Abstract—Permutation codes, in the form of rank modulation, to multipermutation codes, which improves the maximum
1 have shown promise for applications such as flash memory. possible code size [4].
0 One of the metrics recently suggested as appropriate for rank In flash memory devices, permutations or multipermuta-
2 modulation is the Ulam metric, which measures the minimum
tions may be modeled physically by relative rankings of
translocation distance between permutations. Multipermutation
n codeshavealsobeenproposedasageneralizationofpermutation cell charges. The number of possible messages is limited by
a
codes that would improve code size (and consequently the code the number of distinguishable relative rankings. However, it
J
rate). In this paper we analyze the Ulam metric in the context was shown in [4] that multipermutations may significantly
4 of multipermutations, noting some similarities and differences
increase the total possible messages compared to ordinary
1 between the Ulam metric in the context of permutations. We
alsoconsiderspheresizesformultipermutationsundertheUlam permutations. For example, if only k different charge levels

] metric and resulting bounds on code size. are possible, then permutations of length k can be stored.
T
Hence, in r blocks of length k, one may store (k!)r potential
I
s. I. INTRODUCTION messages. On the other hand, if one uses r-regular multiper-
c mutationsinthesamesetofblocks,then(kr)!/(r!)k potential
[ Permutation(andmultipermutation)codeswere inventedas messages are possible.

earlyasthe1960’s,whenSlepianproposedconstructingacode Bounds on permutation codes in the Ulam metric were
1
v by permuting the order of the numbers of an initial sequence studied in [3] and [7]. In [10], the nonexistence of nontrivial
6 [13]. More recently, Jiang et al. proposed permutation codes perfect permutation codes in the Ulam metric was proven by
9 utilizing the Kendall-τ metric for use in flash memoryvia the examining the size of Ulam spheres, spheres comprised of
8 rank modulation scheme [8]. Since then, permutation codes all permutations within a given Ulam distance of a particular
3
and their generalization to multipermutation codes have been permutation. However, no similar study of Multipermutation
0
. a hot topic in the research community with various related Ulamspheresexists,andcurrentlyknownboundsoncodesize
1 schemes being suggested [1], [2], [3], [4], [9], [11]. do not always consider the problem of differing sphere sizes.
0
One scheme of particular interest was the proposal of The current paper examines Ulam sphere sizes in the context
7
1 Farnoud et al. to utilize the Ulam metric in place of the of multipermutations and provides new bounds on code size.
: Kendall-τ metric [3] and subsequent study expounded upon The paper is organized as follows: First, Section II defines
v
codesizebounds[7].TheUlammetricmeasurestheminimum notation and basic concepts used in the paper. Next, Section
i
X numberoftranslocationsneededtotransformonepermutation III compares properties of the Ulam metric as defined for
r into another, whereas the Kendall-τ metric measures the permutationsandmultipermutations,andthenprovidesa sim-
a
minimum adjacent transpositions needed to transform one plificationof the r-regularUlam metric for multipermutations
permutationintoanother.Errorsinflashmemorydevicesoccur (Lemmas 1 and 2). Section IV considers an application of
when cell charges leak or when rewriting operations cause Young Tableaux and the RSK-correspondence to calculate r-
overshoot errors resulting in inaccurate charge levels. While regular Ulam sphere sizes (Lemma 5 and Prop. 6). Section
the Kendall-τ metric is suitable for correctingrelativelysmall V then discusses duplicate translocation sets and a method
errors of this nature, the Ulam metric would be more robust of calculating the size of spheres of radius t = 1 for any
to large charge leakages or overshoot errors within a cell. center (Thm. 13). Section VI follows, demonstrating minimal
However,thereisatrade-offincodesizewhenrankmodula- and maximalsphere sizes (Lemmas14 and 16) and providing
tionisusedinconjunctionwiththeUlammetricinsteadofthe both lower and upper bounds on code size (Lemmas 15, 17,
Kendall-τ metric.TheUlamdistancebetweenpermutationsis and 18). Finally Section VII gives some concluding remarks.
always less than or equal to the Kendall-τ distance between
permutations, which implies that the maximum code size for II. PRELIMINARIES AND NOTATION
a permutation code utilizing the Ulam metric is less than In this section we introduce notation and definitions used
or equal to the maximum code size of a permutation code inthispaper.Unlessotherwisestated,definitionsarebasedon
utilizingthe Kendall-τ metric[3]. Onepossible compensation conventions established in [3], [4], and [10]. Throughout this
for this trade-offis the generalizationfrom permutationcodes papernandrareassumedtobepositiveintegers,rdividingn.
The notation [n] denotes the set {1,2,...,n} and S denotes equivalence class is also contained within C. Thus if C is
n
the set of permutations on [n], i.e. the symmetric group of an MPC(n,r) it can be represented by the set of r-regular
size n!. For σ ∈ S , we write σ = [σ(1),σ(2),...,σ(n)], multipermutations associated with elements of C, i.e. the
n
where for all i ∈ [n], σ(i) is the image of i under σ. set M (C). Moreover, we define the cardinality |C| of an
r r
Throughout this paper we assume σ, π ∈ S . With a slight MPC(n,r)C as|C| :=|M (C)|(thisnotationanddefinition
n r r
abuse of notation, we may also use σ to mean the sequence differs slightly from [4]).
(σ(1),σ(2),...,σ(n)) ∈ Zn associated with σ ∈ S . Multi-
n
plication of permutations is defined by composition so that III. MULTIPERMUTATIONULAMMETRIC
for all i ∈ [n], we have (στ)(i) = σ(τ(i)). The identity
In this section we discuss some similarities and differences
permutation, [1,2,...,n]∈S is denoted by e.
n between the Ulam metric for permutations and the Ulam
An r-regular multiset is a multiset such that each of its
metric for multipermutations.We begin by defining the Ulam
elementsisrepeatedrtimes.Amultipermutationisanordered
metric for permutations.
tuple of the elements of a multiset, and in the instance of
For any two sequences u,v ∈ Zn, ℓ(u,v) denotes the
an r-regularmultiset, is called an r-regular multipermutation.
length of the longest common subsequence of u and v. In
Following the work of [4], this study focuses on r-regular
other words, ℓ(u,v) is the largest integer k ∈ Z such that
multipermutations, although many results are extendible to >0
there exists a sequence (a ,a ,...,a ) where for all p∈[k],
general multipermutations. 1 2 k
For each σ ∈ S we define a corresponding r-regular we have ap = u(ip) = v(jk) with 1 ≤ i1 < i2 < ··· <
n
multipermutationmr asfollows:foralli∈[n]andj ∈[n/r], ik ≤ n and 1 ≤ j1 < j2 < ··· < jk ≤ n. The Ulam
σ distance d (σ,π) between permutations σ,π ∈ S is defined
◦ n
mr(i):=j if and only if (j−1)r+1≤σ(i)≤jr, as d (σ,π):=n−ℓ(σ,π).
