1. For a 1-1 affine cipher to exist in any alphabet, each character must map to another character such

that no two characters may map onto the same character.

a. For an english alphabet of 26 letters, there are unique 1-1 affine ciphers

b. For a Hirangana alphabet, there are unique 1-1 affine ciphers

2. Hypothesis: Assume that B is the encryption of E since it is the most frequent. Numerically, this is expressed as

We get two linear equations and two unknowns:

This system has a unique solution , , and is also valid since , so it is at least a legal key. It remains to compute the decryption function corresponding to the

determined key, , and then decrypt the ciphertext to confirm the validity of the key. The decryption function:

But we need to solve for first…

Now we can verify if the decryption function is correct for :

3. To prove an affine cipher has perfect secrecy, we would need to show that: for all and

If each key is used with equal probability, then the probability that a key is used for a given

transformation from plaintext to cipher text is:

Solving for the event that plaintext x is encrypted as cipher text y:

We would then need to compute the probability distribution on C with and we should get:

Finally, using Bayes’ theorem, and plugging in for

4. The assumption is required because no two distinctly different keys can map a plaintext character to the same cipher text character. If two characters were mapped to the same cipher text

character, then decrypting the cipher text character would violate the invertible encryption

function that must be injective. Otherwise the encrypted information will be lost

5. This system violates the perfect secrecy theorem since the probability of each cipher text is not uniformly distributed amongst the variables such that:

Instead, two examples are calculated to show this violation:

Perfect secrecy has been violated since

.

6. NO SOLUTION

End of Assignment 1

1. We have to find the prime factors, p and q such that . The values of and are these prime factorizations. To find the private key, d we need to use the Euclidean algorithm on with :

2. Syntax:

//when the current exponent bit is set. //on each iteration of bit from right to left

// bit is not set //square base mod 1234 // bit is not set // bit is not set // bit is set // bit is set // bit is not set // bit is set // bit is not set // bit is not set // bit is set

3.

Original RSA defines:

, and

b is defined by . The only thing that differs with the modified RSA is the mod n value. Let a letter of cipher text be defined

by:

Then there exists an inverse function for a letter of plain text defined by:

Combining the two we get:

Since then where If , then

;

; , and

Then , or in English, x is the decrypted plaint text of the encrypted x for the first function, and y is the encryption of the decrypted cipher text of y for the second.

With our example we get: Solving for a in original RSA:

Solving for a in modified RSA:

4. Proof by Example:

Using the RSA values from Question #1:

Define the two plain text characters and as ‘D’ and ‘P’ respectively, or as and .

We will prove true for both sides of the equation of .

Since we have:

is true.

Bonus: For good measure, let us test the decryption method for , and choose and

which is obvious that

We have already calculated . Decrypting this value we get:

This means we know the secret keys used in the decryption of the cipher text. Since the RSA cipher is

one-to-one, no two values can map back to the same pre-image. If they did, there would be no way to

decrypt the cipher text. Just to confirm, we solve for x:

is the correct pre-image of the cipher text of , or that .

5. Euler’s Phi-Function , is defined as the number of positive integers a such that

and . Then, we have that since n is prime. It follows that all integers less

than n are going to be relatively prime to n. If this property did not exist, n would not be prime and

Each of these integers will have the relationship for

, , and The same products will exist for

. The elements of are invertible which means that ,

and since , we have that .

6. Let the plain text be as bits [Image from p.97 of the text]

Since the same process is applied to ,

we have:

7. If is a collision resistant hash function then:

We have where if then , and

. This hash

function takes the bit string which is double the input length of the first hash function, chops it up in half

so that it can be used by the first hash function. shows how the two parts

of chopped up bit strings are fed into the first hash function, , and concatenated, resulting in a bit

string of length 2m. This resulting string is then used to feed again to finally give an m bit string.

Solving for :

And finally, we get . Since is clearly many applications of a hash function

that.

End of Assignment 2

1a. By definition, the Linear Congruential Generator is defined by:

We want to show that is true where:

Assume is true, then such that:

Which implies that it should hold true for .

1b.

Plugging in

into the previous equation that was derived:

We get:

1c. The period of length t may never exceed n where since:

created by the LCM, they each generate a sequence of m different ’s. The linear

congruential generator cannot generate more than m numbers, since , and ,

and . The order of a is the number of factors of the congruential generator. The

value of t may not exceed the number of factors because the generator would lack the prime factors it

would need to generate the pseudo-primes.

2.

Since requires 16 bits to represent this number, then

and . We can calculate the i-th number using the formula:

Using Modular Exponentiation, we can calculate for the 10,000th number:

Subbing in that result, we can raise to that power and calculate using modular exponentiation again:

We have a total of calculations which is less than the limit of multiplications.

3. A key for the affine cipher is represented by where where n is the size of

set of plain text and each key has equal probability of being used. The cipher text is equal probable of

size n characters. denotes the uncertainty of knowing the key if the cipher text is known.

Since each of the keys and cipher texts have equal probability of use, they can be written as:

Then , and ,

4.

5.

Primitive roots of

Alice’s public key

Alice computes:

Bob computes:

6. The points on the curve are any values of x and y for which the equation

holds true. Using the elliptic curve equation, we can calculate some points

that lie on the curve:

Quadratic residue:

x

0 6 no

1 8 no

2 5 yes 3 3 yes 4 8 no

5 4 yes

6 8 No

7 4 yes 8 9 Yes 9 7 No

10 4 Yes

=

. The next multiple would be with the same

mathematical process applied on each round. Repeating this, we get the following points:

which are two points on the curve defined by .

7. Sample output from a programming implementation of the Rho-P Algorithm:

:1 :2 :5 :26 :677 :14915 :10701 :4757 D:5

:14885 :3531 :10112 :13715 D:5 :8806 :11357 :11030 D:5 :17386 D:41

Two factors for 21115 are and A third factor can be figured out from this point by:

8. . If n is odd then and If n is not prime then for some If n is not a power of a prime then and where . Let :

1. 2. 3. 15 is not prime

4. 15 is not a power of a prime,

End of Assignment 3