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�, Xθ also is. Moreover, Xθ and Xθ (cid:48) are independent. In fact, note that E (cid:104) XθXθ (cid:48) T (cid:105) = E [Y cos θ + X sin θ] [−Y sin θ + X cos θ] = 0. T We use Jensen’s again (with respect to the integral now) to get: exp (λ [F (X) − F (Y )]) = exp λ (cid:32) π 2 1 π/2 (cid:90) π/2 (cid:33) ∂ ∂θ F (Xθ)...
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) ≤ λ π θ is a gaussian random variable EX (cid:48) θ exp (cid:16) λ π 2 ∇F (Xθ) · X (cid:48) θ (cid:17) ≤ exp (cid:17) (cid:21) 2 (cid:20) 1 2 (cid:16) λ π 2 σ Taking expectation now in Xθ, and putting everything together, gives which means that E [exp (λF (X))] ≤ exp (cid:20) 1 2 (cid:16) λ π 2 σ (cid:17) (cid:21) 2 ...
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7) ≤ W (1) − W (2) ≤ W (1) − W (2) (cid:12) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:12) (cid:12) (cid:12) (cid:13) (cid:13) (cid:13) F . The symmetry√ factor of 2. of the matrix and the variance 2 of the diagon al entries are responsible for an extra Using Gaussian Concentr...
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4 = 1 n . 4.2.2 Talagrand’s concentration inequality A remarkable result by Talagrand [Tal95], Talangrad’s concentration inequality, provides an analogue of Gaussian concentration to bounded random variables. Theorem 4.9 (Talangrand concentration inequality, Theorem 2.1.13 [Tao12]) Let K > 0, ≤ i ≤ n. Let and let X1, ....
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)(cid:18) p (cid:19)p+ε (cid:18) 1 − p (cid:19)1−p−ε(cid:35)n p + ε 1 − p − ε (cid:41) Xi ≤ p − ε ≤ (cid:34)(cid:18) p p − ε (cid:19)p−ε (cid:18) 1 − p 1 − p + ε (cid:19)1− +ε p (cid:35) n 4.3.2 Multiplicative Chernoff Bound There is also a multiplicative version (see, for example Lemma 2.3.3. in [Dur06]), which is part...
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any t > 0: √ • Prob Z { ≥ 2(cid:107)a(cid:107)2 x + 2(cid:107)a(cid:107)∞x} ≤ exp(−x), 64 • Prob {Z ≤ −2(cid:107)a(cid:107)2 √ x} ≤ exp(−x), (cid:107) (cid:107)2 (cid:80)n where a 2 = k=1 a2 k and (cid:107)a(cid:107) = max ∞ 1≤k ≤n | k| a . Note that if ak = 1, for all k, then Z is a χ2 with n degrees of freedom, so th...
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e σ2 = Note that (cid:107)A(cid:107) denotes the spectral norm of A. (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n (cid:88) k=1 E (cid:0)X 2 k (cid:13) (cid:1)(cid:13) (cid:13) (cid:13) (cid:13) . In what follows we will state and prove various matrix concentration results, similar to Theorem 4.13. Motivated by the de...
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n (cid:88) k=1 n (cid:88) (cid:33) (cid:32) gkAk − n (cid:88) k=1 hkAk (cid:33)(cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (gk − hk)Ak (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) max v: (cid:107)v(cid:107)=1 T v (cid:33) (gk − k)Ak h v (cid:32) n (cid:88) k=1 n (cid:88) (gk − hk) (cid:0)vT Akv (cid:1) = max v: (cid:1...
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ance (see, for example, [Tro15b])) Given A , . . . , A ∈ Rd d metric matrices. We define the weak variance parameter as n 1 × sym- σ2 = max ∗ v: (cid:107)v(cid:107)=1 n (cid:88) (cid:0) k=1 vT Akv 2 (cid:1) . This means that, using Gaussian concentration (and setting t = uσ ), we have ∗ (cid:40)(cid:13) n (cid:13) (cid:...
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17) (cid:116) max v: (cid:107)v(cid:107) =1 n (cid:88) k=1 2 (v Akv) . T We have σ ≤ σ. ∗ Proof. Using the Cauchy-Schwarz inequality, σ2 = max ∗ v: (cid:107)v(cid:107)=1 = max v: (cid:107)v(cid:107)=1 ≤ max (cid:107)v (cid:107)=1 v: = max v: (cid:107)v(cid:107)=1 n (cid:88) (cid:0) vT Akv 2 (cid:1) k=1 n (cid:88) k=1 n...
