Department of Mathematics
University of British Columbia
MATH 342 Practice Final Exam
Family Name:
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I.D. Number:
Problem
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Mark
Out of
1
10
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10
3
10
4
10
5
10
6
10
7
10
8
10
Total
80
CALCULATORS, NOTES OR BOOKS ARE NOT PERMITTED.
THERE ARE 8 PROBLEMS ON THIS EXAM.
JUSTIFY YOUR ANSWERS.
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1.
Find the following principal remainders, (the principal remainder of
a mod q is the unique element b ∈ {0, 1, . . . , q − 1} s.t. a = b mod q).
(a) 103849197491 mod 11
(b) (60)11 mod 11
(c) (1100005)(223347) mod 11
solution:
a. 10 = −1 mod 11. Since the exponent is odd, the answer is −1 = 10
mod 11.
b. By Fermat’s little Theorem, (60)11 = 60 = 5 mod 11.
c. The first factor is 5 mod 11 and the second factor is 3 mod 11 (since
11 divides 22, 33 and 44). Thus the product is equal to 15 = 4 mod 11.
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2.
Let C be the linear code over GF (5) spanned by {01234, 13024, 12340}.
(a) Find bases for C and C ⊥.
(b) Find the length, dimension, and minimum distance of C and C ⊥.
(c) Find the number of cosets of C and C ⊥.
(d) For each coset of C ⊥ with coset leader of weight at least 2, find a
coset leader.
solution:
We first find a generator matrix for C by finding the REF of


0 1 2 3 4
1 3 0 2 4
1 2 3 4 0


1 3 0 2 4
0 1 2 3 4
1 2 3 4 0


1 3 0 2 4
0 1 2 3 4
0 4 3 2 1


1 3 0 2 4
0 1 2 3 4
0 0 0 0 0
So, a generator matrix is
1 3 0 2 4
G=
0 1 2 3 4
We can put this in standard form by reducing to RREF:
1 0 4 3 2
G0 =
0 1 2 3 4
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So a basis for C is {10432, 01234}.
A parity check matrix for C is

 

