MATH447/747 ASSIGNMENT 3 The not to hand ... - People.stat.sfu.ca

The not to hand in questions
MATH447/747 ASSIGNMENT 3
FALL 2012
#13 (a) No column of H is 0 and no column in a multiple of another, while column 1
plus column 2 equals column 3, so the distance of this code is 3 and thus e = 1.
n = 4 and we are over Z3 and so each sphere of radius 1 contains 1 + 2 · 4 = 9
elements. Further k = 2 and so there are 3 2 = 9 codewords. Together this gives
9 · 9 = 3 4 elements in the spheres of radius e around the codewords, and this is
all elements in Z 4 3, so this code is perfect.
(b) (you don’t need to worry about the lexicographic part since we didn’t discuss
that in class, and in fact it makes no difference for this code since it has distance
3 and is perfect)
0000 1021 2012 0111 0222 1102 2201 1210 2120
1000 2021 0012 1111 1222 2102 0201 2210 0120
2000 0021 1012 2111 2222 0102 1201 0210 1120
0100 1121 2112 0211 0022 1202 2001 1010 2220
0200 1221 2212 0011 0122 1002 2101 1110 2020
0010 1001 2022 0121 0202 1112 2211 1220 2100
0020 1011 2002 0101 0212 1122 2221 1200 2110
0001 1022 2020 0112 0220 1100 2202 1211 2121
0002 1020 2021 0110 0221 1101 2200 1212 2122
(c) We might as well use the coset leaders since all coset elements have the same
syndrome and the leaders are easiest to work with. H(0000) T = (00) T . The
other leaders are multiples of individual columns of H according to where they
have their nonzero element. Thus we get the map
00 → 0000
10 → 1000
20 → 2000
01 → 0100
02 → 0200
21 → 0010
12 → 0020
11 → 0001
22 → 0002
(d) Hr T 1 = (21) T which corresponds to 0010, and so we decode to r1 − 0010 =
2211 − 0010 = 2201
1

Hr T 2 = (12) T which corresponds to 0020, and so we decode to r2 − 0020 =
2221 − 0020 = 2201
Hr T 3 = (20) T which corresponds to 2000, and so we decode to r3 − 2000 =
2222 − 2000 = 0222
#21 Let Ci be the coset with syndrome v and write H = [h1 · · · hn]. Suppose Ci contains
a vector x = (x1, . . . , xn) of weight w, then v = Hx T = x1h1 + · · · + xnhn is a linear
combination of w columns of H. Suppose v = αi1hi1 + · · · αiwhiw. Let x be the vector
which is αij in position ij and 0 elsewhere. Then Hx T = αi1hi1 + · · · αiwhiw = v so x
is in Ci and x has weight w.
#26 At most 8 check symbols means that n − k ≤ 8, so k ≥ n − 8. d = 5 so e = 2.
The spheres of radius 2 over F2 contain
n
1 + n + = 1 + n +
2
n(n − 1)
2
elements. Thus together these spheres around codewords contain at least
2 n−8 (1 + n +
n(n − 1)
)
2
elements, and we want this number to be less than or equal to 2 n . That is we want
n−8 1 + n + n(n − 1)/2
2
·2n that is we want
1 + n + n(n − 1)/2
28 ≤ 1
Solving the quadratic equation, taking the positive root, and using that n is an
integer, we get that this happens for n ≤ 22. That’s about all we can say with
the tools we have (our other bounds don’t help much), so next you’d have to actually
construct. In fact the best you can actually do is n = 17, http://www.rz.
uni-karlsruhe.de/~kg11/codetables/BKLC/index.html. This wasn’t such a good
question – I thought we’d be able to get further with the techniques we have – does
someone else see something I’ve missed?
Now for the hand in questions
(1) (a) View det V as a polynomial in xi (with coefficients polynomials in the remaining
variables). Then if we evaluate at xi = xj we see that V has two rows equal,
and so xj is a root of det V viewed as a polynomial in xi. Roots of polynomials
yeild linear factors, thus xi − xj divides det V .
(b) Let P =
1≤i

We have P | det V and deg P = deg(det V ). Thus P and det V must be equal
up to a constant multiple. To check that the constant is correct, calculate the
coefficient of x0 1x1 2 · · · xn−1 n on both sides.
In det V this coefficient arises only by the term in the cofactor expansion which
always chose the diagonal elements. Thus this term has coefficient 1 in det V .
In P to get (n−1) powers of xn we must take xn in every factor it appears in. Of
the remaining factors then we must take xn−1 in every factor it appears in to get
n−2 powers. In all these terms the other variable has smaller index. Continuing
(formally an easy induction) in P the term x0 1x1 2 · · · xn−1 n arises only when we take
the first element of each factor, and thus also appears with coefficient 1.
The result follows.
(c) Consider any k columns of H K =
⎡
⎤
⎣
a i1 · · · a ik
.
. ..
a k(i1) · · · a k(ik)
.
Scale each column of K by the inverse of its first entry and then take a transpose.
The result is a Vandermonde matrix, and so by the previous parts we know its
determinant. In particular
det K = a i1+···+ik
⎦
y

