Secure Communication

Definition:A cryptosystem is a five-tuple (P,C,K,E,D) where the following conditions are satisfied:

1. P is a finite set of possible plaintexts:

2. C is a finite set of possible ciphertexts:

3. K is the keyspace, is the a finite set of possible keys:

4. For each K∈ K there is an encryption rule e_K∈E and a corresponding decryption rule d_K∈D. Where

e_K:P→C and d_K:C→P are functions such that for all M∈P, we have d_K(e_K(M))=M

Caesar cipher- Shift K letters.
P and C are the alphabet. e_K is shift K letters and d_K is shift back K.

Alternative 1: Use permutations rather than shifts.
      26! = 403291461126605635584000000

Alternative 2: Use a fixed sequence of Caesar_cipher shifts. Say (5,7,3).
      The first letter gets shifted by 5, the second by 7 and the third by 3. The fourth is again shifted by 5.
      There are 26ⁿ possibilities for a sequence of n shifts.
      For example, if n=30, there are 2813198901284745919258621029615971520741376 posibilities.

A Useful calculator.

The Tests

If I know P and C and the algorithms e and d but not K∈ K , and I also have captured an encrypted message can I compute K and hence e_K and d_K?

If I know P and C and e and d but not K∈ K , and I also somehow have access to e_K so I can encrypt test messages. But, I do not have access to d_K. Can I compute K and hence d_K?

Secret Key Exchange
To maintain two-way private communication between individuals, assuming both individuals know the plaintext and cybertext, they just have to share a secret key. For the Caesar cipher, simply know the shift number.
An assumption is that observers can see the encrypted conversation, maybe even understand the encryption algorithm, but cannot somehow create "experiments" with the secret key.
The Advanced Encryption Standard (AES)
AES is a symmetric 128-bit block data encryption technology adopted by the U.S government as a standard in 2000.

Public Key Encryption

Sometimes sharing a key is not practical, for example consider taking a credit card over the Web.

Suppose, however, that it were possible to find a cryptosystem for which knowing e , and the general methodology used in its construction, did not lead to an easy computation of d , then we could do the following.

Secure One-Way Communication: (eg Web Form)

1. Publish e for the world to see. Tell the world that, if they want to communicate securely with you, all they need to is apply e to the message before transmitting it. This because there is an acceptably small chance of someone discovering d hence decoding there message.

2. When I received the encrypted message apply d which, presumably only I know.

Secure Two-Way Communication:

Assume that we are dealing with a Cryptosystem such that

$\QTR{bs}{P=C}$

In addition to d(e(M )) $\fallingdotseq$ M, we have e(d(C )) $\fallingdotseq$ C for all C $\in \QTR{bs}{C=P}$ .

Given d(), it is also very hard to compute e().

Suppose that we have two people P and P who want to communicate securely with each other. Each selects their own "one way system", K and K , from a Cryptosystem with the above listed properties .

P and P commmunicate as follows:

1. P gives e to P and P gives e to P.

2. Suppose P wants to send message M to P . P computes C e (d (M )) and transmits it.

3. P computes e (d (C )) e (d (e (d (M )))) e (d (M )) M.

Why does this work?

P knows that the only person who can read the message is P , the owner of K since, presumably P , is the only person who knows d ().

P knows that P sent the message since, presumably P , is the only person who knows d (). We are implicitly assuming that the message P

sees is meaningful.

An Example of a One Way Cryptosystem

RSA ( The RSA algorithm was invented in 1978 by Ron Rivest, Adi Shamir, and Leonard Adleman):

1. Begin by choosing $\QTR{Large}{p}$ and $\QTR{Large}{q}$ to be two very large prime numbers.

2. Next choose $\QTR{Large}{e}$ , , such that $\QTR{Large}{e}$ and are relatively prime.
3. Referring to a topic we will cover , we can find $\QTR{Large}{d}$ such that . ( ed = a(p-1)(q-1) + 1 ) Note that $\QTR{Large}{e}$ and $\QTR{Large}{d}$ are "symmetric" for two way communication.
4. We will also need to know that for any M, M^(p-1)(q-1)=1 mod pq
5. Here is RSA

$\QTR{bs}{P=C}$ is the set of integers between $\QTR{Large}{1}$ and $\QTR{Large}{pq=n}$ and relatively prime to $\QTR{Large}{n}$ .

$\QTR{bf}{K}$ the keyspace, is the set of pairs $\QTR{Large}{e}$ , $\QTR{Large}{d}$ as above:

For each K in $\QTR{bs}{K}$ and all M $\in \QTR{bs}{P}$ , e(M )) $\fallingdotseq$ M $^{\QTR{Large}{e}}$ =C and d(C ) $\fallingdotseq$ C $^{\QTR{Large}{d}}$

Note: d (e (M)) $\fallingdotseq$ M M (M ) $^{\QTR{Large}{a}}$ M $\fallingdotseq$ (1) $^{\QTR{Large}{a}}$ M $\fallingdotseq$ M

And d (e (M))=M.

Example- Let $\QTR{Large}{p=47}$ and $\QTR{Large}{q=53}$ .

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and . So . Choose $\QTR{Large}{e=35}$ . Note

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Somehow Compute

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Suppose M $\QTR{Large}{=25}$ $\$ Check

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Observation:
A reasonable question that could be asked is, while it may be hard to factor $\QTR{Large}{n}$ all we really need to do is find $\QTR{Large}{d}$ such that , so given, , is there a way to compute ?

The answer is that it is as "hard" to compute from $\QTR{Large}{n}$ as it is to factor $\QTR{Large}{n}$ it self. Here is the argument.

1. For the sake of clarity, set . So if we know $\QTR{Large}{p}$ and $\QTR{Large}{q}$ we can quickly compute $\QTR{Large}{m}$ .

Next the important direction.

2. Suppose there was an easy way to compute $\QTR{Large}{m}$ from $\QTR{Large}{n}$ . To factor $\QTR{Large}{n}$ , we would then only have to solve the two simultaneous equations.

in two unknowns $\QTR{Large}{p}$ and $\QTR{Large}{q}$ .

Solving the first equation for p gives.

substituting this into the second equation gives.

or

.

The quadratic formula does the rest.