1. 1IS2150/TEL2810: Introduction to Computer Security Nov 1, 2005 Introduction to Computer Security Lecture 8 Key Management

2. IS2150/TEL2810: Introduction to Computer Security2 Issues Authentication and distribution of keys Authentication and distribution of keys  Session key  Key exchange protocols  Kerberos Mechanisms to bind an identity to a key Mechanisms to bind an identity to a key Generation, maintenance and revoking of keys Generation, maintenance and revoking of keys

3. IS2150/TEL2810: Introduction to Computer Security3 Notation X  Y : { Z || W } k X,Y X  Y : { Z || W } k X,Y  X sends Y the message produced by concatenating Z and W enciphered by key k X,Y, which is shared by users X and Y A  T : { Z } k A || { W } k A,T A  T : { Z } k A || { W } k A,T  A sends T a message consisting of the concatenation of Z enciphered using k A, A’s key, and W enciphered using k A,T, the key shared by A and T r 1, r 2 nonces (nonrepeating random numbers) r 1, r 2 nonces (nonrepeating random numbers)

4. IS2150/TEL2810: Introduction to Computer Security4 Session, Interchange Keys Alice wants to send a message m to Bob Alice wants to send a message m to Bob  Assume public key encryption  Alice generates a random cryptographic key k s and uses it to encipher m To be used for this message only Called a session key  She enciphers k s with Bob’s public key k B k B enciphers all session keys Alice uses to communicate with Bob Called an interchange key  Alice sends { m } k s { k s } k B

5. IS2150/TEL2810: Introduction to Computer Security5 Benefits Limits amount of traffic enciphered with single key Limits amount of traffic enciphered with single key  Standard practice, to decrease the amount of traffic an attacker can obtain Makes replay attack less effective Makes replay attack less effective Prevents some attacks Prevents some attacks  Example: Alice will send Bob message that is either “BUY” or “SELL”.  Eve computes possible ciphertexts {“BUY”} k B and {“SELL”} k B.  Eve intercepts enciphered message, compares, and gets plaintext at once

6. IS2150/TEL2810: Introduction to Computer Security6 Key Exchange Algorithms Goal: Alice, Bob use a shared key to communicate secretly Goal: Alice, Bob use a shared key to communicate secretly Criteria Criteria  Key cannot be sent in clear Attacker can listen in Key can be sent enciphered, or derived from exchanged data plus data not known to an eavesdropper  Alice, Bob may trust third party  All cryptosystems, protocols publicly known Only secret data is the keys, ancillary information known only to Alice and Bob needed to derive keys Anything transmitted is assumed known to attacker

7. IS2150/TEL2810: Introduction to Computer Security7 Classical Key Exchange How do Alice, Bob begin? How do Alice, Bob begin?  Alice can’t send it to Bob in the clear! Assume trusted third party, Cathy Assume trusted third party, Cathy  Alice and Cathy share secret key k A  Bob and Cathy share secret key k B Use this to exchange shared key k s Use this to exchange shared key k s

8. IS2150/TEL2810: Introduction to Computer Security8 Simple Key Exchange Protocol Alice Cathy { request for session key to Bob } k A Alice Cathy { k s }k A, { k s }k B Alice Bob { k s } k B Alice Bob {m}ks{m}ks Eve

9. IS2150/TEL2810: Introduction to Computer Security9 Problems How does Bob know he is talking to Alice? How does Bob know he is talking to Alice?  Replay attack: Eve records message from Alice to Bob, later replays it; Bob may think he’s talking to Alice, but he isn’t  Session key reuse: Eve replays message from Alice to Bob, so Bob re-uses session key Protocols must provide authentication and defense against replay Protocols must provide authentication and defense against replay

10. IS2150/TEL2810: Introduction to Computer Security10 Needham-Schroeder AliceCathy Alice || Bob || r 1 AliceCathy { Alice || Bob || r 1 || k s, { Alice || k s } k B } k A AliceBob { Alice || k s } k B AliceBob { r 2 } k s AliceBob { r 2 – 1 } k s