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:13) (cid:13) n (cid:88) k=1 gkAk (cid:13) 2 (cid:13) (cid:13) (cid:13) (cid:13) = E (cid:32) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) k=1 gkAk(cid:107). (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) kA (cid:33) 2 g k n (cid:88) k=1 = E max v v: (cid:107)v(cid:107)=1 (cid:33) 2 gkAk v (cid:32) n (c...
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.14 gives us i j A2 ≤ 2eieT i . ij = (d + 1)I E(cid:107)W (cid:107) (cid:46) (cid:112)d log d, however, we know that E(cid:107)W case, suboptimal by a logarithmic factor.18 √ (cid:107) (cid:16) d, meaning that the bound given by NCK (Theorem 4.14) is, in this The next example will show that the logarithmic factor is in...
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16) max 1≤i≤d d (cid:88) j=1 b2 ij + max ij |bij| (cid:112)log d. Remark 4.20 X in the theorem above can be written in terms of a Gaussian series by taking Aij = bij eieT j + e T jei (cid:0) (cid:1) , for i < j and Aii = biieieT i . One can then compute σ and σ :∗ σ2 = max 1≤i≤d d (cid:88) j=1 b2 and σ2 ij ∗ (cid:16) b...
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jecture 4.21 is false, no counter example is known, up to date. Open Problem 4.1 (Improvement on Non-Commutative Khintchine Inequality) Prove or disprove Conjecture 4.21. I would also be pretty excited to see interesting examples that satisfy the bound in Conjecture 4.21 while such a bound would not trivially follow fr...
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, is the following true? E(cid:107)X(cid:107) (cid:46) E max (cid:107)Xek(cid:107)2. k The results in [BvH15] answer this in the positive for a large range of variance profiles, but not in full generality. Recently, van Handel [vH15] proved this conjecture in the positive with an extra factor of log log d. More precisel...
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:13)(cid:88) (cid:13) (cid:13) (cid:13) k=1 H 2 k (cid:13) 2 (cid:13) (cid:13) (cid:13) (cid:13) . (40) Before proving this theorem, we take first a small detour in discrepancy theory followed by deriva- tions, using this theorem, of a couple of useful matrix concentration inequalities. 4.6.1 A small detour on discrepan...
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13) (cid:13) (cid:13) n (cid:88) k=1 εkHk (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:46) (cid:112)log n (cid:118) (cid:117) (cid:117) (cid:116) (cid:13) n (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) k=1 H 2 k (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ≤ (cid:112)log n (cid:118) (cid:117) (cid:117) (cid:116) n ...
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orem 4.26 Let T1, . . . , Tn ∈ Rd×d be random independent positive semidefinite matrices, then E (cid:13) n (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) i=1 Ti where  (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) ≤ (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n (cid:88) i=1 ETi 1 2 (cid:13) (cid:13) (cid:13) (cid:13) (c...
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cid:16) (cid:13) (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) i=1 (cid:13) (cid:13) (cid:13)ET (cid:48) (cid:13) (cid:13) Ti − ETi E − Ti (cid:48) n (cid:88) (cid:0)Ti − Ti (cid:48) (cid:1) i =1 (cid:104) (cid:48) − ETi Ti (cid:13) n (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) i=1 (cid:0) (cid:13) (cid:13) (cid:13) ≤ E ...
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3) + (cid:112)C(d)E (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n (cid:88) i=1 T 2 i 1 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) The trick now is to make a term like the one in the LHS appear in the RHS. For that we start by noting (you can see Fact 2.3 in [Tro15a] for an elementary proof) that, since Ti (cid:23)...
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max i (cid:107)Ti(cid:107) (cid:19) 1 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n (cid:88) i=1  1 2  . (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Ti Further applying the Cauchy-Schwarz inequality for E gives, E (cid:13) n (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) i=1 Now that the term E (cid:107)(cid:80)n of...
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13) n (cid:88) i=1 Y 2 i 1 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) . 72 Jensen’s inequality gives E (cid:13) n (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) i=1 Y 2 i 1 2 (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:32) ≤ E (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) n (cid:88) i=1 Y 2 i (cid:33) 1 2 (cid:13...
