Discrete Maths 2003 Lecture 38 3 Slides Pp

8/13/2019 Discrete Maths 2003 Lecture 38 3 Slides Pp

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Lecture 38, 23-October-200Discrete Mathematics 2003

1

Introduction

• Last lecture – we looked at public key encryption

• In public key encryption there are 2 types of keys – a private key & a public key

• The private key is kept by the receiver • The public key is announced by the receiver to

anyone who wants it (e.g. the sender of themessage – or anyone else!)

• Thus there are no security issues associated withthe distribution of keys

• When A wants to send a message to B, Aencrypts the message using B’s public key

• B decrypts the message using their private key

2

The RSA Algorithm• The most common public key encryption

method in use today is the RSA algorithm

• In RSA, keys are generated as follows:

1. A key centre (which generates & distributeskeys) chooses 2 distinct large prime nos p & q,and multiplies them: n = pq

2. The number m = ( p – 1)(q – 1) is then calculated

3. A natural number x (between 1 and m) is chosenso that x & m are coprime

4. Then the centre computes an integer y so that xy = 1 mod m

5. The participant gets the nos x & n as their public key, & y & n as their private key

3

Sending a Message with RSA• Once a participant has their public & private

keys, they receive a message as follows:

1. The sender converts the message to bits, breaksit into substrings of the same length, then eachsubstring (which is an integer a that must be < n)is encrypted using C = a x mod n.

C is the ciphertext – note the formula uses thereceiver’s public key, which is freely available.

2. The receiver then deciphers C back to a plaintextcharacter P by P = C y mod n.

This formula requires knowledge of the privatekey, which is known only to the receiver.




4

Example of Sending a Message

with RSA

• Last lecture – we showed that a valid pair of

keys for RSA is x = 3 & n = 33 (public key),and y = 7 & n = 33 (private key)

• Note that the numbers used in this example are

much smaller than they would be if we were

encrypting information in a practical situation

• Example: Use the above keys to encrypt the

message g (i.e. 7), & show the correct message

is recovered after decryption

5

Why does the RSA Method Work?

• The theoretical basis that ensures the RSA

method always works comes from an area of

mathematics known as number theory (which is

the study of properties of integers)

• Note that the process we described for

generating keys has to be followed closely

• For example, we stated that it was necessary to

start with 2 distinct prime nos p & q, & multiply

them to obtain the number n that appears in both

the public and private keys: n = pq

6

RSA with non-Prime Starting Nos• What happens if p & q are not both prime?

• Suppose we generate keys with p = 3 & q = 8(note that q is not prime)

• Then x = 3 & n = 24 (public), and y = 5 & n = 24

(private) are suitable keys• If the message is the number 3, it is recovered by

the receiver as the number 3 (as we’d expect)

• However, the message 2 is recovered as 8

• Similarly, the message 6 is recovered as 0

• So the RSA method doesn’t necessarily workwith non-prime values of p & q




7

Message Security with RSA

• A message sent using RSA can be deciphered by

an intruder if they are able to determine (e.g.

guess) the number y in the private key• In the earlier example, y was 7, and this would

be guessed easily with a trial-and-error approach

• In practice, though, very large nos are used for x

& y (at least tens of digits), so a trial-and-error

approach to finding y would take months or

years, even with the fastest available computers

• So an intruder is very unlikely to guess y

8

Message Security with RSA (cont)• The other way to break the cipher (other than by

stealing y) is to find the original prime nos p & qon which the public & private keys were based

• Since n = pq, and the value of n is publicallyavailable (in the public key), the cipher will be broken if we can factor n into constituent primes

• While this may sound fairly easy, it is actuallyvery difficult to find factors of large nos (and it issuggested n has at least 200-300 decimal digits)

• To illustrate this difficulty, the RSA Security Co.offers prizes of $10,000 to $200,000 (US) forfactoring certain nos, of length 174 to 617 digits

9

Become Rich by Factoring Numbers

• For example, you can earn $10,000 by factoring188198812920607963838697239461650439807163563379417382700763356422988859715234665485319060606504743045317388011303396

716199692321205734031879550656996221305168759307650257059 (which has 174 digits)

• The company offers these prizes to “encourageresearch into computational number theory andthe practical difficulty of factoring large integers”

• Given the magnitude of the prizes, it is clearly noeasy task to factor large numbers, even with thelatest computers

http://www.rsasecurity.com/rsalabs/challenges/factoring/numbers.html






10

Public Key Encryption – Advantages• The major advantage of public key encryption

over secret key encryption is the number of keys

• An individual can communicate securely with

any number of other people with just 2 keys• So, for 1 million people to communicate

securely, just 2 million keys are needed

• This contrasts with secret key encryption, whichneeded half-a-trillion keys (250,000 times more)

• As well, unlike secret key encryption, there isn’tthe need to find a secure way to distribute keys,since only the public key needs to be distributed – and this can be known to anyone

11

Public Key Encryption –

Disadvantages

• The disadvantage of the public key method isthe time taken to implement the algorithm (i.e.to encrypt and decrypt messages)

• This is because of the very large numbersneeded in the public and private keys

• So public key encryption is not recommendedfor large amounts of text

• Next lecture – we’ll look at an approach thatcombines the advantages of secret key & publickey encryption in a single method

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Discrete Maths 2003 Lecture 38 3 Slides Pp