Weight Distributions
Weight Distribution of a Code
Given an (n,M,d)- q-ary code, let Ai = number of code
words of weight i, for 0 ≤ i ≤ n (can assume that the zero
vector is in the code, so A0 = 1).
The ordered set of Ai is called the weight distribution of
the code.
Clearly we have:
n
M =∑ Ai .
i =0
Generating Function
It is most convenient to represent the weight
distribution as a generating function in two variables,
called the weight enumerator. Thus, for an (n,M,d) code C we would have:
n
W C  x , y=∑ Ai x y
i
n−i
.
i=0
It follows immediately that:
W C  0,1= A0 =1
W C  1,1=∑ Ai =M
W C 1,0= An =0∨1 (binary case)
n−1
n
W C  1,−1= An − An−1⋯−1 A1 −1 A0 .
Examples
The [7,4]- Hamming code has A0 = A7 = 1 and A3 = A4 = 7,
so we obtain:
7
3
4
4
3
7
W [7,4 ]  x , y = y 7x y 7x y  x .
For the ternary code C with generator matrix

2 0 2 1 0
G=
1 1 0 0 1

we obtain:
5
3
2
4
5
W C  x , y= y 4x y 2x y2x .
Problem
Suppose C is a binary code (not necessarily linear) with
length 23, minimum distance 7, having 212 codewords.
Assuming that C contains the zero vector, show that C is
a perfect code.
By counting pairs (x,c) with c ∈ C, wt(x) = 4 and d(x,c) =3
show that there are 253 words of weight 7 in C.
Show that the weight enumerator of C is completely
determined by knowing that C is perfect and contains the
zero vector.
MacWilliams Identity
For a binary linear [n,k]-code C we have:
1
W C  x , y = W C  y −x , y x .
∣C∣
⊥
The identity is not restricted to binary codes, but this is
the “prettiest” form for the identity.
We will rewrite the identity using generating functions
(weight enumerators) of one variable before proving
the result in the binary case.
Alternate Form
The one variable form of the weight enumerator is:
WC(x) = WC(x,1).
We can rewrite the MacWilliams identity using z = x/y to
obtain (do this):
 
1
1−z
n
W C  z = 1z  W C
.
∣C∣
1z
⊥
Note: The non-binary form for the identity can be obtained
by replacing the (1+z) terms above by (1+(q-1)z).
Binary Proof
Let V be vector space [n,2] containing the linear code C.
Define for u ∈ V:
g u= ∑ −1
z
wt v
.
v∈V
Now,
∑ g u= ∑ ∑ −1
u∈C
⟨u , v⟩
u∈C v∈V
⟨u , v⟩
z
wt  v
=∑ z
v∈V
wt v
∑ −1
⟨u ,v ⟩
.
u∈ C
But, <u,v> = 0 if v ∈ C┴ and the inner sum is |C|, while if
v ∉ C┴ half of the terms in this inner sum are +1, and the
rest are -1, so the inner sum is 0. So,
∑ g u=∣C∣W C  z .
⊥
u∈C
Binary Proof
On the other hand we have:
n
g u=
∑
∏ −1
 v 1, v 2, , v n ∈V i =1
n
ui vi
vi
z .
ui
= ∏ 1−1 z
i =1
wt  u
=1−z 
1z 
From which the result follows.
n−wt u
.
Corollary
If we let {Bj} be the weight distribution of the code C┴,
we can equate coefficients in the MacWilliams identity to
prove that:
n
j
  
i
B j =2 ∑ Ai ∑ −1
l
i =0
l =0
−k
l
n−i .
j−l
Example
Consider the [8,4,3] extended binary Hamming code. The
weight distribution is A0 = A8 = 1, A4 = 14 ( & no other wts.)
The weights of the dual code are:

0

1
B4 =
A0 ∑ −1l 0
16
l
l= 0
  
  
0
l 8
8 A
4 4 A
−1
−1
4∑
8∑
−l
l −l
l
l =0
l =0
B 0=
4
0
  
1
l
B 0=
A0 ∑ −1 0
16
l
l =0
  
l
0
−l
1
 A0 A4 A8 =1
16
4
  
4
  
l 4
l 8
8 A
4 A
4 ∑ −1
8 ∑ −1
4−l
l 4−l
l
l =0
l =0
1
B4 = 70A 06A470A 8 =14 .
16
0
4−l
Example
Since we know that this extended Hamming code is selfdual, we will get Bi = Ai for all i. Consider one more
computation:
B 2=

