Midterm Xuenan Xu.md

Midterm Exam

by Xuenan Xu GWID G26980825

 

1

Frequency distribution:

P	C	S	U	R	V	G	H	D	E	Z	M	O	K	X	N	L	I	F	T
1	2	8	4	6	3	4	2	1	1	6	2	2	3	1	2	1	1	1	1

• First we notice there is a word H, and the only character that forms a word itself is a;
• S has a noticeable high frequency than other characters, I guess it's e;
• Three-character word VCS ends with e, the most frequent word of this is the, that is V is t and h stands for C;
• RV ends with t, so we know that R is i;
• RV RG H has been like it i* a, so G could be s;
• GHOS is sa*e now, so O shall be m;
• DESGVRZU looks like **esti**, the first word came to my mind is question, so D is q, E is u, Z is o and U is n;
• ZM is a two-character word going after question and starts with o, the best match shall be question of, that is f represents M;
• OZUSK is now mone*, so K has a great possibility of y;
• I would split SXSNKLZIK into two parts, SXSNK which looks like e*e*y and LZIK which looks like *o*y, so SXSNK shall be every and thus LZIK could be body;
• PCSU is a four-character word like *hen, the most frequent ones are when and then, so in this case P shall be w;
• NSFRTRZU contains the remaining two undecided characters, which are F and T, I did some word search because I couldn't recognize the pattern at first, and the first result is religion that quite makes sense;

Mapping:

P	C	S	U	R	V	G	H	D	E	Z	M	O	K	X	N	L	I	F	T
W	H	E	N	I	T	S	A	Q	U	O	F	M	Y	V	R	B	D	L	G

Result:

When it is a question of money, everybody is of the same religion. -Voltaire

2

a) Pr[1] = Pr[a] * Pr[K₁] + Pr[b] * Pr[K₃] + Pr[c] * Pr[K₂] = 1/2 * 2/3 + 1/3 * 1/6 + 1/6 * 1/6 = 5/12

b) Pr[b|1] = Pr[b] * Pr[K₃] / Pr[1] = 1/3 * 1/6 / 5/12 = 2/15;

c) Pr[c|1] = Pr[c] * Pr[K₂] / Pr[1] = 1/6 * 1/6 / 5/12 = 1/15;

d)

Pr[2] = Pr[a] * Pr[K₂] + Pr[b] * Pr[K₁] + Pr[c] * Pr[K₃] = 1/2 * 1/6 + 1/3 * 2/3 + 1/6 * 1/6 = 1/3

Pr[3] = 1 - Pr[1] - Pr[2] = 1/4

H[C] = -(Pr[1] * log₂Pr[1] + Pr[2] * log₂Pr[2] + Pr[3] * log₂Pr[3]) = -(5/12 * log₂5/12 + 1/3 * log₂1/3 + 1/4 * log₂1/4) = 3/4log₂3 + 2/3log₂4 - 5/12log₂5 = 1.5546

3

According to the cipher process e₂(π(e₁(M))), we shall use the multiplicative decipher with key 5 to decrypt the string first, and permutate the result with π = (2 3 1 4), finally decipher the result with autokey decipher with key 5.

1. First, use the multiplicative decipher to decrypt the string RXJNJPOEHLZGFPBV
   • The reverse of 5 is 21
   • Starting with R, we calculate R * 21 mod 26, and we have T
   • Repeat this process, and we get TPHNHDIGRXFWBDVZ
2. Second, use the permutation decipher with π = (2 3 1 4) to decrypt the result
   • Take the first 4 characters TPHN, put the 2nd char at 1st, 3rd char at 2nd, 1st char at 3rd and 4th at 4th, we have HTPN
   • Repeat this process, and we get HTPNIHDGFRXWVBDZ
3. Finally, use the autokey decipher to decrypt the result
   • Take the first character H, the key is 5, so we calculate H - 5 mod 26 and get C, and use C as the new key
   • Repeat this process, we have CRYPTOPRODUCTIVE as final result

4

1. The formula given can be converted into z_i = z_{i - 6} + z_{i - 4} mod 2
2. We have initial stream 110101
   • When we calculate the 7th number, we take the sum of 1st and 3rd number, mod 2, and get 1
   • Repeat this process, we have result 110101100011101101

5

a)

Key Length Bits	128	192	256
Key Length Bytes	16	24	32
# Rounds	10	12	14

b)

r0	r1	r2	r3
12	67	AB	EF
23	78	BC	1A
45	89	CD	2B
56	9A	DE	3C

c)

r0	r1	r2	r3
12	67	AB	EF
78	BC	1A	23
CD	2B	45	89
3C	56	9A	DE

d)

r0	r1	r2	r3
C9	85	62	DF
BC	65	A2	26
BD	F1	6E	A7
EB	B1	B8	1D

e)

r0	r1	r2	r3
85	62	DF	C9
65	A2	26	BC
F1	6E	A7	BD
B1	B8	1D	EB