Linearized Polynomial Modules and their Application to ...user.math.uzh.ch/trautmann/linearized_poly_mods.pdf · Linearized Polynomial Modules and their Application to Network Coding

Linearized Polynomial Modules and their Application to Network Coding

Linearized Polynomial Modules and theirApplication to Network Coding

Anna-Lena Horlemann-Trautmann

Algorithmics LaboratoryEPF Lausanne, Switzerland

October 18th, 2016Clemson University


Linearized Polynomials

1 Linearized PolynomialsThe Ring Lq(x, q

m)Modules over Lq(x, q

m)

2 Network Coding and Gabidulin CodesIntroductionGabidulin Codes and Interpolation Decoding

3 Summary and Conclusion



The Ring Lq(x, qm)

Brief Recap of Finite Fields

Fq exists (uniquely) for q = pr, p prime

Fp∼= Zp

Fqm∼= Fq[α], α root of irreducible polynomial of degree m

Freshman’s Dream: (a+ b)q = aq + bq for a, b ∈ Fqm

aq = a for a ∈ Fq ≤ Fqm

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The Ring Lq(x, qm)

Definition

A (q-)linearized polynomial is of the form

f(x) =

n∑i=0

aixqi

for ai ∈ Fqm . If an ̸= 0, n is called the q-degree of f(x).

Theorem

The set Lq(x, qm) of all q-linearized polynomials over Fqm forms

a non-commutative ring, equipped with the normal addition +and composition (symbolic multiplication) ◦.

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The Ring Lq(x, qm)

Definition

A (q-)linearized polynomial is of the form

f(x) =

n∑i=0

aixqi

for ai ∈ Fqm . If an ̸= 0, n is called the q-degree of f(x).

Theorem

The set Lq(x, qm) of all q-linearized polynomials over Fqm forms

a non-commutative ring, equipped with the normal addition +and composition (symbolic multiplication) ◦.

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The Ring Lq(x, qm)

Example in L3(x, 32)

F32∼= F3[α], where α2 + 1 = 0

(2x3 + x) + (αx9 + x3) = αx9 + x

(2x3 + x) ◦ (αx9 + x3) = 2(αx9 + x3)3 + (αx9 + x3)= 2α3x27 + (2 + α)x9 + x3 = αx27 + (2 + α)x9 + x3

(αx9 + x3) ◦ (2x3 + x) = α(2x3 + x)9 + (2x3 + x)3

= 2αx27 + (2 + α)x9 + x3

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The Ring Lq(x, qm)

Symbolic division:

f(x) is (symbolically) divisible on the left by g(x) withquotient m(x) if

g(x) ◦m(x) = f(x)

f(x) is (symbolically) divisible on the right by g(x) withquotient m(x) if

m(x) ◦ g(x) = f(x)

left division with remainder: for any f(x), g(x) we have

f(x) = g(x) ◦m(x) + r(x), qdeg(r) < qdeg(g)

right division with remainder: for any f(x), g(x) we have

f(x) = m(x) ◦ g(x) + r(x), qdeg(r) < qdeg(g)

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The Ring Lq(x, qm)

Theorem

A linearized polynomial f(x) ∈ Lq(x, qm) is an Fq-linear map. If

qdeg(f) = k, then ker(f) is a k-dimensional Fq-vector space(over some extension field).


f(x) = x3 + 2α2x

f(λx+µy) = (λx+µy)3+2α2(λx+µy) = λx3+µy3+2α2λx+2α2µy = λ(x3 + 2α2x) + µ(y3 + 2α2y) = λf(x) + µf(y),for λ, µ ∈ F3

ker(f) = {0, α, 2α}

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The Ring Lq(x, qm)

Theorem

A linearized polynomial f(x) ∈ Lq(x, qm) is an Fq-linear map. If

qdeg(f) = k, then ker(f) is a k-dimensional Fq-vector space(over some extension field).


f(x) = x3 + 2α2x

f(λx+µy) = (λx+µy)3+2α2(λx+µy) = λx3+µy3+2α2λx+2α2µy = λ(x3 + 2α2x) + µ(y3 + 2α2y) = λf(x) + µf(y),for λ, µ ∈ F3

ker(f) = {0, α, 2α}

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The Ring Lq(x, qm)

Two special linearized polynomials

Let g = (g1, . . . , gn), r = (r1, . . . , rn) ∈ Fnqm .