σ ◦
It is also known that the Ulam distance d (σ,π) between
and mr := (mr(1),mr(2),...,mr(n)) ∈ Zn. For ex- ◦
σ σ σ σ σ,π ∈ S is equivalent to the minimum number of translo-
ample, if n = 6, r = 2, and σ = [1,5,2,4,3,6], then n
cations needed to transform σ into π [3]. Here, for distinct
mr =(1,3,1,2,2,3).Thisdefinitiondiffersslightlyfromthe
σ i,j ∈[n], the translocation φ(i,j)∈S is defined as follows:
correspondencedefined in [4], which was defined in terms of n
theinversepermutation.ThisissothatpropertiesoftheUlam
metricforpermutationswillcarryovertotheUlammetricfor [1,2,...i−1,i+1,i+2,...,j,i,j+1,...,n]
multipermutations (Lemmas 1 and 2 of Section III). 
lenWceitrheltahteiocnobrreetswpeoenndeenlecmeeanbtosvoef, Swe.mWaeysdayefitnheataσn≡equπivaif- φ(i,j):=[1,2,...j−1,i,j,j+1,...,i−1,i+1i,f.i..<n]j
n r
andonlyifmr =mr.TheequivalenceclassR (σ)ofσ ∈S
Sis ⊆defiSne,dthaesnσRotra(tσio)nπ:=M{π(S∈):=Sn{m: rπ:≡σr ∈σ}Sr.}F,oir.ea. tshuebsseentt The notation φ(i,i) is understood to mean thoetheirdwenisteity
n r σ
of r-regular multipermutations corresponding to elements of permutation, e. When it is not necessary to specify any
S. index, a translocationmay be written simply as φ. Intuitively,
We say that σ ≡r π if and only if mrσ = mrπ. The when multiplied on the right of a permutation σ ∈ Sn, the
equivalence class R (σ) of σ ∈ S is defined by R (σ) := translocationφ(i,j)∈S deletesσ(i)fromtheithpositionof
r n r n
{π ∈ S : π ≡ σ}. For a subset S ⊆ S , the notation σ andtheninsertsitinthenewjthposition(shiftingpositions
n r n
M (S) := {mr : σ ∈ S}, i.e. the set of r-regular between i and j in the process).
r σ
multipermutations corresponding to elements of S. Ther-regularUlamdistancedr(σ,π)betweenpermutations
◦
The following definition is our own. For any m∈Zn, and σ,π ∈ S is defined as the minimum Ulam distance among
n
σ ∈ S , we define the product (a right group action) m·σ all members of R (σ) and R (π). That is, dr(σ,π) :=
n r r ◦
bycomposition,similarlytothedefinitionofmultiplicationof min d (σ′,π′). Notice that the r-regular Ulam
◦
permutations.More precisely,for all i∈[n], let (m·σ)(i):= σ′∈Rr(σ),π′∈Rr(π)
distance is defined over equivalence classes.
m(σ(i)). It is easily confirmed that m·e = m for all m ∈
Although technically a distance between equivalence
Zn. It is also easily confirmed that for all σ,π ∈ S , we
n classes,itisconvenienttothinkofther-regularUlamdistance
have m·(σπ) =(m·σ)·π. With this definition, notice that
instead as a distance between multipermutations. Viewed this
mr ·π = mr . It is possible for different permutations to
σ σπ way, the property of the Ulam metric for permutations, that
correspond to the same multipermutation, but for τ ∈ S , it
n it can be defined in terms of longest common subsequences
is clear that mr =mr implies mr ·τ =mr ·τ.
σ π σ π or equivalently in terms of translocations, carries over to the
We finish this section by defining what a multipermutation
r-regular Ulam distance. The next lemma states that the r-
code is. A subset C ⊆ S is called an r-regular multi-
n regularUlam distance between permutationsσ and π is equal
permutation code if and only if for all σ ∈ C, we also
to n minus the length of the longest common subsequence of
have R (σ) ⊆ C. Such a code is denoted by MPC(n,r),
r their corresponding r-regular multipermutations.
and we say that C is an MPC(n,r). If C is an MPC(n,r)
then whenever a permutation is a member of C its entire Lemma 1. dr(σ,π)=n−ℓ(mr,mr).
◦ σ π
Proof:Wewillfirstshowthatdr(σ,π)≥n−ℓ(mr,mr). as followsso that σ′ ∈R (σ) and π′ ∈R (π): set σ′(2)=1,
◦ σ π r r
By definition of dr(σ,π), there exist σ′ ∈ R (σ) and π′ ∈ σ′(5)=6, π′(1)=1, and π′(6)=6.
◦ r
R (π) such that dr(σ,π)=d (σ′,π′)=n−ℓ(σ′,π′). Hence If two multipermutations mr and mr have a common
r ◦ ◦ σ π
if for all σ′ ∈ Rr(σ) and π′ ∈ Rr(π) we have ℓ(σ′,π′) ≤ subsequenceoflengthk,thenmrσ canbetransformedintomrπ
ℓ(mr,mr), then dr(σ,π) ≥ n − ℓ(mr,mr) (subtracting withn−k(butnofewer)delete/insertoperations.Delete/insert
σ π ◦ σ π
a larger value from n results in a smaller overall value). operationscorrespondto applying(multiplyingontheright)a
Therefore it suffices to show that that for all σ′ ∈ R (σ) translocation. Hence by Lemma 1 we can state the following
r
and π′ ∈ Rr(π), that ℓ(σ′,π′) ≤ ℓ(mrσ,mrπ). This is simple lemma about the r-regular Ulam distance.
to prove because if two permutations have a common subse-
Lemma 2. dr(σ,π) = min{k ∈ Z :
quence, then their corresponding r-regular multipermutations ◦ ≥0
there exists(φ ,φ ,...,φ )s.t.mr ·φ φ ···φ =mr}.
will have a related common subsequence. Let σ′ ∈ R (σ), 1 2 k σ 1 2 k π
r
π′ ∈ Rr(π), and ℓ(σ′,π′) = k. Then there exist indexes Proof: There exists a translocation φ ∈ Sn such that
1≤i1 <i2 <···<ik ≤n and 1≤j1 <j2 <···<jk ≤n ℓ(mr ·φ,mr) = ℓ(mr,mr)+1, since it is always possible
σ π σ π
such that for all p ∈ [k], σ′(ip) = π′(jp). Of course, to arrange one element with a single translocation. This then
wℓ(hσe′n,πev′)er=σk′(i≤) ℓ=(mπrσ′(′,jm),rπth′)en=mℓ(rσm′(rσi,)m=rπ)m. rπ′(j). Therefore iφm1p·l·i·eφsktha=t mminrπ{}k∈≤Zn:−thℓe(rmerσex,mistrπs)(φ=1,.d.r◦.(,σφ,kπ))s..tA.tmtrσhe·
Next, we will show that dr(σ,π) ≤ n−ℓ(mr,mr). Note same time, given ℓ(mr,mr) ≤ n, then for all translocations
◦ σ π σ π
that φ∈S , we have that ℓ(mr ·φ,mr)≤ℓ(mr,mr)+1, since
n σ π σ π
a single translocationcan only arrange one element at a time.
dr(σ,π)= min d (σ′,π′)
◦ σ′∈Rr(σ),π′∈Rr(π) ◦ Therefore min{k ∈ Z : there exists (φ1,...,φk) s.t mrσ ·
= min (n−ℓ(σ′,π′)) φ1···φk = mrπ} ≥ n−ℓ(mrσ,mrπ) = dr◦(σ,π), by Lemma
σ′∈Rr(σ),π′∈Rr(π) 1.
=n− max ℓ(σ′,π′). Lemmas 1 and 2 allow us to view the Ulam metric for
σ′∈Rr(σ),π′∈Rr(π) r-regular multipermutations similarly to the way we view
Here if max ℓ(σ′,π′)≥ℓ(mr,mr), then the Ulam metric for permutations; in terms of longest com-
σ π
dr(σ,π)σ≤′∈Rnr−(σ)ℓ,π(m′∈Rrr,(mπ)r) (subtracting a smaller value from mon subsequences or in terms of the minimum number of
◦ σ π translocations. Another known property of the Ulam metric
n results in a larger overall value). It is enough to show
for permutationsis left invariance,i.e. givenτ ∈S , we have
that there exist σ′ ∈ R (σ) and π′ ∈ R (π) such that n
ℓ(σ′,π′)≥ ℓ(mr,mr). Torprove this fact, wertake a longest d◦(σ,π)=d◦(τσ,τπ).However,leftinvariancedoesnothold
common subseqσuencπe of mr and mr and then carefully in generalformultipermutations,as the nextlemmaindicates.
σ π
choose σ′ ∈ Rr(σ) and π′ ∈ Rr(π) to have an equally long Lemma 3. Let n/r ≥2 and r ≥2. Then there exist σ′,π′ ∈
common subsequence.The next paragraphdescribes how this S such that dr(e,σ′)6=dr(π′e,π′σ′).
n ◦ ◦
can be done.
