Bounds on the weights of vectors

The linear complexity property relates the weight of a vector to the length of the linear recursion that produces the Fourier transform of that vector periodically repeated. By using this property, the Fourier transform of a vector can be constrained to ensure that the vector has a weight at least as large as some desired value d .

The theorems of this section describe how patterns of zeros in the Fourier transform of a vector determine bounds on the weight of that vector. These bounds can also be obtained as consequences of the fundamental theorem of algebra.

Theorem 1.8.1 (BCH bound) The only vector of blocklength n of weight d − 1 or less that has d− 1 (cyclically) consecutive components of its transform equal to zero is the all-zero vector.

Proof: The linear complexity property says that, because the vector v has weight less than d , its Fourier transform V satisfies the following recursion:

V_j = −

d−1

k=1

kV_{(( j−k))}.

This recursion implies that any d− 1 cyclically consecutive components of V equal to zero will be followed by another component of V equal to zero, and so forth. Thus V must be zero everywhere. Therefore v is the all-zero vector.
Theorem 1.8.2 (BCH bound with cyclic permutation) Suppose that b and n are coprime and a is arbitrary. The only vector v of weight d− 1 or less, whose Fourier transform satisfies

V_((a+b
))= 0
= 1, . . . , d − 1, is the all-zero vector.

Proof: The modulation property of the Fourier transform implies that translation of the spectrum V by a places does not change the weight of v. The cyclic permutation property implies that cyclic permutation of the transform V by B= b⁻¹(mod n) places does not change the weight of v. This gives a weight-preserving permutation of v that rearranges the d− 1 given zeros of V so that they are consecutive. The BCH bound

completes the proof.

The BCH bound uses the length of the longest string of zero components in the Fourier transform of a vector to bound the weight of that vector. Theorems1.8.3and1.8.4use other patterns of substrings of zeros. The first of these theorems uses a pattern of evenly spaced substrings of components that are all equal to zero. The second theorem also

31 1.8 Bounds on the weights of vectors

uses a pattern of evenly spaced substrings of components, most of which are zero, but, in this case, several may be nonzero.

Theorem 1.8.3 (Hartmann–Tzeng bound) Suppose that b and n are coprime. The only vector v of blocklength n of weight d − 1 or less, whose spectral components satisfy

V_((a+
1+b
2))= 0
1= 0, . . . , d − 2 − s

2= 0, . . . , s, is the all-zero vector.

Proof: This bound is a special case of the Roos bound, which is given next.
Notice that the Hartmann–Tzeng bound is based on s+1 uniformly spaced substrings of zeros in the spectrum, each substring of length d−1−s. The Roos bound, given next, allows the evenly spaced repetition of these s+ 1 substrings of zeros to be interrupted by some nonzero substrings, as long as there are not too many such nonzero substrings.

The Roos bound can be further extended by combining it with the cyclic decimation property.

Theorem 1.8.4 (Roos bound) Suppose that b and n are coprime. The only vector v of blocklength n of weight d− 1 or less, whose spectral components satisfy

V_((a+
1+b
2))= 0
1= 0, . . . , d − 2 − s,

for at least s+ 1 values of
2in the range 0,. . . , d − 2, is the all-zero vector.

Proof: We only give an outline of a proof. The idea of the proof is to construct a new vector in the transform domain whose cyclic complexity is not smaller and to which the BCH bound can be applied. From V ↔ v, we have the Fourier transform pair [V_{(( j+r))}] ↔ [viω^ir],

where[·] denotes a vector with the indicated components. This allows us to write the Fourier transform relationship for a linear combination such as

[β0V_j+ β1V_{(( j+r))}] ↔ [β0vi+ β1viω^ir],

whereβ0andβ1 are any field elements. The termsvi andviω^ir on the right are both zero or both nonzero. The linear combination of these two nonzero terms can combine to form a zero, but two zero terms cannot combine to form a nonzero. This means that the weights satisfy

wt v≥ wt [β0vi+ β1viω^ir].

V_k 000 000 000

V_{k 1} 000 000 000

V_{k 2}

–

– 000 000 000

Linear

combination 0 0 0 0 0

Newly created zeros

Figure 1.3. Construction of new zeros.

The weight on the right side can be bounded by the zero pattern of the vector[β0V_j+ β1V_{(( j+r))}]. The bound is made large by the choice of β0 andβ1 so as to create a favorable pattern of zeros.

In the same way, one can linearly combine multiple translates of V , as suggested in Figure1.3, to produce multiple new zeros. We then have the following transform pair:



β
V_{(( j+r
))}



↔



β
viω^ir

 .

The coefficients of the linear combination are chosen to create new zeros such that the d− 1 zeros form a regular pattern of zeros spaced by b, as described in Theorem1.8.2.

The new sequence with components V_a+
bfor
= 0, . . . , d − 2 is zero in at least s + 1 components, and so is nonzero in at most d − s − 2 components. The same is true for the sequence V_a+
b+
2 for each of d− s − 1 values of
2, so all the missing zeros can be created. Theorem1.8.2then completes the proof of Theorem1.8.4provided the new vector is not identically zero. But it is easy to see that the new vector cannot be identically zero unless the original vector has a string of d− 1 consecutive zeros in its

spectrum, in which case the BCH bound applies.

The final bound of this section subsumes all the other bounds, but the proof is less transparent and the bound is not as easy to use. It uses the notion of a triangular matrix, which is a matrix that has only zero elements on one side of its diagonal.

Theorem 1.8.5 (van Lint–Wilson bound) Given V ∈ GF(q)^m, define the compo-nents of an n by n matrix M by

M
_j= V_{(( j−
))} j = 0, . . . , n − 1;
= 0, . . . , n − 1,

33 1.8 Bounds on the weights of vectors

and let M be any matrix obtained from M by row permutations and column permutations. Then v, the inverse Fourier transform of V , satisfies

wt v≥ rank T ,

where T is any submatrix of M that is triangular.

Proof: The matrix M can be decomposed as

M = v^T,

where v is an n by n diagonal matrix, whose diagonal elements are equal to the compo-nents of the vector v, and  is the matrix describing the Fourier transform. The matrix

 has the elements ij= ω^ij. Because  has full rank, rank M = rank v = wt v.

Moreover,

rank M = rank M ≥ rank T ,

from which the inequality of the theorem follows.

For an application of Theorem 1.8.5, consider a vector of blocklength 7 whose spectrum is given by(V0, 0, 0, V₃, 0, V₅, V₆). Then

M =

⎡

⎢⎢

⎢⎢

⎢⎢

⎢⎢

⎢⎣

V₀ 0 0 V₃ 0 V₅ V₆ V₆ V₀ 0 0 V₃ 0 V₅ V₅ V₆ V₀ 0 0 V₃ 0

0 V₅ V₆ V₀ 0 0 V₃ V₃ 0 V₅ V₆ V₀ 0 0

0 V₃ 0 V₅ V₆ V₀ 0 0 0 V₃ 0 V₅ V₆ V₀

⎤

⎥⎥

⎥⎥

⎥⎥

⎥⎥

⎥⎦ .

The bottom left contains the three by three triangular submatrix

T =

⎡

⎢⎣

V₃ 0 V₅

0 V₃ 0

0 0 V₃

⎤

⎥⎦ .

Clearly, this matrix has rank 3 whenever V₃is nonzero, so, in this case, the weight of the vector is at least 3. (If V₃is zero, other arguments show that the weight is greater than 3.)