Expressive Power How do the sets of systems that models can describe compare? If HRU equivalent to SPM, SPM provides more specific answer to safety question If HRU describes more systems, SPM applies only to the systems it can describe

HRU vs. SPM SPM more abstract HRU allows revocation Analyses focus on limits of model, not details of representation HRU allows revocation SMP has no equivalent to delete, destroy HRU allows multiparent creates SPM cannot express multiparent creates easily, and not at all if the parents are of different types because can•create allows for only one type of creator April 13, 2004 ECS 235

Multiparent Create Solves mutual suspicion problem In HRU: Create proxy jointly, each gives it needed rights In HRU: command multicreate(s0, s1, o) if r in a[s0, s1] and r in a[s1, s0] then create object o; enter r into a[s0, o]; enter r into a[s1, o]; end April 13, 2004 ECS 235

SPM and Multiparent Create can•create extended in obvious way cc  TS  …  TS  T Symbols X1, …, Xn parents, Y created R1,i, R2,i, R3, R4,i  R Rules crP,i((X1), …, (Xn)) = Y/R1,1  Xi/R2,i crC((X1), …, (Xn)) = Y/R3  X1/R4,1  …  Xn/R4,n April 13, 2004 ECS 235

Example Anna, Bill must do something cooperatively But they don’t trust each other Jointly create a proxy Each gives proxy only necessary rights In ESPM: Anna, Bill type a; proxy type p; right x  R cc(a, a) = p crAnna(a, a, p) = crBill(a, a, p) =  crproxy(a, a, p) = { Anna/x, Bill/x } April 13, 2004 ECS 235

2-Parent Joint Create Suffices Goal: emulate 3-parent joint create with 2-parent joint create Definition of 3-parent joint create (subjects P1, P2, P3; child C): cc((P1), (P2), (P3)) = Z  T crP1((P1), (P2), (P3)) = C/R1,1  P1/R2,1 crP2((P1), (P2), (P3)) = C/R2,1  P2/R2,2 crP3((P1), (P2), (P3)) = C/R3,1  P3/R2,3 April 13, 2004 ECS 235

General Approach Define agents for parents and child Agents act as surrogates for parents If create fails, parents have no extra rights If create succeeds, parents, child have exactly same rights as in 3-parent creates Only extra rights are to agents (which are never used again, and so these rights are irrelevant) April 13, 2004 ECS 235

Entities and Types Parents P1, P2, P3 have types p1, p2, p3 Child C of type c Parent agents A1, A2, A3 of types a1, a2, a3 Child agent S of type s Type t is parentage if X/t  dom(Y), X is Y’s parent Types t, a1, a2, a3, s are new types April 13, 2004 ECS 235

Can•Create Following added to can•create: cc(p1) = a1 cc(p2, a1) = a2 Parents creating their agents; note agents have maximum of 2 parents cc(a3) = s Agent of all parents creates agent of child cc(s) = c Agent of child creates child April 13, 2004 ECS 235

Creation Rules Following added to create rule: crP(p1, a1) =  crC(p1, a1) = p1/Rtc Agent’s parent set to creating parent; agent has all rights over parent crPfirst(p2, a1, a2) =  crPsecond(p2, a1, a2) =  crC(p2, a1, a2) = p2/Rtc  a1/tc Agent’s parent set to creating parent and agent; agent has all rights over parent (but not over agent) April 13, 2004 ECS 235

Creation Rules crPfirst(p3, a2, a3) =  crPsecond(p3, a2, a3) =  crC(p3, a2, a3) = p3/Rtc  a2/tc Agent’s parent set to creating parent and agent; agent has all rights over parent (but not over agent) crP(a3, s) =  crC(a3, s) = a3/tc Child’s agent has third agent as parent crP(a3, s) =  crP(s, c) = C/Rtc crC(s, c) = c/R3t Child’s agent gets full rights over child; child gets R3 rights over agent April 13, 2004 ECS 235

Link Predicates Idea: no tickets to parents until child created Done by requiring each agent to have its own parent rights link1(A1, A2) = A1/t  dom(A2)  A2/t  dom(A2) link1(A2, A3) = A2/t  dom(A3)  A3/t  dom(A3) link2(S, A3) = A3/t  dom(S)  C/t  dom(C) link3(A1, C) = C/t  dom(A1) link3(A2, C) = C/t  dom(A2) link3(A3, C) = C/t  dom(A3) link4(A1, P1) = P1/t  dom(A1)  A1/t  dom(A1) link4(A2, P2) = P2/t  dom(A2)  A2/t  dom(A2) link4(A3, P3) = P3/t  dom(A3)  A3/t  dom(A3) April 13, 2004 ECS 235