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3) 0 (cid:13) 0 T Si Si (cid:21) (cid:13) (cid:13) (cid:13) (cid:13) We defer the details to [Tro15a] = max (cid:13) (cid:8) (cid:13)ST i Si (cid:13) , (cid:13) (cid:13) SiST (cid:13) i (cid:13) (cid:9) . (cid:13) In order to prove Theorem 4.22, we will use an AM-GM like inequality for matrices for which, unlike the on...
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and µ1−θλθ ≤ (1 − θ)µ + θλ. Proof. [of Theorem 4.22] Let X = (cid:80)n k=1 εkHk, then for any positive integer p, E(cid:107)X(cid:107) ≤ (cid:0)E(cid:107)X(cid:107)2p(cid:1) 2p = (cid:0)E(cid:107)X 2p(cid:107)(cid:1) 1 2p ≤ (cid:0)E Tr X 2p(cid:1) 1 1 2p , where the first inequality follows from Jensen’s inequality and ...
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1 2 (cid:104) Tr Hi (cid:16) 2 X −1 p +i − 2X i p − −1 (cid:17)(cid:105) , where the expectation can be taken over εj for j (cid:54)= i. 2p Now we rewrite X+i −1 − 2pX −1 −i as a telescopic sum: X p 2 +i − − − 2p 1 1 X−i = −2 p 2 (cid:88) q=0 q X+i (X+i 2p 2 q − X ) X − − −i −i . Which gives E Tr X 2p = Since X+i − X−i...
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, 2p 2 which gives Applying this inequality, recursively, we get E Tr X 2p ≤ σ2(2p − 1)E Tr X 2p−2. E Tr X 2p ≤ [(2p − 1)(2p − 3) · · · (3)(1)] σ2pE Tr X 0 = (2p − 1)!!σ2pd (46) (47) Hence, E(cid:107)X(cid:107) ≤ (cid:0) 1 E Tr X 2p(cid:1) 2p ≤ [(2p − 1)!!] p 1 2p σd 1 2p . (cid:17) (cid:16) 2p+1 e Taking p = (cid:100)...
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y, it is not uncommon that H 2 is a multiple of the identity i matrix, which would render this step also an equality). (cid:80) In fact, Joel Tropp [Tro15c] recently proved an improvement over the NCK inequality that, essen- tially, consists in replacing inequality (45) with a tighter argument. In a nutshell, the idea ...
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T U = Id×d Prob (cid:8)(cid:13) (ΠU )T (ΠU ) − I ≥ ε < δ, (cid:13) (cid:13) (cid:13) (cid:9) for m ≥ c 1 d+log( 1 δ ) ε2 and s ≥ c2 log( d δ ) ε2 . 2. Same setting as in (1) but conditioning on m (cid:88) r=1 δri = s, for all i, meaning that each column of Π has exactly s non-zero elements, rather than on average. The ...
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k = 1 I m d×d . The conjecture is then that, there exists c1 and c2 positive universal Prob (cid:40)(cid:13) m (cid:13) (cid:13)(cid:88) (cid:13) (cid:13) k=1 (cid:2)zkzT k − Ezkz T k (cid:41) ≥ ε < δ, (cid:13) (cid:13) (cid:3)(cid:13) (cid:13) (cid:13) for m ≥ c1 d+log( 1 δ ) ε2 and s ≥ c2 log( d δ ) ε2 . I think this...
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j ⊗ Πij + e T jei ⊗ ΠT ij (cid:1) , where ⊗ corresponds to the Kronecker product. Note that EA⊗k = A ⊗ (cid:19) J , (cid:18) 1 k where J = 11T is the all-ones matrix. 77 Open Problem 4.5 (Random k-lifts of graphs) Give a tight upperbound to E (cid:13) (cid:13) (cid:13)A⊗k − EA⊗k(cid:13) (cid:13) (cid:13) . Oliveira [Ol...
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4.6 Prove or disprove Conjecture 4.33.21 21We thank Francisco Unda and Philippe Rigollet for suggesting this problem. 78 5 Johnson-Lindenstrauss Lemma and Gordons Theorem 5.1 The Johnson-Lindenstrauss Lemma Suppose one has n points, X = {x1, . . . , xn}, in Rd (with d large). If d > n, since the points have to lie in a...
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map f : Rd → Rk that is an (cid:15)−isometry for X (see (48)). This map can be found in randomized polynomial time. We borrow, from [DG02], an elementary proof for the Theorem. We need a few concentration of measure bounds, we will omit the proof of those but they are available in [DG02] and are essentially the same id...