Annihilator polynomial:

Πg(x) :=∏

β∈⟨g1,...,gn⟩Fq

(x− β)

q-Lagrange polynomial:

Λg,r(x) :=n∑

i=1

(−1)n−iridet(Di(g, x))

det(Mn(g))

Theorem

Πg(x) ∈ Lq(x, qm) and Πg(gi) = 0 for i = 1, . . . , n

qdeg(Πg(gi)) = dim⟨g1, . . . , gn⟩Fq

Λg,r(x) ∈ Lq(x, qm) and Λg,r(gi) = ri for i = 1, . . . , n

qdeg(Λg,r(x)) = dim⟨g1, . . . , gn⟩Fq − 1

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Modules over Lq(x, qm)



m)






Left Module Lq(x, qm)ℓ

for fi(x) ∈ Lq(x, qm), the elements of Lq(x, q

m)ℓ are

f := [f1(x) . . . fℓ(x)] =

ℓ∑i=1

fi(x)ei

for h(x) ∈ Lq(x, qm)

h(x) ◦ f := [h(f1(x)) . . . h(fℓ(x))] =

ℓ∑i=1

h(fi(x))ei.

Definition

A subset M ⊆ Lq(x, qm)ℓ is a (left) submodule of Lq(x, q

m)ℓ if itis closed under addition and composition with Lq(x, q

m) on theleft.

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Definition

Consider f (1), . . . , f (s) ∈ Lq(x, qm)ℓ. f (1), . . . , f (s) are linearly

independent if for any a1(x), . . . , as(x) ∈ Lq(x, qm)

s∑i=1

ai(x) ◦ f (i) = [ 0 . . . 0 ] =⇒ a1(x) = · · · = as(x) = 0.

A generating set of a submodule M ⊆ Lq(x, qm)ℓ is called a

basis of M if all its elements are linearly independent.

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Monomials

Notation: [i] := qi

The monomials of f = [f1(x) . . . fℓ(x)] are of the formx[k]ei for all k such that fik ̸= 0.

A monomial order < on Lq(x, qm)ℓ is a total order on

Lq(x, qm)ℓ, analogous to normal polynomial ring.

The (k1, . . . , kℓ)-weighted term-over-position monomialorder is defined as

x[i1]ej1 <(k1,...,kℓ) x[i2]ej2 : ⇐⇒

i1 + kj1 < i2 + kj2 or [i1 + kj1 = i2 + kj2 and j1 < j2].

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Definition

For any monomial order we define the following:

the leading monomial lm(f) = x[i1]ej1 is the greatestmonomial of f .

the leading position lpos(f) = j1 is the vector coordinate ofthe leading monomial.

the leading term lt(f) = fj1,i1x[i1]ej1 is the complete term of

the leading monomial.

The (k1, . . . , kℓ)-weighted q-degree of [f1(x) . . . fℓ(x)] is definedas max{ki + qdeg(fi(x)) | i = 1, . . . , ℓ}.

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Let f ,h ∈ Lq(x, qm)ℓ and let F = {f (1), . . . , f (s)}. We say that f

reduces to h modulo F in one step if and only if

h = f − ((b1x[a1]) ◦ f (1) + · · ·+ (bkx

[ak]) ◦ f (k))

for some a1, . . . , ak ∈ N0 and b1, . . . , bk ∈ Fqm .We say that f isminimal with respect to F if it cannot be reduced modulo F .

Definition

A module basis B is called minimal if all its elements b areminimal with respect to B\{b}.

Theorem

Let B be a basis of a module M ⊆ Lq(x, qm)ℓ. Then B is a

minimal basis if and only if all leading positions of the elementsof B are distinct.

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Let f ,h ∈ Lq(x, qm)ℓ and let F = {f (1), . . . , f (s)}. We say that f

reduces to h modulo F in one step if and only if

h = f − ((b1x[a1]) ◦ f (1) + · · ·+ (bkx

[ak]) ◦ f (k))

for some a1, . . . , ak ∈ N0 and b1, . . . , bk ∈ Fqm .We say that f isminimal with respect to F if it cannot be reduced modulo F .

Definition

A module basis B is called minimal if all its elements b areminimal with respect to B\{b}.

Theorem

Let B be a basis of a module M ⊆ Lq(x, qm)ℓ. Then B is a

minimal basis if and only if all leading positions of the elementsof B are distinct.

