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Network Security7-1 Chapter 8: Network Security Chapter goals: r understand principles of network security: m cryptography and its many uses beyond “confidentiality”

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Presentation on theme: "Network Security7-1 Chapter 8: Network Security Chapter goals: r understand principles of network security: m cryptography and its many uses beyond “confidentiality”"— Presentation transcript:

1 Network Security7-1 Chapter 8: Network Security Chapter goals: r understand principles of network security: m cryptography and its many uses beyond “confidentiality” m authentication m message integrity m key distribution r security in practice: m firewalls m security in applications m Internet spam, viruses, and worms

2 Network Security7-2 What is network security? Confidentiality: only sender, intended receiver should “understand” message contents m sender encrypts message m receiver decrypts message Authentication: sender, receiver want to confirm identity of each other m Virus email really from your friends? m The website really belongs to the bank?

3 Network Security7-3 What is network security? Message Integrity: sender, receiver want to ensure message not altered (in transit, or afterwards) without detection m Digital signature Nonrepudiation: sender cannot deny later that messages received were not sent by him/her Access and Availability: services must be accessible and available to users upon demand m Denial of service attacks Anonymity: identity of sender is hidden from receiver (within a group of possible senders)

4 Network Security7-4 Friends and enemies: Alice, Bob, Trudy r well-known in network security world r Bob, Alice (lovers!) want to communicate “securely” r Trudy (intruder) may intercept, delete, add messages secure sender secure receiver channel data, control messages data Alice Bob Trudy

5 Network Security7-5 Who might Bob, Alice be? r Web client/server (e.g., on-line purchases) r DNS servers r routers exchanging routing table updates r Two computers in peer-to-peer networks r Wireless laptop and wireless access point r Cell phone and cell tower r Cell phone and bluetooth earphone r RFID tag and reader r.......

6 Network Security7-6 There are bad guys (and girls) out there! Q: What can a “bad guy” do? A: a lot! m eavesdrop: intercept messages m actively insert messages into connection m impersonation: can fake (spoof) source address in packet (or any field in packet) m hijacking: “take over” ongoing connection by removing sender or receiver, inserting himself in place m denial of service: prevent service from being used by others (e.g., by overloading resources) more on this later ……

7 Network Security7-7 The language of cryptography symmetric key crypto: sender, receiver keys identical public-key crypto: encryption key public, decryption key secret (private) plaintext ciphertext K A encryption algorithm decryption algorithm Alice’s encryption key Bob’s decryption key K B

8 Network Security7-8 Classical Cryptography r Transposition Cipher r Substitution Cipher m Simple substitution cipher (Caesar cipher) m Vigenere cipher m One-time pad

9 Network Security7-9 Transposition Cipher: rail fence r Write plaintext in two rows r Generate ciphertext in column order r Example: “HELLOWORLD” HLOOL ELWRD ciphertext: HLOOLELWRD Problem: does not affect the frequency of individual symbols

10 Network Security7-10 Simple substitution cipher substituting one thing for another m Simplest one: monoalphabetic cipher: substitute one letter for another (Caesar Cipher) A B C D E F G H I J K L M N O P Q R S T U V W X Y Z D E F G H I J K L M N O P Q R S T U V W X Y Z A B C Example: encrypt “I attack”

11 Network Security7-11 Problem of simple substitution cipher r The key space for the English Alphabet is very large: 26!  4 x 10 26 r However: m Previous example has a key with only 26 possible values m English texts have statistical structure: the letter “ e ” is the most used letter. Hence, if one performs a frequency count on the ciphers, then the most frequent letter can be assumed to be “ e ”

12 Network Security7-12 Distribution of Letters in English Frequency analysis

13 Network Security7-13 Vigenere Cipher r Idea: Uses Caesar's cipher with various different shifts, in order to hide the distribution of the letters. r A key defines the shift used in each letter in the text r A key word is repeated as many times as required to become the same length Plain text: I a t t a c k Key: 2 3 4 2 3 4 2 (key is “234”) Cipher text: K d x v d g m

14 Network Security7-14 Problem of Vigenere Cipher r Vigenere is easy to break (Kasiski, 1863): r Assume we know the length of the key. We can organize the ciphertext in rows with the same length of the key. Then, every column can be seen as encrypted using Caesar's cipher. r The length of the key can be found using several methods: m 1. If short, try 1, 2, 3,.... m 2. Find repeated strings in the ciphertext. Their distance is expected to be a multiple of the length. Compute the gcd of (most) distances. m 3. Use the index of coincidence.