2
  
1
l
A0 ∑ −1 0
16
l
l =0
2
  
l 4
8 A
−1
4∑
2−l
l
l =0
B 2=
2
   
l 8
4 A
−1
8∑
2−l
l
l =0
1
 28A0−4A 428A8 =0 .
16
0
2−l
Codes from Designs
We have seen how some designs can be contained in (held
by) a code. We now reverse the roles and discuss some
instances where a design gives rise to a code.
The simplest way to do this is to consider the binary code
which is the row space of the block vs. point incidence
matrix of a design.
As a simplification, we will consider the codes that arise this
way from the Steiner designs S(2, n+1, n2+n+1) where
n ≡ 2 mod 4. [These are projective planes of order n].
The Extended Code
The generator matrix of this code has rows with weight
n+1 ≡ 3 mod 4 which is odd. The extended code thus has
generator matrix with all rows having weight a multiple of
4. Furthermore, the inner product of two distinct rows is
exactly 2 (before adding the parity check, these rows
represented lines of a projective plane – which always
meet in a point), and the inner product of a row with itself
is n+2 (even), hence every two rows are orthogonal.
By Theorem 4, the extended code C', is a doubly even
self-orthogonal code. Thus C must have dimension
k ≤ ½(n2 + n + 2).
The Extended Code
We can go further and prove that the code C has
dimension ½(n2 + n + 2), and so C' is in fact self-dual.
Let C have dimension k and define r := n2 + n + 2 – k =
dim C⊥. Let H be the parity check matrix for C, permuted
into the form [Ir A]. Define:
 