Proof: Let n/r≥2 and r ≥2. Define σ′,π′ ∈S by
Let ℓ(mr,mr)= k and let (1 ≤ i < i < ··· < i ≤ n) n
σ π 1 2 k
and (1 ≤ j1 < j2 < ··· < jk ≤ n) be integer sequences σ′ :=[2r,2r−1,...,1,2r+1,2r+2,...,n] and
such that for all p∈[k], mr(i )=mr(j ). The existence of
such sequencesis guaranteeσdbpy the deπfinpitionof ℓ(mr,mr). 2r n−2r
σ π | {z } | {z }
Now for all p ∈ [k], define σ′(ip) to be the smallest integer π′ :=[1,r+1,2,r+2,...,r,2r,2r+1,2r+2,...,n].
l ∈[n] such that m (l)=m (i ) and if q ∈[k] with q <p,
σ σ p
then mr(i ) = mr(i ) implies σ′(i ) < σ′(i ) = l. For all 2r n−2r
p∈[k],σdeqfine π(jπ) spimilarly. Thenqfor all p∈p [k], σ′(i )= First, c|onsider dr({ez,σ′). Note }tha|t for mr{zand mr }, for
p p ◦ e σ′
π′(j ).Theremainingtermsofσ′andπ′mayeasilybechosen any integer i such that 2r < i ≤ n we have e(i) = σ′(i),
p
in such a manner that σ′ ∈ R (σ) and π′ ∈ R (π). Thus which implies mr(i) = mr (i). Meanwhile, the first 2r
r r e σ′
there exist σ′ ∈ R (σ) and π′ ∈ R (π) such that ℓ(σ′,π′) ≥ elements of mr and mr are (1,1,...,1,2,2,...,2) and
r r e σ′
ℓ(mr,mr).
σ π r r
The following example helps to illuminate the choice of (2,2,...,2,1,1,...,1) respectivel|y, so{zthat}th|e lo{nzges}t com-
σ′ and π′ in the proof above. If mr = (2,1,2,1,3,3),
4an,dwmithrπ th=e c(3o,m2m,2o,n1,s3u,b1s)e,qutheenncew(e2σ,h2a,v1e,3ℓ)(mofrσ,mmarπx)im=al cm|oomnp{srrzuisbesde}qouf|ernc1{er’zsoofr}thre2fi’sr.stH2ernceeleℓm(menret,smofrσ′m)=re a(nnd−m2rσr)′ +is
length. Here (1,3,4,6) and (2,3,4,5) are sequences with r =n−r, which by lemma 1 implies that dr◦(e,σ′)=r ≥2.
mrσ(1) = mrπ(2), mrσ(3) = mrπ(3), mrσ(4) = mrπ(4), and Next, consider dr◦(π′e,π′σ′). Multiplying π′ and σ′ yields
mrσ(6) = mrπ(5). Then following the convention outlined in π′σ′ =[2r,r,2r−1,r−1,...,r+1,1,2r+1,2r+2,...,n].
the proof above, σ′(1) = π′(2) = 3, σ′(3) = π′(3) = 4,
σ′(4) = π′(4) = 1, and σ′(6) = π′(5) = 5, so that 2r n−2r
ℓ(σ′,π′)≥4. The other elements of σ′ and π′ can be chosen Forallin|tegersisuchth{zat2r <i≤n,}w|ethenhav{ezπ′e(i)=}
π′(i) = π′σ′(i) =⇒ mr (i) = mr (i). Meanwhile, the As an example, let T be the tableau pictured below on the
π′e π′σ′
first 2r elements of mrπ′e and mπ′σ′ are (1,2,1,2,...,1,2) far left. The following diagrams illustrate the stages of the
and(2,1,2,1,...,2,1)respectively.Thusthelongestcommon Schensted algorithm applied to T to obtain T ←2.
subsequence of the first 2r elements of mr and mr is
π′e π′σ′ T T←2
any length 2r−1 sequence of alternating 1’s and 2’s. Hence 1 1 2 3 1 1 2 2 1 1 2 2 1 1 2 2
ℓ(mrπ′e,mrπ′σ′) = (n−2r)+(2r −1) = n−1, which by 3 4 3 4 3 3 2 3 3
lemma 1 implies that dr(π′e,π′σ′)=1.
◦ 4 4 4 4 4
The fact that left invariance does not hold for the r-regular x=2,i=1 setx:=3,i:=2 setx:=4,i:=3
Ulam metric has implicationson r-regularUlam sphere sizes,
The Schensted algorithm may be applied to the sequence
defined and discussed in the next section. Left invariance im- (mr(1),mr(2),...,mr(n)) of an r-regular multipermuta-
pliesspheresizedoesnotdependuponthecenter.However,we tionσmr toσobtain a uσnique tableau P := (...(mr(1) ←
will demonstrate that in the multipermutation case sizes may mr(2))σ← ...) ← mr(n). Meanwhile, a unique sσtandard
differ depending upon the center, a fact previously unknown. σ σ
tableau results from recording where each new box appears
in the construction of P. This recording is accomplished by
IV. YOUNG TABLEAUX SPHERE SIZE CALCULATION inputing the value i in the new box that appears when mr(i)
σ
In[10],YoungtableauxandtheRSK-Correspondencewere is added (via the Schendsted algorithm) to (...(mr(1) ←
σ
utilized to calculate Ulam Sphere sizes. A similar approach (mr(2) ← ...) ← mr(i − 1). Following conventions,
σ σ
can be applied to r-regular Ulam spheres of arbitrary radius we denote the standard tableau resulting from this recording
centered at mr. It is first necessary to introduce some basic methodbyQ.Asanexample,thetwotableauxpicturedbelow
e
notation and definitionsregardingYoungtableaux. Additional aretherespectiveP andQresultingfromthemultipermutation
informationonthesubjectcanbefoundin[6],[12],and[14]. (2,3,2,1,3,1). Notice that P and Q have the same shape.
A Young diagram is a left-justified collection of cells with Intermediate steps are omitted for brevity.
a (weakly) decreasing number of cells in each row below.
P Q
Listing the number of cells in each row gives a partition 1 1 3 1 2 5
λ = (λ1,λ2,...,λk) of n, where n is the total number of 2 2 3 6
cells in the Young diagram. The notation λ ⊢ n indicates λ
3 4
is a partition of n. Because the partition λ ⊢ n defines a
unique Young diagram and vice versa, a Young diagram may The RSK-correspondence ([6], [14]) provides a bijection
be referred to by its associated partition λ⊢n. For example, between r-regular multipermutations mr and ordered pairs
σ
the partition λ := (4,3,1) ⊢ 8 has the corresponding Young (P,Q) on the same Young diagram λ ⊢ n, where P is a
diagram pictured below. tableauxwhose memberscome from mr and Q is a SYT. A
σ
stronger form of the following lemma appears in [6].
Lemma 4. Let σ ∈ S and P be the tableau resulting from
n
running the Schendsted algorithm on the entries of σ. Then
thenumberofcolumnsinP isequaltoℓ(mr,mr),thelength
e σ
A Young tableau, or simply a tableau, is a filling of a of the longest non-decreasing subsequence of mr.
σ
Young diagram λ⊢n such that values in all cells are weakly
The above lemma, in conjunction with the RSK-
increasing across each row and strictly increasing down each
correspondence, means that for all k ∈ [n], the size of the
column. If each of the integers 1 through n appears exactly
set {mr ∈ M (S ) : ℓ(mr,mr) = k} is equal to the sum
onceinatableauT thatisafillingofaYoungdiagramλ⊢n, σ r n e σ
ofthenumberoforderedpairs(P,Q)oneachYoungdiagram
then we call T a standard Young tableau, abbreviated SYT.
λ⊢nsuchthatλ =k,whereP isatableauxwhosemembers
The Schensted algorithm is an algorithm for obtaining a 1
come from mr and Q is a SYT. The number of SYT on a
tableau T ← x from a tableau T and a real number x. The σ
particular λ ⊢ n is denoted by fλ. We denote by Kλ (our
algorithm may be defined as follows: r
own notation) the number of Young tableaux on λ ⊢ n such
that each i ∈ [n/r] appears exactly r times. The next lemma
1) Set i:=1.
states the relationship between |S(mr,t)|, fλ, and Kλ.