11. IS2150/TEL2810: Introduction to Computer Security11 Argument: Alice talking to Bob Second message Second message  Enciphered using key only she, Cathy know So Cathy enciphered it  Response to first message As r 1 in it matches r 1 in first message Third message Third message  Alice knows only Bob can read it As only Bob can derive session key from message  Any messages enciphered with that key are from Bob

12. IS2150/TEL2810: Introduction to Computer Security12 Argument: Bob talking to Alice Third message Third message  Enciphered using key only he, Cathy know So Cathy enciphered it  Names Alice, session key Cathy provided session key, says Alice is other party Fourth message Fourth message  Uses session key to determine if it is replay from Eve If not, Alice will respond correctly in fifth message If so, Eve can’t decipher r 2 and so can’t respond, or responds incorrectly

13. IS2150/TEL2810: Introduction to Computer Security13 Problem with Needham-Schroeder Assumption: all keys are secret Assumption: all keys are secret Question: suppose Eve can obtain session key. How does that affect protocol? Question: suppose Eve can obtain session key. How does that affect protocol?  In what follows, Eve knows k s EveBob { Alice || k s } k B [Replay] EveBob { r 3 } k s [Eve intercepts] EveBob { r 3 – 1 } k s

14. IS2150/TEL2810: Introduction to Computer Security14 Solution: Denning-Sacco Modification In protocol above, Eve impersonates Alice In protocol above, Eve impersonates Alice Problem: replay in third step Problem: replay in third step  First in previous slide Solution: use time stamp T to detect replay Solution: use time stamp T to detect replay  Needs synchronized clocks Weakness: if clocks not synchronized, may either reject valid messages or accept replays Weakness: if clocks not synchronized, may either reject valid messages or accept replays  Parties with either slow or fast clocks vulnerable to replay  Resetting clock does not eliminate vulnerability

15. IS2150/TEL2810: Introduction to Computer Security15 Needham-Schroeder with Denning-Sacco Modification AliceCathy Alice || Bob || r 1 AliceCathy { Alice || Bob || r 1 || k s || { Alice || T || k s } k B } k A AliceBob { Alice || T || k s } k B AliceBob { r 2 } k s AliceBob { r 2 – 1 } k s

16. IS2150/TEL2810: Introduction to Computer Security16 Otway-Rees Protocol Corrects problem Corrects problem  That is, Eve replaying the third message in the protocol Does not use timestamps Does not use timestamps  Not vulnerable to the problems that Denning- Sacco modification has Uses integer n to associate all messages with a particular exchange Uses integer n to associate all messages with a particular exchange

17. IS2150/TEL2810: Introduction to Computer Security17 The Protocol AliceBob n || Alice || Bob || { r 1 || n || Alice || Bob } k A CathyBob n || Alice || Bob || { r 1 || n || Alice || Bob } k A || { r 2 || n || Alice || Bob } k B CathyBob n || { r 1 || k s } k A || { r 2 || k s } k B AliceBob n || { r 1 || k s } k A

18. IS2150/TEL2810: Introduction to Computer Security18 Argument: Alice talking to Bob Fourth message Fourth message  If n matches first message, Alice knows it is part of this protocol exchange  Cathy generated k s because only she, Alice know k A  Enciphered part belongs to exchange as r 1 matches r 1 in encrypted part of first message

19. IS2150/TEL2810: Introduction to Computer Security19 Argument: Bob talking to Alice Third message Third message  If n matches second message, Bob knows it is part of this protocol exchange  Cathy generated k s because only she, Bob know k B  Enciphered part belongs to exchange as r 2 matches r 2 in encrypted part of second message

20. IS2150/TEL2810: Introduction to Computer Security20 Replay Attack Eve acquires old k s, message in third step Eve acquires old k s, message in third step  n || { r 1 || k s } k A || { r 2 || k s } k B Eve forwards appropriate part to Alice Eve forwards appropriate part to Alice  Alice has no ongoing key exchange with Bob: n matches nothing, so is rejected  Alice has ongoing key exchange with Bob: n does not match, so is again rejected If replay is for the current key exchange, and Eve sent the relevant part before Bob did, Eve could simply listen to traffic; no replay involved