Filter Functions f1(a2, a1) = a1/t  c/Rtc f1(a3, a2) = a2/t  c/Rtc f2(s, a3) = a3/t  c/Rtc f3(a1, c) = p1/R4,1 f3(a2, c) = p2/R4,2 f3(a3, c) = p3/R4,3 f4(a1, p1) = c/R1,1  p1/R2,1 f4(a2, p2) = c/R1,2  p2/R2,2 f4(a3, p3) = c/R1,3  p3/R2,3 April 13, 2004 ECS 235

Construction Create A1, A2, A3, S, C; then P1 has no relevant tickets A1 has P1/Rtc A2 has P2/Rtc  A1/tc A3 has P3/Rtc  A2/tc S has A3/tc  C/Rtc C has C/R3 April 13, 2004 ECS 235

Construction Only link2(S, A3) true  apply f2 A3 has P3/Rtc  A2/t  A3/t  C/Rtc Now link1(A3, A2) true  apply f1 A2 has P2/Rtc  A1/tc  A2/t  C/Rtc Now link1(A2, A1) true  apply f1 A1 has P2/Rtc  A1/tc  A1/t  C/Rtc Now all link3s true  apply f3 C has C/R3  P1/R4,1  P2/R4,2  P3/R4,3 April 13, 2004 ECS 235

Finish Construction Now link4s true  apply f4 P1 has C/R1,1  P1/R2,1 P2 has C/R1,2  P2/R2,2 P3 has C/R1,3  P3/R2,3 3-parent joint create gives same rights to P1, P2, P3, C If create of C fails, link2 fails, so construction fails April 13, 2004 ECS 235

Theorem The two-parent joint creation operation can implement an n-parent joint creation operation with a fixed number of additional types and rights, and augmentations to the link predicates and filter functions. Proof: by construction, as above Difference is that the two systems need not start at the same initial state April 13, 2004 ECS 235

Theorems Monotonic ESPM and the monotonic HRU model are equivalent. Safety question in ESPM also decidable if acyclic attenuating scheme April 13, 2004 ECS 235

Expressiveness Graph-based representation to compare models Graph Vertex: represents entity, has static type Edge: represents right, has static type Graph rewriting rules: Initial state operations create graph in a particular state Node creation operations add nodes, incoming edges Edge adding operations add new edges between existing vertices April 13, 2004 ECS 235

Example: 3-Parent Joint Creation Simulate with 2-parent Nodes P1, P2, P3 parents Create node C with type c with edges of type e Add node A1 of type a and edge from P1 to A1 of type e´ April 13, 2004 ECS 235

Next Step A1, P2 create A2; A2, P3 create A3 Type of nodes, edges are a and e´ April 13, 2004 ECS 235

Next Step A3 creates S, of type a S creates C, of type c P3 P1 P2 A3 April 13, 2004 ECS 235

Last Step Edge adding operations: P1A1A2A3SC: P1 to C edge type e P2A2A3SC: P2 to C edge type e P3A3SC: P3 to C edge type e P3 P1 P2 A2 A3 A1 S C April 13, 2004 ECS 235

Definitions Scheme: graph representation as above Model: set of schemes Schemes A, B correspond if graph for both is identical when all nodes with types not in A and edges with types in A are deleted April 13, 2004 ECS 235

Example Above 2-parent joint creation simulation in scheme TWO Equivalent to 3-parent joint creation scheme THREE in which P1, P2, P3, C are of same type as in TWO, and edges from P1, P2, P3 to C are of type e, and no types a and e´ exist in TWO April 13, 2004 ECS 235

Simulation Scheme A simulates scheme B iff every state B can reach has a corresponding state in A that A can reach; and every state that A can reach either corresponds to a state B can reach, or has a successor state that corresponds to a state B can reach The last means that A can have intermediate states not corresponding to states in B, like the intermediate ones in TWO in the simulation of THREE April 13, 2004 ECS 235

Expressive Power If scheme in MA no scheme in MB can simulate, MB less expressive than MA If every scheme in MA can be simulated by a scheme in MB, MB as expressive as MA If MA as expressive as MB and vice versa, MA and MB equivalent April 13, 2004 ECS 235