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dimension k is the same as understanding the norm of the projection of a (uniformly) 79 random point on Sd−1 the unit sphere in Rd on a specific k−dimensional subspace, let’s say the one generated by the first k canonical basis vectors. This means that we are interested in the distribution of the norm of the first k entri...
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:20) ∃i,j : Pr (cid:107)f (xi) − f (xj)(cid:107)2 (cid:107)xi − xj(cid:107)2 −dimensional subspace is an Therefore, choosing f as a properly scaled projection onto a random k (cid:15)− isometry on X (see (48)) with probability at least . We can achieve any desirable constant probability of success by trying O(n) such r...
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and Diffusion Maps are, in some sense, data adaptive). Using Lemma 5.3 you can just choose a projection at random in the beginning of the process (all ones needs to know is an estimate of the log of the size of the data set) and just map each point using this projection matrix which can be done online – we don’t need to...
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the lower bound by Alon [Alo03]). The only hope is to speed up the matrix-vector multiplication. If we were able to construct a sparse matrix M then we would definitely speed up the computation of M x but sparse matrices tend to distort sparse vectors, and the data set may contain. Another option would be to exploit the...
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projection approximately preserves the norm of every point in a set S, in this case that, in order to approximately the set of differences between pairs of n points. What we showed is preserve the norm of every point in S it is enough to project to Θ (cid:0)(cid:15)−2 log |S| dimensions. The question this section is mea...
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1 be a closed subset of the unit sphere in d dimensions. Then E max x∈S (cid:13) 1 (cid:13) (cid:13) (cid:13) ak Gx (cid:13) (cid:13) (cid:13) (cid:13) ≤ 1 + ω(S) ak , 82 and E min x∈S (cid:13) 1 (cid:13) (cid:13) (cid:13) ak Gx (cid:13) (cid:13) (cid:13) (cid:13) ≥ 1 − ω(S) ak , where ak = E(cid:107)gk(cid:107) and ω(...
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:107) x∈S (cid:107)| ≤ ≤ max |(cid:107)G1x(cid:107) − (cid:107)G2x max x∈S = (cid:107)G1 − G2(cid:107) ≤ (cid:107)G1 − G2(cid:107)F . (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) (cid:13) Concentration to get: and (cid:26) (cid:27) (cid:18) Prob max (cid:107)Gx(cid:107) ≥ ak + ω(S) + t ∈ x S ≤ exp − (cid:26) Prob min x...
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13) ≤ (1 + ε)(cid:107)x(cid:107), Recall that k − k k+1 ≤ 2 a k ≤ k. Proof. This is readily obtained by taking ε = ω(S)+t ak , using (49), (50), and recalling that a2 k ≤ k. (cid:50) Remark 5.8 Note that a simple use of a union bound23 shows that ω(S) (cid:46) (cid:112)2 log |S|, which means that taking k to be of the ...
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� S (cid:54)= ∅} ≤ (cid:18) exp − 1 18 7 2 2 (ak − ω(S)) (cid:19) , where ω(S) is the gaussian width of S and ak = E(cid:107)gk(cid:107) where gk ∼ N (0, Ik×k). 5.2.2 Proof of Gordon’s Theorem In order to prove this Theorem we will use extensions of the Slepian’s Comparison Lemma. Slepian’s Comparison Lemma, and the cl...
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The following extension is due to Gordon [Gor85, Gor88]. Theorem 5.11 [Theorem A in [Gor88]] Let {Xt,u} bounded) centered Gaussian processes indexed by the same (compact) sets T and U . t1, t2 ∈ T and u1, u2 ∈ U : }(t,u)∈T ×U { and Yt,u (t,u)∈T ×U be two (almost surely If, for every E [Xt1,u1 − 2 Xt1,u2] ≤ E 2 [Yt1,u1 ...
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),u(cid:48) = (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) 0 2 This means that w e can use Theorem 5.11 with X = A and Y = B, to get E min max Av,u v∈S u∈Sk−1 ≤ E min max Bv,u. v∈S u∈Sk−1 Noting that E min max B v∈S u Sk−1 ∈ v,u = E min max uT v∈S u Sk−1 ∈ Gv = E min (cid:107)Gv v∈S (cid:107) , and (c...