21. IS2150/TEL2810: Introduction to Computer Security21 Kerberos Authentication system Authentication system  Based on Needham-Schroeder with Denning-Sacco modification  Central server plays role of trusted third party (“Cathy”) Ticket (credential) Ticket (credential)  Issuer vouches for identity of requester of service Authenticator Authenticator  Identifies sender Alice must Alice must 1.Authenticate herself to the system 2.Obtain ticket to use server S

22. IS2150/TEL2810: Introduction to Computer Security22

23. IS2150/TEL2810: Introduction to Computer Security23 Overview User u authenticates to Kerberos server User u authenticates to Kerberos server  Obtains ticket T u,TGS for ticket granting service (TGS) User u wants to use service s: User u wants to use service s:  User sends authenticator A u, ticket T u,TGS to TGS asking for ticket for service  TGS sends ticket T u,s to user  User sends A u, T u,s to server as request to use s Details follow Details follow

24. IS2150/TEL2810: Introduction to Computer Security24 Ticket Credential saying issuer has identified ticket requester Credential saying issuer has identified ticket requester Example ticket issued to user u for service s Example ticket issued to user u for service s T u,s = s || { u || u’s address || valid time || k u,s } k s where:  k u,s is session key for user and service  Valid time is interval for which the ticket is valid  u’s address may be IP address or something else Note: more fields, but not relevant here

25. IS2150/TEL2810: Introduction to Computer Security25 Authenticator Credential containing identity of sender of ticket Credential containing identity of sender of ticket  Used to confirm sender is entity to which ticket was issued Example: authenticator user u generates for service s Example: authenticator user u generates for service s A u,s = { u || generation time || k t } k u,s where:  k t is alternate session key  Generation time is when authenticator generated Note: more fields, not relevant here

26. IS2150/TEL2810: Introduction to Computer Security26 Protocol userAS user || TGS { k u,TGS } k u || T u,TGS userTGS service || A u,TGS || T u,TGS userTGS user || { k u,s } k u,TGS || T u,s userservice A u,s || T u,s userservice { t + 1 } k u,s Authentication server userAS

27. IS2150/TEL2810: Introduction to Computer Security27 Analysis First two steps get user ticket to use TGS First two steps get user ticket to use TGS  User u can obtain session key only if u knows key shared with Cathy Next four steps show how u gets and uses ticket for service s Next four steps show how u gets and uses ticket for service s  Service s validates request by checking sender (using A u,s ) is same as entity ticket issued to  Step 6 optional; used when u requests confirmation

28. IS2150/TEL2810: Introduction to Computer Security28 Problems Relies on synchronized clocks Relies on synchronized clocks  If not synchronized and old tickets, authenticators not cached, replay is possible Tickets have some fixed fields Tickets have some fixed fields  Dictionary attacks possible  Kerberos 4 session keys weak (had much less than 56 bits of randomness); researchers at Purdue found them from tickets in minutes

29. IS2150/TEL2810: Introduction to Computer Security29 Public Key Key Exchange Here interchange keys known Here interchange keys known  e A, e B Alice and Bob’s public keys known to all  d A, d B Alice and Bob’s private keys known only to owner Simple protocol Simple protocol  k s is desired session key Alice Bob { k s } e B

30. IS2150/TEL2810: Introduction to Computer Security30 Problem and Solution Vulnerable to forgery or replay Vulnerable to forgery or replay  Because e B known to anyone, Bob has no assurance that Alice sent message Simple fix uses Alice’s private key Simple fix uses Alice’s private key  k s is desired session key Alice Bob { { k s } d A } e B

31. IS2150/TEL2810: Introduction to Computer Security31 Notes Can include message enciphered with k s Can include message enciphered with k s Assumes Bob has Alice’s public key, and vice versa Assumes Bob has Alice’s public key, and vice versa  If not, each must get it from public server  If keys not bound to identity of owner, attacker Eve can launch a man-in-the-middle attack (next slide; Peter is public server providing public keys)

