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Last Time: Summary. Hypercube Eigenvalues of hypercube. V = { 0 , 1 - PowerPoint PPT Presentation

Last Time: Summary. Hypercube Eigenvalues of hypercube. V = { 0 , 1 } d ( x , y ) E when x and y differ in one bit. Graph G = ( V , E ) , assume regular graph of degree d . | V | = 2 d | E | = d 2 d 1 . | E ( S , V S ) | Edge


  1. Last Time: Summary. Hypercube Eigenvalues of hypercube. V = { 0 , 1 } d ( x , y ) ∈ E when x and y differ in one bit. Graph G = ( V , E ) , assume regular graph of degree d . | V | = 2 d | E | = d 2 d − 1 . | E ( S , V − S ) | Edge Expansion. h ( S ) = d min | S | , | V − S | , h ( G ) = min S h ( S ) Anyone see any symmetry? M = A / d adjacency matrix, A Coordinate cuts. +1 on one side, -1 on other. Eigenvector: a vector v where Mv = λ v ( Mv ) i = ( 1 − 2 / d ) v i . Spectral theorem: Eigenvectors form basis: v 1 ,..., v n . Eigenvalue 1 − 2 / d . d Eigenvectors. Why orthogonal? x = α 1 v 1 + α 2 v 2 + ··· α n v n . Mx = α 1 λ 1 v 1 + α 2 λ 2 v 2 + ··· α n λ n v n Next eigenvectors? Good cuts? “Coordinate cut”: d of them. Highest eigenvalue: λ 1 = 1. Proof: Plug in 1 . 2 d − 1 d 2 d − 1 = 1 Delete edges in two dimensions. Edge expansion: Second Eigenvalue: λ 2 < 1 if connected. Proof: v 2 is not v 1 . d Four subcubes: bipartite. Color ± 1 Ball cut: All nodes within d / 2 of node, say 00 ··· 0. Eigenvalue gap: µ = λ 1 − λ 2 . � d Cheeger: µ � � d � Vertex cut size: bit strings with d / 2 1’s. � 2 ≤ h ( G ) ≤ = 2 µ Eigenvalue: 1 − 4 / d . eigenvectors. d / 2 2 ≈ 2 d √ Proof of LHI: Plug in “cut” vector, x , into Rayleigh Quotient. � d � Eigenvalues: 1 − 2 k / d . eigenvectors. d k µ = 1 − max x ⊥ 1 x t Mx 1 x t x . Vertex expansion: ≈ √ d . This expression ’counts’ edges in cut ’ x ’ plus scales by volume. 1 Edge expansion: d / 2 edges to next level. ≈ √ Yields h ( S ) . 2 d √ Worse by a factor of d Back to Cheeger. Cycle Find x ⊥ 1 with Rayleigh quotient, x T Mx x T x close to 1. Coordinate Cuts: � i − n / 4 if i ≤ n / 2 Eigenvalue 1 − 2 / d . d Eigenvectors. x i = 3 n / 4 − i if i > n / 2 Tight example for Other side of Cheeger? 2 = 1 − λ 2 µ � � ≤ h ( G ) ≤ 2 ( 1 − λ 2 ) = 2 µ 2 µ 2 = 1 − λ 2 � � ≤ h ( G ) ≤ 2 ( 1 − λ 2 ) = 2 µ Hit with M . For hypercube: h ( G ) = 1 2 d λ 1 − λ 2 = 2 / d . Cycle on n nodes. Left hand side is tight.  − n / 4 + 1 / 2 if i = 1 , n Will show other side of Cheeger is tight.  Note: hamming weight vector also in first eigenspace.  ( Mx ) i = n / 4 − 1 if i = n / 2 Edge expansion:Cut in half.  x i otherwise Lose “names” in hypercube, find coordinate cut?  | S | = n / 2, | E ( S , S ) | = 2 → h ( G ) = 2 Find coordinate cut? n . → x T Mx = x T x ( 1 − O ( 1 → λ 2 ≥ 1 − O ( 1 n 2 )) n 2 ) Show eigenvalue gap µ ≤ 1 Eigenvector v maps to line. n 2 . µ = λ 1 − λ 2 = O ( 1 n 2 ) Cut along line. Find x ⊥ 1 with Rayleigh quotient, x T Mx x T x close to 1. n = Θ( √ µ ) h ( G ) = 2 Eigenvector algorithm gets a linear combination of coordinate cuts. Something like ball cut. µ 2 = 1 − λ 2 � � ≤ h ( G ) ≤ 2 ( 1 − λ 2 ) = 2 µ 2 Find coordinate cut? Tight example for upper bound for Cheeger.

