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UVA CS 588 - Security through complexity

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Security through complexity Ana Nora SovarelProjectsTuring MachineDefinitionNondeterministic Turing MachineSlide 6More Power?Complexity ClassesReductionMore definitions …How do we prove a problem is hard?Cook’s Theorem (‘71)Subset SumSlide 14Subset Sum is in NPSlide 161. Algorithm1. Algorithm (cont.)Slide 192. Reduction ‘’3. Reduction ‘’4. PolynomialBack to cryptology Knapsack CipherDecryptionLinear Time DecryptionHow to build the keys?Slide 28Trade offsEvaluationKnapsack Cipher - SummaryShamir’s break (’82)Shamir’s break (cont.)Slide 34A little bit of historyConclusion1Security through complexityAna Nora Sovarel2ProjectsPlease fill one slot on the signup sheet.One meeting for each group.All members must agree.3Turing MachineFinite Control0 0 1 1 0 0 1 0 0 04DefinitionA Turing Machine is a 7-tuple (Q, ∑, Γ, δ, q0, qaccept, qreject) where Q, ∑, Γ are finite sets and1. Q is the set of states2. ∑ is the input alphabet3. Γ is the tape alphabet4. δ : Q X Γ  Q X Γ X {L,R} is the transition function5. q0 is the start state6. qaccept is the accept state7. qreject is the reject state, where qaccept ≠ qreject5Nondeterministic Turing MachineFinite Control0 0 1 1 0 0 1 0 0 0Finite Control0 0 1 1 0 0 1 0 00Finite Control0 0 0 1 0 0 1 0 006DefinitionA Turing Machine is a 7-tuple (Q, ∑, Γ, δ, q0, qaccept, qreject) where Q, ∑, Γ are finite sets and1. Q is the set of states2. ∑ is the input alphabet3. Γ is the tape alphabet4. δ : Q X Γ P(Q X Γ X {L,R}) is the transition function5. q0 is the start state6. qaccept is the accept state7. qreject is the reject state, where qaccept ≠ qreject7More Power?Does nondeterminism affect the power of Turing Machine?NO – more power means it recognizes more languagesBut, maybe it can do things faster …8Complexity Classes•P = decidable in polynomial time by a deterministic TM•NP = decidable in polynomial time by a nondeterministic TM9Reductionf – polynomial time transformation What we know about A and B?A is at most as hard as B ( can be easier if we find another way to solve it )B is at least as hard as A.A’s Inputf(A) BYes/NoB’s Input10More definitions …•NP-Hard = the set of problems Q such that any problem Q’ in NP is polynomial reducible to it.•NP-complete = the problems Q such that Q is in NP-Hard and Q is in NP11How do we prove a problem is hard?•Let A be a known hard problem•Find a polynomial transformation from A’s input to your problem’s input•Why it works? –If your problem is easy ( P ) then we can solve A easy ( P ).–So A is not hard. Contradiction•Need a hard problem to start with ….12Cook’s Theorem (‘71)SAT is NP-complete. ( SAT = given a boolean formula, is it satisfiable? )3SAT is NP-complete.Example: Ф(x1,x2,x3,x4)=(x1+x2+x3)(x’1+x3+x4)13Subset SumGiven a set {x1,x2,…,xn} of integers and an integer t, find {y1,y2,…,yk} a subset of {x1,x2,…,xn} such that:kiiyt114Subset SumTo prove NP-complete:1. Prove is in NP•Verifiable in polynomial time•Give a nondeterministic algorithm 2. Reduction from a known NP-complete problem to subset sum•Reduction from 3SAT to subset sum15Subset Sum is in NPsum = 0A = {x1,x2,…,xn}for each x in Ay  choice(A)sum = sum + yif ( sum = t ) then successA  A – {y}donefail16ReductionGoal: Reduce 3SAT to SUBSET-SUM.How: Let Ф be a 3 conjunctive normal formformula. Build an instance of SUBSET-SUMproblem (S, t) such that Ф is satisfiable if and only if there is a subset T of S whoseelements sum to t.Prove the reduction is polynomial.171. AlgorithmInput: Ф - 3 conjunctive normal form formulaVariables: x1, x2, …, xlClauses: c1,c2,…,ck.Output: S, t such that Ф is satisfiable iff there is T subset of Swhich sums to t.181. Algorithm (cont.)x1x2…. xlc1c2…. cky11 0 0 1 0 0z11 0 0 0 1 0y21 0 0 0 1z21 0 0 0 0…yl1 0 0 0zl1 0 0 0g11 0 0h11 0 0g21 0h21 0…gk1hk1t 1 1 … 1 3 3 … 3191. Algorithm (cont.)(yi,xj), (zi,xj) – 1 if i=j, 0 otherwise (yi,cj) – 1 if cj contains variable xi, 0 otherwise(zi,cj) – 1 if cj contains variable x’i, 0 otherwise(gi,xj), (hi,xj) – 0(gi,cj), (hi,cj) – 1 if i=j, 0 otherwiseEach row represents a decimal number.S={y1,z1,..,yl,zl,g1,h1,…,gk,hk}t is the last row in the table.202. Reduction ‘’Given a variable assignment which satisfiesФ, find T.1. If xi is true then yi is in T, else zi is in T2. Add gi and/or hi to T such all last k digits of T to be 3.213. Reduction ‘’Given T a subset of S which sums to t, find avariable assignment which satisfies Ф.1. If yi is in T then xi is true2. If zi is in T then xi is false224. PolynomialTable size is (k+l)2O(n2)23Back to cryptology •P=NP is still an open question•factorization is not known to be NP-complete•cipher based on a known NP-complete problem24Knapsack Cipher•Public Key: {a1,a2,…,an} set of integers•Plain Text: x1…xn•Cipher Text: [Merkle and Hellman, ’78]niiiaxs125Decryption•Based on an easier problem•{a1,a2,…,an} is a superincreasing sequence11ijjiaa26Linear Time Decryption•xn = 1 iff •Solve it recursively on {a1,a2,…,an-1} and s - xnanniias127How to build the keys?•Modular multiplication (Merkle and Hellman)•Starts with superincreasing sequence {b1,b2,…,bn} •Choose M and W such that•Compute {a1,a2,…,an} such that1),(,1WMGCDaMniiMWbaiimod)(28Decryption•C = (s W-1) mod M, where (W-1W) mod M = 1•Solve subset sum problem with superincreasing sequence {b1,b2,…,bn} and sum c.29Trade offs•bi large  M large  n bits encoded with log2M bits•bi small  easy to break–If bi = 1  aj = W.–Break O(n)•Merkle and Hellman recommended: b1 ≈ 2n, , bn ≈ 22n 12,11nibbijji30Evaluation+ speed ( 100 times faster than RSA )-needs twice the communication capacity (m bits encoded into approximate 2m bits)-larger public key(2n2 bits, 20,000 for n=100, RSA - 500)? security31Knapsack Cipher - Summary•Secret –superincreasing sequence {b1,b2,…,bn}–M–W•Public–{a1,a2,…,an}Remember:MWba ii mod)(32Shamir’s break (’82)•based on the choice of superincreasing sequence• linear transformation to generate public key•What do we need to guess ?(Only one of W and M is enough)33Shamir’s break (cont.)Given the public key {a1,a2,…,an} find M and W such that (ai W) mod M is a superincreasing sequence.b1 = (ai W) mod M  b1 = ai W + k1Mb1/(Mai) =


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