32. IS2150/TEL2810: Introduction to Computer Security32 Man-in-the-Middle Attack Alice Peter send me Bob’s public key Eve Peter send me Bob’s public key Eve Peter eBeB Alice eEeE Eve Alice Bob { k s } e E Eve Bob { k s } e B Eve intercepts request Eve intercepts message

33. IS2150/TEL2810: Introduction to Computer Security33 Key Generation Goal: generate difficult to guess keys Goal: generate difficult to guess keys Problem statement: given a set of K potential keys, choose one randomly Problem statement: given a set of K potential keys, choose one randomly  Equivalent to selecting a random number between 0 and K–1 inclusive Why is this hard: generating random numbers Why is this hard: generating random numbers  Actually, numbers are usually pseudo-random, that is, generated by an algorithm

34. IS2150/TEL2810: Introduction to Computer Security34 What is “Random”? Sequence of cryptographically random numbers: a sequence of numbers n 1, n 2, … such that for any integer k > 0, an observer cannot predict n k even if all of n 1, …, n k–1 are known Sequence of cryptographically random numbers: a sequence of numbers n 1, n 2, … such that for any integer k > 0, an observer cannot predict n k even if all of n 1, …, n k–1 are known  Best: physical source of randomness Electromagnetic phenomena Characteristics of computing environment such as disk latency Ambient background noise

35. IS2150/TEL2810: Introduction to Computer Security35 What is “Pseudorandom”? Sequence of cryptographically pseudorandom numbers: sequence of numbers intended to simulate a sequence of cryptographically random numbers but generated by an algorithm Sequence of cryptographically pseudorandom numbers: sequence of numbers intended to simulate a sequence of cryptographically random numbers but generated by an algorithm  Very difficult to do this well Linear congruential generators [n k = (an k–1 + b) mod n] broken (a, b and n are relatively prime) Polynomial congruential generators [n k = (a j n k–1 j + … + a 1 n k–1 a 0 ) mod n] broken too Here, “broken” means next number in sequence can be determined

36. IS2150/TEL2810: Introduction to Computer Security36 Best Pseudorandom Numbers Strong mixing function: function of 2 or more inputs with each bit of output depending on some nonlinear function of all input bits Strong mixing function: function of 2 or more inputs with each bit of output depending on some nonlinear function of all input bits  Examples: DES, MD5, SHA-1  Use on UNIX-based systems: (date; ps gaux) | md5 where “ps gaux” lists all information about all processes on system

37. IS2150/TEL2810: Introduction to Computer Security37 Cryptographic Key Infrastructure Goal: bind identity to key Goal: bind identity to key Classical Crypto: Classical Crypto:  Not possible as all keys are shared Public key Crypto: Public key Crypto:  Bind identity to public key  Crucial as people will use key to communicate with principal whose identity is bound to key  Erroneous binding means no secrecy between principals  Assume principal identified by an acceptable name

38. IS2150/TEL2810: Introduction to Computer Security38 Certificates Create token (message) containing Create token (message) containing  Identity of principal (here, Alice)  Corresponding public key  Timestamp (when issued)  Other information (perhaps identity of signer) signed by trusted authority (here, Cathy) C A = { e A || Alice || T } d C C A is A’s certificate

39. IS2150/TEL2810: Introduction to Computer Security39 Use Bob gets Alice’s certificate Bob gets Alice’s certificate  If he knows Cathy’s public key, he can decipher the certificate When was certificate issued? Is the principal Alice?  Now Bob has Alice’s public key Problem: Bob needs Cathy’s public key to validate certificate Problem: Bob needs Cathy’s public key to validate certificate  Problem pushed “up” a level  Two approaches: Merkle’s tree, signature chains

