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On the metrical rigidity of binary codes
S.V. Avgustinovich, a;1 F.I. Solov'eva a;2
aSobolev Institute of Mathematics, pr. Koptyuga 4, Novosibirsk 630090, Russia
Abstract
A code C in the n-dimensional vector space En over GF (2) is called metrically rigid
if every isometry I : C ! En with respect to the Hamming metric is extendable
to an isometry of the whole space En. A code C is reduced if it contains the all-
zero vector. For n large enough the metrical rigidity of length n reduced binary
codes containing a 2-(n; k; �)-design is proved. The class of such codes includes
all the families of uniformly packed codes of su�ciently large length satisfying the
condition d� � � 2; where d is the code distance and � is the covering radius.
Key words: Metrically Rigid Codes; Strongly Rigid Codes; 2-(n; k; �)-design;
Uniformly Packed Codes
1 Introduction
The aim of this paper is to investigate the problem of metrical rigidity of binary
codes. A notion of the metrical regidity is closely and naturally connected with
a well known in geometry notion of rigidity (see, for example, [1{3]). It should
be noted that the term rigid code is used in [4,5] in a di�erent context for
codes with the trivial automorphism group.
Let En be the n-dimensional vector space over GF (2) with respect to the
Hamming metric. A binary code C of length n is a subset of En. In what
follows, we omit the term binary because we only deal with such codes. A
mapping I : C ! En is called an isometry from the code C to the code
1 This research was partially supported by the Russian Foundation for Basic Re-
search under the grant 00-01-00916 and by INTAS 97-1001.
E-mail address: [email protected] (S.V. Avgustinovich)2 This research was supported by the Russian Foundation for Basic Research under
the grant 00-01-00822 and partially supported by the grant FZP \Integration" 274.
E-mail address: [email protected] (F.I. Solov'eva)
Preprint submitted to Elsevier Preprint 4 April 2001
I(C) if d(x; y) = d(I(x); I(y)) for all codewords x; y 2 C: A code C in En
is called metrically rigid if every isometry I : C ! En with respect to theHamming metric is extendable to an isometry of the whole space En (see [6]).The automorphism group of En is transitive on En, therefore without loss ofgenerality for an isometry I : C ! En we can put 0n 2 C and I(0n) = 0
n,where 0n is the all-zero vector of length n. We shall call an isometry I fromthe code C into the code I(C) reduced if it preserves weights of codewords.It is clear that an isometry I such that I(0n) = 0
n is reduced. Two codesC;D � En are equivalent if there exists an isometry of En which transformsthe code C into the code D. It is known that every isometry is de�ned by amapping Av
�: x ! �(x) + v; where � is a permutation of coordinates and
v 2 En:
Let R be any metric space with the automorphism group rich enough, for ex-ample double transitive. Two metric subspaces R1 and R2 in R are equivalentif there exists an automorphism I from the automorphism group of R suchthat I(R1) = R2: Quite often all problems concerning metric spaces are inves-tigated up to equivalence. Let us consider the situation when only metricalstructure of the metric subspace is important, but not the way how a metricsubspace puts in R. The investigation of the situation from equivalence pointof view is not possible. The isometric approach permits the di�culties to beovercome. The situation described above takes place both in geometry (see[1-3, 7]) and discrete mathematics (see [6, 8-11]).
If two codes are equivalent they are isometric, but the inverse is not true inevery case. For example the codes
C = f(0000); (1100); (1010); (0110)g
and
D = f(0000); (1100); (1010); (1001)g
are isometric but are not equivalent. All Hadamard codes obtained fromHadamard matrices by replacing -1 to 1 and 1 to 0 are mutually isometricbut there are a lot of nonequivalent Hadamard matrices (see [15]) and there-fore Hadamard codes.
A code C is called strongly rigid if it is metrically rigid and every code ofthe same length containing C is also metrically rigid. In the paper we onlyconsider reduced codes, i.e. codes with 0 2 C. We call a 2-(n; k; �)-designwith the joined all-zero word 0
nreduced design. It is proved that any reduced
2-(n; k; �)-design is strongly rigid for n large enough. The class of stronglyrigid codes containing reduced 2-(n; k; �)-designs includes all the families ofuniformly packed codes of su�ciently large length satisfying the conditiond � � � 2; where d is the code distance and � is the covering radius. It hasbeen shown in [12-14] that uniformly packed codes contain (d��)-designs and
2
include BCH-codes of distance 5 and 7, Preparata codes, Goethals codes of
distance 7, extended perfect codes.
