Question 1.
(n,l) = (5,3),p = 507997
(a) x = {1, 2, 3, 4, 5}
fll = 3«95lT mod 101021 = 71631 a2 = 540a517 mod 101021 = 78075 S = oo = 469517
Next, we can construct the polynomial:
/(#) = ao'x0 + aix1 + aix1 f(x) = 469517 + 71631a: + 78075x2
Now, we can compute the shares:
Sl =/(])= 619223 mod 567997 = 51226 s2 = /(2) = 925079 mod 567997 = 357082 s3 = /(3) = 1387085 mod 567997 = 251091 Si = /(4) = 2005241 mod 567997 = 301250 «5 = /(5) = 2779547 mod 567997 = 507559
The shares are:
(1,51226), (2,357082), (3,251091), (4,301250), (5,507559)
(b) The shares create a system of equations:
5' + ai + a2 = 438605 5 + 2ai + 4o2 = 273820 S + 3ai + 9a2 = 133642 S = 627997 mod 567997 = 60000
The secret from the shares is 60000.
Question 2.
(a) Yes, such orthogonal array exists. Following the condition that any pair of symbols n2 occurs in exactly A rows, we can find the following array
fO 0 0 0 0\ 0 0 1 0 1 0 1 0 0 0 0 110 1 10 0 10 10 111 110 10
Vi l l l i/
(b) The upper bound is computed as
(c) Number of rows is computed as
r = \n"
where r is the number of rows
(d) We need to find an orthogonal array, where each subset of length 3 occurs only once. There are 1 • 23 = 8 rows in the OA.
	0	0	
0	I)	1	0
0	1	0	0
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	1
\l	1	1	0
(e) We take the t — (n. k. A) array and choose an element e in it. We then select rows that have this element on the first position and delete the first position to get a subarray. This subarray is the (t — 1) — (n, k — 1, A) orthogonal array.
Question 3.
(a) a = av * vr mod p v = 47
a = 169 (13, 23, 11)
169u * 4723 mod 311 = 13 = a correct (U3, 17, 3)
1693 * 4717 mod 311 = 113 = a correct (113, 19, 15)
16915 * 4719 mod 311 = 105 ^ a incorrect (13, 27, 15)
1691E * 4727 mod 311 = 260 f a incorrect
(b) If random k was used in (113.17,3) and pseudo-random in (13, 23.11) it must hold that: a\ = a   mod p
da = q3**+4 mod p <J2 — <r' * 1694 mod p <7] = 113
13 = a2 = 113s * 1694 mod 311 Therefore:
y1 = k\ + a * ri mod q
D2 = 3 * k\ + 4 + a * T2 mod q
3 = ki + 17 *a mod 31
11 = 3 * fci + 4 -I- 23 * a mod 31
9 - 7 = 3 * fei - 3 * fei + 51 *a - 23 * a mod 31 2 = 28 * a mod 31 = 20 Secret key is 20,
Question 4.
Peggy will take every row and every column as a lists, together there will be 2n such lists. Every list will contains a numbers. Every number will be twice in all the lists. Sum of every list will be m, which will be a prove that she knows the solution. However she will mix the numbers in every list, so he doesn't know the exact combination. She wouldn't tell him which list is a row and which is a column.
Victor will see 2n lists, where sum of every list is in. He can see that every number is twice in all the lists and that every list is different, which means that she knows the solution. However it wouldn't help him to find the solution.
Question 5.
(a) Zi can be expressed as Zi = h{i), where h(i) = f(i) + g(i). f{x) = x + 2>=i aj a;-*'(mod p)
g(x) = y +        bjx1 (mod p)
h(x) = z + Ejli(aj + bj)xi(mod p), where z — x + y
h(x) is not polynomial of degree t — 1 if at-\ + 6j-i = 0 mod p and therefore f < t.
(b) t' = t only when at-\ + bt-i ^ 0 mod p.
Question 6.
(a) Victor would have at the end of every round w, c. z, Victor accepts with:
bz — u * ac mod p V+Ct* = V * U"c mod p rJrc*x = r + c*x mod p
(b) Victor can only change r in this protocol.
He can e.g. precompute a large number of results for k = bv mod p, where y < p.
Then he sets c = 1 every time. Because there is p/2 rounds, it is probable, that in some point, he find u = fe. Therefore, he found r for that specific round.
Now, he can compute z = r + c* x, because he know z, r, c. Question 7.
Using script to xor pixels in given qr codes I have found:
Question 8.
Inside the dot at the end of the text there is text: microdot
microdot
Figure 2: Zoomed at the dot