40. IS2150/TEL2810: Introduction to Computer Security40 Merkle’s Tree Scheme Keep certificates in a file Keep certificates in a file  Changing any certificate changes the file  Use crypto hash functions to detect this (data integrity) Define hashes recursively Define hashes recursively  h is hash function  Y i is certificate I (identity, Public key) Hash of file (h(1,4) in example) known to all Hash of file (h(1,4) in example) known to all h(1,4) h(1,2) h(3,4) h(1,1) h(2,2) h(3,3) h(4,4) Y 1 Y 2 Y 3 Y 4

41. IS2150/TEL2810: Introduction to Computer Security41 Details f: D  D  D maps bit strings to bit strings f: D  D  D maps bit strings to bit strings h: N  N  D maps integers to bit strings h: N  N  D maps integers to bit strings  if i ≥ j, h(i, j) = f( Y i, Y j )  if i < j, h(i, j) = f(h(i,  (i+j)/2  ), h(  (i+j)/2  +1, j))

42. IS2150/TEL2810: Introduction to Computer Security42 Y 1 Y 2 Y 3 Y 4 Validation To validate 1 : To validate Y 1 :  Compute h(1, 1)  Obtain h(2, 2)  Compute h(1, 2)  Obtain h(3, 4)  Compute h(1,4)  Compare to known h(1, 4) Need to know hashes of children of nodes on path that are not computed Need to know hashes of children of nodes on path that are not computed h(1,4) h(1,2) h(3,4) h(1,1) h(2,2) h(3,3) h(4,4)

43. IS2150/TEL2810: Introduction to Computer Security43 Problem File must be available for validation File must be available for validation  Otherwise, can’t recompute hash at root of tree  Intermediate hashes would do Not practical in most circumstances Not practical in most circumstances  Too many certificates and users  Users and certificates distributed over widely separated systems

44. IS2150/TEL2810: Introduction to Computer Security44 Certificate Signature Chains Create certificate Create certificate  Generate hash of certificate  Encipher hash with issuer’s private key Validate Validate  Obtain issuer’s public key  Decipher enciphered hash  Recompute hash from certificate and compare Problem: Problem:  Validating the certificate of the issuer and getting issuer’s public key

45. IS2150/TEL2810: Introduction to Computer Security45 X.509 Chains Key certificate fields in X.509v3: Key certificate fields in X.509v3:  Version  Serial number(unique)  Signature algorithm identifier  Issuer’s name; uniquely identifies issuer  Interval of validity  Subject’s name; uniquely identifies subject  Subject’s public key  Signature: Identifies algorithm used to sign the certificate Signature (enciphered hash)

46. IS2150/TEL2810: Introduction to Computer Security46 X.509 Certificate Validation Obtain issuer’s public key Obtain issuer’s public key  The one for the particular signature algorithm Decipher signature Decipher signature  Gives hash of certificate Recompute hash from certificate and compare Recompute hash from certificate and compare  If they differ, there’s a problem Check interval of validity Check interval of validity  This confirms that certificate is current

47. IS2150/TEL2810: Introduction to Computer Security47 Issuers Certification Authority (CA): entity that issues certificates Certification Authority (CA): entity that issues certificates  Multiple issuers pose validation problem  Alice’s CA is Cathy; Bob’s CA is Dan; how can Alice validate Bob’s certificate?  Have Cathy and Don cross-certify Each issues certificate for the other

48. IS2150/TEL2810: Introduction to Computer Security48 Validation and Cross-Certifying Certificates: Certificates:  Cathy > represents the certificate that C has generated for A  Dan  Cathy >  Dan > Alice validates Bob’s certificate Alice validates Bob’s certificate  Alice obtains Cathy >  Alice uses (known) public key of Cathy to validate Cathy >  Alice uses Cathy > to validate Dan > Cathy > Dan > is a signature chain  How about Bob validating Alice?