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, (cid:107) (cid:107) v∈S u∈Sk−1 v∈S (cid:20) E (cid:21) max max Av,u v∈S u∈Sk−1 = E max gT u + E max hT v = ak + ω(S), v∈S u∈Sk−1 concludes the proof of the Theorem. 5.3 Sparse vectors and Low-rank matrices In this Section we illustrate the utility of Gordon’s theorem by undertanding which projections are expected to ...
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injective on sparse vectors. Let us assume that x is s-sparse, meaning that x has at most s non-zero entries (often written as (cid:107)x(cid:107)0 ≤ s, where (cid:107) · (cid:107)0 is called the 0-norm and counts the number of non-zero entries in a vector25). It is also intuitive that, in order for reconstruction to b...
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s where g ∼ N (0, IN N ). We have × ω (Ss) = max Γ⊂[N ], |Γ|=s (cid:107) gΓ , (cid:107) where gΓ is the restriction of g to the set of indices Γ. Given a set Γ, Theorem 4.12 gives Prob (cid:110) gΓ (cid:107) √ (cid:107)2 ≥ s + 2 s √ (cid:111) t + t ≤ exp(−t). Union bounding over all Γ ⊂ [N ], |Γ| = s gives (cid:26) Pro...
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Γ(cid:107)2. This suggests that ≈ 2s log (cid:0) N measurements suffice to identify a 2s-sparse vector. As we’ll see, ts suffices to identify a sparse vector but also for certain efficient 2s (cid:1) not only such a number of measuremen algorithms to do so. 5.3.2 The Restricted Isometry Property and a couple of open problems...
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BFMM14, BMM14, B+11, Mix14a]. The only known unconditional construction that is able to break this bottleneck is due to Bourgain et al. [B+11] that achieves for a small, but positive, ε. There is a conditional construction, based on the Paley s ≈ Equiangular Tight Frame, that will be briefly described in the next Lectur...
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more on this problem and recent advances. Note that checking whether a matrix has RIP or not is, in general, NP-hard [BDMS13, TP13]. 5.3.3 Gaussian width of rank-r matrices Another structured set of interest is the set of low rank matrices. Low-rank matrices appear in countless applications, a prime example being the N...
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rank( =1 X)≤r Let X = U ΣV T be the SVD decomposition of X, then ω (cid:0)(cid:8)X : X ∈ Rn1×n2, rank(X) ≤ r(cid:9)(cid:1) = E max U T U =V T V =Ir×r Σ∈Rr×r diagonal (cid:107)Σ(cid:107)F =1 Tr(Σ V T GU ). (cid:1) (cid:0) This implies that ω (cid:0)(cid:8)X : X ∈ Rn1×n2, rank(X) ≤ r(cid:9)(cid:1) ≤ (Tr Σ) (E(cid:107)G(c...
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naturally embeds into the bi-infinite path whose vertices are the set of integers Z with edges between consecutive integers. By an abuse of notation we denote the bi-infiite path as Z. It is intuitive to say that Pn converges to Z as these paths can be embedded into Z in a nested manner such that they exhaust Z. Clearly,...
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see the red vertices in Figure 1). Pick exactly one vertex from each cluster, say the median vertex. Connect each such vertex, say (n, y), to the 2 − 1, y), which is the unique edge connecting (n, y) to vertex (n − 1, y) via the edge (n, y) ↔ (n Z[−n + 1, n − 1] . If a vertex (n, y(cid:48)) on the right hand side is no...
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on Z2, which in an intuitively sense is the limit of the grids Z[−n, n]2. We will in fact calculate the tree entropy. − The tree entropy of a sequence of bounded degree connected graphs {Gn} is the limiting value of log sptr(Gn)/|Gn| provided it exists. It measures the exponential rate of growth of the number of spanni...
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CONVERGENCE OF GRAPHS AND ENUMERATION OF SPANNING TREES Borel σ-algebra of G under this metric; it is generated by sets of the from A(H, y, r) = {(G, x) ∈ G : Nr(G, x) ∼= Nr(H, y)}. A random rooted graph (G, ◦) is a probability space (G, F , µ); we think of (G, ◦) as a G-valued random variable such that P(cid:2) (G, ◦)...
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ery r > 0 and any finite connected For ev rooted graph (H, y) with that Nr(H, y) = (H, y) we require (cid:0) (cid:1) that |x ∈ V (Gn) : Nr(G, x) ∼= (H, y)| /|Gn| converges as n → ∞. Using tools from measure theory G, ◦) ∈ G such that that there is a random rooted graph ( and compactness of G it can be shown (cid:3) if t...
