Dining Cryptographers
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Presentation Transcript
Dining Cryptographers R. Newman
Topics • Defining anonymity • Need for anonymity • Defining privacy • Threats to anonymity and privacy • Mechanisms to provide anonymity • Metrics for Anonymity • Applications of anonymity technology
Dining Cryptographer Problem • Three cryptographers go out for dinner • They are told that the bill has been paid • Benefactor wishes to remain anonymous • Could be one of them, or a fourth party (e.g., NSA) • Want to know if one of them paid • Respect desire to remain anonymous • But want to find this piece of information
Dining Cryptographer Protocol • Each pair of cryptographers flips a fair coin • This is done in secret, so only the pair can see it • Each cryptographer states XOR of coins • States whether the pair of coins they saw were same • If one paid, reports the opposite result • Each computes answer • Odd number of differences => a cryptographer paid • Even number of difference => someone else paid
Dining Cryptographer Protocol • Why does this work? • Assuming all cryptographers are honest • 1. First, consider case where NSA paid • 1a. All heads or all tails => no differences • Even number of differences, • So conclude NSA paid
Case 1a: NSA Paid, all same “Same” All same All report same Zero (even) diffs A heads B heads “Same” C heads “Same”
Dining Cryptographer Protocol • Why does this work? • Assuming all cryptographers are honest • 1. First, consider case where NSA paid • 1a. All heads or all tails => no differences • 1b. Two of one and one of the other => two differences • Either way, even number of differences!
Case 1b: NSA Paid, one different “Different” One different All report same Two (even) diffs A heads B tails “Same” C heads “Different”
Dining Cryptographer Protocol • Why does this work? • Assuming all cryptographers are honest • 2. Now what if one cryptographer inverts report? • 2a. All same => two say same, one says different • => one difference
Case 2a: Alice Paid, all same “Different” All same B and C report same Alice inverts report One (odd) diff A heads B heads “Same” C heads “Same”
Dining Cryptographer Protocol • Why does this work? • Assuming all cryptographers are honest • 2. Now what if one cryptographer inverts report? • 2a. All same => one difference • 2b/c. Two of one, one of other: • 2b. Payer sees equal pair => says different, and other two see different pairs, say different => 3 differences (odd)
Case 2b: Alice Paid, sees same “Different” One different Alice sees same B, C report different Alice reports different Three (odd) diffs A heads B heads “Different” C tails “Different”
Dining Cryptographer Protocol • Why does this work? • Assuming all cryptographers are honest • 2. Now what if one cryptographer inverts report? • 2a. All same => 1 difference • 2b/c. Two of one, one of other: • 2b. Payer sees equal pair => 3 differences • 2c. Payer sees unequal pair => says same One of the others sees equal, one sees unequal Hence 1 difference reported
Case 2c: Alice Paid, sees different “Same” One different Alice sees different B reports same C reports different A reports same One (odd) diff A heads B tails “Same” C heads “Different”
Dining Cryptographer Protocol • Why does this work? • 1. No cryptographers invert report • 1a. All heads or all tails => 0 differences • 1b. Two of one and one of the other => 2 differences • Either way, even number of differences! • 2. One cryptographer inverts report • 2a. All same => 1 difference • 2b/c. Two of one, one of other: • 2b. Payer sees equal pair => 3 differences • 2c. Payer sees unequal pair => 1 difference • Always odd number of differences reported
Dining Cryptographer Protocol • How does it preserve anonymity? • View of non-paying cryptographer: • If even difference, there is nothing to discover • If odd difference, two cases: • Cryptographer sees equal values • One of the others said ”same”, other said ”different” • Hidden coin is same => one who said ”different” paid • Hidden coin different => one who said ”same” paid • Each is equally likely! (Fair coin)
Dining Cryptographer Protocol • How does it preserve anonymity? • View of non-paying cryptographer: • If even difference, there is nothing to discover • If odd difference, two cases: • Cryptographer sees unequal values • Both of the others said ”different” => payer closest to coin that is same as hidden coin • Both of the others said ”same” => payer closest to coin different from hidden coin • Each is equally likely! (Fair coin)
Dining Cryptographer Protocol • OK – so what? • Now can send one bit anonymously • Extend protocol to anonymously transmission • Repeat protocol in rounds • Each round, act like non-payer unless you have msg • When you have message, start sending bits • Invert report when sending 1’s, not when 0’s • What about collisions? • Use collision detection, backoff protocol • CSMA/CD with backoff – like Ethernet!