49. IS2150/TEL2810: Introduction to Computer Security49 PGP Chains Pretty Good Privacy: Pretty Good Privacy:  Widely used to provide privacy for electronic mail  Sign files digitally OpenPGP certificates structured into packets OpenPGP certificates structured into packets  One public key packet  Zero or more signature packets Public key packet: Public key packet:  Version (3 or 4; 3 compatible with all versions of PGP, 4 not compatible with older versions of PGP)  Creation time  Validity period (not present in version 3)  Public key algorithm, associated parameters  Public key

50. IS2150/TEL2810: Introduction to Computer Security50 OpenPGP Signature Packet Version 3 signature packet Version 3 signature packet  Version (3)  Signature type (level of trust)  Creation time (when next fields hashed)  Signer’s key identifier (identifies key to encipher hash)  Public key algorithm (used to encipher hash)  Hash algorithm  Part of signed hash (used for quick check)  Signature (enciphered hash using signer’s private key) Version 4 packet more complex Version 4 packet more complex

51. IS2150/TEL2810: Introduction to Computer Security51 Signing Single certificate may have multiple signatures Single certificate may have multiple signatures Notion of “trust” embedded in each signature Notion of “trust” embedded in each signature  Range from “untrusted” to “ultimate trust”  Signer defines meaning of trust level (no standards!) All version 4 keys signed by subject All version 4 keys signed by subject  Called “self-signing”

52. IS2150/TEL2810: Introduction to Computer Security52 Validating Certificates Alice needs to validate Bob’s OpenPGP cert Alice needs to validate Bob’s OpenPGP cert  Does not know Fred, Giselle, or Ellen Alice gets Giselle’s cert Alice gets Giselle’s cert  Knows Henry slightly, but his signature is at “casual” level of trust Alice gets Ellen’s cert Alice gets Ellen’s cert  Knows Jack, so uses his cert to validate Ellen’s, then hers to validate Bob’s Bob Fred Giselle Ellen Irene Henry Jack Arrows show signatures Self signatures not shown

53. IS2150/TEL2810: Introduction to Computer Security53 Stream and Block Cipher Block cipher (E k is encryption) Block cipher (E k is encryption)  m = b 1 b 2 …., where b i is of a fixed length  E k (m) = E k (b 1 )E k (b 2 )…  DES is a block cipher (64 bit blocks) Stream cipher (E k is encryption) Stream cipher (E k is encryption)  m = b 1 b 2 …., where b i is of a fixed length  k = k 1 k 2 ….  E k (m) = E k1 (b 1 )E k2 (b 2 )…  Vinegere cipher, one-time pad

54. IS2150/TEL2810: Introduction to Computer Security54 Digital Signature Construct that authenticates origin, contents of message in a manner provable to a disinterested third party (“judge”) Construct that authenticates origin, contents of message in a manner provable to a disinterested third party (“judge”) Sender cannot deny having sent message (service is “nonrepudiation”) Sender cannot deny having sent message (service is “nonrepudiation”)  Limited to technical proofs Inability to deny one’s cryptographic key was used to sign  One could claim the cryptographic key was stolen or compromised Legal proofs, etc., probably required;

55. IS2150/TEL2810: Introduction to Computer Security55 Common Error Classical: Alice, Bob share key k Classical: Alice, Bob share key k  Alice sends m || { m }k to Bob  Does this satisfy the requirement for message authentication? How?  Does this satisfy the requirement for a digital signature? This is not a digital signature This is not a digital signature  Why? Third party cannot determine whether Alice or Bob generated message

56. IS2150/TEL2810: Introduction to Computer Security56 Classical Digital Signatures Require trusted third party Require trusted third party  Alice, Bob each share keys with trusted party Cathy The judge must trust the trusted party Cathy The judge must trust the trusted party Cathy To resolve dispute, judge gets { m }k Alice, { m }k Bob, and has Cathy decipher them; if messages matched, contract was signed, else one is a forgery To resolve dispute, judge gets { m }k Alice, { m }k Bob, and has Cathy decipher them; if messages matched, contract was signed, else one is a forgery Alice Bob Cathy Bob { m }k Alice { m }k Bob

57. IS2150/TEL2810: Introduction to Computer Security57 Public Key Digital Signatures (RSA) Alice’s keys are d Alice, e Alice Alice’s keys are d Alice, e Alice Alice sends Bob Alice sends Bob m || { m }d Alice In case of dispute, judge computes In case of dispute, judge computes { { m }d Alice }e Alice and if it is m, Alice signed message and if it is m, Alice signed message  She’s the only one who knows d Alice !

