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Coding Theory

Ruud Pellikaang.r.pellikaan@tue.nl

MasterMath 2MMC30

Lecture 5.1March 17 - 2016

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Introduction

Apart from the minimum Hamming weighta code has other important invariants

We introduce the weight enumerator andthe extended weight enumerator of a code

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Content lecture 5

1. Weight enumerator• Weight enumerator• Average weight enumerator

2. Error probability• Probability of undetected error• Probability of decoding error

3. Weight enumerator• Arrangement of hyperplanes• Weight enumerator of MDS codes

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Weight enumerator

Let C be a code of length nThe weight spectrum, also called the weight distributionis the following set

{(w,Aw) | w = 0, 1, . . . , n}

where Aw denotes the number of codewords in C of weight wThe weight enumerator of C is defined as the polynomial

WC (Z ) =

n∑w=0

AwZw

The homogeneous weight enumerator of C is defined as

WC (X , Y ) =

n∑w=0

AwX n−wYw

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REMARK - 1

WC (Z ) and WC (X , Y ) determine each other uniquely:

WC (Z ) = WC (1, Z )

andWC (X , Y ) = X nWC (X−1Y )

Given the weight enumerator or the homogeneous weight enumeratorthe weight spectrum is determined completely by the coefficients

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REMARK - 2

Clearly, the weight enumerator and homogeneous weight enumeratorcan be written in another form, that is

WC (Z ) =∑c∈C

Zwt(c)

andWC (X , Y ) =

∑c∈C

X n−wt(c)Ywt(c)

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REMARK - 3

Let C be a linear codeThen A0 = 1

The minimum distance d (C ) which is equal to the minimum weightis determined by the weight enumerator as follows:

d (C ) = min{ i | Ai 6= 0, i > 0 }

It also determines the dimension k (C ), since

WC (1, 1) =

n∑w=0

Aw = qk (C )

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Trivial codes

The zero code has one codeword, and its weight is zeroHence the homogeneous weight enumerator of this code is

W{0}(X , Y ) = X n

The number of words of weight w in the trivial code Fnq is

Aw =

(nw

)(q − 1)w

So

WFnq (X , Y ) =

n∑w=0

(nw

)(q − 1)wX n−wYw

= (X + (q − 1)Y )n

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Repetition code

The n-fold repetition code C has homogeneous weight enumerator

WC (X , Y ) = X n+ (q − 1)Y n

In the binary case its dual is the even weight codeHence it has homogeneous weight enumerator

WC⊥(X , Y ) =

bn/2c∑t=0

(n2t

)X n−2tY2t

=12

((X + Y )n + (X − Y )n

)

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[7, 4, 3] Hamming code

The nonzero entries of the weight distribution of the [7,4,3] binaryHamming code are given by

A0 = 1, A3 = 7, A4 = 7, A7 = 1

as is seen by inspecting the weights of all 16 codewordsHence its homogeneous weight enumerator is

X7+ 7X4Y3

+ 7X3Y4+ Y7

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Simplex code

The simplex code Sr(q) is a constant weight code with parameters

[(q r− 1)/(q − 1), r, q r−1

]

Hence its homogeneous weight enumerator is

WSr (q)(X , Y ) = X n+ (q r

− 1)X n−q r−1Yq r−1

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Recurrence relation for Hamming code

The Hamming code over Fq of length n = (q r− 1)/(q − 1)

has parameters[n, n − r, 3]

and is perfect with covering radius 1

The following recurrence relation holds(nw

)(q−1)w = Aw−1(n−w+1)(q−1)+Aw(1+w(q−2))+Aw+1(w+1)

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PROOF

Every word y of weight w is at distance at most 1 to a unique codeword cand such a codeword has possible weights w − 1, w or w + 1

If wt(c) = w − 1 then there are n − w + 1 possible positions jin the complement of the support of c where cj = 0 could be changedinto a nonzero element in order to get the word y of weight w

If wt(c) = w then either y = cor there are w possible positions j in the support of cwhere cj could be changed into another nonzero element to get y

If wt(c) = w + 1, then there are w + 1 possible positions jin the support of c where cj could be changed into zero to get y

