Princeton University COS 433 Cryptography Fall 2005 Boaz Barak COS 433: Cryptography Princeton University Fall 2005 Boaz Barak Lecture 3: Computational Hardness

2 “Impatient Cryptography” Last Lecture: Simple algorithm (~10 lines of C code) to break every encryption with key-length < message-length. Caveat: Takes roughly 2 key length steps to run. For 1KB key will take more than galaxy’s lifetime to finish. This holds even if we run in parallel one processor for each atom of the universe each running in light speed. Encryption that takes this long to break seems to provide “pretty good privacy”.

3 Measuring Computational Steps Church-Turing thesis: every physically feasible computation can be described roughly as follows memory in initial state s Step 1Step 2Step iStep i+1Final state t Captures every computer or other computing device ever built, including human brain. Bit in state i+1 depends on constant (e.g. 3) bits of state i. W.l.o.g, each bit depends on · 2 bits, using either Æ, Ç or : functions. We call this a t ¢ s –sized Boolean circuit.

4 Computational Hardness Fact 1: 8f:{0,1} n  {0,1} n, 9 n 2 2 n -sized Boolean circuit C s.t. C computes f. Def: Let f:{0,1} n  {0,1} m be a function. We say that f is (T,  )-hard if for every T-sized circuit C Pr x  { 0,1 } n [ C(x) = f(x) ] <  Thm: 8n, there exists a function f:{0,1} n  {0,1} n that is (2 n/10,2 -n/10 )-hard We know: If f:{0,1} n  {0,1} n then f can’t be (n 2 2 n,1)- hard and can’t be (T,0)-hard for any T (where T>n).

5 Asymptotic Hardness Def: A function h:N  N is super-polynomial if 8 polynomial q(¢), 9n s.t. h(n)>q(n). In other words h(n)=n  (1). Def: Let { f n } be family of functions, where f n :{0,1} n  {0,1} n. We say that { f n } is hard if 9 super-poly h(¢) s.t. for every n, f n is (h(n), 1/h(n) ) –hard. Examples: h(n) = 2 n, h(n)=n log n, h(n) = 2 n, h(n) = 1/3 1 n< n n¸10 6 Def: Let f:{0,1} n  {0,1} m be a function. We say that f is (T,  )-hard if for every T-sized circuit C Pr x  { 0,1 } n [ C(x) = f(x) ] <  We know: There exists a hard family of functions.

6 Asymptotic Easyness Def: Let { f n } be family of functions, where f n :{0,1} n  {0,1} m. We say that { f n } is in the class P/poly if 9 polynomial q(¢) and circuit family { C n } s.t. 8n |C n | · q(n) and We say that a circuit family {C n } is uniform if there’s a (constant-size) Turing machine that on input n,i outputs the i th gate of C n in · |C n | 2 steps. 8 x 2 {0,1} n, C(x) = f(x) If the family {C n } is uniform then { f n } is in the class P. Def: Let { f n } be family of functions, where f n :{0,1} n  {0,1} n. We say that { f n } is hard if 9 super-poly h(¢) s.t. for every n, f n is (h(n), 1/h(n) ) –hard.

7 World View P - easy for poly-size uniform circuits P /poly - easy for poly-size circuits All function families Mildly hard func. families Hard function families

8 Functions Known to be in P Arithmetic functions: f 1 (a,b) = a+b, f 2 (a,b) = a¢ b, etc.. (elementary school algorithms) Arithmetic mod n: f 1 (a,b,n) = a+b (mod n), f 2 (a,b,n) = a¢ b (mod n) GCD(a,b) = largest c s.t. c|a, c|b Inverse(a): If GCD(a,n)=1 find a -1 (mod n) EXP(a,b) = a b (mod n) Maximal Ind. Set: Given graph G(V,E), we say that Sµ V is independent if 8 u,v 2 S, (u,v)  E. Find S s.t. forall v  S, S[{ v } is not independent.

9 Functions Believed Outside of P/poly Factoring: f 1 (a) = prime factors of a (can be computed in time 2 |a|, even in 2 |a| ) 1/3 Maximum Ind. Set: Given graph G(V,E), find S of maximum size s.t. S independent (i.e., 8u,v2 S, (u,v)  E) Believed to be a hard function family 3SAT: Given 3CNF Boolean formula  on n variables, find (if exist) x 1,…,x n such that  (x 1,…,x n ) is true.  = (x 1 Ç : x 7 Ç x 9 ) Æ (: x 2 Ç x 9 Ç : x 12 ) Æ … Æ (x 1 Ç x 2 Ç x 100 ) For all these functions, it is easy to verify a given solution. Such functions are in the class NP If either max-ind-set or 3SAT is in P/poly then every problem in NP is in P/poly Such functions are called NP-complete

10 Proving NP - Completeness Maximum Ind. Set: Given graph G(V,E), find S of maximum size s.t. S independent (i.e., 8u,v2 S, (u,v)  E) 3SAT: Given 3CNF Boolean formula  on n variables, find (if exist) x 1,…,x n such that  (x 1,…,x n ) is true.  = (x 1 Ç : x 7 Ç x 9 ) Æ (: x 2 Ç x 9 Ç : x 12 ) Æ … Æ (x 1 Ç x 2 Ç x 100 ) If either max-ind-set or 3SAT is in P/poly then every problem in NP is in P/poly This is proven using a reduction. For example. Circuit to compute IND-SET Thm: Assume that 9 poly(n) circuit family for IND-SET. Then 9 poly(n) circuit family for 3SAT. Circuit to compute 3SAT GGSS  xx

11 World View P - easy for poly-size uniform circuits P /poly - easy for poly-size circuits All function families Mildly hard func. families Hard function families add, mult, GCD, EXP, maximal IS 3SAT, Ind-SET, NP-Complete Factoring

12 Randomization Possible objection to our model: Can circuits get more power by tossing coins? Input Random bits Probabilistic circuit Output In this context, answer is NO. Thm: If f:{0,1} n  {0,1} m is (T,  )-hard for standard (deterministic) circuits, then it is (T/10,  )-hard for probabilistic circuits.