Example Scheme A in model M Scheme B in model N Nodes X1, X2, X3 2-parent joint create 1 node type, 1 edge type No edge adding operations Initial state: X1, X2, X3, no edges Scheme B in model N All same as A except no 2-parent joint create 1-parent create Which is more expressive? April 13, 2004 ECS 235

Can A Simulate B? Scheme A simulates 1-parent create: have both parents be same node Model M as expressive as model N April 13, 2004 ECS 235

Can B Simulate A? Suppose X1, X2 jointly create Y in A Edges from X1, X2 to Y, no edge from X3 to Y Can B simulate this? Without loss of generality, X1 creates Y Must have edge adding operation to add edge from X2 to Y One type of node, one type of edge, so operation can add edge between any 2 nodes April 13, 2004 ECS 235

No All nodes in A have even number of incoming edges 2-parent create adds 2 incoming edges Edge adding operation in B that can edge from X2 to C can add one from X3 to C A cannot enter this state B cannot transition to a state in which Y has even number of incoming edges No remove rule So B cannot simulate A; N less expressive than M April 13, 2004 ECS 235

Theorem Monotonic single-parent models are less expressive than monotonic multiparent models ESPM more expressive than SPM ESPM multiparent and monotonic SPM monotonic but single parent April 13, 2004 ECS 235

Typed Access Matrix Model Like ACM, but with set of types T All subjects, objects have types Set of types for subjects TS Protection state is (S, O, , A), where :OT specifies type of each object If X subject, (X) in TS If X object, (X) in T – TS April 13, 2004 ECS 235

Create Rules Subject creation Object creation create subject s of type ts s must not exist as subject or object when operation executed ts  TS Object creation create object o of type to o must not exist as subject or object when operation executed to  T – TS April 13, 2004 ECS 235

Create Subject Precondition: s  S Primitive command: create subject s of type t Postconditions: S´ = S  { s }, O´ = O  { s } (y  O)[´(y) =  (y)], ´(s) = t (y  O´)[a´[s, y] = ], (x  S´)[a´[x, s] = ] (x  S)(y  O)[a´[x, y] = a[x, y]] April 13, 2004 ECS 235

Create Object Precondition: o  O Primitive command: create object o of type t Postconditions: S´ = S, O´ = O  { o } (y  O)[´(y) =  (y)], ´(o) = t (x  S´)[a´[x, o] = ] (x  S)(y  O)[a´[x, y] = a[x, y]] April 13, 2004 ECS 235

Definitions MTAM Model: TAM model without delete, destroy MTAM is Monotonic TAM (x1:t1, ..., xn:tn) create command ti child type in  if any of create subject xi of type ti or create object xi of type ti occur in  ti parent type otherwise April 13, 2004 ECS 235

Cyclic Creates command havoc(s1 : u, s2 : u, o1 : v, o2 : v, o3 : w, o4 : w) create subject s1 of type u; create object o1 of type v; create object o3 of type w; enter r into a[s2, s1]; enter r into a[s2, o2]; enter r into a[s2, o4] end April 13, 2004 ECS 235

Creation Graph u, v, w child types u, v, w also parent types Graph: lines from parent types to child types This one has cycles u v w April 13, 2004 ECS 235

Theorems Safety decidable for systems with acyclic MTAM schemes Safety for acyclic ternary MATM decidable in time polynomial in the size of initial ACM “ternary” means commands have no more than 3 parameters Equivalent in expressive power to MTAM April 13, 2004 ECS 235

Key Points Safety problem undecidable Limiting scope of systems can make problem decidable Types critical to safety problem’s analysis April 13, 2004 ECS 235

Overview Overview Policies Trust Nature of Security Mechanisms Policy Expression Languages Limits on Secure and Precise Mechanisms April 13, 2004 ECS 235

Security Policy Policy partitions system states into: Secure system Authorized (secure) These are states the system can enter Unauthorized (nonsecure) If the system enters any of these states, it’s a security violation Secure system Starts in authorized state Never enters unauthorized state April 13, 2004 ECS 235

Confidentiality X set of entities, I information I has confidentiality property with respect to X if no x in X can obtain information from I I can be disclosed to others Example: X set of students I final exam answer key I is confidential with respect to X if students cannot obtain final exam answer key April 13, 2004 ECS 235