2) If row i (of T) is empty or if x is greater or equal to each e r
of the entries in the ith row then input x in a new box at the Lemma5. Lett∈[0,n−1],andΛ:={λ⊢n : λ ≥n−t}.
1
end of row i and terminate the algorithm.Otherwise, proceed Then |S(mr,t)|= (fλ)(Kλ).
e r
to step 3. λP∈Λ
3) Find the minimum entry y in row i such that y > x and Proof: Assume t ∈ [0,n−1]. Let Λ := {λ ⊢ n : λ ≥
1
swap y and x. That is, replace y with x in its box and set n−t}. Furthermore, let Λ(l) := {λ ⊢ n : λ = l}, the set
1
x:=y. of all partitions of n having exactly l columns. By the RSK-
4) Set i:=i+1, and return to step 2. Correspondence, and Lemma 4, there is a bijection between
the set {mr : ℓ(mr,mr) = l} and the set of ordered pairs
σ e σ
(P,Q) where both P and Q have exactly l columns. This Proposition 6 demonstrates how Young Tableaux may be
implies that |{mr : ℓ(mr,mr) = l}| = (fλ)(Kλ). used to calculate r-regular Ulam spheres centered at mr.
σ e σ r e
λ∈PΛ(l)
Note that by Lemma 1,
V. r-REGULAR ULAMSPHERES AND DUPLICATION SETS
|S(mr,t)| = |{m : dr(e,σ)≤t}|
e σ ◦ In the previous section we showed how multipermutation
= |{m : ℓ(mr,mr)≥n−t}|. Ulam spheres may be calculated when the center is mr. In
σ e σ e
thissectionweprovideawaytocalculatespheresizesforany
Hence it follows that |S(mr,t)|= (fλ)(Kλ).
e r center when the radius is t = 1. The r-regular Ulam sphere
λP∈Λ
The formula below, known as the hook length formula, is sizesplayanimportantroleinunderstandingthepotentialcode
due to Frame, Robinson, and Thrall [5], [6]. In the formula, size for a given minimum distance.
the notation (i,j) ∈ λ is used to refer to the cell in the ith For example, the well-known sphere-packing and Gilbert-
row and jth column of a Young diagram λ⊢n. The notation Varshamov bounds rely on calculating, or at least bounding
h(i,j) denotes the hook length of (i,j)∈λ, i.e., the number sphere sizes. In the case of permutations, recall that the
of boxes below or to the right of (i,j), including the box Ulam sphere S(σ,t) centered at σ of radius t was defined
(i,j) itself. More formally, h(i,j) := |{(i,j∗) ∈ λ : j∗ ≥ as S(σ,t) := {π ∈ S : d (σ,π) ≤ t}, which is equivalent
n ◦
j}∪{(i∗,j)∈λ : i∗ ≥i}|. The formula is as follows: by definition to the set {π ∈S : n−ℓ(σ,π)≤t}.
n
In the case of r-regularmultipermutations,for t∈Z , we
>0
n!
fλ = . introduce the following analogous definition of a sphere.
Π h(i,j)
(i,j)∈λ Definition. Define
Thus by applying Lemma 5, it is possible to calculate r- S(mr,t):={mr ∈M (S ) : dr(σ,π)≤t}
regularspheresize byusingthe hooklengthformula.We will σ π r n ◦
use this strategy to treat the sphere of radius r =1. However, We call S(mr,t) the r-regular Ulam sphere centered at mr
σ σ
because sphere sizes are calculated recursively, we must first of radius t.
calculate the sphere size when r =0. By Lemma 1, S(mr,t) = {mr ∈ M (S ) : n −
σ π r n
Remark. |S(mr,0)=1|.Althoughthisisanobviousfact,we ℓ(mr,mr)≤t}.
e σ π
wish to considerwhy it is true from the perspective ofLemma It should be noted, however, that the notation mr is a bit
π
5. Note first that there is only one partition λ ⊢ n such that misleading because given mrπ ∈M(Sn), we cannot uniquely
λ =n, namely λ′ :=(n) with the associated Young diagram determine π. The r-regular Ulam sphere definition can also
1
below. be viewed in terms of translocations. Lemma 2 implies that
n S(mr,t) is equivalentto {mr ∈M (S ) : there exists k ∈
σ π r n
... [t] and (φ ,...,φ ) s.t. mr ·φ ···φ =mr}.
1 k σ 1 k π
z }| {
Lemma 5 provided a way to calculate r-regular Ulam
It is clear that there is only one possible Young tableau on λ′ spheres centered at mr. Unfortunately, the choice of center
so that (fλ′)=1, and thus by Lemma 5 |S(mre,0)| = 1. hasan impactonthe sieze ofthesphere,asis easilyconfirmed
bycomparingProposition6toLemma16(SectionVI).Hence
The following proposition is an application of Lemma 5.
the applicability of Lemma 5 is limited.
Proposition 6. |S(mr,1)|=1+(n−1)(n/r−1). We begin to address the issue of differing sphere sizes by
e
considering the radius t = 1 case. To aid with calculating
Proof: First note that |S(mr,0)|=1. There is only one
e such sphere sizes, we introduce (as our own definition) the
possible partition λ ⊢ n such that λ = n−1, namely λ :=
1 following subset of the set of translocations.
(n−1,1), with its Young diagram pictured below.
Definition. Let n∈Z . Define
n−1 >0
... Tn :={φ(i,j)∈Sn : i−j 6=1}.
z }| {
We call T the unique set of translocations.
n
Therefore by Lemma 5, |S(mr,1)| = 1 + (fλ′)(Kλ′). Bydefinition,T isthesetofalltranslocationsinS ,except
e r n n
Applying the well-known hook length formula ([5], [6]), we translocationsoftheformφ(i,i−1).Weexcludetranslocations
obtainfλ′ =n−1.ThevalueKλ′ ischaracterizedbypossible ofthisformbecausetheycanbemodeledbytranslocationsof
r
fillings of row 2 with the stipulation that each i∈[n/r] must the form φ(i−1,i), and are therefore redundant.
appear exactly r times in the diagram. In this case, since WeclaimthatthesetT ispreciselythesetoftranslocations
n
there is only a single box in row 2, the possible fillings are needed to obtain all unique permutations within the Ulam
i ∈ [n/r−1], each of which yields a unique Young tableau sphere of radius 1 via multiplication. Moreover, there is no
ofthe desiredtype.Hence Kλ′ =n/r−1,whichimpliesthat redundancy in the set, that is, there is no smaller set of
r
|S(mr,1)|=1+(n−1)(n/r−1). translocationsyieldingtheentireUlamsphereofradius1when
e
multiplied with a given center permutation. These facts are first variety corresponds to the first condition of the D(m)
stated in the next lemma. definition, when m(i) = m(j). For example, if σ ∈ S such
6
that m2 = (1,3,2,2,3,1), then we have m2 · φ(1,5) =
Lemma 7. S(σ,1)={σφ∈S : φ∈T } and σ σ
n n (3,2,2,3,1,1) = m2 ·φ(1,6), since m2(2) = 3 = m2(4).
|T |=|S(σ,1)|. σ σ σ
n Thisis because movingthe first 1 to the left or to the rightof
Proof: We will first show that S(σ,1) = {σφ ∈ S : the last 1 results in the same tuple.
n
φ∈T }. Note that The second variety corresponds to the second condition of
n
the D(m) definition above, when m(i) = m(i − 1). For
S(σ,1) = {π ∈S : d (σ,π)≤1}
n ◦ example, if m2 = (1,3,2,2,3,1) as before, then for all
= {σφ(i,j)∈Sn : i,j ∈[n]}. j ∈ {1,2,3,4,σ5,6}, we have m2 · φ(3,j) = m2 ·φ(4,j).