Dining Cryptographer Protocol • OK – so what? • Now can send one bit anonymously • For three senders • Extend protocol to multiple senders • Complete graph for N senders • Each edge represents a fair coin • Report XOR of all coins (or invert it for 1) • Note that with N=2, only non-participants don’t know the sender (not secret from participants)
Dining Cryptographer Protocol • Why does this work? • Each bit appears in two sums • In sum of sums, these cancel each other out • If one cryptographer inverts, then odd number of sum of reports is 1, otherwise it is 0 • Replace coin flips with key bits • Each participant shares a key with each other participant • Same number of bits in key as rounds of protocol • Use key bits as coin values in protocol
Modeling DC Nets • Two kinds of secret per participant: • Secret pairwise keys shared with other participants • Message bits • Consider collusion later.... • Remaining information: • Which pairs share a key (not required to be secret) • What each participant outputs each round • Compute sum of outputs
Modeling DC Nets • Model with graph: • Each participant is a node • Each key is represented by an edge • Edge is incident on participants sharing key • Graph is connected, may not be complete
Modeling DC Nets Originally coin flips Replace with random bit Which is “key bit” A Tails 0 B Heads 1 C Heads 1
Modeling DC Nets • Model with graph • Anonymity Set seen by a set of keys • AS = Set of vertices in a connected component remaining in graph after removing edges corresponding to keys in set • Two participants connected by non-compromised keys are in same AS, and are indistinguishable – only parity of report can be determined
Examples • Non-participant observer • All participants in same CC are in same AS • (Graph remains connected after removing 0 edges) • Complete key compromise • All edges are removed • All nodes are singletons • No anonymity: Sent bit = XOR of key bits with report
Modeling DC Nets Distribute keys Alice has message Others report sums Alice inverts her sums Sums_A = 100 Report_A = 101 Msg_A = 001 A Kab=010 Sums_B = 110 B Sum of sums: 101 110 100 100 010 001 Kac=110 Kbe=001 C Kbc=101 D Sums_C = 100 Sums_D = 100 Kce=111 E Kde=101 Sums_E = 010
Modeling DC Nets B and C collude Alice has message All report as before B and C know what A should have sent Sums_A = 100 Report_A = 101 Msg_A = 001 A Kab=010 Sums_B = 110 B Kac=110 Kbe=001 Sum of A keys: Kab=010 Kac=110 100 What A reported: 101 What A said: 100 101 001 C Kbc=101 D Sums_C = 100 Sums_D = 100 Kce=111 E Kde=101 Sums_E = 010
Modeling DC Nets B and C collude Notice that B and C Do not have to share All keys (Kce or Kbe) To attack Alice Sums_A = 100 Report_A = 101 Msg_A = 001 A Kab=010 Sums_B = 110 B Kac=110 Kbe=001 C Kbc=101 D Sums_C = 100 Sums_D = 100 Kce=111 E Kde=101 Sums_E = 010
Modeling DC Nets B and C collude Ed has message All report B and C know that A reported honestly, So D or E sent msg Sums_A = 100 A Kab=010 Sums_B = 110 B Kac=110 Kbe=001 C Kbc=101 D Sums_C = 100 Sums_D = 100 Kce=111 E Kde=101 Sums_E = 010 Msg_E = 100 Report_E = 110
Modeling DC Nets Bob by himself Cannot reduce AS Sums_A = 100 A Kab=010 Sums_B = 110 B Kac=110 Kbe=001 C Kbc=101 D Sums_C = 100 Sums_D = 100 Kce=111 E Kde=101 Sums_E = 010 Msg_E = 100 Report_E = 110
Examples • Biconnected graph • All pairs of participants are connected by at least two node-disjoint paths • No single participant can reduce AS size of other participants by itself • Requires collusion to learn anything! • All collusion buys is parity of sum of inversions of each connected component • Inversions hidden by one or more key bits
Formal Model • Connected component C: m nodes and n edges • m x n incidence matrix M • nodes = rows and edges = columns • Stochastic variable keys K over GF(2n) • One per edge, uniform random • Stochastic variable msg bits I over GF(2m) • One per vertex, uniform random • A = (MK) + I = reports of the vertices • Parity(A) = parity(I) • since columns of M have even parity Nota bene!
Formal Model edges keys sums 1 2 3 4 5 6 A1 1 0 0 0 0 B 1 0 1 1 0 0 C 0 1 1 0 1 0 D 0 0 0 1 0 1 E 0 0 0 0 1 1 K 0 1 1 0 1 1 S 1 1 1 1 0 = nodes Info_A = 1 Report_A = 0 Sum_A = 1 X A K1=0 Incidence Matrix M 1 Sum_B = 1 S 1 1 1 1 0 I 1 0 0 0 0 A 0 1 1 1 0 2 B K2=1 + = K4=0 4 K3=1 3 C D sums msg bits reports Sum_C = 1 Info_C = 0 Report_C = 1 K5=1 Sum_D = 1 6 5 K6=1 E Sum_E = 0
Formal Model • Thm: Let a be in GF(2n). For each i in GF(2n), which is assumed by I with non-zero probability, and which has the same parity as a, Prob(A=a | I=i) = 21-m. hence Prob(I=i | A=a) = prob(I=i) a priori. • Prf: Since every proper subset of rows of M is is linearly independent, the rank of M is m-1, and any zero parity vector in GF(2n) can be written as a linear combination of the columns of M. So the system of linear equalities MK+i = a is solvable, since MK = a+i has zero parity. The system has exactly 2n-m+1 solutions. Since K and I are mutually independent and K is uniformly distributed, the theorem follows.
