Network Security Section 3: Public Key, Digital Signature.

Network Security

Section 3: Public Key,

Digital Signature

New Algorithm Requirements

• Definitions:– E = Encryption Key– D = Decryption Key

• New Requirements:1. D(E(P)) = P

2. E #> D

3. E not crack-able by “known texts” attack.

Example

AliceEa, Da

BobEb,Db

Eb

Ea

ABCDEFGHI

ABCDEFGHI

P Eb(P) ******************

******************

Send to Bob

******************

******************

Eb(P)D(Eb(P))ABCDEFGHI

ABCDEFGHI

Bob reads P

Rivest

ShmirAdelman

RSA

RSA

• One of the public key algorithms• RSA Algorithm:

1. Chose two number p & q (1024bit)2. n=p×q and z=(p-1)×(q-1)3. Choose a number d that is relatively prime to z4. e: e×d mod z = 15. Divde P to blocks, 0 <= block length < n6. C = Pe mod n7. Exit.

• OK. Where is the security location?

RSA Example

• p = 3, q = 11

• n = 33, z = 20, d = 7, e = 3OK!

Lets finish cryptography algorithmsHOOORAY!

No more cryptography

algorithms PLEASE!

Digital Signatures

• Why do we use signatures?

Authorization and Validity

• What is the problem of signature in digital world?

1. Authorize sender

2. Message must be undeniable from sender’s prospective.

3. Receiver can not produce fake messages.

Symmetric-Key Signatures

• Store signature on valid institute (BB: Big Brother).

• What is the problem of this method?

Do you trust

Big Brother?

Public-Key SignaturesE(D(P)) = PD(E(P)) = P

Can Alice evade message P?

Yes! How?

No! Why?

1. Be lost Da!!2. Changing Da!!

Message Digests

• Digital Signature do both authorization and confidentiality of message

• Message Digests only authorize messages.

• MD features:– Easy Calculation of MD(P)– MD(P) #> P– MD(P) # MD(Q)– MD(P) # MD(P+1)

MD5• MD5: 5th Message Digest. 128bit buffer• md5(apple) = 1f3870be274f6c49b3e31a0c6728957f

SHA-1

• Secure Hash Algorithm

• Developed by NSA

• 160bit buffer

The Birthday attack

• Problem: If it is easy to find two random messages that map to the same signature then a birthday attack is easy

• Example: the probability of 2 people having the same birthday in a group of 23 people is more than 0.5