15 Network Security7-15 One-time Pad r Extended from Vigenere cipher r Key is as long as the plaintext r Key string is random chosen m Pro: Proven to be “perfect secure” m Cons: How to generate Key? How to let bob/alice share the same key pad? m Code book

16 Network Security7-16 Symmetric key cryptography symmetric key crypto: Bob and Alice share know same (symmetric) key: K r e.g., key is knowing substitution pattern in mono alphabetic substitution cipher r Q: how do Bob and Alice agree on key value? plaintext ciphertext K A-B encryption algorithm decryption algorithm A-B K plaintext message, m K (m) A-B K (m) A-B m = K ( ) A-B

17 Network Security7-17 Symmetric key crypto: DES DES: Data Encryption Standard r US encryption standard [NIST 1993] r 56-bit symmetric key, 64-bit plaintext input r How secure is DES? m DES Challenge: 56-bit-key-encrypted phrase (“Strong cryptography makes the world a safer place”) decrypted (brute force) in 4 months m no known “backdoor” decryption approach r making DES more secure (3DES): m use three keys sequentially on each datum m use cipher-block chaining

18 Network Security7-18 Symmetric key crypto: DES initial permutation 16 identical “rounds” of function application, each using different 48 bits of key final permutation DES operation

19 Network Security7-19 AES: Advanced Encryption Standard r new (Nov. 2001) symmetric-key NIST standard, replacing DES r processes data in 128 bit blocks r 128, 192, or 256 bit keys r brute force decryption (try each key) taking 1 sec on DES, takes 149 trillion years for AES

20 Network Security7-20 Block Cipher r one pass through: one input bit affects eight output bits 64-bit input T1T1 8bits 64-bit scrambler 64-bit output loop for n rounds T2T2 T3T3 T4T4 T6T6 T5T5 T7T7 T8T8 r multiple passes: each input bit affects most output bits r block ciphers: DES, 3DES, AES

21 Network Security7-21 Cipher Block Chaining r cipher block: if input block repeated, will produce same cipher text: t=1 m(1) = “HTTP/1.1” block cipher c(1) = “k329aM02” … r cipher block chaining: XOR ith input block, m(i), with previous block of cipher text, c(i-1) m c(0) transmitted to receiver in clear m what happens in “HTTP/1.1” scenario from above? + m(i) c(i) t=17 m(17) = “HTTP/1.1” block cipher c(17) = “k329aM02” block cipher c(i-1)

22 Network Security7-22 Public Key Cryptography symmetric key crypto r requires sender, receiver know shared secret key r Q: how to agree on key in first place (particularly if never “met”)? public key cryptography r radically different approach [Diffie- Hellman76, RSA78] r sender, receiver do not share secret key r public encryption key known to all r private decryption key known only to receiver

23 Network Security7-23 Public key cryptography plaintext message, m ciphertext encryption algorithm decryption algorithm Bob’s public key plaintext message K (m) B + K B + Bob’s private key K B - m = K ( K (m) ) B + B -

24 Network Security7-24 Public key encryption algorithms need K ( ) and K ( ) such that B B.. given public key K, it should be impossible to compute private key K B B Requirements: 1 2 RSA: Rivest, Shamir, Adelson algorithm + - K (K (m)) = m B B - + + -

25 Network Security7-25 RSA: Choosing keys 1. Choose two large prime numbers p, q. (e.g., 1024 bits each) 2. Compute n = pq, z = (p-1)(q-1) 3. Choose e (with e<n) that has no common factors with z. (e, z are “relatively prime”). 4. Choose d such that ed-1 is exactly divisible by z. (in other words: ed mod z = 1 ). 5. Public key is (n,e). Private key is (n,d). K B + K B -

26 Network Security7-26 RSA: Encryption, decryption 0. Given (n,e) and (n,d) as computed above 1. To encrypt bit pattern, m, compute c = m mod n e (i.e., remainder when m is divided by n) e 2. To decrypt received bit pattern, c, compute m = c mod n d (i.e., remainder when c is divided by n) d m = (m mod n) e mod n d Magic happens! c

27 Network Security7-27 RSA example: Bob chooses p=5, q=7. Then n=35, z=24. e=5 (so e, z relatively prime). d=29 (so ed-1 exactly divisible by z. letter m m e c = m mod n e l 12 1524832 17 c m = c mod n d 17 481968572106750915091411825223071697 12 c d letter l encrypt: decrypt: Computational extensive

28 Network Security7-28 RSA: Why is that m = (m mod n) e mod n d (m mod n) e mod n = m mod n d ed Useful number theory result: If p,q prime and n = pq, then: x mod n = x mod n yy mod (p-1)(q-1) = m mod n ed mod (p-1)(q-1) = m mod n 1 = m (using number theory result above) (since we chose ed to be divisible by (p-1)(q-1) with remainder 1 )

29 Network Security7-29 RSA: another important property The following property will be very useful later: K ( K (m) ) = m B B - + K ( K (m) ) B B + - = use public key first, followed by private key use private key first, followed by public key Result is the same!


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