It should be noted that it is more di�cult to prove rigidity for codes with
large code distance than for codes with small distances.
The metrical rigidity of the following classes of codes has been established in
[9,6,10]:
1) perfect q-ary codes for q � 2; with the exception of the binary Hamming
code of length 7 and the ternary Hamming code of length 4;
2) the binary even-weight code of length n with the exception of the case of
n = 4;
3) q-ary (n; n � 1) MDS codes with the exception of several codes of small
length;
4) full constant-weight codes.
2 Metrical rigidity of codes
Let N = f1; 2; : : : ; ng. A subset D � En of weight k vectors is called a 2-
(n; k; �)-design if the number of vectors in D with ones in the ith and jth
coordinates is equal to exactly � for all di�erent i; j 2 N . The set supp(x) =
fk : xk = 1g is called the support of a vector x = (x1; : : : ; xn) 2 En.
Let i 2 N . We call a set Si � En a �-homogeneous i-star if the following
conditions hold:
1) wt(v) = k for every v 2 Si;
2) i 2 supp(v) for every v 2 Si;
3) the number of vectors in Si with ones in the jth position is equal to � for
every j 2 N; j 6= i.
Is it easy to see from the de�nition of a 2-(n; k; �)-design and the above de�-
nition that the design contains a �-homogeneous i-star for every i 2 N . The
i-star is uniquely de�ned by the design.
We call a set S�
i� E
n a u�y i-star if the following conditions hold:
1) wt(v) = k for every v 2 Si;
3
2) supp(v) \ supp(u) = fig for all v; u 2 S�
i; v 6= u;
3) jS�
ij > k
2 � k + 1:
Note that the condition 2 implies d(v; u) = 2k � 2.
An incidence matrix of a vector set S is a rectangular array whose lines are
vectors from S.
Using the de�nition of a 2-(n; k; �)-design it is not di�cult to prove the fol-
lowing
Proposition 1 For every �-homogeneous i-star Si; i 2 N; we have
jSij = �(n� 1)=(k � 1):
Proposition 2 For all n � k4 and i 2 N , every �-homogeneous i-star Si
contains a u�y i-star S�
i:
Proof. Let us consider a 2-(n; k; �)-design and for every i 2 N take the
maximal cardinality set S�
iof its of vectors with ones in ith coordinate and
pairwise distances 2k � 2. We should prove that S�
iis the u�y i-star, i.e.,
that jS�
ij > k
2 � k + 1.
Using the de�nition of �-homogeneous i-star Si we conclude that S�
i� Si:
Let us take the incidence matrix of Si and delete the ith colomn, all colomns
containing zero-colomns of the submatrix S�
iand all rows containing the sub-
matrix S�
iin it. On the one hand, the number of ones in all colomns of the
matrix obtained equals to jS�
ij(k� 1)(�� 1); and on the other hand, there are
not less than jSij � jS�
ij ones in the rows of the matrix . Therefore
jSij � jS�
ij � jS
�
ij(k � 1)(�� 1):
>From this inequality, we get using Proposition 1 that
�(n� 1)=(k � 1) � jS�
ij((k � 1)(�� 1) + 1);
i.e.
n �jS�
ij(k � 1)((k � 1)(�� 1) + 1)
�+ 1:
If jS�
ij � k
2 � k + 1; then
n �(k2 � k + 1)(k � 1)((k � 1)(�� 1) + 1)
�+ 1 < k
4;
4
a contradiction.
}
Proposition 3 If n � k4, then every reduced isometry I takes every u�y
i-star into the u�y i0-star for some i
0 2 N:
Proof. Suppose that the proposition is not true and the image I(S�
i) of the
u�y i-star S�
iis not a u�y i
0-star for any i
0 2 N:
Without loss of generality, � = (1; : : : ; 1; 0; : : : ; 0) 2 I(S�
i). Since a reduced
isometry preserves weights, every vector in I(S�
i) has weight k. Suppose that
there are is � 0 vectors having ones in the sth coordinate, s = 1; : : : ; k: The
assumption that I(S�
i ) is not a u�y i-star implies existence of at least two
is > 0.
