Computational and Statistical Learning Theory

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1 Computational and Statistical Learning Theory TTIC Prof. Nati Srebro Lecture 6: Computational Complexity of Learning Proper vs Improper Learning

2 Learning Using FIND-CONS For any family of hypothesis classes consider: FINDCONS H : Given a sample S, either return h H n consistent with the sample (i.e. s.t. L S h = 0), or declare that none exists. Decision problem CONS H S = 1 iff h Hn s.t. L S h = 0. Theorem: If H n over X n = R n is efficiently properly PAC learnable, then VCdim H n poly(n), and CONS H RP 3 TERM DNF n = T 1 T 2 T 3 T i CONJ n E.g. h x = x 1 x 7 x 23 x 75 x 2 x 3 x 5 x 6 x 7

3 Hardness of 3-TERM-DNF Claim: CONS 3 TERM DNF is NP-hard Proof: Reduction from graph 3-colorability 3COLOR = { undirected graph G(V, E) c:v 1,2,3 u v E c u c(v) } Map G V, E S G s.t. G 3COLOR S G CONS 3 T DNF S G = x i = 1,1,1,, 1,0,1,, 1,1, y = 1 i = 1.. V } { x ij = 1,1, 1,0,1,1, 1,0,1,1, 1,1, y = 0 i j E i i Claim: G 3COLOR S G CONS 3 T DNF S G satisfied by T = T 1 T 2 T 3 where T k = c i k x i Claim:S G CONS 3 T DNF G 3COLOR c(i) = min k s.t. T k satisfies x i j

4 Hardness of Learning 3-TERM-DNF Conclusion: If NP RP, then 3 TERM DNF is not efficiently properly PAC learnable Proof: 3 TERM DNF efficiently properly PAC learnable CONST 3 TERM DNF RP 3COLOR RP RP = NP Theorem: Assuming NP RP, H n over X n = 0,1 n of X n = R n is efficiently properly PAC learnable iff: VCdim H n poly(n), and Required for learnability There is a poly-time algorithm for CONS H (polynomial in size of input) What about improper learning? Required for efficient proper learnability

5 Hardness of CONSISTENT Axis-aligned rectangles in n dimensions Halfspaces in n dimensions Conjunctions on n variables 3-term DNF s DNF formulas of size poly(n) Generic logical formulas of size poly(n) Python programs of size at most poly(n) Decision trees of size at most poly(n) Neural nets with at most poly(n) units Functions computable in poly(n) time CONSISTENT: Poly-time What does this imply? CONSISTENT: NP-Hard What does this imply?

6 Another Look at 3-TERM-DNF 3 TERM DNF n = T 1 T 2 T 3 T i CONJ n E.g. h x = x[1] x 7 x 23 x 2 x 3 x 6 x 7 Can also write: h x = x 1 x 2 x 6 x 1 x 2 x 7 x 1 x 3 x 6 x 1 x 3 x 7 x 7 x 2 x 6 x 7 x 2 x 7 x 7 x 3 x 6 x 7 x 3 x 7 x 23 x 2 x 6 x 23 x 2 x 7 x 23 x[3] x 6 x 23 x[3] x 7 Claim: 3 TERM DNF n 3CNF n r 3CNF n = T i T i = of three literals i=1 But 3CNF n CONJ disjunctions of three literals Can be efficiently learned using FINDCONS CONJ, treating each 2n 3 3-disjunction as a single variable I.e. map x ቀx 1 x 2 x 3, x 1 x 2 x 4,, x 1 x 2

7 Efficiently Learning 3-TERM-DNF Sample complexity of learning 3-TERM-DNF using FINDCONS 3CNF : O Compare with cardinality bound: O Why the gap? n Τε n 3 Τε 3CNF Relax statistically easy but computationally hard class to larger class that s statistically harder but computationally easier: 3-TERM-DNF CONS 3 T DNF CONS 3CNF Sample complexity O n Τε O n 3 ε Runtime 2 O n /ε O n 6 ε x 1 x 2 x 3 x 4

8 Hardness of CONSISTENT Axis-aligned rectangles in n dimensions Halfspaces in n dimensions Conjunctions on n variables CONSISTENT: Poly-time Efficiently Properly Learnable 3-term DNF s DNF formulas of size poly(n) Generic logical formulas of size poly(n) Decision trees of size at most poly(n) Neural nets with at most poly(n) units Functions computable in poly(n) time Python programs of size at most poly(n) CONSISTENT: NP-Hard Not Efficiently Properly Learnable Improperly Learnable?

9 Proper vs Improper Learning 3-TERM-DNF is not efficiently properly learnable, but is efficiently (improperly) learnable Are there classes that are not efficiently learnable, even improperly? What about the class of all functions computable in time poly(n)? Universal class: any H n where the predictors are poly-time computable is H n TIME poly n VCdim = O poly n CONS is NP-Complete: If P = NP we could actually learn this universal class! If RP NP, we can t efficiently properly learn it But can we improperly learn it? How do we show that a class cannot be learned, no matter what hypothesis we are allowed to output?

plaintext message s 0,1 n f K ( ) Learning and Crypto How to attack the cryptosystem: Usually assume you know the form

you know target is using a GSM phone, or ssh, or file format specifies encryption, or you captured an Enigma machine

g. s i is an English word, and so Pr s i 1 = "t"

