Equivalencia de Codigos

8/14/2019 Equivalencia de Codigos

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Equivalence of codes: sufficient

conditions

Jay A. Wood

Department of MathematicsWestern Michigan University

http://homepages.wmich.edu/jwood/

Algebra for Secure and Reliable Communications

ModelingMorelia, Michoacan, Mexico

October 12, 2012
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Introduction

The next three lectures will address the topic ofcode equivalence.

If two linear codes are isomorphic via an

isomorphism that preserves weight, does theisomorphism extend to a weight-preservingmonomial transformation?

In the first two lectures I will concentrate on linear

codes over rings with the Hamming weight. More general weights and alphabets are the subject

of the final lecture.

JW (WMU) Sufficient conditions October 12, 2012 2 / 28
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Monomial transformations

Let Rbe a finite associative ring with 1.

A monomial transformation T :Rn Rn is ahomomorphism of left R-modules of the form

(a1, . . . , an)T = (a(1)u1, . . . , a(n)un),

where the uiare units in R and is a permutation

of{1, . . . , n}. Homomorphisms of left modules are written on the

right.

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Weights and symmetry groups

A weighton R is a function w :R C withw(0) = 0.

The left symmetry groupisGl :={u U(R) :w(ua) =w(a), aR}.

The right symmetry groupisGr :={v U(R) :w(av) =w(a), aR}.

U(R) denotes the group of units in R. For Hamming weight wt, Gl=Gr=U(R).

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Gr-monomial transformations

If the units of a monomial transformation T belongto the subgroup Gr, then T is a Gr-monomial

transformation. Gr-monomial transformations preserve weight:

w(aT) =w(a), for all aRn.

The Gr-monomial transformations are isometriesfor

the weight w. (Alonzo Sepulvedas talk.)

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Equivalence of codes

Suppose C1, C2Rn are two left R-linear codes,

where Rhas weight w.

The codes C1, C2 are equivalentif there exists aGr-monomial transformationTsuch thatC1T =C2.

If so, the restriction ofT to C1, T :C1C2, is anR-linear isomorphism that preserves the weight w

(an isometry).

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The extension problem

Suppose f :C1C2 is an R-linear isomorphismthat preserves the weight w. Does fextend to aGr-monomial transformation?

For R= Fqand Hamming weight wt, the answer isyes: MacWilliams, 1961-1962.

Also true for Frobenius rings, 1999, by character

theoretic methods; Greferath and Schmidt, 2000, bycombinatorial methods; Greferath, 2002.



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Main Theorem

TheoremLet R be a finite Frobenius ring equipped with the

Hamming weightwt. Suppose C1, C

2Rn are two left

R-linear codes. If f :C1C2 is an R-linearisomorphism that preserveswt, then f extends to amonomial transformation.

For w= wt, Gr=U(R).



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Idea

Express weight-preservation as an equation ofcharacters. (This idea comes from Ward and was

used to prove the extension theorem for finite fields,1996.)

Use linear independence of characters and a partialordering to match up terms and produce the

monomial transformation.



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Vanishing formula for characters

IfG is a finite abelian group, then

G(g) = |G|, g= 0,

0, g= 0.

For a= (a1, a2, . . . , an)Rn,

wt(a) =n 1|R|

ni=1

R

(ai).



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Re-cast the problem

Re-phrase the hypothesis slightly. Let Mbe a finiteleft R-module (the module underlying the linearcode C1).

Suppose there are two embeddings g, h:M Rnsuch that wt(mg) = wt(mh) for all mM. (Viewgas the inclusion ofC1R

n, and h asg f :MC2.)

Write the components ofg, h as g= (g1, . . . , gn),h= (h1, . . . , hn).

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Weight preservation condition Weight-preservation means wt(mg) = wt(mh), for

all mM. By vanishing formula, for all mM,

n

1

|R|

ni=1R(mgi) =n

1

|R|

ni=1R(mhi).

Simplifying, and changing summation variables, for

all mM,n

i=1

R

(mgi) =

nj=1

R

(mhj).



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Using Frobenius hypothesis

Because Ris Frobenius, there exists a (left)generating character on R. Every character Ris of the form rforrR.

Remember that (r)(a) =(ar), for a, rR. Weight-preservation becomes, for all mM,

n

i=1 rR (mgir) =n

j=1 sR (mhjs). This is an equation of characters on the module M.



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Using linear independencetrial run

The weight-preservation condition, again, is, for allmM,

n

i=1 rR (mgir) =n

j=1 sR (mhjs). This is an equation of characters (with all

coefficients equal to 1 C).

By linear independence of characters, the terms onthe left must match the terms on the right.

But how to get units?