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Let (G, x) be a rooted graph. The simple random walk (SRW) on (G, x) (started as x) is a V (G)-valued stochastic process X0 = x, X1, X2 . . . k−1. The such that Xk is a uniform random neighbour of Xk 1 picked independently of X0, . . . , X SRW is a Markov process given by the transition matrix P (u, v) = 1u∼v where u ∼...
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of (G, ◦), n) ∼= Nr(G(cid:48), ◦(cid:48))) → 1 as n → ∞. This common and for every r ≥ 0 the probability µ(Nr(G(cid:48) probability space where all the graphs can be jointly defined and satisfy the stated claim follows n) ∼= Nk/2(G(cid:48), ◦(cid:48))} we have from Shorokhod’s representation theorem. On the event {Nk/2(...
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n) (cid:29) Nk/2(G(cid:48), ◦(cid:48)) as n → ∞ . (cid:12) (cid:3)(cid:12) (cid:12) 2.2. Local weak limit of random regular graphs. In this section we will show a classical result that random d-regular graphs converge to the d-regular tree Td in the local weak sense (see Bollob´as [4]). There are a finite number of d-re...
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− 2d e 4 . Also, conditioned on Gn,d being simple its distribution is a uniform random d-regular graph on It follows from these observations that any sequence of graph properties An whose n vertices. probability under Gn,d tends to 1 as n → ∞ also tends to 1 under the uniform random d-regular graph model. In particular...
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,d of length at most 2r. It follows from this where C 2r is the ≤ (cid:3) (cid:2) wing lemma argument that (cid:2) E |v ∈ V (Gn,d) : Nr(Gn,d, v) = Nr(Td, ◦)| ≤ 2rdrE C (cid:2) , E if d ≥ 3, and more precisely (cid:3) ≤ 2r(3d − 3) shows that E C follo ∼ 2r (cid:2) ≤2r (cid:3) ≤2r . The (cid:3) C≤2 r erges to converges t...
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6)−1)!(d( ( d−1)) 2(nd−1)!! } v1, . . . , v(cid:96) { conclude that (cid:2) (cid:3) E C(cid:96) = (cid:88) {v1 ,...,v(cid:96)} (cid:2) P {v1, . . . , v(cid:96)} forms an (cid:96) − cycle = (cid:3) (cid:18) (cid:19) n ((cid:96) (cid:96) − 1)!(d(d − 2(nd 1))(cid:96)(nd − 1)!! − − 2(cid:96) 1)!! . Note that (cid:0)n(cid:1...
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G. As we will see, this expression it turn can be written in terms of the return probabilities of the SRW on G. This is good for our purposes because if a sequence of bounded degree graphs Gn converges in the local weak limit to n) in terms of the expected a random rooted graph (G, ◦ ) then we will be able to express l...
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L(x, y) equals negative of the number of edges from x to y). Exercise 3.1. The Laplacian L of a graph G is a matrix acting on the vector space RV (G). Let (f, g) = (cid:80) x V (G) f (x)g(x) denote the standard inner product on RV (G). Prove each of the following statements. ∈ (1) (Lf, g) = 1 (cid:80) 2 (x,y)(f (x) − f...
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obtained from L by removing its i-th row and column; Li is called the (i, i)-cofactor of L. The Matrix-Tree Theorem states that det(Li) = sptr(G) for every i. To derive (2) we consider the characteristic polynomial det(L − tI) of L and note that the coefficient of t is − i det(Li) = −nsptr(G). On the other hand, if we wr...
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) = π(y)P (y, x). The latter condition is equivalent to (P f, g)π = (f, P g)π for all f, g ∈ RV (G), which means that P is self-adjoint w.r.t. the inner product (·, ·)π. Due to being self-adjoint it has n real eigenvalues and an orthonormal basis of eigenvector w.r.t. the new inner product. Notice that the eigenvalues ...
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≥ log(1−x) = − −1 n i = TrP k i=1 µk Now, − 1 since we exclude the eigenvalue 1 Note that TrP k = (cid:80) x V (G) pk ∈ started from x. Consequently, we conclude that G(x) where pk x < 1 we see that (cid:80) n 1 − n − k 1 i=1 log(1−µi) = − i=1 µi /k. of P that occurs with y one. G(x) in the k-step return probability of...
