1. 1 Cryptography on weak BSS model of computation Ilir Çapuni ilir@cs.bu.edu

2. 2 Tripling an angle with ruler and compass X 3X If x is an angle, then we define f(x) := 3x

3. 3 Can we invert this function using the same tools? Algebra: “NO” Important assumption: we are working with straightedge and compass with infinite precision

4. 4 Identification using this function Initialization phase  Alice generates a secret angle X A, computes Y A =3 * X A and publishes Y A Protocol  Alice generates an angle S, and sends a copy of the it’s triple value R to Bob  Bob tosses a coin and sends a response to Alice  If Bob said “head” Alice will send a copy of S and Bob will verify if 3S=R  If Bob said “tail” Alice will send a copy of S+X A and Bob will check if Y A +R == 3*(S + X A )

5. 5 The structure Introduction of BSS model of computation Algebra recap Auxiliary results Cryptography with ruler and compass

6. 6 State space Computation node Output space …0x0x0 x1x1 x2x2 … x k-2 x k-1 xkxk... Input node 1 Input space Branch node Output node N Shifting node x l =0 otherwise Program is a finite directed graph Lin. map. I Lin. map. O Legend Polynomial (rational) function

7. 7 What if R = Z 2 ? … we have a Turing machine! State space Computation node Output space …0010… 110... Input node 1 Input space Branch node Output node N Shifting node x l =0 otherwise Program is a finite directed graph Lin. map. I Lin. map. O

8. 8 Some facts BSS model provides a framework for algorithms of Numerical Analysis Gives new perspective and adds additional (algebraic) flavor to P vs NP question  In the weak BSS model, there is unconditional separation between these two classes

9. 9 Discrepancies of this model Overly realistic Cheating … and a couple of other problems

10. 10 735,661.59 euros worth problem + 2 more 59.6 million Serbian dinars Is P = NP ? Is P R = NP R ? Is P C = NP C ? Transfer results  Theorem. P C = NP C if and only if P K = NP K where K is any algebraically closed field of characteristic 0 (say algebraic numbers)  Theorem. If P C = NP C then BPP contains NP Solve 1, get 2 for free!!!

11. 11 Talk progress Introduction of BSS model of computation Algebra recap Auxiliary results Cryptography with ruler and compass

12. 12 Algebraic preliminaries Element t is algebraic over the field F if it is a root of a polynomial over F[X] F(t) is the intersection of all fields containing F and t F(t)/F could be viewed as a vector space over F The dimension of this vector space is the degree of the extension

13. 13 Some previous work All parties start with 0 and 1 and can perform finitely many operations +, -, * and / Parties can sample real numbers from [0,1] State of knowledge of each party is the field that he/she can generate

14. 14 Talk progress Introduction of BSS model of computation Algebra recap Definitions and auxiliary results Cryptography with ruler and compass

15. 15 Algebraic one-way functions Easy to compute, but hard to invert Alice samples a real number r and computes r 2 It is impossible to deduce r from r 2 with infinite precision in finitely many steps P [ Q (t 1, t 2, …, t n, r 2 )  Q( r ) = Q] =1

16. 16 PK Encryption Alice samples a real number SK then she computes PK which is in Q (SK) m is a real number that Bob wants to send to Alice and c is its encryption using PK We have

17. 17 Who knows what? c, PK Q(PK), Q(SK), Q(SK,c) Q(PK), Q(PK,c), Q(PK,m) Q(PK), Q(PK,c)

18. 18 Results PKE is not possible since Q(PK,m)=Q(PK,c) Secure signature schemes are impossible Secret key exchange is impossible

19. 19 Talk progress Introduction of BSS model of computation Algebra recap Auxiliary results Cryptography with ruler and compass

20. 20 Constructability OA is a unit segment in complex plane O(0,0), A(0,1) Point M(x,y) is constructible if it can be constructed in finitely many steps using ruler and compass from OA

21. 21 Axioms of constructability Points O and A are constructible If B and C are constructible, then segment BC and the line defined by them are constructible Circle with constructible center and radius is constructible Intersection of 2 constructible rays is a constructible point Intersection of 2 constructible circles are constructible points Intersections of constructible circle and constructible ray are constructible points

22. 22 Algebraic facts Set of all constructible points on C is called Pitaghorean plane If M(x,y) is constructible, then x and y are constructible real numbers The set of all constructible real numbers is a subfield of the field of real numbers

23. 23 Computing vs constructing If K=Q(S), S = set of coordinates of the points from the set which contains at least O and A Every line has an equation of the form Every circle has an equation

24. 24 Facts Theorem: If M(x,y) is constructible in one step, then K(x,y) = K or to a quadratic extension of K Theorem: a) For every constructible point M(x,y) there exists a finite sequence of subfields K i, i=0,1,…, m each of which is quadratic extension of the previous one such that K 0 =K, and K m subset of R and x,y are elements of K m b) x and y are algebraic over K and their degrees over K are powers of 2 c) Every point with coordinates in K or any of its quadratic extensions is constructible

25. 25 Computational model We use BSS model over the field of complex numbers Each party can sample random points from unit circle Each party can also toss a coin The state of knowledge of each party is the field he/she can generate

26. 26 Is our computational system complete? State space Computation node -10 Output space …0x0x0 x1x1 x2x2 … x k-2 x k-1 xkxk... Input node 1 Input space If -10=0 Output node N Computation node Sqrt(-10) x l =0 otherwise Program is a finite directed graph

27. 27 PK Encryption Euclid before publishing his Elements has sampled a point SK=(SK x,SK y ) and then he has computed PK=(PK x,PK y ) and published in page 655 of the XIV book Archimedes wants to send him a secret point M(x,y). Using Euclid’s PK he computes the ciphertext C(x c, y c ). Archimedes sends this point to Euclid

28. 28 But… Using previous results over the field K, we will have Malicious Romans that have copied C, enumerate all points and using encryption machine PK and X they obtain some C x. If C=C x then M=X

29. 29 So We have given a partial answer to Rivest, Shamir and Burmester’s question if the secure encryption could be performed with the ruler and compass  In the weak algebraic model, where operations are done with ruler and compass with infinite precision, “algebraic OWFs” exist, ZK identification protocols do exist… but, secure PK encryption is impossible