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Dana Moshkovitz MIT Joint work with Subhash Khot, NYU 1

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Best Approximations Known For 2CSP vs. 3CSP 23 OR [FG95] 7/8 XOR [GW92] 1/2 AND [FG95] 1/2 any better approximation: NP-hard [Håstad97] 2 MAX-CUT: If exists cut with (1- ± ) fraction of edges, efficiently cut at least (1-c ± ) fraction of edges.

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1 2 (n) n O(1) exponential polynomial running- time Exponential hardness: time 2 n 1-o(1) Sharp threshold: Drop at ½+o(1) Assuming it takes 2 (n) time to solve 3SAT exactly on size n inputs. 1/2 The Complexity of Approximating 3XOR [=3LIN(GF(2))] [ AroraSafra92, AroraLundMotwaniSudanSzegedy92, Raz94, BellareGoldreichSudan95, Håstad97, MRaz08 ] Approx. factor 3

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23 OR [FG95] 7/8 XOR [GW92] 1/2 AND [FG95] 1/2 Best Approximations Known For 2CSP vs. 3CSP any better approximation: NP-hrd [Håstad97] Any better approximation: NP-hard assuming the Unique Games Conjecture [K02,KKMO04,R08] 4 Hard time showing easy problems are hard!

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Proving Hardness of Constraint Satisfaction Problems 5 The Bellare-Goldreich-Sudan-Håstad paradigm 2CSP(GF(q)) desired problem composition with long- code/dictator code [Khot02]: Start with 2LIN(GF(q)). The Unique Games Conjecture: 2LIN(GF(q)) is extremely hard to approximate.

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The Unique Games Conjecture (UGC) - The Hardness of 2LIN Over Large Alphabets 6 Unique Games Conjecture [Khot02, formulation by KKMO04] : For all ,δ>0, for sufficiently large q, given a set of linear equations of the form x i – x j = c (mod q) where 1-δ fraction of the equations can be satisfied, it is NP-hard to find an assignment that satisfies fraction of the equations. [Raghavendra08]: Assuming the Unique Games Conjecture, the best approximation for all CSPs is obtained by rounding a natural SDP.

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Alphabet/Hardness Rules of Thumb: As alphabet gets larger, problem gets harder. Important Exception: problem can (but not necessarily) become easy if alphabet is R : linear/semidefinite programming can be used. Typically, Easy, if allow fractional solutions Hard, if encode large discrete alphabets (e.g., exact 3LIN( R ) [GR07]) 7

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A natural optimization problem: find a balanced assignment for real homogeneous linear equations: SDP-based algorithm: Best approximation known obtained by rounding natural SDP (when 1- equations exactly satisfiable, get margin O( ) for (1) of equations), – Unlike GF(q): Same for 2 or 3 variables per equation! We show NP-hardness for 3 vars per equation! Approach to proving the Unique Games Conj. 8 x 15 - x x 37 = 0 x 1 - 2x 3 + x 89 = 0 ... Work with Subhash Khot, 2010: Margin of equation ax+by+cz = 0 is |ax+by+cz|

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Main Theorem Theorem [KM10]: For any ± >0, it is NP-hard, given a system of linear equations over reals where (1- ± ) fraction can be exactly satisfied by a balanced assignment, to find an assignment where 0.99 fraction of the equations have margin at most ¢ ±. Tight: Efficient algorithm gives margin O( ) for (1) of equations. Blow up: Reduction from SAT has blow-up n n poly(1/ ), matching the recent result of Arora, Barak, Steurer. 9

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Approach to Proving The Unique Games Conjecture 1.Show that approximate 3LIN over the reals is NP-hard. 2.Show that approximate 2LIN over the reals is NP-hard, possibly by reduction from 3LIN. 3.Deduce the Unique Games Conjecture using parallel repetition. 10

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Approximation Algorithm Min E[| ax + by + cz | 2 ] s.t. assignment is balanced Analysis & Rounding. Assume ( 1- ) equations exactly satisfiable. Then the SDP minimum is at most O( ). Solve the SDP to get vector assignment. Pick random Gaussian ³, get real assignment x i = with similar value. The margin is ( ) for constant fraction of equations. SDP 11

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Hardness of Approximately Solving Real Linear Equations with Three Variables Per Equation 12

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Proving Hardness of Approximate Real Equations 1.Dictatorship test over reals. 2.Adapting the paradigm 13 The Bellare-Goldreich-Sudan-Håstad paradigm 2CSP(GF(q)) desired problem composition with long- code/dictator code dictatorship test for desired problem

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Real Functions Variables correspond to points in R n. Assignments correspond to functions R n R. We consider the Gaussian distribution over R n, where each coordinate has mean 0 and variance 1. Fact: If x,y are independent Gaussians, then px+qy is Gaussian where p 2 +q 2 = 1. 14

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F F Dictator Testing Tester Function F: R n R. Two possibilities for F. A tester queries three positions x,y,z R n, tests aF(x)+bF(y)+cF(z) = 0 ? If dictator, equation satisfied exactly with probability 1- . If not approximated by linear junta, there is a margin with probability