P=
Ir
0
At
Ik
.
With N the incidence matrix of the design, we have
½(n2+n)
det (NP) = det N = (n+1)n
Since all entries in the first r columns of NP are even,
The Extended Code
det N must have a factor of 2r (since every term in this
determinant must have such a factor). But then, since n
only has 2 as a factor, we must have r ≤ ½(n2+n), and
hence k = n2 + n + 2 – r ≥ ½(n2 + n + 2) and the equality
now follows.
Minimum Weight Codewords
The code C we are examining has minimum weight n+1
and every codeword of minimum weight corresponds to a
line in the projective plane of order n.
Recall that the lines of the projective plane give
codewords in the extended code C' that have a 1 for a
parity check.
Let c be a codeword of C with weight d. Since C' is
self-dual, when d is odd c must meet every line at least
once (an odd number of times) and if d is even then
every line which meets c (thought of as a set of points)
must meet c a second time.
Minimum Weight Codewords
The code C we are examining has minimum weight n+1
and every codeword of minimum weight corresponds to a
line in the projective plane of order n.
When d is even, as there are n+1 lines through a point,
we must have d > n+1 (because the second points are
all distinct on the lines through a given point). If d is
odd, we have d(n+1) ≥ n2+n+1, so d ≥ n+1.
Now suppose wt(c) = n+1, which is odd. The line L
determined by two points of c must intersect c in at
least one more point. If P is a point of L not in c, then
the n lines through P other than L must meet c, so wt(c)
≥ n + 3 →← Thus c must be the line L.
Hyperovals
In the code C, the codewords of weight n+2 are precisely
the hyperovals in the projective plane.
Let c be a codeword of C of weight n+2. As the weight is
even, every line meets c in an even number of points. Let
L be a line meeting c in 2a points. Each one of the n lines
other than L through one of these points must meet c at
least once more. Thus 2a+n ≤ n+2, so a = 0 or 1, and c is
a hyperoval.
Let H be a hyperoval. Each point not on H is on ½(n+2)
secant lines (an even number since n≡2 mod 4), and each
point on H is on n+1 (an odd number) of secants. Thus,
the algebraic sum of all the secants (codewords) is the
characteristic function of H, and so, a codeword.
Weight Distribution
The weight of any codeword of C is ≡ 0,3 mod 4.
Since C', the extended code, is doubly even, the code
words of C which have even weight must have weight
divisible by 4, while the odd weight ones (which would
have a parity check of 1) have weight one less than a
multiple of 4.
Ai = An-i for all 0 ≤ i ≤ n.
As C' is self-dual, it contains the all 1 vector. But then C
must contain the all 1 vector of length n, from which the
result follows.
Looking for S(2,11,111)
Consider the case n = 10.
It can be shown that the weight distribution of the code C
is completely determined from the values of A12, A15 and
A16 (A0 and A11 are already known).
In 1973, MacWilliams, Sloane and Thompson showed that
A15 = 0 by a theoretical argument with a little computer
assist.
In 1983, Lam et.al, carried out an extensive computer
search to show that A12 = 0 (at which point I lost interest in
the problem.)
Looking for S(2,11,111)
In 1986, Lam et.al, finished a computer search started by
Carter for his Ph.D. thesis which showed that W16 = 0.
The weight distribution
was now known for C.
Finally, in 1989, Lam et.
al., with another massive
computer search showed
that A19 = 24,675 did not
lead to a solution.
nd
Stirling Numbers - 2 kind
The number of partitions of an n-set is denoted by B(n)
and is called a Bell number. The number of partitions of
an n-set into k (non-empty) pieces (blocks, boxes, etc.) is
denoted by S(n,k) and called a Stirling number of the
second kind. Clearly,
n
B n= ∑ S  n , k 
n1.
k =1
Example: Let n = 3, then B(3) = 5.
{1,2,3}, {1}{2,3}, {2}{1,3}, {3}{1,2}, {1}{2}{3} so
S(3,1) = 1, S(3,2) = 3 and S(3,3) = 1.
nd
Stirling Numbers - 2 Kind
We shall define S(0,0) = 1, but S(n,0) = 0 for n > 0. Also
we have S(n,k) = 0 if k > n.
The Stirling numbers of the 2nd kind satisfy the recursion:
S(n+1,k) = kS(n,k) + S(n,k-1) for n ≥ k ≥ 1.
Pf: Distinguish an element of the (n+1)-set. This element
is in precisely one of the partition pieces for any partition
into k pieces. There are S(n,k-1) partitions in which this
element is in a singleton piece. Otherwise, removal of this
element from any of the k pieces it could be in would
leave a partition of an n-set into k pieces, i.e, kS(n,k) ways
that this could happen.
nd
Stirling Numbers - 2 Kind
S(n,1) = S(n,n) = 1 for all n ≥ 1 is clear since there is only
one partition in these cases. Together with the recursion
relation, we can now construct a “Pascal” - like triangle for
these Stirling numbers.
1
1
1
1 3 1
1 7 6 1
1 15 25 10 1
1 31 90 65 15 1
nd
Stirling Number - 2 Kind
We also have a direct formula for the computation of the
Stirling Numbers of the 2nd Kind.
k
1
k− i k
n
S  n , k = ∑ −1
i .
k ! i =0
i

Pless Power Moments
Let C be a binary [n,k] code with weight distribution {Ai }
and let the weight distribution of C⊥ be {Bj} then:
n
∑
j= 0
n
j A j =∑ −1 B j
r
j
j =0
n
r

