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What Crypto Can Do for You: Solutions in Search of Problems Anna Lysyanskaya Brown University

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Systemic Risk from Local Information

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M.C.Escher, Belvedere

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Who Puts Together the Big Picture?

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The government?

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Who Puts Together the Big Picture? The government?

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Who Puts Together the Big Picture? An independent trustworthy party?

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Who Puts Together the Big Picture? An independent trustworthy party?

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Who Puts Together the Big Picture? The data owners (financial institutions) themselves?

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Who Puts Together the Big Picture? The data owners (financial institutions) themselves?

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Who Puts Together the Big Picture? Cryptography tells us: For any efficiently computable function F, there is an “efficient” interactive algorithm that n data owners, P 1 (x 1 ),…,P n (x n ), can run together such that: 1.They learn F(x 1,x 2,…,x n ) 2.Other than that, P i learns nothing about x j, j≠i [Yao, GMW, BGW, …]

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Example: Set Intersection 12 18 5 6 31 42 5 24 12 3 Alice’s set Bob’s set 5 12 Intersection

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How to compute the intersection w/o learning the rest of each other’s sets? [FMP04,…,BCCKLS09,…,KMRS14]

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Step 1: Alice’s set becomes a polynomial 12 18 5 6 31 Alice’s set p(x) = (x-12)(x-18)(x-5)(x-6)(x-31) mod q = x 5 + c 4 x 4 + c 3 x 3 + c 2 x 2 + c 1 x + c 0 c4c4 c3c3 c2c2 c1c1 c0c0

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Step 1: Alice’s set becomes a polynomial Alice’s polynomial p(x) c4c4 c3c3 c2c2 c1c1 c0c0

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Step 2: Alice encrypts her polynomial Alice’s polynomial p(x) c4c4 c3c3 c2c2 c1c1 c0c0

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Step 2: Alice encrypts her polynomial Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 )

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Step 2: Alice encrypts her polynomial… Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 ) …using an “additive” encryption scheme E(x) * E(y) = E(x+y) [Paillier’99]

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Step 2: Alice encrypts her polynomial… Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 ) …using an “additive” encryption scheme …for which she holds the decryption key

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Step 3: Alice sends the encrypted Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 ) polynomial to Bob

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Step 4: Bob evaluates the encrypted Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 ) polynomial on his set 42 5 24 12 3 Bob’s set p(42) = 42 5 + c 4 42 4 + c 3 42 3 +c 2 42 2 +c 1 42+c 0 mod q E(p(42)) = E(42 5 ) * E(c 4 )42 4 * E(c 3 )42 3 * E(c 2 )42 2 * E(c 1 )42 * E(c 0 )

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Step 4: Bob evaluates the encrypted Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 ) polynomial on his set 42 5 24 12 3 Bob’s set p(x) evaluated on Bob’s set E(p(42))E(p(5))E(p(24))E(p(12))E(p(3))

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Step 4: Bob evaluates the encrypted Alice’s encrypted polynomial p(x) E(c 4 )E(c 3 )E(c 2 )E(c 1 )E(c 0 ) polynomial on his set p(x) evaluated on Bob’s set E(p(42))E(0)E(p(24))E(0)E(p(3)) Note: p(y) = 0 iff y is in Alice’s set

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Step 5: Bob randomizes the result E(p(42))R 1 E(0)R 2 E(p(24))R 3 E(0)R 4 E(p(3))R 5

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Step 5: Bob randomizes the result E(u 1 )E(0)E(u 3 )E(0)E(u 5 )

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Step 6: Bob sends the result to Alice E(u 1 )E(0)E(u 3 )E(0)E(u 5 )

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Step 7: Alice decrypts it... E(u 1 )E(0)E(u 3 )E(0)E(u 5 ) u1u1 0u3u3 0u5u5

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Step 7: Alice decrypts it... and sends the locations of 0’s to Bob u1u1 0u3u3 0u5u5

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Step 7: Alice decrypts it... and sends the locations of 0’s to Bob ?0?0?

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Step 8: Bob derives the intersection ?0?0? 42 5 24 12 3

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Step 8: Bob derives the intersection and sends it to Alice 5 12

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A More General Solution for Two Parties: Yao’s Encrypted Circuit Alice’s logical circuit C Bob’s input x 0 1 1 Encrypted circuit Oblivious transfer of keys

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A More General Solution for N Parties: Secure Multi-Party Computation Split the computation into logical steps (ANDs, ORs, NOTs) or algebraic steps (ADD, MULT) Securely evaluate step by step [GMW, BGW, …]

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Conclusion Tell me how you could detect systemic risk given complete information… …and I will tell you how to do it via a privacy- preserving protocol!

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