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Example in L3(x, 32)2

consider (1, 2)-weighted TOP order and

f = [x3 + x αx3], g = [2x9 + x3 2x]

qdeg(1,2)(f) = max(1 + 1, 1 + 2) = 3,qdeg(1,2)(g) = max(2 + 1, 0 + 2) = 3

lpos(f) = 2, lpos(g) = 1

lm(f) = [0 x3], lm(g) = [x9 0]

lt(f) = [0 αx3], lt(g) = [2x9 0]

=⇒ {f ,g} is a minimal basis of

rowspan

[x3 + x αx3

2x9 + x3 2x

]

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consider (1, 2)-weighted TOP order and

f = [x3 + x αx3], g = [2x9 + x3 2x]

qdeg(1,2)(f) = max(1 + 1, 1 + 2) = 3,qdeg(1,2)(g) = max(2 + 1, 0 + 2) = 3

lpos(f) = 2, lpos(g) = 1

lm(f) = [0 x3], lm(g) = [x9 0]

lt(f) = [0 αx3], lt(g) = [2x9 0]

=⇒ {f ,g} is a minimal basis of

rowspan

[x3 + x αx3

2x9 + x3 2x

]

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Computation of minimal bases (EEA)

Start with a basis {f (1), f (2), . . . , f (s)} ⊆ Lq(x, qm)ℓ of the

module.

Consider all elements with leading position 1 and reduce allbut one of them such that they have leading position > 1.

Consider all elements with leading position 2 and reduce allbut one of them such that they have leading position > 2.

etc.

Receive another basis with all distinct leading positions.=⇒ minimal basis!

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f = [2x27 + x9 + x3 + x (α+ 2)x3], g = [2x9 + x3 2x]

both have leading position 1 in (1, 2)-weighted order

compute h = f − x3 ◦ g = [x3 + x αx3]

h has leading position 2

{g,h} is minimal basis of ⟨f ,g⟩

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Theorem (Predictable Leading Monomial Property)

Let M be a module in Lq(x, qm)ℓ with minimal basis

B = {b(1), . . . ,b(L)}. Then for any 0 ̸= f ∈ M , written as

f = a1(x) ◦ b(1) + · · ·+ aL(x) ◦ b(L),

where a1(x), . . . , aL(x) ∈ Lq(x, qm), we have

lm(f) = max1≤i≤L;ai(x) ̸=0

{lm(ai) ◦ lm(b(i))}

where lm(ai(x)) is the term of ai(x) of highest q-degree.

Corollary

The leading positions and weighted q-degrees of elements of twodistinct minimal bases for the same module in Lq(x, q

m)ℓ haveto be the same. =⇒ cardinalities of minimal bases are equal

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Theorem (Predictable Leading Monomial Property)

Let M be a module in Lq(x, qm)ℓ with minimal basis

B = {b(1), . . . ,b(L)}. Then for any 0 ̸= f ∈ M , written as

f = a1(x) ◦ b(1) + · · ·+ aL(x) ◦ b(L),

where a1(x), . . . , aL(x) ∈ Lq(x, qm), we have

lm(f) = max1≤i≤L;ai(x) ̸=0

{lm(ai) ◦ lm(b(i))}

where lm(ai(x)) is the term of ai(x) of highest q-degree.

Corollary

The leading positions and weighted q-degrees of elements of twodistinct minimal bases for the same module in Lq(x, q

m)ℓ haveto be the same. =⇒ cardinalities of minimal bases are equal

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The Interpolation Module


Remember: annihilator Πg(x), q-Lagrange Λg,r(x)

Define the interpolation module

M(g, r) = rowspan

[Πg(x) 0

−Λg,r(x) x

]

Theorem

M(g, r) ⊆ Lq(x, qm)2 contains exactly all [f1(x) − f2(x)] with

f1(gi)− f2(ri) = 0.

=⇒ minimal basis of M(g, r) can be computed in Oqm(n3)

operations with EEA

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M(g, r) = rowspan

[Πg(x) 0

−Λg,r(x) x

]

Theorem


f1(gi)− f2(ri) = 0.


operations with EEA

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M(g, r) = rowspan

[Πg(x) 0

−Λg,r(x) x

]

Theorem


f1(gi)− f2(ri) = 0.


operations with EEA

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Network Coding and Gabidulin Codes

Introduction



m)





Introduction

Classical coding theory (simple channel):

codewords are vectors of length n over a finite field Fq

received word = codeword + error vector (r = c+ e ∈ Fnq )

most likely sent codeword ∼= closest codeword w.r.t.Hamming distance:

dH((u1, . . . , un), (v1, . . . , vn)) := |{i | ui ̸= vi}|.