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Noise Sensitivity Approach to Dictator Testing Pick Gaussian x 2 R n. Perturb x by re-sampling each coordinate with probability ±. Obtain x’. Check F(x)=F(x’). 16

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Problem with Noise Sensitivity Approach For F half-space, – With probability 1-c ±, equality holds. – With probability c ±, constant margin. We want: with probability ¸ 0.01, margin ¸ ±. 17

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Linear Non-Juntas Are Noise-Sensitive Observation: If F = i 2I a i x i for | I | À 1/ ± 2, then (F(x) – F(x’)) is Gaussian with variance ¸ ± . Hence |F(x)-F(x’)| is ¸ ± with prob ¸ Basic Idea: test linearity and noise sensitivity. 18

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Hermite Analysis – Fourier Anlysis for Real Functions Fact: For any F: R n R with bounded |f| 2, can write F = c i 1,…, i n ¢ H i 1 (x 1 ) ¢ … ¢ H i n (x n ), where H d is a polynomial (“Hermite polynomial”) of degree d. Notation: F · 1 is the linear part of F, and F >1 (x) is its non-linear part. 19

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Linearity Testing Pick Gaussian x,y 2 R n, Pick p 2 [0,1]; set q= (1-p 2 ). Check F(px+qy)=pF(x)+qF(y). Lemma: |F-F · 1 | 2 2 · E[Margin 2 ]. Proof: Via Hermite analysis. 20 Equation depends on three variables!

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Decompose into linear and non-linear part: E x |F(x)-F(x’)| 2 = E x |F · 1 (x)-F · 1 (x’)| 2 + E x |F >1 (x)-F >1 (x’)| 2 Non junta ) with prob 0.1 margin ¸ 0.1 Linearity test ) | F >1 (x )| 2 · Cancellation problem: On average | F >1 (x )- F >1 (x ’)| · , but it can be much larger when |F · 1 (x)-F · 1 (x’)| is large and cancel |F · 1 (x)-F · 1 (x’)| ! 21

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Cancelations Don’t Arise! coordinate-wise perturbation x » c x’: x’ is obtained from x by re-sampling with prob ±. random perturbation x » r x’: x’ is obtained from x as px+qy for p=(1- ± ), p 2 +q 2 =1 Cancelation: P x » c x’ [|F >1 (x)-F >1 (x’)| 2 ¸ 0.1 ] ¸ 0.1 By linearity testing: P x » r x’ [|F >1 (x)-F >1 (x’)| 2 · 0.01 ] ¸ 0.99 We’ll show that they cannot co-exist! 22

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Cancelations Don’t Arise! coordinate-wise perturbation x » c x’: x’ is obtained from x by re-sampling with prob ±. random perturbation x » r x’: x’ is obtained from x as px+qy for p=(1- ± ), p 2 +q 2 =1 Cancelation: P x » c x’ [|F >1 (x)-F >1 (x’)| 2 ¸ 0.1 ] ¸ 0.1 By linearity testing: P x » r x’ [|F >1 (x)-F >1 (x’)| 2 · 0.01 ] ¸ 0.99 We’ll show that they cannot co-exist! 23 Claim: For any G: R n R, E x » r x’ |G(x)-G(x’)| 2 ¸ E x » c x’ |G(x)-G(x’)| 2.

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Main Lemma: From Small Difference to Constant Difference Main Lemma: If for H: R n R, P x » c x’ [|H(x)-H(x’)| 2 ¸ A ¢ T] ¸ 10 ®, P x » r x’ [|H(x)-H(x’)| 2 · T] ¸ 1- ®, Then, there is Boolean B: R n {0,1} where P x » c x’ [|B(x)-B(x’)| 2 = 1] ¸ 0.01, P x » r x’ [|B(x)-B(x’)| 2 = 0] ¸ 1-1/A, 24 Proof: Combinatorial, by considering “perturbation graphs”, and constructing an appropriate cut.

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Thank you! 25

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Proof of Main Lemma Perturbation graphs: Our task: Find a cut in C µ R n that cuts: E c edges of weight ¸ E r edges of weight ·

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Cutting The Gaussian Space Claim: There is a distribution D over cuts s.t.: Every edge e 2 E c is cut with prob ¸ 0.1 . Every edge e 2 E r is cut with prob · 0.01 -1/21/2 H(x)H(x’) 27

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From Small Cuts To Large Cuts In previous lemma, cut only ¼ weight. Want to cut constant weight. 28

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Larger Cut Let m=11/ . Sample from D cuts C 1,…,C m. Pick random Iµ [m]. Let C = i 2I C i. Claim: Edge e 2 E c is cut with prob ¸ 0.5 ¢ (1-1/e 2 ). Edge e 2 E r is cut with prob 0.01 ¢ 10/ 0.1 Corr: A cut with 0.2 ¢ 0.1 weight of E c and 0.8 ¢ 0.99 weight of E r. 29

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