r
∑ s ! Sr ,s2
k−s
s=0
n
∑  n− j  A j =∑ B j
j =0
j =0

r
∑ s! S r ,s2
s=0
 
 
n− j
n−s
k −s
n− j
n−s
and
.
where S(r,s) is a Stirling number of the 2nd kind, and r ≥ 0.
These equations are known as the Pless Power Moments.
Pless Power Moments
The Pless Power Moment equations can be derived from
the MacWilliams identity and some properties of the
Stirling numbers of the 2nd kind.
These equations are useful if not all the Bi are known,
since they lead to sets of linear equations which may be
solved and often have unique solutions (when the
determinant of the system is a Vandermonde
determinant).
Pless Power Moments
Example: Consider a [24,12,8] doubly even binary code
C. As doubly even codes are self dual (by definition), we
have that B1 = B2 = 0 because C = C⊥. Writing the first
three Pless Power Moment equations (recall A0 = A24 =1,
A4 = A20 = 0):
A8 + A12 + A16 = 212 – 2 = 4094
8A8 + 12A12 + 16A16 = 211(24) – 24
64A8 + 144A12 + 256A16 = 210(24)(25)- (24)2
as the determinant of this system is Vandermonde, there
is a unique solution : A8 = A16 = 759, A12 = 2576.
Gleason Polynomials
In 1970 A. Gleason showed that the weight enumerators
of certain classes of self-dual codes can be generated by
fixed pairs of polynomials, now called Gleason
polynomials. [Recent work by Nebe, Rains & Sloane
(2006) has increased the number of classes of self-dual
codes that can be dealt with in this way].
A binary self-dual code C has its weight enumerator in the
ring ℂ[f,g] where f(x,y) = x2 + y2 and g(x,y) =x2y2(x2-y2)2.
A binary doubly even code C has its weight enumerator in
the ring ℂ[f,g] where f(x,y) = x8 + 14x4y4 + y8 and
g(x,y) = x4y4(x4-y4)4
Proof of Assmus-Mattson
Theorem 95: Let C be an [n,k,d] binary code, and let 0 < t < d. Let Bi be the
number of vectors of weight i in C⊥, and let
s = | {i: Bi ≠ 0 and 0 < i ≤ n-t} |.
If s ≤ d-t, then the vectors of weight d in C hold a t-Design and the vectors of
any weight i in C⊥ such that i ≤ n-t hold a t-Design.
Pf: Let T be a set of t coordinate positions and I the
remaining n-t positions. Let C(I) be the code obtained from
C by removing these t positions. Now C(I) is an [n-t,k] code
since t < d (so n-t ≥ n-d+1) and any n-d+1 columns of C
contain k independent columns. The minimum weight of
C(I) is ≥ d – t. Consider the subcode D of C⊥ which is 0 in all
the positions of T. C(I) and D(I) are dual codes (clearly, D(I)
⊆ C(I)⊥ , now take d' in C(I)⊥ and extend it with 0's to get a
vector in D).
Proof of Assmus-Mattson
Theorem 95: Let C be an [n,k,d] binary code, and let 0 < t < d. Let Bi be the
number of vectors of weight i in C⊥, and let
s = | {i: Bi ≠ 0 and 0 < i ≤ n-t} |.
If s ≤ d-t, then the vectors of weight d in C hold a t-Design and the vectors of
any weight i in C⊥ such that i ≤ n-t hold a t-Design.
Pf(cont): Let Bi' be the number of vectors of weight i in D(I).
Then Bi' ≤ Bi and if s' is the number of i so that Bi' ≠ 0, then
s' ≤ s. So, s' ≤ d-t and thus s' is less than or equal to the
minimum weight of C(I). In this situation the Pless Moment
Equations give a unique solution to the weight enumerator
which depends only on the dimensions of the codes. This
means that the enumerator does not depend on the choice
of T, just its size.
Proof of Assmus-Mattson
Theorem 95: Let C be an [n,k,d] binary code, and let 0 < t < d. Let Bi be the
number of vectors of weight i in C⊥, and let
s = | {i: Bi ≠ 0 and 0 < i ≤ n-t} |.
If s ≤ d-t, then the vectors of weight d in C hold a t-Design and the vectors of
any weight i in C⊥ such that i ≤ n-t hold a t-Design.
Pf (cont.): Consider the vectors of weight d in C, all of
which have 1's on a fixed T of size t. Each of these gives
rise to a vector of weight d-t in C(I). Conversely, every
vector of weight d-t in C(I) comes from a weight d vector in
C in the obvious way since d is the minimum weight. Since
the number of vectors of weight d-t in C(I) is independent of
the set T, the vectors of weight d in C hold a t-design.
Proof of Assmus-Mattson
Theorem 95: Let C be an [n,k,d] binary code, and let 0 < t < d. Let Bi be the
number of vectors of weight i in C⊥, and let
s = | {i: Bi ≠ 0 and 0 < i ≤ n-t} |.
If s ≤ d-t, then the vectors of weight d in C hold a t-Design and the vectors of
any weight i in C⊥ such that i ≤ n-t hold a t-Design.
Pf (cont): Let {ci} be the complements of vectors of weight
w in C⊥. Consider the ci's that are 1 throughout T. Each
such ci is the complement of a vector of weight w in D. The
number of these ci's is therefore independent of the
particular choice of T since the weight distribution of D is
the same as the weight distribution of D(I), and that is
independent of T. The {ci} thus hold a t-design, and so their
complements hold a t-design.