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Introduction





dH((u1, . . . , un), (v1, . . . , vn)) := |{i | ui ̸= vi}|.

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Introduction





dH((u1, . . . , un), (v1, . . . , vn)) := |{i | ui ̸= vi}|.

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Introduction

Multicast Channel

All receivers want to receive the same information.

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Introduction

Example (Butterfly Network)

Linearly combining is better than forwarding:

R2

R1

a

a

a

aa

b b

b

a

R1 receives only a, R2 receives a and b.

Forwarding: need 2 transmissions to transmit a, b to bothreceivers

Linearly combining: need 1 transmission to transmit a, b toboth receivers

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Introduction

Example (Butterfly Network)

Linearly combining is better than forwarding:

R2

R1

a

a

a

a+ba+b

b b

b

a+b

R1 and R2 can both recover a and b with one operation.

Forwarding: need 2 transmissions to transmit a, b to bothreceivers

Linearly combining: need 1 transmission to transmit a, b toboth receivers

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Introduction

It turns out that linear combinations at the inner nodes are“sufficient” to reach capacity:

Theorem

One can reach the capacity of a single-source multicast networkchannel with linear combinations at the inner nodes.

Problem: Hamming metric does not reflect the number oferrors in the network.Solution: Use metric space (Fm×n

q , dR),

dR(A,B) = rank(A−B).

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Introduction


Theorem


Problem: Hamming metric does not reflect the number oferrors in the network.

Solution: Use metric space (Fm×nq , dR),


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Introduction


Theorem


Problem: Hamming metric does not reflect the number oferrors in the network.Solution: Use metric space (Fm×n

q , dR),


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Introduction

Definition

A rank-metric code is a subset of Fm×nq . The minimum distance

of C ⊆ Fm×nq is defined as

dR(C) := min{dR(A,B) | A,B ∈ C,A ̸= B}.

For a codeword A ∈ C, error matrix E ∈ Fm×nq and channel

operation matrix Mc, the channel output is

McA+ E.

Decoding a received word ∼= finding the closest codewordw.r.t. dR.

The error correction capability of C is (dR(C)− 1)/2.

The transmission rate of C is logq(|C|)/(mn).

We can use Fm×nq

∼= Fnqm and define C ⊆ Fn

qm .

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Introduction

Definition






McA+ E.




We can use Fm×nq


qm .

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Introduction

Definition






McA+ E.




We can use Fm×nq


qm .

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Gabidulin Codes and Interpolation Decoding



m)






Definition

Let g1, . . . , gn ∈ Fqm be linearly independent over Fq. Then

CGab := {(f(g1), . . . , f(gn)) | f(x) ∈ Lq(x, qm), qdeg < k}

is called a Gabidulin code in Fnqm of dimension k.

Theorem

A Gabidulin code C ⊆ Fnqm of dimension k has minimum rank

distance n− k + 1, which is optimal (Singleton bound).Therefore, it is called a maximum rank distance (MRD) code.

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Definition

Let g1, . . . , gn ∈ Fqm be linearly independent over Fq. Then

CGab := {(f(g1), . . . , f(gn)) | f(x) ∈ Lq(x, qm), qdeg < k}

is called a Gabidulin code in Fnqm of dimension k.

Theorem

A Gabidulin code C ⊆ Fnqm of dimension k has minimum rank

distance n− k + 1, which is optimal (Singleton bound).Therefore, it is called a maximum rank distance (MRD) code.

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Example:Consider F22

∼= F2[α], with α2 + α+ 1 = 0. Fix g1 = 1, g2 = α(linearly independent). The Gabidulin code of length n = 2,dimension k = 1 and minimum distance n− k + 1 = 2 is

(0 0) 7→(

0 00 0

)f(x) = 0

(1 α) 7→(

1 00 1

)f(x) = x

(α α+ 1) 7→(

0 11 1

)f(x) = αx

(α+ 1 1) 7→(

1 11 0

)f(x) = (α+ 1)x

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Interpolation Decoding:

c = (f(g1), . . . , f(gn)) ∈ Fnqm codeword;

r = c+ e ∈ Fnqm received word.

