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Nuray Kutlu

#1: Small sample of 91-end of Frank Garvan's awesome Maple booklet

with(LinearAlgebra);
A := Matrix(3, 3, [1, 4, 5, 6, 3, 6, 7, 3, 4]);
Determinant(A);
                               51
B := Transpose(A);
                              [1  6  7]
                         B := [4  3  3]
                              [5  6  4]
Rank(A);
                               3
combine(x*x);
                                2
                               x 
floor(3.5);
                               3
#2: form a matrix with 0-1 whose number of rows and number of columns are both prime numbers and "draw" your initials
10001001001
11001001010
10101001100
10011001010
10001001001

Then turn it into a 1-dimensional vector:
1000100100111001001010101010011001001100101010001001001

#3:Use RC(q,n,d,K) with q=2, d=3,5,7, and 5 ≤ n ≤16 and K very big (say 10000), and each time run it several times and pick the one with the most elements (to get M as big as possible). How do the codes that you got compare to the best possible values in table 2.4 on p.14 of Hill's book?

After running the lines multiple times, I got the following results. Compared to the values in Hill's book, the small numbers are pretty similar but Hill's numbers are a lot greater when it comes to larger numbers.

temp := RC(2, 5, 3, 10000);
nops(temp);
                               4
temp := RC(2, 6, 3, 10000);
nops(temp);
                               6
temp := RC(2, 7, 3, 10000);
nops(temp);
                               9
temp := RC(2, 8, 3, 10000);
nops(temp);
                               16
temp := RC(2, 9, 3, 10000);
nops(temp);
                               29
temp := RC(2, 10, 3, 10000);
nops(temp);
                               47
temp := RC(2, 11, 3, 10000);
nops(temp);
                               88
temp := RC(2, 12, 3, 10000);
nops(temp);
                              156
temp := RC(2, 13, 3, 10000);
nops(temp);
                              264
temp := RC(2, 14, 3, 10000);
nops(temp);
                              453
temp := RC(2, 15, 3, 10000);
nops(temp);
temp := RC(2, 16, 3, 10000);
nops(temp);
                              743

                              1179
temp := RC(2, 5, 5, 10000);
nops(temp);
temp := RC(2, 6, 5, 10000);
nops(temp);
temp := RC(2, 7, 5, 10000);
nops(temp);
temp := RC(2, 8, 5, 10000);
nops(temp);
temp := RC(2, 9, 5, 10000);
nops(temp);
temp := RC(2, 10, 5, 10000);
nops(temp);
temp := RC(2, 11, 5, 10000);
nops(temp);
temp := RC(2, 12, 5, 10000);
nops(temp);
temp := RC(2, 13, 5, 10000);
nops(temp);
temp := RC(2, 14, 5, 10000);
nops(temp);
temp := RC(2, 15, 5, 10000);
nops(temp);
temp := RC(2, 16, 5, 10000);
nops(temp);
                               2
                               2
                               2
                               4
                               4
                               8
                               10
                               19
                               28
                               45
                               66
                              106
temp := RC(2, 5, 7, 10000);
nops(temp);
temp := RC(2, 6, 7, 10000);
nops(temp);
temp := RC(2, 7, 7, 10000);
nops(temp);
temp := RC(2, 8, 7, 10000);
nops(temp);
temp := RC(2, 9, 7, 10000);
nops(temp);
temp := RC(2, 10, 7, 10000);
nops(temp);
temp := RC(2, 11, 7, 10000);
nops(temp);
temp := RC(2, 12, 7, 10000);
nops(temp);
temp := RC(2, 13, 7, 10000);
nops(temp);
temp := RC(2, 14, 7, 10000);
nops(temp);
temp := RC(2, 15, 7, 10000);
nops(temp);
temp := RC(2, 16, 7, 10000);
nops(temp);
                               1
                               1
                               2
                               2
                               2
                               2
                               4
                               4
                               6
                               10
                               13
                               19


#4:The binary Hamming codes (to be studied later) have n=2r-1 (where r is a positive integer), and minimal distance 3. They are known to be perfect. How big are they? (i.e. what is there M?) (Hint: use the Sphere Packing Bound)

since we are using the binary hamming code, q =2

since n= 2^r -1,
and we have d = 3 so t=1

use sphere packing bound, 


<= (2^(2^r -1))/ (sum i form 0 to 1 (bi(2^r -1, i) * (2-1)^i))

= 

(2^2^r/2)/ (1 + 2^r -1)

<=2^(2^r -1 -r)