Invariants Risi Kondor

Transcription

1 Risi Kondor

2 Let g R, and f : R C The real numbers, as a group acts on the space of functions on R by translation: g : f f where f (x) = f(x g) Question: How do we construct functionals Υ[f] that are invariant to this action, ie, for which Υ[f] = Υ[f ] for any f and any g? Many applications in signal processing, image analysis, etc 2 2/25

3 The of f is a(x) = f(x + y)f(y)dy Tells us how much f changes when we translate it by an amount y Clearly invariant to translation: 3 3/25

4 The of f is â(ω) = f(ω) f(ω) dx = f(ω) 2 dx Literally measures the amount of energy in each Fourier mode Clearly invariant to translation: â f (ω) = (e 2πiωg f(ω)) (e 2πiωg f(ω)) dx = f(ω) f(ω) dx = âf (ω) In fact, the power spectrum is just the Fourier transform of the autocorrelation Their common limitation: lose all the information in the phase 4 4/25

5 The of f is b(x 1, x 2 ) = f(y x 1 ) f(y x 2 ) f(y) dy The of f is b(k1, k 2 ) = f(k 1 ) f(k2 ) f(k1 + k 2 ) 5 5/25

6 f b b(k1, k 2 ) = f(k 1 ) f(k2 ) f(k1 + k 2 ) Use the following algorithm to recover f from b: 1 f(0) = b(0, 0) 1/3 2 f(1) = e iϕ b(0, 1)/ f(0) indeterminacy in ϕ inevitable 3 f(k + 1) = b(1, k) k = 2, 3, f(1) f(k) If f(k) 0 for any k, then b uniquely determines f up to translation 6 6/25

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8 G is a group acting on a set X transitively This means that each g G is a map g : X X, sending g : x gx L(X) is a space of functions on X The action of G on X extends to functions in L(X) by g : f f g where (f g )(x) = f(g 1 x) (Assume that gf L(X) for all f L(X) and g G) A functional Υ[f] is an to this action if Υ[f] = Υ[f g ] f L(X), g G 8 8/25

9 The rotation group SO(3) acts on the sphere S 2 by x Rx Consider L 2 (S 2 ) The symmetric group acts on the adjacency matrix of a graph by (i, j) (σ(i), σ(j)) 9 9/25

10 Restrict ourselves for now to finite X and finite G Recall that f induces a function f G (g) = f(gx 0 ), and the Fourier transform of f is f(ρ) = g G f(g) ρ(g) ρ R G Moreover, if f t (g) = f(t 1 g), then f t (ρ) = ρ(t) f(ρ) 10 10/25

11 The of a function f : X C is Clearly invariant because â(ρ) = f(ρ) f(ρ) f τ (ρ) f τ (ρ) = (ρ ρ (t) f(ρ)) (ρ ρ (t) f(ρ)) = f(ρ) f(ρ) The power spectrum is the FT of the (flipped) autocorrelation function a(h) = g G f(gh 1 )f(g) 11 11/25

12 Recall the Clebsch-Gordan decomposition ρ 1 (σ) ρ 2 (σ) = C ρ1,ρ 2 [ ρ R ρ1,ρ 2 c(ρ 1,ρ 2,ρ) i=1 [ bf (ρ 1, ρ 2 ) = C ρ 1,ρ 2 f(ρ1 ) f(ρ ] 2 ) Cρ1,ρ 2 The bispectrum is the FT of the b(h 1, h 2 ) = g G ] ρ(σ) C ρ 1,ρ 2 ρ Λ ρ1,ρ 2 f(gh 1 1 )f(gh 1 2 )f(g) c(ρ 1,ρ 2,ρ) i=1 f(ρ) 12 12/25

13 Let f and f be a pair of complex valued integrable functions on a compact group G Assume that f(ρ) is invertible for each ρ R Then f = f z for some z G if and only if b f (ρ 1, ρ 2 ) = b f (ρ 1, ρ 2 ) for all ρ 1, ρ 2 R Generalizes to any Tatsuuma duality group (ISO(n), ihomog Lorentz Group, etc) 13 13/25

14 The of f : S n C is the collection of matrices q h (ρ) = r h (ρ) f(ρ), ρ R G, G, with r h (g) = f(gh)f(g) Unitarily equivalent to the bispectrum, but sometimes easier to compute [K, 2007] 14 14/25

15 Risi Kondor

16 Given A, A R n n, the is to solve Equivalently, maximize σ S n f(σ) f(σ) = n A σ(i),σ(j) A i,j i,j=1 maximize σ S n tr(p σ AP σ A ) [P σ ] i,j = { 1 if σ(j) = i 0 else Can interpret A and A as the adjacency matrices of two (weighted) graphs The QAP aims to find the best way to match A to A Most graph-to-graph matching problems reduce to special cases of the QAP: traveling salesman, (sub)-graph isomorphism, graph partitioning, etc NP-hard Also hard in practice No PTAS until recently Sometimes written in minimization form 16 16/25

17 Subdivide in the form of a tree based on where vertex 1 is mapped, where vertex 2 is mapped, etc Each node corresponds to a coset τ i1,,i k S n k Define bounds B i1,,i k max σ τ i1,,i k S n k f(σ) Do a directed depth-first search: follow most promising branch, go down to leaf, then backtrack, but never follow branches which are guaranteed to be worse that optimum so far Hard to come up with theoretical performance guarantees Empirical performance depends critically on the tightness of the bounds 17 17/25

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19 Consider the Fourier transform of the objective function: f(λ) = σ Sn f(σ) ρ λ (σ) { If f is the QAP objective} function, then f(λ) = 0 unless λ,,, f( ) and f( ) are rank one matrices Question: Why is this? 19 19/25

20 S n acts transitively on the off-diagonal entries of A, so it induces a function g A (σ) = A σ(n),σ(n 1) with the property π (g A ) = g (π A) (note g A (στ) = g A (σ) τ S n 2 ) 20 20/25

21 The QAP objective function can be written in the form f(σ) = n A σ(i), σ(j) A i,j = i,j=1 1 (n 2)! 1 (n 2)! 1 (n 2)! τ S n g A (στ) g A (τ) π S n A σπ(n), σπ(n 1) A π(n),π(n 1) = π S n g A (σπ)g A (π) 21 21/25

22 The QAP objective function can be written in the form f(σ) = 1 (n 2)! τ S n g A (στ) g A (τ) The Fourier transform of the QAP objective is of the form f(λ) = 1 (n 2)! ĝ A (λ) (ĝ A (λ)), λ n (1) In particular, f(λ) is identically zero unless λ = (n), (n 1, 1), (n 2, 2) or (n 2, 1, 1), and f((n 2, 2)) and f((n 2, 1, 1)) are dyadic (rank one) matrices Question: Is this useful for anything? 22 22/25

23 First compute ĝ A and ĝ A O(n 2 ) space, O(n 3 ) time Compute f O(n 2 ) space, O(n 3 ) time Do branch and bound on the same coset tree as before Compute bounds directly in Fourier space 23 23/25

24 f i (λ) = d λ nλ λ λ n [ ρ λ (in) f(λ) ] λ 24 24/25

25 For any f : S n R max σ S n f(σ) 1 n! d λ f(λ) tr λ n f(σ) = 1 d λ tr [ f(λ) ρλ (σ) 1] n! λ n For any orthogonal matrix O, tr(mo) M tr This bound is computed in time O(n 3 ) (requires an SVD) 25 25/25