Integrity X set of entities, I information I has integrity property with respect to X if all x in X trust information in I Types of integrity: trust I, its conveyance and protection (data integrity) I information about origin of something or an identity (origin integrity, authentication) I resource: means resource functions as it should (assurance) April 13, 2004 ECS 235

Availability X set of entities, I resource I has availability property with respect to X if all x in X can access I Types of availability: traditional: x gets access or not quality of service: promised a level of access (for example, a specific level of bandwidth) and not meet it, even though some access is achieved April 13, 2004 ECS 235

Policy Models Abstract description of a policy or class of policies Focus on points of interest in policies Security levels in multilevel security models Separation of duty in Clark-Wilson model Conflict of interest in Chinese Wall model April 13, 2004 ECS 235

Types of Security Policies Military (governmental) security policy Policy primarily protecting confidentiality Commercial security policy Policy primarily protecting integrity Confidentiality policy Policy protecting only confidentiality Integrity policy Policy protecting only integrity April 13, 2004 ECS 235

Integrity and Transactions Begin in consistent state “Consistent” defined by specification Perform series of actions (transaction) Actions cannot be interrupted If actions complete, system in consistent state If actions do not complete, system reverts to beginning (consistent) state April 13, 2004 ECS 235

Trust Administrator installs patch Trusts patch came from vendor, not tampered with in transit Trusts vendor tested patch thoroughly Trusts vendor’s test environment corresponds to local environment Trusts patch is installed correctly April 13, 2004 ECS 235

Trust in Formal Verification Gives formal mathematical proof that given input i, program P produces output o as specified Suppose a security-related program S formally verified to work with operating system O What are the assumptions? April 13, 2004 ECS 235

Trust in Formal Methods Proof has no errors Bugs in automated theorem provers Preconditions hold in environment in which S is to be used S transformed into executable S’ whose actions follow source code Compiler bugs, linker/loader/library problems Hardware executes S’ as intended Hardware bugs (Pentium f00f bug, for example) April 13, 2004 ECS 235

Types of Access Control Discretionary Access Control (DAC, IBAC) individual user sets access control mechanism to allow or deny access to an object Mandatory Access Control (MAC) system mechanism controls access to object, and individual cannot alter that access Originator Controlled Access Control (ORCON) originator (creator) of information controls who can access information April 13, 2004 ECS 235

Question Policy disallows cheating Includes copying homework, with or without permission CS class has students do homework on computer Anne forgets to read-protect her homework file Bill copies it Who cheated? Anne, Bill, or both? April 13, 2004 ECS 235

Answer Part 1 Bill cheated Policy forbids copying homework assignment Bill did it System entered unauthorized state (Bill having a copy of Anne’s assignment) If not explicit in computer security policy, certainly implicit Not credible that a unit of the university allows something that the university as a whole forbids, unless the unit explicitly says so April 13, 2004 ECS 235

Answer Part #2 Anne didn’t protect her homework Not required by security policy She didn’t breach security If policy said students had to read-protect homework files, then Anna did breach security She didn’t do so April 13, 2004 ECS 235

Mechanisms Entity or procedure that enforces some part of the security policy Access controls (like bits to prevent someone from reading a homework file) Disallowing people from bringing CDs and floppy disks into a computer facility to control what is placed on systems April 13, 2004 ECS 235

Overview Key exchange Cryptographic key infrastructure Key storage Session vs. interchange keys Classical, public key methods Key generation Cryptographic key infrastructure Certificates Key storage Key escrow Key revocation Digital signatures May 13, 2004 ECS 235

Notation X  Y : { Z || W } kX,Y A  T : { Z } kA || { W } kA,T X sends Y the message produced by concatenating Z and W enciphered by key kX,Y, which is shared by users X and Y A  T : { Z } kA || { W } kA,T A sends T a message consisting of the concatenation of Z enciphered using kA, A’s key, and W enciphered using kA,T, the key shared by A and T r1, r2 nonces (nonrepeating random numbers) May 13, 2004 ECS 235

Session, Interchange Keys Alice wants to send a message m to Bob Assume public key encryption Alice generates a random cryptographic key ks and uses it to encipher m To be used for this message only Called a session key She enciphers ks with Bob;s public key kB kB enciphers all session keys Alice uses to communicate with Bob Called an interchange key Alice sends { m } ks { ks } kB May 13, 2004 ECS 235