0 =
% I actually only need to correct errors in positions 3,5,6,7
% as those are the data positions in the original word and we have assumed
% no more than 1 error
c = [r(3),r(5:7)];
if (s == [1;1;0])
c = c - [1 0 0 0];
elseif (s == [1;0;0])
c = c - [0 1 0 0];
elseif (s == [0;1;0])
c = c - [0 0 1 0];
elseif (s == [0;0;1])
c = c - [0 0 0 1];
endif
Then doing the source decoding by hand I get
(a) DECODE B
(b) TEARS FOR FEARS
(c) SEE YOU SOON
(3)#36: (a) There are q q+1 vectors total. There are q 2 codewords. Thus there are q q−1
cosets. Each coset other than C can contribute at most one vector to the
sphere of radius e around any codeword c. This is true for any code C as
if x and y were both in the sphere of radius e around c then x − y is a
codeword, but x = c + x ′ , y = c + y ′ with x and x ′ of weight ≤ e, and then
w(x − y) ≤ w(x ′ ) + w(y ′ ) ≤ 2e < d(C).
First note that if q is even then the code cannot be perfect, as if a code C
has even distance 2d, then there are two codewords c, c ′ with d(c, c ′ ) = 2d,
so moving from one to the other changing one coordinate at a time we
obtain a vector v with d(c, v) = d and d(c ′ , v) = d. This v is not in the
sphere of radius e = d − 1 around c or c ′ , and if it was in the sphere of
radius e around some other codeword c ′′ then d(c, c ′′ ) ≤ d(c, v)+d(c ′′ , v) ≤
d + d − 1 < 2d contradicting the distance, so C is not perfect.
Now suppose q is odd so e = (q −1)/2. If q = 3 then e = 1 and the spheres
contain 1 + 2 · 4 = 9 elements and there are 9 spheres, giving 81 = 3 4
elements in the sphere so the code is perfect.
For larger odd q we would need that
q 2
1 + (q − 1)(q + 1) + (q − 1) 2
q + 1
+ · · · + (q − 1)
2
e
q + 1
= q
e
q+1
dividing by q2 and rewriting we’d need
e
(2e) i
2e + 2
= (2e + 1)
i
2e
i=0
Rewriting with the binomial theorem we’d need
e
(2e) i
2e + 2 2e
− (2e)
i
i
e
2e 2e
=
(2e)
i
i
i
(2e + 2)(2e + 1)
− 1
(2e + 2 − i)(2e + 1 − i)
i=0
i=0
i=0
4
− (2e) 2e−i

the i = 0 term is 0, and otherwise the terms are strictly decreasing with e,
so since we have seen equality at e = 1 there will be no further equalities.
(b) Fix γ ∈ Fq and indices i = j.
If γ = 0 then the zero codeword is γ in positions i and j, and no other
codeword can be, as any such codeword would have weight ≤ q − 1 which
contradicts that the distance of the code is q.
Now suppose γ = 0.
Suppose there were two codewords c and c ′ which were both γ in positions
i and j. Then c − c ′ would be 0 in positions i and j, and so c − c ′ would
have weight ≤ q − 1 which contradicts that the distance of the code is q.
Take any two linearly independent codewords c and c ′ . These form a basis
for C. Project c and c ′ onto F 2 q by projecting onto postions i and j. Call the
projected vectors v and v ′ . If v and v ′ are linearly independent, then some
linear combination of them is [γ, γ] and so the same linear combination of
c and c ′ is γ in positions i and j. Suppose v and v ′ are linearly dependent.
Then some linear combination of them is 0. The same linear combination
of c and c ′ is 0 in positions i and j, but this codeword would have weight
≤ q − 1 which contradicts that the distance of the code is q.
Thus for every γ and i = j there is exactly one codeword which is γ in
positions i and j.
(c) As remarked above no nonzero codeword can have two or more zero entries.
Suppose c is a codeword with no zero entries.
As remarked above no linearly independent codewords can have a projection
onto two coordinates such that the projections are linearly dependent.
That is the generator matrix has no two columns which are multiples of
each other. Use c along with another codeword c ′ to make a basis for
C. Let G be the matrix with first row c and second row c ′ . Whether
two columns of G are linearly dependent or not is unchanged by scaling
columns of G. So scale all columns of G to have 1 in the first position. This
is always possible since c has no zero entries. Then since no two columns
which are multiples of each other all the entries in the secodn row (now
that we have scaled) must be distinct. But there are only q distinct values
and q + 1 columns.
This is a contradiction and so all nonzero codewords have exactly 1 zero
entry.
#69: (a) An (n, k, d)-code has a parity check matrix of size n−k×n. So in particular
any set of more than n − k columns must be linearly dependent. Thus by
a result from class d ≤ n − k + 1.
(b) Since C is capable of correcting t errors we must have 2t+1 ≤ d. Using the
result from the previous part we get 2t + 1 ≤ n − k + 1, so 2t ≤ n − k. But
n − k is the number of check symbols, so C has at least 2t check symbols.
#70: (a) Consider the code with parity check matrix [1 1 1 1] (This is a simple
parity bit code). This has no zero column, but all pairs of columns are
linearly dependent, so it has distance 2. n = 4, and n − k = 1, so k = 3.
Thus this code satisfies the singleton bound. To make a second example I
5

could do something similar over Z3 with the parity check matrix [1 2 1 2].
The statistics are the same by the same calculations. This code is a sort of
modified parity bit code where different input bits have different weights.
(b) Hamming codes have n = (q r − 1)/(q − 1), k = n − r and d = 3. So
d − n + k − 1 = 3 − r − 1 = 2 − r
which is 0 iff r = 2. So Hamming codes are MDS codes iff they are order
2.
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