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A partial ordering

Let M := HomR(M, R) be the collection of all leftR-homomorphisms from M to R.

M is itself a right R-module under m(gr) = (rm)g.

Let Pbe the set of principal right R-submodules ofM, gRM, partially ordered by inclusion.

Fact: it follows from work of Bass that gR=hR iffthere exists a unit u U(R) such that h=gu.

In their work, Greferath-Schmidt use a similar posetconstruction directly on R.

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Using linear independence

The weight-preservation condition, again, is, for allmM,

n

i=1rR

(mgir) =

n

j=1sR

(mhjs).

This is an equation of characters (with all

coefficients equal to 1 C). By linear independence of characters, the terms on

the left must match the terms on the right.

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Matching

From the finite list of principal submodulesg1R, g2R, . . . , gnR, h1R, . . . , hnR, choose one that ismaximal in the partial ordering. (Say, h1R.)

Consider the term on the right with j= 1 ands= 1R. By linear independence, there exists i0 andrRon the left so that (mgi0r) =(mh1) for allmM.

This says that M(h1 gi0 r)ker is a leftR-submodule of ker .

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Frobenius condition, again

But is a left generating character on R, soM(h1 gi0 r) = 0; that is, h1=gi0r.

Thus, h1Rgi0R.

But h1Rwas chosen to be maximal. Thush1R=gi0R.

By Bass, there exists a unit u1 U(R) such thath1=gi0u1.

Begin to define a permutation of{1, . . . , n}with(1) =i0.

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Inner sums

In the weight-preservation condition

n

i=1 rR(mgir) =

n

j=1 sR(mhjs),

we now examine the inner sums for h1 and g(1).

Because h1=g(1)u1, h1s=g(1)u1s.

As svaries over all ofR, so does u1s. Thus

sR(mh1s) =

rR(mg(1)r), for all

mM.

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Induction step

SubtractsR(mh1s) =rR(mg(1)r) fromboth sides ofn

i=1 rR(mgir) =

n

j=1 sR(mhjs).

This reduces the size of the outer sums (in i andj) by one.

Do the same process repeatedly. This inductively builds a permutation of

{1, 2, . . . , n}and produces units ui U(R) suchthat hi=g(i)ui, as desired.

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Other alphabets

Greferath, Nechaev, Wisbauer, 2004, proved theextension theorem for homogeneous and Hammingweights with A=

Rfor ANY finite ring R

(Frobenius or not). The Frobenius ring case is then a special case.

Most general, 2008: any ring R, alphabet AwithHamming weight, provided Ais pseudo-injective andembeds intoR (A admits a left generatingcharacter).

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Converses?

If a ring Ror an R-module alphabet Asatisfy theextension property for Hamming weight, what canwe say about R orA?

In the ring case, Rmust be Frobenius (2008).(Next lecture.)

In the module case, Amust be pseudo-injective and

embed inR(2008).JW (WMU) Sufficient conditions October 12, 2012 22 / 28
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Other weights?

For weights other than the Hamming weight and

the homogeneous weight, much less is known. This topic is the subject of the final lecture.

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Parametrized codes

Setting: ring R, left R-module alphabet A.

A left R-linear parametrized codeover A is pair(M, ), where M is a finite left R-module, and

:MAn is a homomorphism of left R-modules. is given by an n-tuple (1, 2, . . . , n) of

homomorphisms i :MA. (Evaluation codes.)

Suppose A has weight wand symmetry groupsGl U(R) and GrAut(A).

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Equivalence of parametrized codes

Two parametrized codes over the same module M,(M, ), (M, ), are equivalentif there is aGr-monomial transformation T on A

n such that

= T. Then w(m) =w(mT) =w(m) for all mM.

Up to equivalence, what matters is the number of

coordinate functionals (the i) that belong to eachGr-scale class in HomR(M, A).

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Multiplicity functions

Given a finite left R-module M, let O be the set ofGr-scale classes in HomR(M, A).

Up to equivalence, a parametrized code on M iscompletely determined by a multiplicity function:O N, i.e., an element F(O,N).

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Weight mapping

Define an additive mappingW :F(O,N)F(M,C) by

(m

[]Ow(m)([])).

[] denotes the Gr-scale class ofHomR(M, A).

This calculates the weight w(m) for every mM,

where is determined by the multiplicity function . This mapping is well-defined and additive. (Addition

in F(O,N) corresponds to concatenation ofgenerator matrices.)

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Extension problem re-cast

The extension problem for the weight wover thealphabet Ais satisfied iff the mapping W is injectivefor every module M.

I.e., if two parametrized codes based on Myield thesame weights (weight preservation), then the codesare equivalent (differ by a Gr-monomialtransformation).

This formulation will be used in the next lecture onnecessary conditions.

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