Theorem

dR(c, r) = t if and only if there exists a D(x) ∈ Lq(x, qm), such

that qdeg(D(x)) = t and

D(f(gi)) = D(ri) ∀i ∈ {1, . . . , n}.

This D(x) is called the error span polynomial.

Set N(x) := D(f(x)). Then N(gi)−D(ri) = 0.

=⇒ [N(x) −D(x)] ∈ M(g, r) (interpolation module)!

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Interpolation Decoding:

c = (f(g1), . . . , f(gn)) ∈ Fnqm codeword;

r = c+ e ∈ Fnqm received word.

Theorem

dR(c, r) = t if and only if there exists a D(x) ∈ Lq(x, qm), such

that qdeg(D(x)) = t and

D(f(gi)) = D(ri) ∀i ∈ {1, . . . , n}.

This D(x) is called the error span polynomial.

Set N(x) := D(f(x)). Then N(gi)−D(ri) = 0.

=⇒ [N(x) −D(x)] ∈ M(g, r) (interpolation module)!

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Theorem

The elements [N(x) −D(x)] of M(g, r) that fulfill

1 qdeg(N(x)) ≤ t+ k − 1,

2 qdeg(D(x)) = t,

3 N(x) is symbolically divisible on the left by D(x), i.e. thereexists f(x) ∈ Lq(x, q

m) such that D(f(x)) = N(x),

are in one-to-one correspondence with the codewords of rankdistance t to r.

The quotient is the respective message polynomial in Lq(x, qm).

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Theorem

The elements [N(x) −D(x)] of M(g, r) that fulfill

1 qdeg(N(x)) ≤ t+ k − 1,

2 qdeg(D(x)) = t,

3 N(x) is symbolically divisible on the left by D(x), i.e. thereexists f(x) ∈ Lq(x, q

m) such that D(f(x)) = N(x),

are in one-to-one correspondence with the codewords of rankdistance t to r.

The quotient is the respective message polynomial in Lq(x, qm).

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The Parametrization:

Goal: Find all [N(x) −D(x)] ∈ M(r) with

1 qdeg(N(x)) ≤ t+ k − 1,

2 qdeg(D(x)) = t,

Solution: If {b1, b2} is minimal basis (ordered by leadingpositions) with ℓ1, ℓ2 the respective weighted row q-degrees, then

λ(x) ◦ b1 + µ(x) ◦ b2 with

1 qdeg(λ(x)) ≤ t− ℓ1 + k − 1,

2 qdeg(µ(x)) = t− ℓ2 + k − 1 and µ(x) is monic

yield all the desired elements.

Proof: Based on the Predictable Leading Monomial Property.

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1 qdeg(N(x)) ≤ t+ k − 1,

2 qdeg(D(x)) = t,


λ(x) ◦ b1 + µ(x) ◦ b2 with

1 qdeg(λ(x)) ≤ t− ℓ1 + k − 1,




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1 qdeg(N(x)) ≤ t+ k − 1,

2 qdeg(D(x)) = t,


λ(x) ◦ b1 + µ(x) ◦ b2 with

1 qdeg(λ(x)) ≤ t− ℓ1 + k − 1,




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The Decoding Algorithm:

Precomputed and stored: Πg(x) ∈ Lq(x, qm) (annihilator)

Input: received word r ∈ Fnqm

Compute q-Lagrange polynomial Λg,r(x).

Compute a minimal basis {b1, b2} of the interpolationmodule

M(g, r) = rowspan

[Πg(x) 0

−Λg,r(x) x

].

Check all elements of the form λ(x) ◦ b1 + µ(x) ◦ b2 withrestricted degrees of λ(x), µ(x) ∈ Lq(x, q

m) for divisibility.

Output: the symbolic quotients (where existent)

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The Decoding Algorithm:

Precomputed and stored: Πg(x) ∈ Lq(x, qm) (annihilator)

Input: received word r ∈ Fnqm

Compute q-Lagrange polynomial Λg,r(x).

Compute a minimal basis {b1, b2} of the interpolationmodule

M(g, r) = rowspan

[Πg(x) 0

−Λg,r(x) x

].

Check all elements of the form λ(x) ◦ b1 + µ(x) ◦ b2 withrestricted degrees of λ(x), µ(x) ∈ Lq(x, q

m) for divisibility.