58. IS2150/TEL2810: Introduction to Computer Security58 RSA Digital Signatures Use private key to encipher message Use private key to encipher message  Protocol for use is critical Key points: Key points:  Never sign random documents, and when signing, always sign hash and never document Mathematical properties can be turned against signer  Sign message first, then encipher Changing public keys causes forgery

59. IS2150/TEL2810: Introduction to Computer Security59 Attack #1 Example: Alice, Bob communicating Example: Alice, Bob communicating  n A = 95, e A = 59, d A = 11  n B = 77, e B = 53, d B = 17 26 contracts, numbered 00 to 25 26 contracts, numbered 00 to 25  Alice has Bob sign 05 and 17: c = m d B mod n B = 05 17 mod 77 = 3 c = m d B mod n B = 17 17 mod 77 = 19  Alice computes 05  17 mod 77 = 08; corresponding signature is 03  19 mod 77 = 57; claims Bob signed 08  Note: [(a mod n) × (b mod n)] mod n = (a × b) mod n  Judge computes c e B mod n B = 57 53 mod 77 = 08 Signature validated; Bob is toast!

60. IS2150/TEL2810: Introduction to Computer Security60 Attack #2: Bob’s Revenge Bob, Alice agree to sign contract 06 Bob, Alice agree to sign contract 06 Alice enciphers, then signs: Alice enciphers, then signs:  Enciper: c = m e B mod n B = (06 53 mod 77) 11  Sign: c d A mod n A = (06 53 mod 77) 11 mod 95 = 63 Bob now changes his public key Bob now changes his public key  Bob wants to claim that Alice singed N (13)  Computes r such that 13 r mod 77 = 6; say, r = 59  Computes r.e B mod  (n B ) = 59  53 mod 60 = 7  Replace public key e B with 7, private key d B = 43 Bob claims contract was 13. Judge computes: Bob claims contract was 13. Judge computes:  (63 59 mod 95) 43 mod 77 = 13  Verified; now Alice is toast Solution: sign first and then enciher!! Solution: sign first and then enciher!!

61. IS2150/TEL2810: Introduction to Computer Security61 El Gamal Digital Signature Relies on discrete log problem Relies on discrete log problem Choose p prime, g, d < p; Choose p prime, g, d < p; Compute y = g d mod p Compute y = g d mod p Public key: (y, g, p); private key: d Public key: (y, g, p); private key: d To sign contract m: To sign contract m:  Choose k relatively prime to p–1, and not yet used  Compute a = g k mod p  Find b such that m = (da + kb) mod p–1  Signature is (a, b) To validate, check that To validate, check that  y a a b mod p = g m mod p

62. IS2150/TEL2810: Introduction to Computer Security62 Example Alice chooses p = 29, g = 3, d = 6 Alice chooses p = 29, g = 3, d = 6 y = 3 6 mod 29 = 4 Alice wants to send Bob signed contract 23 Alice wants to send Bob signed contract 23  Chooses k = 5 (relatively prime to 28)  This gives a = g k mod p = 3 5 mod 29 = 11  Then solving 23 = (6  11 + 5b) mod 28 gives b = 25  Alice sends message 23 and signature (11, 25) Bob verifies signature: g m mod p = 3 23 mod 29 = 8 and y a a b mod p = 4 11 11 25 mod 29 = 8 Bob verifies signature: g m mod p = 3 23 mod 29 = 8 and y a a b mod p = 4 11 11 25 mod 29 = 8  They match, so Alice signed

63. IS2150/TEL2810: Introduction to Computer Security63 Attack Eve learns k, corresponding message m, and signature (a, b) Eve learns k, corresponding message m, and signature (a, b)  Extended Euclidean Algorithm gives d, the private key Example from above: Eve learned Alice signed last message with k = 5 Example from above: Eve learned Alice signed last message with k = 5 m = (da + kb) mod p–1 = 23 =(11d + 5  25) mod 28 So Alice’s private key is d = 6