σ σ
This is because any translocation that deletes and inserts the
It is trivial that
second of the two adjacent 2’s does not result in a different
Tn = {φ(i,j)∈Sn : i−j 6=1} tuple when compared to deleting and inserting the first of the
⊆ {φ(i,j)∈S : i,j ∈[n]}. two adjacent 2’s.
n
Therefore {σφ∈Sn : φ∈Tn}⊆S(σ,1). Lemma 8. S(mrσ,1) = {mrσ · φ ∈ Mr(Sn) : φ ∈
To see why S(σ,1) ⊆ {σφ ∈ Sn : φ ∈ Tn}, consider Tn\D(mrσ)}.
any σφ(i,j) ∈ {σφ(i,j) ∈ Sn : i,j ∈ [n]} = S(σ,1). If Proof: Notice S(mr,1) = {mr ·φ ∈ M (S ) : φ ∈
i−j 6=1, then φ(i,j)∈Tn, and thus σφ(i,j)∈{σφ∈Sn : T }. Hence it suffices toσshow that σfor all φ(i,rj) ∈n D(mr),
n σ
φ ∈ Tn}. Otherwise, if i−j = 1, then σφ(i,j) = σφ(j,i), there exists some i′,j′ ∈[n] such that φ(i′,j′)∈T \D(mr)
n σ
and i−j = 1 =⇒ j−i =−16= 1, so φ(j,i) ∈Tn. Hence and mr ·φ(i,j)=mr ·φ(i′,j′). We proceed by dividing the
σφ(i,j)=σφ(j,i)∈{σφ∈Sn : φ∈Tn}. proofinσtotwomaincaσses.CaseIiswhen(mr(i)6=mr(i−1)
σ σ
Next we show that |Tn| = |S(σ,1)|. By Proposition 6 (in or i=1. Case II is when (mr(i)=mr(i−1).
the case that r = 1), |S(σ,1)| = 1+(n−1)2. On the other Case I (when (mr(i) 6= mσr(i−1) σor i = 1) can be split
hand, |T | = |{φ(i,j) ∈ S : i−j 6= 1}|. If i = 1, then σ σ
n n into two subcases:
there are n values j ∈ [n] such that i−j 6= 1. Otherwise,
if i ∈ [n] but i 6= 1, then there are n −1 values j ∈ [n] Case IA: i<j
such that i− j 6= 1. However, for all i,j ∈ [n], φ(i,i) = Case IB: i>j.
φ(j,j) = e so that there are n−1 redundancies. Therefore
|T |=n+(n−1)(n−1)−(n−1)=1+(n−1)2. We can ignore the instance when i = j, since φ(i,j) ∈
n D(mr) implies i 6= j. For case IA, if for all p ∈ [i,j] (for
Inthecaseofpermutations,thesetTn hasnoredundancies. a,b ∈σZ with a < b, the notation [a,b] := {a,a+1,...,b})
If φ ,φ ∈ T , then σφ = σφ implies φ = φ .
1 2 n 1 2 1 2 we have mr(i) = mr(p), then mr ·φ(i,j) = mr ·e. Thus
Alternatively,in the case of multipermutations,the set Tn can σ σ σ σ
setting i′ = j′ = 1 yields the desired result. Otherwise, if
generally be shrunken to exclude redundancies. Notice that
S(mrσ,1) = {mrπ ∈ Mr(Sn) : there exists φ s.t. mrσ ·φ = there exists p∈[i,j] such that mrσ(i)6=mrσ(p), then let
mπ}, which is equal to {mrσ ·φ ∈ Mr(Sn) : φ ∈ Tn}. j∗ :=j−min{k ∈Z>0 : mrσ(i)6=mrσ(j−k)}.
However, it is possible that there exist φ ,φ ∈ T such that
1 2 n
φ1 6=φ2, but mrσ·φ1 =mrσ·φ2. In such an instance we may Then φ(i,j∗)∈Tn\D(mrσ) and mrσ·φ(i,j)=mrσ·φ(i,j∗).
refer to either φ or φ as a duplicate translocationfor mr. Thussetting i′ =i and j′ =j∗ yields the desired result. Case
1 2 σ
If we remove all duplicate translocations for mr from IB is similar to Case IA.
Tn, then the resulting set will have the same cardinσality as Case II (when mrσ(i) = mrσ(i−1)), can also be divided
the r-regular Ulam sphere of radius 1 centered at mr. The into two subcases.
σ
next definition (our own) is the set of standard duplicate
Case IIA: i<j
translocations. For the remainder of the paper, assume that
m is an n-length integer tuple, i.e. m∈Zn. Case IIB: i>j.
As in Case I, we can ignore the instance when i = j. For
Definition. Define Case IIA, if for all p ∈ [i,j] we have mr(i) = mr(p), then
D(m):={ φ(i,j)∈Tn\{e} : (m(i)=m(j)) mdesrσir·eφd(rie,sju)lt=. Omtherσrw· eis,e,soifstehtteirnegeixi=stsjσp=∈ 1[i,ajc]hσsieuvcehstthhaet
or (m(i)=m(i−1) or i=1) } mr(i)6=mr(p), then let
σ σ
We call D(m) the standard duplicate translocation set for i∗ :=i−min{k∈Z>0 : (mrσ(i)6=mrσ(i−k−1)) or (i−k=1)}.
m. For each i ∈ [n], also define D (m) := {φ(i,j) ∈ D :
i n Then mr · φ(i,j) = mr · φ(i∗,j) and either one of the
j ∈[n]}. σ σ
following is true: (1) φ(i∗,j) ∈/ Di∗(mrσ) =⇒ φ(i∗,j) ∈/
Ifwetakeanr-regularmultipermutationmr,thenremoving D(mr), so set i′ = i∗ and j′ = j; or (2) by Case IA
σ σ
D(mr) from T equates to removing a set of duplicate there exist i′,j′ ∈ [n] such that φ(i′,j′) ∈ T \D(mr) and
σ n n σ
translocations. These duplications come in two varieties. The mr · φ(i′,j′) = mr · φ(i∗,j) = mr · φ(i,j). Case IIB is
σ σ σ
similar to Case IIA. Moreover,forallp∈/ [k,k+3],wehave(m·φ(i,k+3))(p)=
While Lemma 8 shows that D(mr) is a set of duplicate m(p) = (m·φ(k +3,k))(p). Therefore m·φ(i,k +3) =
σ
translocationsfor mr, we have notshown that T \D(mr) is m·φ(k +3,k). Also notice that m(i) 6= m(i−1) implies
σ n σ
the set of minimal size having the quality that S(mrσ,1) = that mφ(i,k+3)∈/ D(m). Hence φ(i,k+3)∈Ei(m).
{mrσ ·φ ∈ Mr(Sn) : φ ∈ Tn\D(mrσ)}. In fact it is not We now prove the second half of the lemma. That is,
minimal. In some instances it is possible to remove further we assume that E (m) 6= ∅ and then show that 1) and
i
duplicate translocations to reduce the set size. We will define 2) necessarily hold. Suppose that E (m) is nonempty. Then
i
another set of duplicate translocations, but a few preliminary m(i) 6= m(i−1), since otherwise there would not exist any
definitions are first necessary. φ(i,j)∈T \D(m).
n
We say thatm isalternatingif foralloddintegers1≤i≤
Let j ∈ [i+1,n] and k ∈ [i,j −2] such that φ(j,k) ∈
n,m(i)=m(1)andforallevenintegers2≤i′ ≤n,m(i′)=
T \D(m) and m · φ(i,j) = m(j,k). Existence of such j,
m(2)butm(1)6=m(2).Inotherwords,anyalternatingtuple n
k, and φ(j,k) is guaranteed by definition of E (m) and the
is of the form (a,b,a,b,...,a,b) or (a,b,a,b,...,a) where i
fact that E (m) was assumed to be nonempty. Then for all
a,b∈Zanda6=b.Anysingletonisalsosaidtobealternating. i
p ∈ [i,k −1], we have m(p) = m(p+1) and for all p ∈
Now for integers 1≤i≤n and 0≤k ≤n−i, the substring
[k,j−2], we have m(p)=m(p+2). Hence either m[k,j] is
m[i,i+k] of m is defined as m[i,i+k] := (m(i),m(i+
alternating,orelseforallp,q∈[k,j],wehavem(p)=m(q).