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of graphs Gn converging to G in the local weak limit. To prove this theorem let ◦n be a uniform random vertex of Gn. Then from (3) we get that log sptr(Gn) |Gn | = 2e(Gn) |Gn| (cid:2) + E log deg(◦n) (cid:3) − 1 k (cid:88) k ≥1 (cid:16) (cid:2) E pk Gn ◦ ( n) (cid:3) − | Gn | (cid:17) −1 . As Gn has degree bounded by ∆...
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(cid:80) k as required. con vergence theorem that (cid:80) 1 k k≥1 (E(cid:2) pk G (◦ n n) (cid:3) − |Gn 1 |− ) con- Lemma 3.4. G(x) denote the k-step return probability of the SRW on G starting at x. Let π(x) = deg(x)/2e for x ∈ V (G), where e is te connected graph of maximum degree ∆. Let pk Let G be a fini the number ...
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= z in G from x0 maxx V (G) |f (x)|, and by replacing f with −f if necessary we may assume that f (x such that f (z) ≤ 0. Then ||f || ≤ |f (x0) − f (z)|. There is a path x0, x1, . . . , x to z. Therefore, ∞ ||f ||∞ ≤ t (cid:88) i=1 |f (xi−1) − f (xi)| ≤ 1 2 (cid:88) |f (x) − f (y)|. (x,y)∈V (G)×V (G) x∼y Let K(x, y) = ...
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f (y)| and (cid:112)K(x, y)(|f (x)|+ |f (y)|) then we deduce that ||f ||4 ∞ ≤ e2(cid:0)(I − P )f, f (cid:1) π · (cid:0)(I + P )|f |, |f | (cid:1) .π Notice that (cid:0) | (I + P )|f |, |f | ≤ 2( f m (P f, f )π ≤ (f, f )π because all eigenvalues of P lies in (cid:1) |, |f |)π. If (f, f )π ≤ 1 then we see that π ||f ||4 ...
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π √ this to the function f (y) = √1x(y)−π(x) π(x)(1−π(x)) || ≤ 2e(k + 1)−1/4 if f ∈ U and (f, f )π ≤ 1. Let us now apply ∞ | ≤ 2e(k + 1)−1/4. The value of P kf (x) is . Then |P kf (x) √ P k(x, x)(1 − π(x)) − π(x) (cid:80) (cid:112) π(x)(1 − π(x)) y=x P k(y, x) P k(x, x)(1 − (cid:112) = π(x)) − π(x)(1 P − k(x, x)) π(x)(...
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12) (cid:12)(cid:0) (cid:12) (cid:88) pk G(x) (cid:1) − 1 (cid:12) (cid:12) (cid:12) (cid:12) (cid:88) = |G|−1(cid:12) (cid:12) (cid:0)pk (cid:12) (cid:1) (cid:12) G(x) − π(x) (cid:12) x∈ V (G) ≤ |G|− x∈V (G) pk 1 (cid:88) (cid:12) π(x) G(x) (cid:12) (cid:12) π(x) x∈V (G) − 1 (cid:12) (cid:12) (cid:12) This proves that...
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as well.) There is a classical problem called the Ballot Box problem that allows us to compute p2k (Td ◦) explicitly. It turns out that p2k Td to find a closed form of F (t) (see Wo are called the Catalan numbers. From this it is possible ( ) (d 1) − − d − 2k 1 k+1 2k ( ) k k+1 ess [9 ] Lemma 1.24): . The numbers (◦) = ...
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(G)/n = (1/n) (log n)/n. There is an infinitary version of this representation for the tree entropy of Zd. If L is the Laplacian on Zd, which acts on (cid:96)2(Zd), then one can define an operator log L on (cid:96)2(Zd) that satisfies (cid:16) h(Zd) = (log L) 1o, 1o (cid:17) . 99 LOCAL CONVERGENCE OF GRAPHS AND ENUMERATIO...
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. Open problems One can consider the space of (isomorphism classes of) doubly rooted graphs (G, x, y) of bounded degree, analogous to the space G. It is also a compact metric space where the distance between (G, x, y) and (H, u, v) is 1/(1 + R) where R is the minimal r such that the r-neighborhood of (x, y) in G is iso...
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unimodular trees (see Aldous and Lyons [1]). This is a major open problem in the field. Here is a problem on tree entropy. Bernoulli bond percolation on Td at density p is a random forest of Td obtained by deleting each edge independently with probability 1 − p. Let ◦ be a fixed vertex and denote by Cp(◦) the component o...