Benefits Limits amount of traffic enciphered with single key Standard practice, to decrease the amount of traffic an attacker can obtain Prevents some attacks Example: Alice will send Bob message that is either “BUY” or “SELL”. Eve computes possible ciphertexts { “BUY” } kB and { “SELL” } kB. Eve intercepts enciphered message, compares, and gets plaintext at once May 13, 2004 ECS 235

Key Exchange Algorithms Goal: Alice, Bob get shared key 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 May 13, 2004 ECS 235

Classical Key Exchange Bootstrap problem: how do Alice, Bob begin? Alice can’t send it to Bob in the clear! Assume trusted third party, Cathy Alice and Cathy share secret key kA Bob and Cathy share secret key kB Use this to exchange shared key ks May 13, 2004 ECS 235

Simple Protocol { request for session key to Bob } kA Alice Cathy { ks } kA || { ks } kB Alice Cathy { ks } kB Alice Bob May 13, 2004 ECS 235

Problems 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 May 13, 2004 ECS 235

Needham-Schroeder Alice || Bob || r1 Alice Cathy { Alice || Bob || r1 || ks || { Alice || ks } kB } kA Alice Cathy { Alice || ks } kB Alice Bob { r2 } ks Alice Bob { r2 – 1 } ks Alice Bob May 13, 2004 ECS 235

Argument: Alice talking to Bob Second message Enciphered using key only she, Cathy know So Cathy enciphered it Response to first message As r1 in it matches r1 in first 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 May 13, 2004 ECS 235

Argument: Bob talking to Alice 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 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 r2 and so can’t respond, or responds incorrectly May 13, 2004 ECS 235

Denning-Sacco Modification Assumption: all keys are secret Question: suppose Eve can obtain session key. How does that affect protocol? In what follows, Eve knows ks { Alice || ks } kB Eve Bob { r2 } ks Eve Bob { r2 – 1 } ks Eve Bob May 13, 2004 ECS 235

Solution In protocol above, Eve impersonates Alice Problem: replay in third step First in previous slide Solution: use time stamp T to detect replay 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 May 13, 2004 ECS 235

Needham-Schroeder with Denning-Sacco Modification Alice || Bob || r1 Alice Cathy { Alice || Bob || r1 || ks || { Alice || T || ks } kB } kA Alice Cathy { Alice || T || ks } kB Alice Bob { r2 } ks Alice Bob { r2 – 1 } ks Alice Bob May 13, 2004 ECS 235

Otway-Rees Protocol Corrects problem Does not use timestamps That is, Eve replaying the third message in the protocol Does not use timestamps Not vulnerable to the problems that Denning-Sacco modification has Uses integer n to associate all messages with particular exchange May 13, 2004 ECS 235

The Protocol n || Alice || Bob || { r1 || n || Alice || Bob } kA Alice { r2 || n || Alice || Bob } kB Cathy Bob n || { r1 || ks } kA || { r2 || ks } kB Cathy Bob n || { r1 || ks } kA Alice Bob May 13, 2004 ECS 235

Argument: Alice talking to Bob Fourth message If n matches first message, Alice knows it is part of this protocol exchange Cathy generated ks because only she, Alice know kA Enciphered part belongs to exchange as r1 matches r1 in encrypted part of first message May 13, 2004 ECS 235

Argument: Bob talking to Alice Third message If n matches second message, Bob knows it is part of this protocol exchange Cathy generated ks because only she, Bob know kB Enciphered part belongs to exchange as r2 matches r2 in encrypted part of second message May 13, 2004 ECS 235

Replay Attack Eve acquires old ks, message in third step n || { r1 || ks } kA || { r2 || ks } kB 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 May 13, 2004 ECS 235

Kerberos Authentication system Ticket Authenticator Based on Needham-Schroeder with Denning-Sacco modification Central server plays role of trusted third party (“Cathy”) Ticket Issuer vouches for identity of requester of service Authenticator Identifies sender May 13, 2004 ECS 235

Idea User u authenticates to Kerberos server Obtains ticket Tu,TGS for ticket granting service (TGS) User u wants to use service s: User sends authenticator Au, ticket Tu,TGS to TGS asking for ticket for service TGS sends ticket Tu,s to user User sends Au, Tu,s to server as request to use s Details follow May 13, 2004 ECS 235