1),...m(i+k)). Given a substring m[i,j] of m, the length
However, the latter case is impossible, since it would imply
|m[i,j]| of m[i,j] is defined as |m[i,j]| := j −i+1. As
that for all p,q ∈ [i,j] that m(p) = m(q), which would
an example, if m′ := (1,2,2,4,2,4,3,1,3), then m′[3,6] =
mean φ(j,k)∈/ T \D(m), a contradiction. Therefore m[k,j]
(2,4,2,4) is an alternating substring of m′ of length 4. n
is alternating.
Definition. Next define It remains only to show that |m[k,j]| ≥ 4. Since k ∈
E(m):={ φ(i,j)∈T \D(m) : [i,j −2], it must be the case that |m[k,j]| ≥ 3. However,
n if |m[k,j]| = 3 (which occurs when k = j − 2), then
i<j and there exists k ∈[i,j−2] s.t.
(m·φ(i,j))(j)=m(i)=m(k)6=m(k+1)=(m·φ(j,k)(j),
(φ(j,k)∈Tn\D(m)) which implies that m·φ(i,j) 6= m·φ(j,k), a contradiction.
and (m·φ(i,j)=m·φ(j,k)) }. Hence |m[k,j]|≥4.
We call E(m) the alternating duplicate translocation set One implication of Lemma 9 is that there are only two
for m because it is only nonempty when m contains an possible forms for m[i,j] where φ(i,j) ∈ Ei(m). The first
possibilityisthatm[i,j]isanalternatingsubstringoftheform
alternating substring of length at least 4. For each i ∈ [n],
also define E (m):={φ(i,j)∈E(m) : j ∈[n]}. (a,b,a,b,...,a,b)(herea,b∈Z),sothatm[i,j]·φ(i,j)isof
i theform(b,a,b,a...,b,a).Inthiscase,aslongasm[i,j]|≥
In the example of m′ := (1,2,2,4,2,4,3,1,3) above, 4, then setting k =i implies that k ∈[i,j−2], that φ(j,k)∈
m∗·φ(2,6)=m′·φ(6,3) and φ(2,6),φ(6,3)∈T9\D(m′), Tn\D(m), and that m[i,j]·φ(i,j)=m[i,j]·φ(j,k).
implyingthatφ(2,6)∈E(m′).Infact,itcaneasilybeshown The other possibility is that m[i,j] is of the form
that E(m′)={φ(2,6)}. (a,a,a,...,a,b,a,b,...,a,b) (again a,b ∈ Z), so that
Lemma 9. Let i∈[n]. Then E (m)6=∅ if and only if
i k n−k
1) m(i)6=m(i−1) m|[i,j]{·zφ(i,j})|isofth{ezform}(a,...,a,b,a,b,...,b,a).Again
2) There exists j ∈[i+1,n] and k ∈[i,j−2] such that
k−1 n−k+1
i) For all p∈[i,k−1], m(p)=m(p+1) in this case, as long as |m[i|,j]{|z≥}4|, then{zk ∈ [}i,j − 2],
ii) m[k,j] is alternating φ(j,k)∈Tn\D(m), and m[i,j]·φ(i,j)=m[i,j]·φ(j,k).
iii) |m[k,j]|≥4.
Remark. If
Proof: Let i∈[n]. We will first assume 1) and 2) in the 1) m is alternating and
lemma statementandshow thatEi(m) is notempty.Suppose 2) n is even
m(i)6=m(i−1), and that there exists j ∈[i+1,n] and k ∈ then m·φ(1,n)=m·φ(n,1).
[i,j−2]suchthatforallp∈[i,k−1],wehavem(p)=m(p+
1).Supposealsothatm[k,j]isalternatingwith|m[k,j]|≥4. Remark. If
For ease of notation, let a := m(k) = m(k+2) and b := 1) m is alternating
m(k+1)=m(k+3) so that m[k,k+3]=(a,b,a,b)∈Z4. 2) n≥3
Then 3) n is odd,
(m·φ(i,k+3))[k,k+3]= (m·φ(k,k+3))[k,k+3] then m·φ(1,n)6=m·φ(n,1).
= (b,a,b,a)
To calculate |E(mr|, we define a set of equal size that is
= (m·φ(k+3,k))[k,k+3]. σ
easier to calculate.
Definition. Define to be alternating,
E∗(m):={ (i,j)∈[n]×[n] : (m[i,j] is alternating) | E∗(m)|
i
and (|m[i,j]|≥4) i∈[[n]
and (|m[i,j]| is even) }. = |{(i,j)∈[n]×[n] : |m[i,j]|≥4 and |m[i,j]| is even}|
= |{(i,j)∈[n]×[n] : j−i+1∈K}|,
For each i ∈ [n], also define E∗(m) := {(i,j) ∈ E∗(m) :
i
j ∈[n]}. Notice that E∗(m)= E∗(m). where K is the set of even integers between 4 and n, i.e.
i
i∈S[n] K :={k∈[4,n] : k is even}. For each k ∈K, we have
Lemma 10. |E(m)|=|E∗(m)| |{(i,j)∈[n]×[n] : j−i+1=k}|
Proof: The idea of the proof is simple. Each element = |{i∈[n] : i∈[1,n−k+1]}|
φ(i,j) ∈ E(m) involves exactly one alternating sequence of = n−k+1.
lengthgreaterorequalto4,sothesetsizesmustbeequal.We
formalize the argument by showing that |E(m)| ≤ |E∗(m)| Therefore |E(m)|= (n−k+1). In the case that n is
and then that |E∗(m)|≤|E(m)|. even, then kP∈K
To see why |E(m)| ≤ |E∗(m)|, we define a mapping f :
n/2
[n]→[n], whichmapsindexvalueseitherto thebeginningof (n−k+1) = (n−2i+1)
the nearest alternating subsequence to the right, or else to n. kX∈K Xi=2
For all i∈[n], let
(n−2)/2 2
n−2
= (2i−1)= .
(cid:18) 2 (cid:19)
i+min{p∈Z≥0 : (m(i)6=m(i+p+1))∨(i+p=n)} Xi=1
f(i):= (ifm(i)6=m(i−1)ori=1) In the case that n is odd, then
n (otherwise)
 (n−1)/2
Notice by definition of f, if i,i′ ∈ [n] such that i 6= i′, (n−k+1) = (n−2i+1)
and if m(i) 6= m(i − 1) or i = 1 and at the same time kX∈K Xi=2
m(i′)6=m(i′−1) or i′ =1, then f(i)6=f(i′). (n−3)/2 n−3 n−1
Now for each i ∈ [n], if m(i) 6= m(i − 1) or i = 1, = 2i=(cid:18) 2 (cid:19)(cid:18) 2 (cid:19).
then|E (m)|= |E∗ (m)|byLemma9andthetwoprevious Xi=1
i f(i)
remarks. Otherwise, if m(i) = m(i − 1), then |E (m)| =
|E∗ (m)| = 0. Therefore |E (m)| ≤ |E∗(m)|. Thiis is true Notice that by Lemma 11, it suffices to calculate |E(m)|
forf(ail)l i∈[n], so |E(m)|≤|Ei∗(m)|. i for locally maximal length alternating substrings of m. An
alternating substring m[i,j] is of locally maximal length if
The argument to show that |E∗(m)| ≤ |E(m)| is similar, and only if 1) m[i−1] is not alternating or i = 1; and 2)
except it uses the following function g : [n] →[n] instead of m[i,j+1] is not alternating or j =n.
f. For all i∈[n], let
Finally, we define the general set of duplications. The
lemma that follows the definition also shows that removing
i−min{p∈Z≥0 : (m(i)6=m(i−p−1))∨(i−p=1)} the set D∗(mrσ) from Tn removesall duplicate translocations
g(i):= (ifm(i)6=m(i−1)ori=n) associated with mrσ.
n (otherwise) Definition (D∗(m), duplication set). Define
D∗(m):=D(m)∪E(m).