Output: the symbolic quotients (where existent)

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Example:Consider the Gabidulin code C over F23

∼= F2[α] (withα3 = α+ 1) with generator matrix

G =

(1 α α2

1 α2 α4

).

=⇒ dR = n− k + 1 = 3− 2 + 1 = 2=⇒ C is no-error correcting

Received word:r = ( α+ 1 0 α ).

Interpolation module:

M(r) = rowspan

[Πg(x) 0Λg,r(x) x

]= rowspan

[x8 + x 0

α2x4 + α5x x

].

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G =

(1 α α2

1 α2 α4

).

=⇒ dR = n− k + 1 = 3− 2 + 1 = 2=⇒ C is no-error correctingReceived word:

r = ( α+ 1 0 α ).


M(r) = rowspan

[Πg(x) 0Λg,r(x) x

]= rowspan

[x8 + x 0

α2x4 + α5x x

].

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G =

(1 α α2

1 α2 α4

).

=⇒ dR = n− k + 1 = 3− 2 + 1 = 2=⇒ C is no-error correctingReceived word:

r = ( α+ 1 0 α ).


M(r) = rowspan

[Πg(x) 0Λg,r(x) x

]= rowspan

[x8 + x 0

α2x4 + α5x x

].

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Finding the minimal basis with the Euclidean algorithm:

x8 + x = (α3x2) ◦ (α2x4 + α5x) + (α6x2 + x).

Since qdeg(α3x2) + k − 1 = 2 ≥ 1 = qdeg(α6x2 + x), thealgorithm terminates and a minimal basis (w.r.t. the(0, 1)-weighted 2-degree) of this module is[

0 xx α3x2

]◦[

x8 + x 0α2x4 + α5x x

]︸︷︷︸

original basis

=

[α2x4 + α5x xα6x2 + x α3x2

]︸︷︷︸

minimal basis

.

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Parametrization λ(x) ◦ b1 + µ(x) ◦ b2:

Use all λ(x) ∈ L2(x, 23) with 2-degree 0 and all monic

µ(x) ∈ L2(x, 23) with 2-degree 0.

=⇒ λ(x) = a0x, a0 ∈ F23 , and µ(x) = x

E.g. for a0 = α:

(αx) ◦ [α2x4 + α5x x] + [α6x2 + x α3x2]

= [α3x4 + α6x αx] + [α6x2 + x α3x2]

= [α3x4 + α6x2 + α2x α3x2 + αx]

Symbolic division:

α3x4 + α6x2 + α2x = (α3x2 + αx) ◦ ( x2 + αx︸︷︷︸message polynomial

)

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=⇒ λ(x) = a0x, a0 ∈ F23 , and µ(x) = x

E.g. for a0 = α:

(αx) ◦ [α2x4 + α5x x] + [α6x2 + x α3x2]

= [α3x4 + α6x αx] + [α6x2 + x α3x2]

= [α3x4 + α6x2 + α2x α3x2 + αx]

Symbolic division:


)

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=⇒ λ(x) = a0x, a0 ∈ F23 , and µ(x) = x

E.g. for a0 = α:

(αx) ◦ [α2x4 + α5x x] + [α6x2 + x α3x2]

= [α3x4 + α6x αx] + [α6x2 + x α3x2]

= [α3x4 + α6x2 + α2x α3x2 + αx]

Symbolic division:


)

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We get divisibility for all a0 ∈ F23\{0}. Thus our list decodingalgorithm finds the following list of message polynomials andcorresponding codewords:

m1(x) = x2 + αx c1 = ( α+ 1 0 α2 + 1),m2(x) = α5x2 + α2x c2 = ( α+ 1 α α),m3(x) = α3x2 + α4x c3 = ( α2 + 1 0 α2),m4(x) = α4x2 c4 = ( α2 + α α2 + 1 α),m5(x) = α6x2 + α6x c5 = ( 0 α+ 1 1),m6(x) = α2x2 + α3x c6 = ( α2 + α+ 1 0 α),m7(x) = αx2 + x c7 = ( α+ 1 1 α+ 1).

All these codewords are rank distance 1 away from

r = ( α+ 1 0 α ).

Note: Hamming distance can be 1, 2 or 3.