10 plaintext message s 0,1 n f K ( ) Learning and Crypto How to attack the cryptosystem: Usually assume you know the form of f (but not K) e.g. you know target is using a GSM phone, or ssh, or file format specifies encryption, or you captured an Enigma machine Collect messages for which you know the plaintext, i.e. pairs s i, c i or have partial or statistical information on the plaintext s i, e.g. s i is an English word, and so Pr s i 1 = "t" = 0.17 Search for a key K s.t. c i = f K (s i ) Information Theoretically: Need m = log K n H(s) encrypted message c = f K (s) message pairs Only way to be secure: One Time Pad, i.e. K = 0,1 m n f K 1 ( ) Encryption key K K Claude Shannon decoded message s = f K 1 ( c)

11 Learning and Crypto plaintext message s 0,1 n f K ( ) encrypted message f 1 K ( ) c = f K (s) Encryption key K K decoded message s = f K 1 ( c) Reducing Code Breaking to Machine Learning: Hypothesis class f K 1 K K over domain X = {encrypted messages c} and targets plaintext messages s Training set c i, s i i=1..m, possibly noisy s i Learn decoder c s Proper learning: find decoder of the form f K 1 But really, finding any decoder that decodes most messages is fine Improper learning Statistically, enough to have m log H log K messages So it better be computationally hard Learning is computationally easy Breaking code is easy No crypto Crypto is possible learning is computationally hard

12 Discrete Cube Root Cryptosystem Choose primes p, q s.t. 3 (p 1)(q 1) Public (encryption) key: K = p q 2 n Private (decryption) key: D = 1 Τ3 mod p 1 q 1 f K s = s 3 mod K easy to compute from K and s in time O n 2 f K 1 c = c D mod K Proof: 3D = a p 1 q for some integer a and so mod K we have c D 3 = c 3D = c p 1 a c q 1 a c = 1 a 1 a c = c easy to compute from D and c in time O n 3 Hard to compute from K and c Hard in what sense? Worst case: no poly time alg that works for every c maybe only hard on crazy c, but we can decrypt most messages We would like: no poly time alg that works for any c impossible always return 3487, you ll be right on some c Enough: no poly time alg that works for random c can make c random by sending f K s r, r for random r

13 Cryptographic Assumption 1. There exists a poly-time algorithm ENC K, s = f K s = s 3 mod K 2. For every K K, there exists D(K) and a poly-time algorithm DEC D(K), c = f K 1 c = s D mod K, i.e. s.t. s S DEC D K, f K (s) = s 3. There is no poly-time (randomized) algorithm BREAK(K, c) s.t. K K Pr BREAK K, f 1 K s = s s~unif S,randomness in B poly n K, S 0,1 n, i.e. size of inputs is O(n) For discrete cube root: K = K = p 1 q 1 p, q prime, 2 n 2 K < 2 n S = 0,1 n 2 = s 0 s < 2 n 2 } For a useable public-key cryptosystem also need: A poly(n) algorithm for generating K, D K -- we won t use this Pr BREAK K, f k s = s 1 rather then K,s poly n K Pr 1 s ( K enough for us) poly n

14 Hardness of Learning Discrete Cube Root H n = c, i f K 1 c [i] K K, i = 1.. n 2 VCdim H n log H n log 0,1 n = n All predictors in H n are computable in time O n 3 Claim: Subject to discrete cube root assumption, H n is not efficiently PAC learnable. Proof: Given poly-time learning algorithm A, construct BREAK(K, c): For i=1..n, h i A D i, ε = 1 4n, δ = 1 4n D i x, y : a Unif 0,1 n 2 x = (f K a, i), y = a[i] Return [h 1 c, h 2 c,, h n 2 (c)] W.p. 1 n 2 > 3 (over randomness in BREAK) learning worked for all i, and in that 4n 4 case, the probability to err on some bit is n 2 < 1 (over the choice of c), and so overall 4n 4 w.p.>1/2, BREAK K, c = f 1 K (c).

15 Crypto Hardness of Discrete Cube Root Hardness of Learning We identified a family of functions that is provably not efficiently PAC-learnable Despite having polynomial VC-dimension Despite containing only easily computable functions Subject to an assumption about security of a cryptosystem Since H n TIME O n 3 efficiently PAC-learnable, this implies poly-time computable functions are not In fact, if there is any secure public-key crypto-system, then poly-time computable functions are not efficiently PAC-learnable Even stronger: if we can efficiently learn poly-time computable functions then no cipher is secure to known-plaintext attacks

16 Learning and Crypto plaintext message s 0,1 n f( ) encrypted message c = f(s) f 1 ( ) Any poly-time encryption function f with a poly-time computable decryption function f 1 decoded message s = f 1 ( c) Reducing Breaking an unknown code to learning poly-time functions: Hypothesis class poly time computable h: X Y over domain X = {encrypted messages c} and targets plaintext messages s Training set c i, s i i=1..m, possibly noisy s i Learn decoder c s Conclusion: if we can efficiently learn poly-time computable functions, we can break any cipher with a known-plaintext attack If there exists any secure cipher system, then poly-time functions are not efficiently PAC learnable

17 Hardness of Learning via Crypto Public-key crypto is possible (one way functions exist) hard to learn poly-time functions Hardness of Discrete Cube Root hard to learn log(n)-depth logic circuits hard to learn log(n)-depth poly-size neural networks Hardness of breaking RSA hard to learn poly-length logical formulas hard to learn poly-size automata hard to learn push-down automata, ie regexps for some depth d, hard to learn poly-size depth-d threshold circuits (output of unit is one iff number of input units that are one is greater than threshold) Hardness of lattice-shortest-vector based cryptography hard to learn intersection of n r halfspaces (for any r > 0)

Computational and Statistical Learning Theory

Computational and Statistical Learning Theory TTIC 31120 Prof. Nati Srebro Lecture 6: Computational Complexity of Learning Proper vs Improper Learning Efficient PAC Learning Definition: A family H n of