Problem 1. Consider GF(33) defined by the primitive polynomial p(x) = x3 + 2x + 1, and let ksi = x mod p(x). Find the minimum polynomial m5(x) of ksi5. You may assume the following theorems:
You may use the following
table for you calculations:
GF(33) defined by the primitive polynomial p(x) = x3 + 2x + 1 |
|||
Antilog |
Log |
Antilog |
Log |
000 |
-INF |
|
|
100 |
0 |
200 |
13 |
010 |
1 |
020 |
14 |
001 |
2 |
002 |
15 |
210 |
3 |
120 |
16 |
021 |
4 |
012 |
17 |
212 |
5 |
121 |
18 |
111 |
6 |
222 |
19 |
221 |
7 |
112 |
20 |
202 |
8 |
101 |
21 |
110 |
9 |
220 |
22 |
011 |
10 |
022 |
23 |
211 |
11 |
122 |
24 |
201 |
12 |
102 |
25 |
Problem 2.
You will find below the Antilog/Log table of GF(26) based on the primitive polynomial
p(x) = x6 + x + 1
Use this table to compute the minimum
polynomial m5(x) of ksi5
, where ksi is
the primitive element defined by p(x).
Antilog |
Log |
|
Antilog |
Log |
|
Antilog |
Log |
|
Antilog |
Log |
000000 |
-INF |
|
000101 |
15 |
|
101001 |
31 |
|
111001 |
47 |
100000 |
0 |
|
110010 |
16 |
|
100100 |
32 |
|
101100 |
48 |
010000 |
1 |
|
011001 |
17 |
|
010010 |
33 |
|
010110 |
49 |
001000 |
2 |
|
111100 |
18 |
|
001001 |
34 |
|
001011 |
50 |
000100 |
3 |
|
011110 |
19 |
|
110100 |
35 |
|
110101 |
51 |
000010 |
4 |
|
001111 |
20 |
|
011010 |
36 |
|
101010 |
52 |
000001 |
5 |
|
110111 |
21 |
|
001101 |
37 |
|
010101 |
53 |
110000 |
6 |
|
101011 |
22 |
|
110110 |
38 |
|
111010 |
54 |
011000 |
7 |
|
100101 |
23 |
|
011011 |
39 |
|
011101 |
55 |
001100 |
8 |
|
100010 |
24 |
|
111101 |
40 |
|
111110 |
56 |
000110 |
9 |
|
010001 |
25 |
|
101110 |
41 |
|
011111 |
57 |
000011 |
10 |
|
111000 |
26 |
|
010111 |
42 |
|
111111 |
58 |
110001 |
11 |
|
011100 |
27 |
|
111011 |
43 |
|
101111 |
59 |
101000 |
12 |
|
001110 |
28 |
|
101101 |
44 |
|
100111 |
60 |
010100 |
13 |
|
000111 |
29 |
|
100110 |
45 |
|
100011 |
61 |
001010 |
14 |
|
110011 |
30 |
|
010011 |
46 |
|
100001 |
62 |
Problem 3.
h(x) = 1 + x3 +x6
g(x) = 1 + x2 + x4
+ x6 + x7
Problem 4. Draw an LSC which takes as inputs polynomials a(x)
and b(x) and then produces the output h(x)a(x) + k(x)b(x),
where h(x) and k(x) are the polynomials:
h(x) = 1 + x4 + x10
and k(x)
= x + x2 + x4 + x7 + x9