  2. Eigenvalues of cycle? Random Walk. p - probability distribution. Probability distrubtion after choose a random neighbor. Eigenvalues: cos 2 π k n . Mp . x i = cos 2 π ki Converge to uniform distribution. n Eigenvalues, random walks, volume estimation, counting. � � � � � � � � 2 π k ( i + 1 ) 2 π k ( i − 1 ) 2 π k 2 π ki Power method: M t x goes to highest eigenvector. ( Mx ) i = cos + cos = 2cos cos n n n n M t x = a 1 λ t 1 v 1 + a 2 λ 2 v 2 + ··· Eigenvalue: cos 2 π k n . λ 1 − λ 2 - rate of convergence. Eigenvalues: Ω( n 2 ) steps to get close to uniform. vibration modes of system. Fourier basis. Start at node 0, probability distribution, [ 1 , 0 , 0 , ··· , 0 ] . Takes Ω( n 2 ) to get n steps away. Recall drunken sailor. Sampling. Convex Bodies. Convex Body Graph. S ⊂ [ k ] n is set of grid points inside Convex Body. S ⊂ [ k ] n is grid points inside Convex Body. Sample Space: S . Ex: Numerically integrate convex function in d dimensions. Compute ∑ i v i Vol ( f ( x ) > v i ) where v i = i δ . Graph on grid points inside P or on Sample Space. Example: P defined by set of linear inequalities. One neighbor in each direction for each dimension Or other “membership oracle” for P (if neighbor is inside P .) Sampling: Random element of subset S ⊂ { 0 , 1 } n or { 0 ,..., k } k . Degree: 2 d . S is set of grid points inside Convex Body. Related Problem: Approximate | S | within factor of 1 + ε . Grid points that satisfy linear inequalities. How big is graph? Big! Random walk to do both for some interesting sets S . or “other” membership oracle. So big it ..it INSERT JOKE HERE. O ( k n ) if coordinates in [ k ] . Choose a uniformly random elt? Easy to choose randomly from [ k ] n which is big. That’s a big graph! For convex body? How to find a random node? Choose random point in [ k ] n and check if in P . Start at a grid point, and take a (random) walk. Works. But P could be exponentially small compared to | [ k ] n | . When close to uniform distribution...have a sample point. Takes a long time to even find a point in P . How long does this take? More later. But remember power method...which finds first eigenvector.

  3. Spanning Trees. Spin systems. Sampling structures and the BIG GRAPH Each element of S may have associated weight. Sample element proportional to weight. Problem: How many? Example? Another Problem: find a random one. 2 or 3 dimensional grid of particles. Particle State ± 1. System State {− 1 , + 1 } n . Algorithm: Start with spanning tree. Sampling Algorithms ≡ Random walk on BIG GRAPH. Small degree. Energy on local interactions: E = ∑ ( i , j ) − σ i σ j . Repeat: Vertices Neighbors Degree (ish) Grid points in convex body. Change one dimension 2 d “Ferromagnetic regime”: same spin is good. ≤ | V | 2 neighbors per node Swap a random nontree edge with a random tree edge. Spanning Trees. Change two edges. Gibbs distribution ∝ e − E / kT . Spin States. Change one spin O ( n ) neighbors. How long? Physical properties from Gibbs distribution. Sample space graph (BIG GRAPH) of spanning trees. Metropolis Algorithm: Node for each tree. Neighboring trees differ in two edges. At x , generate y with a single random flip. Go to y with probability min ( 1 , w ( y ) / w ( x )) Algorithm is random walk on BIG GRAPH (sample space graph.) Random walk in sample space graph (BIG GRAPH ALERT) (not random walk in 2d grid of particles.) Markov Chain on statespace of system. Analyzing random walks on graph. Fix-it-up chappie! Rapid mixing, volume, and surface area.. “Lazy” random walk: With probability 1 / 2 stay at current vertex. Start at vertex, go to random neighbor. Evolution Matrix: I + M Recall volume of convex body. For d -regular graph: eventually uniform. 2 Grid graph on grid points inside convex body. if not bipartite. Odd /even step! Eigenvalues: 1 + λ i 2 Recall Cheeger: µ 2 ( v i + λ i v i ) = 1 + λ i � How to analyse? 2 ( I + M ) v i = 1 1 2 ≤ h ( G ) ≤ 2 µ . 2 v i Eigenvalues in interval [ 0 , 1 ] . Random Walk Matrix: M . Lower bound expansion → lower bounds on spectral gap µ Spectral gap: 1 − λ 2 = µ → Upper bound mixing time. M - normalized adjacency matrix. 2 . 2 h ( G ) ≈ Surface Area Symmetric, ∑ j M [ i , j ] = 1. Uniform distribution: π = [ 1 N ,..., 1 Volume N ] M [ i , j ] - probability of going to j from i . Distance to uniform: d 1 ( v t , π ) = ∑ i | ( v t ) i − π i | Isoperimetric inequality. Probability distribution at time t : v t . “Rapidly mixing”: d 1 ( v t , π ) ≤ ε in poly ( log N , log 1 Vol n − 1 ( S , S ) ≥ min ( Vol ( S ) , Vol ( S )) ε ) time. v t + 1 = Mv t Each node is average over neighbors. diam ( P ) When is chain rapidly mixing? Evolution? Random walk starts at 1, distribution e 1 = [ 1 , 0 ,..., 0 ] . Edges ∝ surface area, Assume Diam ( P ) ≤ p ′ ( n ) Another measure: d 2 ( v t , π ) = ∑ i (( v t ) i − π i ) 2 . M t v 1 = 1 N v 1 + ∑ i > 1 λ t i α i v i . √ → h ( G ) ≥ 1 / p ′ ( n ) Note: d 1 ( v t , π ) ≤ Nd 2 ( v t , π ) v 1 = [ 1 N ,..., 1 → µ > 1 / 2 p ′ ( n ) 2 N ] → Uniform distribution. 1 n – “size” of vertex, µ ≥ p ( n ) for poly p ( n ) , t = O ( p ( n ) log N ) . → O ( p ′ ( n ) 2 log N ) convergence for Markov chain on BIG GRAPH. Doh! What if bipartite? � 2 t d 2 ( v t , π ) = | A t e 1 − π | 2 ≤ � 2 p ( n ) ) 2 t ≤ → Rapidly mixing chain: ( 1 + λ 2 ) 1 1 Negative eigenvalues of value -1: (+ 1 , − 1 ) on two sides. ≤ ( 1 − 2 poly ( N ) Side question: Why the same size? Assumed regular graph. 1 Rapidly mixing with big ( ≥ p ( n ) ) spectral gap.

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