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We get divisibility for all a0 ∈ F23\{0}. Thus our list decodingalgorithm finds the following list of message polynomials andcorresponding codewords:

m1(x) = x2 + αx c1 = ( α+ 1 0 α2 + 1),m2(x) = α5x2 + α2x c2 = ( α+ 1 α α),m3(x) = α3x2 + α4x c3 = ( α2 + 1 0 α2),m4(x) = α4x2 c4 = ( α2 + α α2 + 1 α),m5(x) = α6x2 + α6x c5 = ( 0 α+ 1 1),m6(x) = α2x2 + α3x c6 = ( α2 + α+ 1 0 α),m7(x) = αx2 + x c7 = ( α+ 1 1 α+ 1).

All these codewords are rank distance 1 away from

r = ( α+ 1 0 α ).

Note: Hamming distance can be 1, 2 or 3.

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Complexity:

Computing Λg,r(x): Oqm(n2)

Computing the minimal basis of M(r) with Euclideanalgorithm: Oqm(n

3)

Computing the minimal basis of M(r) iteratively (not inthis talk): Oqm(n

2)

Parametrization: Oqm((qm(2t+k−n) + q)n2)

If t ≤ (n− k)/2 (unique decoding), then polynomial.

If t > (n− k)/2, then exponential.

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Complexity:

Computing Λg,r(x): Oqm(n2)

Computing the minimal basis of M(r) with Euclideanalgorithm: Oqm(n

3)

Computing the minimal basis of M(r) iteratively (not inthis talk): Oqm(n

2)

Parametrization: Oqm((qm(2t+k−n) + q)n2)

If t ≤ (n− k)/2 (unique decoding), then polynomial.

If t > (n− k)/2, then exponential.

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Idea to improve parametrization (open problem):

When finding all [N(x) −D(x)] ∈ M(r) with the degreerestrictions we can impose extra condition that the errorspan polynomial D(x) has only distinct roots.

Open problem: How to parametrize this condition?

In Reed-Solomon case it can be done with a curve fittingalgorithm (based on Wu’s algorithm). This idea does notwork in the linearized case.

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Idea to improve parametrization (open problem):

When finding all [N(x) −D(x)] ∈ M(r) with the degreerestrictions we can impose extra condition that the errorspan polynomial D(x) has only distinct roots.

Open problem: How to parametrize this condition?

In Reed-Solomon case it can be done with a curve fittingalgorithm (based on Wu’s algorithm). This idea does notwork in the linearized case.

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Summary and Conclusion



m)






Lq(x, qm) is non-commutative ring with Euclidean

algorithm

Lq(x, qm)ℓ, as left module, can be equipped with monomial

order (e.g. (k, . . . , kℓ)-weighted TOP order)

Minimal basis = basis with all distinct leading positions

Predictable Leading Monomial Property holds for minimalbases

PLMP gives rise to parametrizations

Remark: Linearized polynomials are skew polynomials, manystatements can be proven in this setting, as well.

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Lq(x, qm) is non-commutative ring with Euclidean

algorithm

Lq(x, qm)ℓ, as left module, can be equipped with monomial

order (e.g. (k, . . . , kℓ)-weighted TOP order)

Minimal basis = basis with all distinct leading positions

Predictable Leading Monomial Property holds for minimalbases

PLMP gives rise to parametrizations

Remark: Linearized polynomials are skew polynomials, manystatements can be proven in this setting, as well.

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Decoding in the network coding setting (Gabidulin codes)can be translated to a parametrization in the interpolationmodule.Set up algebraic decoding algorithm, that finds allcodewords within the ball of radius t around a givenreceived word, for any decoding radius t.Can easily be altered to find all closest codewords to agiven received word.Complexity is exponential in code length iff the decodingradius is beyond the unique decoding radius.Nonetheless, it is still feasible for radii close to the uniquedecoding radius.

Thank you for your attention!

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Decoding in the network coding setting (Gabidulin codes)can be translated to a parametrization in the interpolationmodule.Set up algebraic decoding algorithm, that finds allcodewords within the ball of radius t around a givenreceived word, for any decoding radius t.Can easily be altered to find all closest codewords to agiven received word.Complexity is exponential in code length iff the decodingradius is beyond the unique decoding radius.Nonetheless, it is still feasible for radii close to the uniquedecoding radius.

Thank you for your attention!

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Linearized Polynomial Modules and their Application to ...user.math.uzh.ch/trautmann/linearized_poly_mods.pdf · Linearized Polynomial Modules and their Application to Network Coding