By definition, calculating |E∗(m)| equates to calculating
the number of alternating substrings m[i,j] of m such that We call D∗(m) the duplication set for m. For each i ∈ [n],
thelengthofthesubstringisbothevenandlongerthan4.The we also define D∗(m):={φ(i,j)∈D∗(m) : j ∈[n]}.
i
following lemma helps to simplify this calculation further.
Lemma 12. Let φ ,φ ∈T \D∗(mr). Then φ =φ if and
1 2 n σ 1 2
Lemma 11. Let m be an alternating string. Then only if mrσ·φ1 =mrσ·φ2.
1) If n is even then |E∗(m)|= n−2 2 φ =Prmoorf:·Lφettφri1v,iφal2ly∈. ITtnr\emDa∗i(nmsrσto).pIrfoφve1 =thaφt2mthren·φmrσ=·
(cid:18) 2 (cid:19) 1 σ 2 σ 1
mr·φ =⇒ φ =φ .Weproceedbycontrapositive.Suppose
n−3 n−1 σ 2 1 2
2) If n is odd then |E∗(m)|=(cid:18) 2 (cid:19)(cid:18) 2 (cid:19) that φ1 6= φ2. We want to show that mrσ ·φ1 6= mrσφ2. Let
φ := φ(i ,j ) and φ := φ(i ,j ). The remainder of the
1 1 1 2 2 2
Proof:Assumemisanalternatingstring.ByLemma10, proof can be split into two main cases: Case I is if i1 = i2
|E(m)| = |E∗(m)| = | Ei∗(m)|. Since m was assumed andCaCsaeseI I(Iwihseinf ii1 6==i2i. ), can be further divided into two
i∈S[n] 1 2
subcases: Lemma 12 implies that we can calculate r-regular Ulam
Case IA: mrσ(i1)=mrσ(i1−1) sphere sizes of radius 1 whenever we can calculate the
appropriateduplication set. This calculation can be simplified
Case IB: mrσ(i1)6=mrσ(i1−1). bynotingthatforasequencem∈ZnthatD(m)∩E(m)=∅
CaseIAiseasytoprove.WehaveD∗(mr)=D∗(mr)= (by the definition of E(m)) and then decomposingthe dupli-
i1 σ i2 σ
{φ(i ,j) ∈ T \{e} : j ∈ [n]}, so φ = e = φ , a cation set into these components. This idea is stated in the
1 n 1 2
contradiction. For Case IB, we can first assume without loss next theorem
of generality that j < j and then split into the following
1 2
smaller subcases: Theorem13. |S(mr,1)|=1+(n−1)2−|D(mr)|−|E(mr)|.
σ σ σ
i) (j1 <i1) and (j2 >i1) Proof: By the definition of D∗(mr) and lemma 8,
σ
ii) (j1 <i1) and (j2 ≤i1)
{mr ·φ∈M (S ):φ∈T \D∗(mr)}
iii) (j1 >i1) and (j2 >i1) σ r n n σ
= {mr ·φ∈M (S ) : φ∈T \D(mr)}
iv) (j1 >i1) and (j2 ≤i1). σ r n n σ
= S(mr,1).
However, subcase iv) is unnecessary since it was assumed σ
that j1 < j2, so j1 > i1 =⇒ j2 > j1 > i1. Subcase ii) can This implies |Tn\D∗(mrσ)| ≥ |S(mrσ,1)|. By lemma 12, for
alsobereducedto(j1 <i1) and (j2 <i1)sincej2 6=i2 =i1. φ1,φ2 ∈ Tn\D∗(mrσ), if φ1 6= φ2, then mrσ ·φ1 6= mrσ ·φ2.
Each of the remaining subcases is proven by noting that Hence we have |Tn\D∗(mrσ)| ≤ |S(mrσ,1)|, which implies
there is some element in the multipermutationmrσ·φ1 that is that |Tn\D∗(mrσ)| = |S(mrσ,1)|. It remains to show that
necessarily different from mrσ ·φ2. For example, in subcase |Tn\D∗(mrσ)|=(n−1)2+1−|D(mrσ)|−|E(mrσ)|. This is
i), we have mrσ·φ1(j1)=mrσ(i1)6=mrσ(j1)=mrσ·φ2(j1). animmediateconsequenceofthefactthat|Tn|=(n−1)2+1
Subcases ii) and iii) are solved similarly. and D(mr)∩E(mr)=∅.
σ σ
Theorem13reducesthecalculationof|S(mr,1)| to calcu-
σ
Case II (when i 6=i ) can be divided into three subcases: lating |D(mr)| and |E(mr|. It is an easy matter to calculate
1 2 σ σ
Case IIA: (mrσ(i1)=mrσ(i1−1) and mrσ(i2)=mrσ(i2−1)), o|Df(im∈rσ[)n|,]ssinuccehitthiastemxarc(tliy)e=qumalrt(oi(−n−1)2)pltuimse(srt−he1n)utmimbeesr
Case IIB: either σ σ
the number of i∈[n] such that mr(i)6=mr(i−1). We also
(mrσ(i1)=mrσ(i1−1) and mrσ(i2)6=mrσ(i2−1)) showed how to calculate |E(m)| eσarlier. Theσnext example is
or (mrσ(i1)6=mrσ(i1−1) and mrσ(i2)=mrσ(i2−1)), an application of Theorem 13
Case IIC: (mrσ(i1)6=mrσ(i1−1) and mrσ(i2)6=mrσ(i2−1)).
Example. Suppose σ := [1,2,3,4,9,6,7,11,5,10,12,8].
Case IIA is easily solved by mimicking the proof Thenm3 :=(1,1,1,2,3,2,3,2,4,4,3,4).There are 3 values
σ
of Case IA. Case IIB is also easily solved as follows. of i∈[12] such that m3(i)=m3(i−1), which implies that
σ σ
First, without loss of generality, we assume that |D(m3|=(3)(12−2)+(12−3)(3−1)=48.Meanwhile,by
σ
mrσ(i1) = mrσ(i1 − 1) and mrσ(i2) 6= mrσ(i2 − 1). Lemmas 10 and 11, |E(m3σ)|=((5−3)/2)((5−1)/2))=2.
TThheernefDori∗1e(mwerσ)h=av{eφm(i1r,j·)φ∈(Tjn)\{=e}m: rj(j∈)[n6=]},msorφ(i1 =) =e. By Theorem 13, |S(m3σ),1|=(12−1)2−48−2=71.
σ 1 2 σ 2 σ 2
mrφ (i −1). VI. MIN/MAX SPHERES AND CODE SIZE BOUNDS
σ 2 2
In this section we show choices of center achieving mini-
Finally, for Case IIC, without loss of generality we may
mumandmaximumr-regularUlamspheresizesfortheradius
assume that i < i and then split into the following four
1 2 t=1 case. The minimum and maximum valuesare explicitly
subcases:
given.We thendiscussresultingboundsoncodesize.Firstlet
i) (j1 <i2) and (j2 ≥i2) us consider the r-regular Ulam sphere of minimal size.
ii) (j1 <i2) and (j2 <i2) Lemma 14. |S(mr,1)|≤|S(mr,1)|
iii) (j1 ≥i2) and (j2 ≥i2) e σ
iv) (j1 ≥i2) and (j2 <i2). Proof:In the case thatn/r=1, then mre =e andmrσ =
σ,sothat|S(mr,1)|=|S(mr,1)|.Thereforewemayassume
. that n/r ≥2. Bey Theorem 13σ, min(|S(mr,1)|)=1+(n−
However, since φ(i2,j2) ∈ Tn\D∗(mrσ) implies i2 6= j2, 1)2−max(|D(mr)|+|E(mr)σ|∈).SnSince nσ/r ≥ 2, we know
siu)bcaansdes(ji) a≥ndi i)iia)ncdan(jbe>reid)ucreedspteoct(ivj1ely<. Fio2r)saunbdca(jse2 i>), that nσ−∈S2n> r−σ1, which imσplies that for all σ ∈ S , that
2 1 2 2 2 n
we have mr ·φ (j ) = mr(i ) 6= mr(j ) = mr ·φ (j ). |D(mr)| ismaximizedbymaximizingthenumberofintegers
σ 1 1 σ 1 σ 1 σ 2 1 σ
Subcasesii)andiii)aresolvedinasimilarmanner.Forsubcase i∈[n]suchthatmr(i)=mr(i−1).Thisisaccomplishedby
σ σ
iv), if j > i , then mr · φ (j ) = mr(i ) 6= mr(j ) = choosingσ =e,andhenceforallσ ∈S ,wehave|D(mr)|≥
1 2 σ 1 1 σ 1 σ 1 n e
mr ·φ (j ). Otherwise, if j = i , then φ = φ(i ,i ) and |D(mr)|.