Eliminating the vector � in I(S�
i ), we have
kX
s=1
is = jI(S�
i )j � 1:
Compare a vector having one in the sth coordinate, s � k, with a vector �
having one in the jth coordinate, s 6= j; where is; ij > 0: A reduced isometry
preserves an intersection of vectors, i.e. preserves a number of common 1th
coordinates in these vectors. Then
jsupp(�) \ supp( )j = jsupp(�) \ supp(�)j = jsupp( ) \ supp(�)j = 1:
Hence only coordinates s and �j are equal to 1 among the �rst k coordinates
of vectors and �, and there is exactly one coordinate �t such that t > k and
t = �t = 1 among coordinates of the vector �. Therefore is � k�1, ij � k�1
and
jI(S�
i )j � 1 =
kX
s=1
is � k(k � 1):
Hence
jI(S�
i )j � k2 � k + 1:
By the de�nition of an isometry we have jI(S�
i )j = jS�
i j. But jS�
i j > k2�k+1;
a contradiction.
}
Proposition 4 If n � k4, then every reduced isometry I takes every �-
homogeneous i-star into a �-homogeneous i0-star for some i0 2 N:
5
Proof. Let us consider the p � n incidence matrices M and M0 of the stars
Si and I(Si) respectively, where Si is a �-homogeneous i-star and p = jI(Si)j:
By the de�niton of an i-star, every row of the matrix M contains k ones and
every column except the ith one contains � ones. Every reduced isometry I
preserves weigths, therefore there are k ones in every row of the matrix M0.
Using Proposition 3 and contradiction arguments, it is easy to prove that there
exists some i0 2 N such that the i0th column in the matrix M0 contains only
ones.
Let �1; �2; : : : ; �i0�1; �i0+1; : : : ; �n be numbers of ones in all columns except the
i0th column of the matrixM 0. We shall prove that �j = � if j 6= i
0: Let us take
the sum � of all pairwise distances between vectors from Si: It is clear that
a reduced isometry preserves the sum of the distances. Taking into account
that every \zero-one" pair in a column of the matrix M (and similarly of the
matrix M0) adds one into the sum �, we obtain that
� = (n� 1)�(p� �) =X
j 6=i0
�j(p� �j) = pX
j 6=i0
�j �X
j 6=i0
�2
j :
ThusP
j 6=i0 �j = (n� 1)�. According to the Cauchy-Bunyakowsky inequality,
we have X
j 6=i0
�2
j �1
(n� 1)(X
j 6=i0
�j)2
and the equality holds only if �j = � for all j 2 N; j 6= i0.
}
Let us consider a 2-(n; k; �)-design and its reduced isometry I. According
to Proposition 4, the pair (Si; I) de�nes a mapping � : N ! N such that
�(i) = i0.
Using contradiction arguments and Proposition 4, it is not di�cult to prove
the following
Proposition 5 The mapping � is a permutation.
Proposition 6 Let a code C contain a 2-(n; k; �)-design. Let u be a vector in
C. Then for every reduced isometry I we have
1) if i 2 supp(u) then �(i) 2 supp(I(u));
2) if i =2 supp(u) then �(i) =2 supp(I(u)):
Proof. Let J� : En ! En be the isometry generated by the permutation
�. For both cases let us count the sum of distances between any vector u =
(u1; : : : ; un) 2 C and all vectors of the �-homogeneous i-star Si(C). If the
6
sums are di�erent it means according to Propositions 4 and 5 that a reduced
isometry I is extendable to the isometry J�.
There are two cases.
1. Let i 2 supp(u) and w(u) = t. Then ui= 1 and the vector u gives zero to the
sum of the distances. There are t� 1 coordinates of the vector u equal to one.
Every such coordinate of u together with every vector v 2 Si(C) having zero in
the ith coordinate gives one in the sum of the distances. There are jSi(C)j��
such vectors in Si(C). There are n � t zero coordinates of the vector u and
every such coordinate together with every vector v 2 Si(C) having one in the
ith coordinate gives one in the sum of the distances. There are � such vectors
in Si(C). Summing up all distances we get
X
v2Si(C)
d(u; v) = (t� 1)(jSi(C)j � �) + (n� t)�: (1)
2. In the case i =2 supp(u) considerations analogous to those in the case 1 allow
to obtain
X
v2Si(C)
d(u; v) = jSi(C)j+ t(jS
i(C)j � �) + (n� t� 1)�: (2)
Comparing (1) and (2), we see that the numbers are di�erent.
}
It follows from Propositions 2-6 that the automorphism � coincides with the
reduced isometry I. Hence every reduced 2-(n; k; �)-design is metrically rigid.
It is not di�cult to prove that every reduced code of length n; n � k4; contain-
ing 2-(n; k; �)-design is metrically rigid. Therefore every reduced 2-(n; k; �)-
design is strongly rigid and we obtain the main result of the paper.
Theorem 7 For n � k4 every reduced 2-(n; k; �)-design is a strongly rigid
code.
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