Department of Mathematics
University of British Columbia
MATH 342 Practice Midterm 1
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100
CALCULATORS, NOTES OR BOOKS ARE NOT PERMITTED.
THERE ARE 5 PROBLEMS ON THIS EXAM.
JUSTIFY YOUR ANSWERS.
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1.
Let C be a code with minimum distance d(C) = 7.
(a) Find the maximum number of errors that can be corrected, using an
incomplete nearest neighbour decoder.
(b) Find the maximum number of errors that can be detected.
(c) Explicitly describe a hybrid decoder which is guaranteed to correct
up to 2 errors and to detect 3 errors and to detect 4 errors. Prove
that your decoder works.
Solution:
a. As shown in class, the maximum number of errors that can be corrected is b d(C)−1
c = 3.
2
b. As shown in class, the maximum number of errors that can be detected
is d(C) − 1 = 6.
c. Let x be the received word and c0 be the nearest neighbour codeword.
If d(x, c0) ≤ 2, decode x to c0. If d(x, c0) ≥ 3, declare an error.
Proof that this works:
Let c be the transmitted codeword.
Case 1:
1 or 2 channel errors.
Then d(x, c) ≤ 2 and so d(x, c0) ≤ 2. Thus, decoder decodes to c0. And
d(c, c0) ≤ d(c, x) + d(x, c0) ≤ 4.
Since d(C) = 7, c = c0, and so decoder (correctly) decodes to c.
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Case 2:
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3 or 4 channel errors.
Then 3 ≤ d(x, c) ≤ 4. So, the decoder will correctly detect an error
unless d(x, c0) ≤ 2. But then
d(c, c0) ≤ d(c, x) + d(x, c0) ≤ 6.
Since d(C) = 7, c = c0. But this is impossible since d(x, c0) 6= d(x, c).
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2.
True or False (if True, give a brief proof; if False, give a counter-example)
a) Every binary code C of length 8 is equivalent to a binary code which
contains the codeword 01010101.
b) Every binary code C of length 8 with d(C) = 8 is equivalent to a
binary code which contains the codewords 00000000 and 11111111.
c) If two binary codes have the same n, M, and d parameters, then they
are equivalent.
Solution:
a. True: Let c ∈ C. For each of i = 1, 3, 5, 7, if ci = 0, do nothing;
if xi = 1, flip the bit in position i in each codeword. For each of i =
2, 4, 6, 8, if ci = 1, do nothing; if xi = 0, flip the bit in position i in
each codeword. This yields an equivalent code where x is transformed
to 01010101.
b. True: Let c, c0 ∈ C such that d(c, c0) = 8. Then c and c0 differ in
each position. Apply bit flips in each position i in all codewords such
that ci = 1. This yields an equivalent code where c is transformed to
00000000. Since c and c0 differ in each position, c0 is transformed to
11111111.
c. False: from HW2, K1 = {000, 100, 010}, K2 = {000, 100, 111}; even
though the two codes have the same (n, M, d) parameters, K2 has a pair
of codewords with distance = 3, and K1 does not.
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3.
Find A2(10, 7).
Solution:
Let C be a (10, M, 7) code. Up to equivalence, we may assume that
C contains the all 0’s word 0. Then all other codewords must have
weight at least 7. If there were two distinct such codewords c, c0, then
d(c, c0) ≤ d(c, 1) + d(1, c0) ≤ 3 + 3 = 6, a contradiction (here, 1 denotes
the all 1’s word). Thus, A2(10, 7) ≤ 2.
Letting x denote any word of length 10 with exactly 7 ones, we see that
the code {x, 0} has minimum distance 7. Thus, A2(10, 7) ≥ 2, and so
A2(10, 7) = 2.
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4. Let G = {I, R1, R2, F1, F2, F3} be the group of rigid motions that preserve an equilateral triangle centered at the origin, with composition of
mappings as addition. Here, I is the identity map, R1 is rotation by
2π/3 counter-clockwise, R2 is rotation by 4π/3 counter-clockwise, and
Fi, i = 1, 2, 3 is the reflection that fixes vertex i (with the vertices in
counter-clockwise order).
Recall that G is non-abelian.
(a) Show that H = {I, F1} is a subgroup of G.
(b) Find all the cosets, a + H, of H in G.
(c) Show that K = {I, R1, R2} is a subgroup of G.
(d) Find all the cosets, a + K, of K in G.
(e) Find the smallest subgroup of G that contains both F1 and F2.
Solution:
a. Addition table for H:
+ I F1
I I F1
F1 F1 I
Thus, H is closed under addition.
Since, I −1 = I and F1−1 = F1, H is closed under inverses.
So H is a subgroup.
b. There are 3 cosets:
H = {I, F1}, R1 + H = {R1, R1 ◦ F1} = {R1, F3},
R2 + H = {R2, R2 ◦ F1} = {R2, F2}
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c.
Addition table for K:
+
I
R1
R2
I
I
R1
R2
R1
R1
R2
I
R2
R2
I
R1
Thus, K is closed under addition.
Since, I −1 = I, R1−1 = R2, and R2−1 = R1, K is closed under inverses.
Thus, K is a subgroup.
d.
There are 2 cosets:
K = {I, R1, R2}, F1 + K = {F1, F1 ◦ R1, F1 ◦ R2} = {F1, F2, F3}
e. The smallest such subgroup L must contain F1 ◦ F2 = R1 and so also
contains R1 ◦ R1 = R2. And then it must contain F1 ◦ R2 = F3. Since
any subgroup contains the identity, L = G.
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5.
Let q and m be positive integers. Show that Zm, as a group with addition
modulo m, has a subgroup of size q iff q divides m.
Solution:
Only if: Let H be a subgroup of size q. By Lagrange’s theorem, H has
|G|
|G|
cosets
and
thus
|H|
|H| must be an integer. Thus, q = |H| must divide
m = |G|.
If: Let H = {im/q : 0 ≤ i ≤ q − 1}, a subset of Zm of size q.
We can view H as {im/q mod m : i ∈ Z}.
To see that H is a subgroup of Zm, we must show that it is closed under
addition and inverses.
Addition: im/q + jm/q = (i + j)m/q mod m ∈ H
Inverses: −im/q = (−i)m/q mod m ∈ H
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