σ 2 1 1 2 1 1 2 σ
φ =φ(i ,j ).Thusifmr·φ =mr·φ thenφ ∈D∗(mr), We next will show that for any increase in the size of
w1hich im2plie2s that φ ∈/ Tσ \D1 ∗(mrσ), a2contrad1ictioni.1 σ |E(mr)| compared to |E(mr)|, that |D(mr)| is decreased
1 n σ σ e σ
by a larger value compared to |D(mr)|, so that (|D(mr)|+ completing the proof of the first statement in the lemma.
e σ
|E(mr)|) is maximized when σ =e. We now prove that |S(mr,1)| ≤ |S(mr,1)|. Recall that
σ σ ω
Suppose σ ∈ S . By Lemmas 10 and 11, |E(mr| is |D(mr)| is equalto (n−2)times the numberof i∈[n] such
n σ σ
characterizedby the lengthsof its locally maximalalternating that mr(i) = mr(i−1) plus (r −1) times the number of
σ σ
substrings. For every locally maximal alternating substring i ∈ [n] such that mr(i) 6= mr(i−1). But n/r > 2 implies
σ σ
mr[a,a+k−1] of mr of length k, there are at least k−2 thatr−1<n−2,whichimplies min|D(mr,1)|=(r−1)n.
fewσer instances where mσ r = mr(i−1) when compared to π∈Sn π
σ σ Therefore
instanceswheremr(i)=mr(i−1).Thisisbecauseforalli∈
e e
[a+1,a+k−1],mr(i)6=mr(i−1).Hence foreachlocally
σ σ
maximalalternatingsubstringmr(a,a+k−1),then|D(mr)| |S(mr,1)| ≤ 1+(n−1)2− min|D(mr,1)|
isdecreasedbyatleast(k−2)(n−σ 2−(r−1))≥(k−2)(r−σ1) σ π∈Sn π
−min|E(mr,1)|
whencomparedto|D(mre)|.Meanwhile,|E(mrσ)|isincreased π∈Sn π
by the same locally maximal alternating substring by at most ≤ 1+(n−1)2− min|D(mr,1)|
(k −2)((k −2)/4) by Lemma 11. However, since k ≤ 2r, π∈Sn π
we have (k−2)((k−2)/4)≤(k−2)(r−1)/2, which is of = 1+(n−1)2−(r−1)n
course less than (k−2)(r−1). = |S(mr,1)|.
ω
Lemma 14, along with Proposition 6 implies that the r-
regular Ulam sphere size of radius t=1 is bounded (tightly)
below by (1+(n −1)(n/r −1)). This in turn implies the Extending the concept of perfect permutation codes dis-
following sphere-packing type upper bound on any single cussed in [10], we define a perfect multipermutation code.
error-correcting code. Let C be an MPC(n,r) code. Then C is a perfect t-error
correcting code if and only if for all σ ∈ S , there exists
Lemma 15. Let C be a single-error correcting MPC◦(n,r) a unique mr ∈ M (C) such that mr ∈ S(mnr,t). We call
c r σ c
code. Then such C a perfect t-error correcting MPC(n,r). With this
n! definitiontheupperboundoflemma16impliesalowerbound
|C| ≤ .
r (r!)n/r(1+(n−1)(n/r−1)) on a perfect single-error correcting MPC(n,r).
Proof: Let C be a single-error correcting MPC◦(n,r) Lemma 17. Let n/r6=2 and let C be a perfect single-error
code.A standardsphere-packingboundargumentimpliesthat correcting MPC(n,r).Then
|C| ≤ n! . The remainder of the proof
follrows fr(orm!)n/Prr(oσmp∈ioSnns(i|tSio(mnσ6,1a)|n)d Lemma 14. |C|r ≥ (r!)n/r((1+(n−n!1)2)−(r−1)n).
We have seen that |S(mr)| is minimized when σ = e.
σ Proof: Suppose n/r 6= 2 and C is a perfect single-error
We now discuss the choice of center yielding the max-
iωm(ia)l s:p=here((isiz−e.1L)etmoωd (∈n/rS)n)r b+e ⌈dierfi/nne⌉d aansdfoωllow:=s: cmoerarencsting MPC(n,r). Then cP∈C|S(mrc,1)| = (r!n)n!/r. This
[ω(1),ω∗(2),...ω∗(n)]. With this definition, for all i ∈ [n], n!
we have mrω(i) = i mod (n/r) For example, if r = 3 (|C|r)·(cid:18)mc∈aCx(|S(mrc,1)|)(cid:19)≥ (r!)n/r,
and n = 12, then ω = [1,4,7,10,2,5,8,11,3,6,9,12] and
which by Lemma 16 implies the desired result.
mr = (1,2,3,4,1,2,3,4,1,2,3,4). We can use Theorem
13ωto calculate |S(mr,1)|, and then show that this is the A more generallowerboundis easily obtainedbyapplying
ω Lemma 16 with a standard Gilbert-Varshamov bound argu-
maximalr-regularUlamspheresize(exceptforthecasewhen
ment. In the lemma statement, C is an MPC (n,r,d) if and
n/r=2). ◦
only if C is an MPC(n,r) such that min dr(σ,π)=d.
◦
Lemma 16. Letn/r6=2.Then|S(mr,1)|≤|S(mr,1)| and σ,π∈C,σ6=π
σ ω
if n/r>2, then |S(mr,1)|=(1+(n−1)2)−(r−1)n. Lemma 18. Let n/r6=2 andC be an MPC (n,r,d) codeof
ω ◦
maximal cardinality. Then
Proof: Assume n/r 6= 2. First notice that if n/r = 1
then for any π ∈ S (including π = ω), the sphere n!
n |C| ≥
S(mr,1) contains exactly one element (the tuple of the r (r!)n/r(1+(n−1)2−(r−1)n)d−1
π
form (1,1,...,1)). Hence the lemma holds trivially in this
Proof: Suppose that n/r 6= 2 and that C is an
instance. Next, assume that n/r >2. We will first prove that MPC (n,r,d) code of maximal cardinality. For all σ ∈ S ,
|S(mr,1)|=(1+(n−1)2)−(r−1)n. ◦ n
ω there exists c ∈C such that dr(σ,c) ≤d−1. Otherwise, we
Since n/r > 2, it is clear that mr contains no alternating ◦
ω could add σ ∈/ C (and its entire equivalence class R (σ)) to
r
subsequences of length greater than 2. Thus by Lemma 9,
C while maintaining a minimum distance of d, contradicting
E(mr) = ∅ and therefore by Theorem 13, |S(mr,1)| =
ω ω the assumption that |C| is maximal.
1+(n−1)2 −|D(mr)|. Since there does not exist i ∈ [n] r
ω Therefore S(mr,d−1)=M (S ).Thisinturnimplies
such that mr(i)=mr(i−1), we have |D(mr)|=(r−1)n, c r n
ω ω ω cS∈C