−4 −2 1 0 0
1 3 1 0 0
H =  −3 −3 0 1 0  =  2 2 0 1 0 
−2 −4 0 0 1
3 1 0 0 1
So a basis for C ⊥ is {13100, 22010, 31001}.
b. Using the fact that H is a parity check matrix for C and G0 is a
parity check matrix for C ⊥, we see that C is a [5, 2, 4] code and C ⊥ is
a [5, 3, 3] code (for the latter, observe that the first 3 columns of G0 are
linearly dependent; for the former observe that the determinant of each
matrix consisting of three columns of H is nonzero – argue by cases).
c. C has 55−2 = 125 cosets and C ⊥ has 55−3 = 25 cosets.
d. Since d(C ⊥) = 3, all words of weight 0 or 1 are coset leaders. There
are 21 such words. The 4 remaining coset leaders correspond to syndromes that are not multiples of single columns of G0 and not the zero
vector. These syndromes are the multiples of the vector [41]. The vector
x := [41000] has syndrome [41] and since x has weight two, it can be
chosen as a coset leader for its coset. The other coset leaders are the
nonzero multiples of x, namely [32000], [23000], [14000].
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3.
A subgroup H of a group G is proper if H 6= G.
(a) Find all proper subgroups of Z4 (with addition modulo 4).
(b) Find all proper subgroups of GF (4) (with addition).
Recall the addition table for GF (4):
+
0
1
a
b
0
0
1
a
b
1
1
0
b
a
a
a
b
0
1
b
b
a
1
0
solution:
The identity element must be an element of every subgroup.
a. Z4 = {0, 1, 2, 3}. So any subgroup H must contain 0. If H contains
1, then it must also contain 2 and 3 and thus is not a proper subgroup.
Similarly, if it contains 3, it must also contain 1 and 2. The only other
possibility is H = {0, 2} which is a proper subgroup. So the only proper
subgroups are {0} and {0, 2}.
b. GF (4) = {0, 1, a, b}. Since every element is its own inverse and
the sum of any element and itself is 0, each of {0, 1}, {0, a} and {0, b}
is a proper subgroup. If a subgroup contains two of the three nonzero
elements it must contain the third nonzero element. So the only proper
subgroups are {0}, {0, 1}, {0, a}, {0, b}.
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4.
Which of the following polynomials is (are) irreducible over GF (2)?
(a) f (x) = x5 + x2 + 1
(b) g(x) = x5 + x + 1
solution:
a. Irreducible: There are no degree-one factors since neither x = 0 nor
x = 1 is a root. So, it is irreducible iff it has a quadratic factor. The
quadratic polynomials are {x2, x2 + 1, x2 + x, x2 + x + 1}. The first and
third of these are divisible by x, but f (x) is not. So, the only possible
factors are x2 + 1 and x2 + x + 1. By long division, the first gives the
non-zero remainder, x, and the second gives the non-zero remainder, 1.
b. Reducible: Using the same procedure, g(x) is irreducible iff x2 + 1
and x2 +x+1 do not divide it. But x2 +x+1 does divide it: x5 +x+1 =
(x3 + x2 + 1)(x2 + x + 1).
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5.
Recall the definition of characteristic of a finite field F : the smallest
positive integer q such that q · 1 = 0 (recall that for an element a ∈ F ,
q · a denotes the sum of q copies of a).
(a) Show that if F is a finite field with characteristic q, then for all
nonzero elements a ∈ F , q is the smallest positive integer such that
q·a=0
(b) For any prime p and positive integer n, what is the characteristic of
GF (pn)?
solution:
a. Let a be a nonzero element of F . Let q be a positive integer. If
q · a = 0, then (q · 1)a = q · a = 0. Since F is a field and a 6= 0,
(q · 1) = 0. Conversely, if (q · 1) = 0, then (q · a) = 0. Thus, (q · a) = 0
iff (q · 1) = 0. So, the characteristic of F is the smallest positive integer
q s.t. (q · a) = 0.
b. We proved in class that the characteristic of GF (p) (which is the
same as Zp) is p. Now GF (pn) = GF (p)[x]/(f (x)) where f (x) is an
irreducible polynomial over GF (p). The multiplicative identity element
of GF (pn) is the same as the multiplicative identity element, 1, of GF (p).
For a positive integer q, the expression q · 1 means the same in GF (pn)
as in GF (p). So, the characteristic of GF (pn) is p.
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6.
Let a be a primitive element of GF (27).
(a) Find all positive integers n such that an = a3.
(b) Find all powers of a that are primitive.
solution:
a. Since a is primitive it is nonzero. So, a26 = 1. Since the powers ai,
i = 1, . . . , 26 exhaust all of GF (27) \ {0} and a26 = 1, we have that
k = 26 is the only positive integer in the range [1, 26] such that ak = 1.
Thus, am = 1 iff m is a multiple of 26. Since an = a3 iff an−3 = 1, we
have an = a3 iff n = 3 mod 26.
b. Let g := gcd(m, 26). If g 6= 1, then 26/g < 26 and since (am)26/g =
(a26)m/g = 1, am is not primitive (because (am)26/g = 1 and (am)26 =
(a26)m = 1 and so the first 26 powers of am cannot exhaust all of
GF (27) \ {0}).
If g = 1, then by Bezout, there exist integers u, v s.t. mu + 26v = 1.
Then
aum = auma26v = aum+26v = a
We claim that am is primitive. If not, then for some 1 ≤ i < 26,
(am)i = 1. But then ai = (aum)i = (ami)u = 1u = 1, a contradiction to
the fact that a is primitive.
So, am is primitive iff g = 1 iff m is not congruent to an even integer or
13 mod 26.
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7.
In this problem, note the “round brackets” (n, M, d) (not square brackets [n, k, d]).
(a) i. Find a (3,4,2) binary code.
ii. Find a (5,4,3) binary code.
iii. Usse parts i and ii to find a (8,4,5) binary code.
(b) Show that A2(8, 5) = 4.
solution:
a.
i. {000, 011, 101, 110}
ii. {00000, 01101, 10110, 11011}
iii. {00000000, 01101011, 10110101, 11011011}
b. By part a, we know that A2(8, 5) ≥ 4. So, it suffices to show that
there does not exist a binary code of length 8, minimum distance 5 with
5 codewords.
Suppose that C is an (8, 5) code with 5 codewords. Up to equivalence,
we may assume that C contains 0. Thus, the other four codewords must
have weight at least 5.
If C contains a word x of weight 7 or 8, then any other word y of
weight at least 5 would share at least 4 1’s in the same positions and so
d(x, y) ≤ 4, a contradiction. So, the remaining words of C must have
weight 5 or 6.
Suppose that C contains two words u and v of weight 6. They must share
at least 4 1’s in the same positions and so d(u, v) ≤ 4, a contradiction.
So, C contains at most one word of weight 6. Thus, it must contain at
least three words of weight 5.
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Up to equivalence, we may assume that one of these words is z =
11111000. If z 0 is another word of weight 5, then z and z 0 must share
at least 2 1’s in the same positions (otherwise z 0 would not have weight
5). Suppose that they share at least 3 1’s in the same positions. Then
they must differ in the other 5 positions. But this is impossible since
they both have weight 5. Thus, z 0 must share exactly 2 1’s in the same
positions as z. If z 00 were another such word then it would also share
exactly 2 1’s in the same positions as z. Since z, z 0 have weight 5, they
must both have 1’s in the last 3 positions. Since they differ in 5 positions
they must be of the form: z 0 = x1x2x3x4x5111 and z 00 = xc1xc2xc3xc4xc5111.
Since z 0 has weight 5, z 00 would have weight 6, a contradiction.
Thus, there can be only two words of weight 5, a contradiction.
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8.
Let C be a binary code of length n with at least two codewords. Let t be
a positive integer such that the Hamming balls of radius t, centered at
the codewords, are pairwise disjoint and completely cover V (n, 2). Show
that d(C) = 2t + 1.
solution:
To show that d(C) = 2t + 1, we must show:
i. for all pairs of distinct codewords c, c0, d(c, c0) ≥ 2t + 1.
ii. there exists a pair of codewords c, c0 such that d(c, c0) = 2t + 1.
Proof of (i): Suppose not. Then there exists a pair of distinct codewords
c, c0 such that d(c, c0) ≤ 2t. If d(c, c0) ≤ t, then c0 ∈ Bt(c), contrary
to the assumption that the Hamming balls of radius t centered at the
codewords are pairwise disjoint. If not, then t < d(c, c0) ≤ 2t. Let x
be obtained from c by swapping 0 and 1 in exactly t positions where c
and c0 disagree. Then d(c, x) = t and d(c0, x) = 2t − d(c, c0) < t, again
contrary to he assumption that the hamming balls of radius t centered
at the codewords are pairwise disjoint.
Proof of (ii): Let c ∈ C and let x be obtained from c by swapping 0
and 1 in exactly t + 1 bit positions. Then x 6∈ Bt(c). But since the
Hamming balls of radius t cover V (n, 2), x ∈ Bt(c0) for some c0 ∈ C.
Clearly c 6= c0. Then
d(c, c0) ≤ d(c, x) + d(x, c0) ≤ (t + 1) + t = 2t + 1.
By part (i), d(c, c0) = 2t + 1.
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