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The recurrence relation implies that W (Z ) =∑

w AwZw

satisfies the ordinary first order differential equation:

((q−1)Z2−(q−2)Z−1)W ′(Z )−(1+(q−1)nZ )W (Z )+(1+(q−1)Z )n = 0

The corresponding homogeneous differential equation is separableAs a result

W (Z ) =1q r

(1+ (q − 1)Z )n +q r− 1q r

(Z − 1)qr−1

((q − 1)Z + 1)n−qr−1

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REMARK

The computation of the weight enumerator of a given code ismost of the time hard

For the perfect codes such as the Hamming codes andthe binary and ternary Golay codes this is left as exercises to the readerand can be done using recurrence relations between the Aw

The weight distribution of MDS codes will be treatedThe weight enumerator of only a few infinite families of codes is known

On the other hand the average weight enumerator of a class of codesis very often easy to determine

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Average weight enumerator

Let C be a nonempty class of codes over Fq of the same lengthThe average weight enumerator of C is defined asthe average of all WC with C ∈ C

WC(Z ) =1|C|

∑C∈C

WC (Z )

and similarly for thehomogeneous average weight enumerator WC(X , Y ) of this class

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Balanced collection of codes

A class C of [n, k ] codes over Fq is called balancedif there is a number N (C) such that

N (C) = |{ C ∈ C | y ∈ C }|

for every nonzero word y in Fnq

The prime example of a class of balanced codesis the set C[n, k ]q of all [n, k ] codes over Fq

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LEMMA

Let C be a balanced class of [n, k ] codes over Fq

Then

N (C) = |C|qk− 1

qn − 1

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PROOF

Compute in two ways the number of elements of the set of pairs

{ (y,C ) | y 6= 0, y ∈ C ∈ C }

First by keeping a nonzero y in Fnq fixed

and letting C vary in C such that y ∈ CThis gives the number (qn

− 1)N (C), since C is balanced

Secondly by keeping C in C fixedand letting the nonzero y in C varyThis gives the number |C|(qk

− 1)

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PROPOSITION

Let f be a function on Fnq with values in a complex vector space

Let C be a balanced class of [n, k ] codes over Fq

Then1|C|

∑C∈C

∑c∈C∗

f (c) =qk− 1

qn − 1

∑v∈(Fnq )∗

f (v)

where C ∗ denotes the set of all nonzero elements of C

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PROOF

By interchanging the order of summation we get∑C∈C

∑v∈C∗

f (v) =∑

v∈(Fnq )∗

f (v)∑

v∈C∈C

1

The last summand is constant and equals N (C), by assumptionNow the result follows by the computation of N (C) in the Lemma

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COROLLARY

Let C be a balanced class of [n, k ] codes over Fq

Then

WC(Z ) = 1+qk− 1

qn − 1

n∑w=1

(nw

)(q − 1)wZw

PROOFApply the Proposition to the function f (v) = Zwt(v)

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MacWilliams identity

Although there is no apparent relation between theminimum distances of a code and its dualthe weight enumerators satisfy the MacWilliams identity

Let C be an [n, k ] code over Fq

ThenWC⊥(X , Y ) = q−kWC (X + (q − 1)Y , X − Y )

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Trivial codes

The zero code C has homogeneous weight enumerator

X n

and its dual Fnq has homogeneous weight enumerator

(X + (q − 1)Y )n

which is indeed equal to

q0WC (X + (q − 1)Y , X − Y )

and confirms MacWilliams identity

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Repetition code and its dual - 1

The n-fold repetition code C has homogeneous weight enumerator

X n+ (q − 1)Y n

If q = 2 the homogeneous weight enumerator of its dual code is

12

((X + Y )n + (X − Y )n

)which is equal to

2−1WC (X + Y , X − Y )

confirming the MacWilliams identity for q = 2

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Repetition code and its dual - 2

For arbitrary q we have

WC (X , Y ) = X n+ (q − 1)Y n

HenceWC⊥(X , Y ) =

q−1WC (X + (q − 1)Y , X − Y ) =

q−1((X + (q − 1)Y )n + (q − 1)(X − Y )n) =

n∑w=0

(nw

)(q − 1)w + (q − 1)(−1)w

qX n−wYw