Tu,s = s || { u || u’s address || valid time || ku,s } ks Ticket Credential saying issuer has identified ticket requester Example ticket issued to user u for service s Tu,s = s || { u || u’s address || valid time || ku,s } ks where: ku,s is session key for user and service Valid time is interval for which ticket valid u’s address may be IP address or something else Note: more fields, but not relevant here May 13, 2004 ECS 235

Au,s = { u || generation time || kt } ku,s Authenticator 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 Au,s = { u || generation time || kt } ku,s where: kt is alternate session key Generation time is when authenticator generated Note: more fields, not relevant here May 13, 2004 ECS 235

Protocol user || TGS user Cathy { ku,TGS } ku || Tu,TGS Cathy user service || Au,TGS || Tu,TGS user TGS user || { ku,s } ku,TGS || Tu,s user TGS Au,s || Tu,s user service { t + 1 } ku,s user service May 13, 2004 ECS 235

Analysis 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 Service s validates request by checking sender (using Au,s) is same as entity ticket issued to Step 6 optional; used when u requests confirmation May 13, 2004 ECS 235

Problems Relies on synchronized clocks Tickets have some fixed fields If not synchronized and old tickets, authenticators not cached, replay is possible 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 May 13, 2004 ECS 235

Public Key Key Exchange Here interchange keys known eA, eB Alice and Bob’s public keys known to all dA, dB Alice and Bob’s private keys known only to owner Simple protocol ks is desired session key { ks } eB Alice Bob May 13, 2004 ECS 235

Problem and Solution Vulnerable to forgery or replay Because eB known to anyone, Bob has no assurance that Alice sent message Simple fix uses Alice’s private key ks is desired session key { { ks } dA } eB Alice Bob May 13, 2004 ECS 235

Notes Can include message enciphered with ks 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; Cathy is public server providing public keys) Solution to this (binding identity to keys) discussed later as public key infrastructure (PKI) May 13, 2004 ECS 235

Man-in-the-Middle Attack send Bob’s public key Eve intercepts request Alice Cathy send Bob’s public key Eve Cathy eB Eve Cathy eE Alice Eve { ks } eE Eve intercepts message Alice Bob { ks } eB Eve Bob May 13, 2004 ECS 235

Key Generation Goal: generate difficult to guess keys 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 Actually, numbers are usually pseudo-random, that is, generated by an algorithm May 13, 2004 ECS 235

What is “Random”? Sequence of cryptographically random numbers: a sequence of numbers n1, n2, … such that, for any integer k > 0, an observer cannot predict nk even if all of n1, …, nk–1 are known Best: physical source of randomness Random pulses Electromagnetic phenomena Characteristics of computing environment such as disk latency Ambient background noise May 13, 2004 ECS 235

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 Very difficult to do this well Linear congruential generators [nk = (ank–1 + b) mod n] broken Polynomial congruential generators [nk = (ajnk–1j + … + a1nk–1 a0) mod n] broken too Here, “broken” means next number in sequence can be determined May 13, 2004 ECS 235

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 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 May 13, 2004 ECS 235

Cryptographic Key Infrastructure Goal: bind identity to key Classical: not possible as all keys are shared Use protocols to agree on a shared key (see earlier) Public key: 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 May 13, 2004 ECS 235

Certificates 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) CA = { eA || Alice || T } dC May 13, 2004 ECS 235

Use 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 pushed “up” a level Two approaches: Merkle’s tree, signature chains May 13, 2004 ECS 235

Merkle’s Tree Scheme h(1,4) h(1,2) h(3,4) h(1,1) h(2,2) h(3,3) h(4,4) Keep certificates in a file Changing any certificate changes the file Use crypto hash functions to detect this Define hashes recursively h is hash function Ci is certificate i 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) C1 C2 C3 C4 May 13, 2004 ECS 235

Validation h(1,4) h(1,2) h(3,4) h(1,1) h(2,2) h(3,3) h(4,4) To validate C1: 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 h(1,4) h(1,2) h(3,4) h(1,1) h(2,2) h(3,3) h(4,4) C1 C2 C3 C4 May 13, 2004 ECS 235

Details f: DDD maps bit strings to bit strings h: NND maps integers to bit strings if i ≥ j, h(i, j) = f(Ci, Cj) if i < j, h(i, j) = f(h(i, (i+j)/2), h((i+j)/2+1, j)) May 13, 2004 ECS 235