Formal Model • Thm: Let a be in GF(2n). For each i in GF(2n), which is assumed by I with non-zero probability, and which has the same parity as a, Prob(A=a | I=i) = 21-m. hence Prob(I=i | A=a) = prob(I=i) a priori. • Prf: Since the rank of M is m-1, The system has exactly 2n-m+1 solutions. Since K and I are mutually independent and K is uniformly distributed, the theorem follows.
Building Graphs • Complete graphs do not scale • Can use a ring • But any two colluders can partition ring • If colluders surround a target node • It is compromised!
Ring B C A D Ring is binconnected – removal of any one node does not partition graph H E G F
Ring B C A D But any two nodesthat collude can partition graph and possibly compromise a single participant (C) H E G F
Building Graphs • ”Trusted not to collude” clique – • Subset of participants whom all believe will not collude • Subset forms a clique • All others share a key with each member of clique • All members of clique must collude to compromise
Trusted not to Collude Clique B A, B, and C are mutually hostile Hence trusted not to collude They form a “root clique” All others nodes connect to each member of root clique A C D E F G H
Trusted not to Collude Clique B Size of clique = K Number of keys = K(K-1)/2 for clique Plus for N total nodes K(N-K) for others And the total is … K[(K-1)/2 + (N-K)] Example here: K=3 3[2/2 + (8-3)] = 18 Compared to N(N-1)/2 = 28 for complete graph A C D E F G H
Trusted not to Collude Clique B All members of root clique must collude to compromise any node Suppose B and C collude… A C D E F G H
Trusted not to Collude Clique B All members of root clique must collude to compromise any node Suppose B and C collude… Then A still connects all other nodes The AS is maximal! A C D E F G H
Preventing Disruption • Well, can’t really prevent it ... • But can detect it and weed out disrupters • Requires: • Key-sharing graph is publically agreed on • Each participant’s outputs are publically agreed on such that no participant can change their output for a round based on the other participant’s outputs for that round • Some rounds contain inversions that would not compromise the untraceability of any non-disrupter
Preventing Disruption • Key-sharing graph is publically agreed on • Distributed consensus • Participantd can’t change outputs • Simultaneous broadcast channels • Commitment protocols • Contestable rounds that do not compromise the untraceability of any non-disrupter • Slot reservation protocol
Slot Reservation Protocol • Messages sent in two blocks • Reservation block with one bit per msg slot • Message block with multiple message slots • Sender reserves one or more slots • Sets corresponding bit(s) in reservation block • Sends message in corresponding slots • For contestable rounds, all N participants must always make one reserveration each round • Requires quadratic slots due to Birthday Paradox • Disrupted reservation block likely to have Hamming weight unequal to N • All bits of reserving block can be safely contested
Single Disrupter • If it tells the truth about shared keys bits for contested bit, or lies about an even number of key bits, it implicates itself • The sum of the claimed key bit values is not what it reported (apart from allowed inversion) • If it lies about an odd number... • Values it claims will differ from values claimed by those who share the keys it lies about • Casting suspicion on itself and each of them • But all disputed bits point to disrupter • And falsely accused participants know disrupter • And can refuse to share edge with disrupter in future
Multiple Disrupters • At least one inversion revealed as illegit or at least one key bit disputed • Since parity of outputs does not correspond to parity of legit inversions • Result of each contested round • Remove at least one edge, or • Remove at least one vertex from agree graph • If every disruption has non-zero probability of being contested • Then bounded amount of disruption possible before disrupters excluded • Removed (vertex) or • Share no keys (edges)
Tracing by Consent • Deter antisocial use of network by... • Allowing trace of any message by cooperation of most participants • Example: court orders all participants to reveal their shared key bits for a round of the message • Sender may try to spread blame by lying about and odd number of shared bits • Digital signatures on shared bits can stop this • Allow contested rounds to be fully resolved • Allow accused senders to exonerate themselves • Allow colluders to convince each other to trust them • But allow sender self-incrimination: non-repudiation! • Variant prevents self-incrimination
Split-bit Signatures • Variant prevents self-incrimination • Each participant in a pair signs a differnt bit whose sum is the actual shared bit • Sharers can tell if the signatures are good • Others can’t tell what bit is if one is lying • Helps resolve contested rounds • Contester of a bit shows signature of other party • Other party must reveal contester’s signature... or be considered a disrupter