Problem File must be available for validation Otherwise, can’t recompute hash at root of tree Intermediate hashes would do Not practical in most circumstances Too many certificates and users Users and certificates distributed over widely separated systems May 13, 2004 ECS 235

Certificate Signature Chains Create certificate Generate hash of certificate Encipher hash with issuer’s private key Validate Obtain issuer’s public key Decipher enciphered hash Recompute hash from certificate and compare Problem: getting issuer’s public key May 13, 2004 ECS 235

X.509 Chains Some certificate components in X.509v3: Version Serial number Signature algorithm identifier: hash algorithm Issuer’s name; uniquely identifies issuer Interval of validity Subject’s name; uniquely identifies subject Subject’s public key Signature: enciphered hash May 13, 2004 ECS 235

X.509 Certificate Validation Obtain issuer’s public key The one for the particular signature algorithm Decipher signature Gives hash of certificate Recompute hash from certificate and compare If they differ, there’s a problem Check interval of validity This confirms that certificate is current May 13, 2004 ECS 235

Issuers Certification Authority (CA): entity that issues certificates Multiple issuers pose validation problem Alice’s CA is Cathy; Bob’s CA is Don; how can Alice validate Bob’s certificate? Have Cathy and Don cross-certify Each issues certificate for the other May 13, 2004 ECS 235

Validation and Cross-Certifying Certificates: Cathy<<Alice>> Dan<<Bob> Cathy<<Dan>> Dan<<Cathy>> Alice validates Bob’s certificate Alice obtains Cathy<<Dan>> Alice uses (known) public key of Cathy to validate Cathy<<Dan>> Alice uses Cathy<<Dan>> to validate Dan<<Bob>> May 13, 2004 ECS 235

PGP Chains OpenPGP certificates structured into packets One public key packet Zero or more signature packets 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 May 13, 2004 ECS 235

OpenPGP 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) Version 4 packet more complex May 13, 2004 ECS 235

Signing Single certificate may have multiple signatures 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 Called “self-signing” May 13, 2004 ECS 235

Validating Certificates Arrows show signatures Self signatures not shown Alice needs to validate Bob’s OpenPGP cert Does not know Fred, Giselle, or Ellen Alice gets Giselle’s cert Knows Henry slightly, but his signature is at “casual” level of trust Alice gets Ellen’s cert Knows Jack, so uses his cert to validate Ellen’s, then hers to validate Bob’s Jack Henry Ellen Irene Giselle Fred Bob May 13, 2004 ECS 235

Storing Keys Multi-user or networked systems: attackers may defeat access control mechanisms Encipher file containing key Attacker can monitor keystrokes to decipher files Key will be resident in memory that attacker may be able to read Use physical devices like “smart card” Key never enters system Card can be stolen, so have 2 devices combine bits to make single key May 13, 2004 ECS 235

Key Escrow Key escrow system allows authorized third party to recover key Useful when keys belong to roles, such as system operator, rather than individuals Business: recovery of backup keys Law enforcement: recovery of keys that authorized parties require access to Goal: provide this without weakening cryptosystem Very controversial May 13, 2004 ECS 235

Desirable Properties Escrow system should not depend on encipherment algorithm Privacy protection mechanisms must work from end to end and be part of user interface Requirements must map to key exchange protocol System supporting key escrow must require all parties to authenticate themselves If message to be observable for limited time, key escrow system must ensure keys valid for that period of time only May 13, 2004 ECS 235

Components User security component Key escrow component Does the encipherment, decipherment Supports the key escrow component Key escrow component Manages storage, use of data recovery keys Data recovery component Does key recovery May 13, 2004 ECS 235

Example: EES, Clipper Chip Escrow Encryption Standard Set of interlocking components Designed to balance need for law enforcement access to enciphered traffic with citizens’ right to privacy Clipper chip prepares per-message escrow information Each chip numbered uniquely by UID Special facility programs chip Key Escrow Decrypt Processor (KEDP) Available to agencies authorized to read messages May 13, 2004 ECS 235

User Security Component Unique device key kunique Nonunique family key kfamily Cipher is Skipjack Classical cipher: 80 bit key, 64 bit input, output blocks Generates Law Enforcement Access Field (LEAF) of 128 bits: { UID || { ksession } kunique || hash } kfamily hash: 16 bit authenticator from session key and initialization vector May 13, 2004 ECS 235