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, Identities and inequalities for tree entropy, Combin. Probab. Comput. 19 (2010), pp. 303-313. [7] R. Lyons and Y. Peres, Probability on Trees and Networks, Cambridge University Press, 2014, in preparation. Current version available at http://mypage.iu.edu/~rdlyons/(cid:856) . [8] B. D. McKay, Spanning trees in regula...
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:107)x(cid:107)0 of a vector x is the number of non-zero entries of x. This means that, when we take a picture, our camera makes N measurements (each corresponding to a pixel) but then, after an appropriate change of basis, it keeps only s (cid:28) N non-zero coefficients and drops the others. This motivates the question...
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this intuition is indeed correct. e to (cid:1) s Since the system is underdetermined and we know x is sparse, the natural thing to try, in order to recover x, is to solve min (cid:107)z(cid:107)0 s.t. Az = y, (55) and hope that the optimal solution z corresponds to the signal in question x. Unfortunately, (55) is known...
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�+ as the positive part of x and ω− as the symmetric of the negative part of it, meaning that x = ω+ − ω− and, for each i, either ωi is zero. Note that, in that case, − or ω+ i (cid:107)x(cid:107)1 = N (cid:88) i=1 Motivated by this we consider: ω+ i + ωi − = 1 (cid:0)ω+ + ω− T (cid:1) . min 1T (ω+ + ω−) s.t. A (ω+ − ω...
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v+ T ω+ − (cid:0) v− T (cid:1) min max u + ω v+ 0 ≥ ω− −≥0 v (cid:1)(cid:1) ω− + uT y − A ω+ − ω− . (cid:0) (cid:0) (58) Indeed, if the primal player (picking ω+ and ω− and attempting to minimize the objective) picks variables that do not satisfy the original constraints, then the dual player (picking u, v+, and v− and...
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0) is equivalent to uT y max u v+ 0 ≥ −≥0 v 1−v+−AT u=0 1−v−+AT u=0 or equivalently, maxu uT y s.t. −1 ≤ AT u ≤ 1. (61) The linear program (61) is known as the dual program to (57). The discussion above shows that (61) ≤ (57) which is known as weak duality. More remarkably, strong duality guarantees that the optimal va...
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we must is non-zero (and be +1 when it is positive and −1 when it is negative), in other words (cid:1) (cid:0) tries of AT u have the en (cid:1) (cid:0)AT u(cid:1) = sign (xS) , S where S = supp(x), and (cid:13) (cid:13)AT u(cid:13) (cid:13) ≤ 1 (in order to be dual feasible). ∞ (cid:13) < 1 any optimal primal (cid:13)...
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Corollary. S − 1 (cid:1) AT S AS is the Moore Penrose pseudo-inverse of AT S . This gives the following Corollary 6.3 Consider the problem of sparse recovery discussed this lecture. Let S = supp(x), if AS is injective and (cid:16) (cid:13) (cid:13) (cid:13) AT ScAS (cid:0) AT S AS − 1 (cid:1) sign (xS) (cid:17) (cid:13...
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≤ s one has (cid:107)AS(cid:107) ≤ 3 , where (cid:107)x(cid:107)=1 (cid:107) 2 1 + 1 (cid:107) Bx . (cid:1) entries then, for any |S ≤ s| we have (cid:107) · (cid:107) denotes the operator norm (cid:107)B(cid:107) = max This means that, if we take A random with i.i.d. N (cid:0)0, 1 M (cid:1)−1 (cid:107) ≤ (cid:0)1 − 1 ...
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) (cid:13) ∞ 1 ≥ √ M √ √ 3 (cid:19) (cid:18) st ≤ 2N exp − (cid:19) , t2 2 which implies Prob (cid:16)(cid:13) (cid:13)AT (cid:13) S AS (cid:0)AT S AS (cid:1)−1 sign (xS) (cid:17) ≥ 1 (cid:13) (cid:13) (cid:13)∞ ≤ 2N exp   − (cid:16) √ M√ 3s 2 (cid:17)2    = exp (cid:18) − (cid:20) M s 3 1 2 (cid:21)(cid:19) − 2 ...
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1 ≥ (cid:107)v+x(cid:107)1 = (cid:107) (v + x)S (cid:107)1+(cid:107)vSc(cid:107)1 ≥ (cid:107)x(cid:107)S −(cid:107)vS(cid:107)1+(cid:107)v(cid:107)Sc, where the last inequality follows by triangular inequality. This means that (cid:107)vS(cid:107)1 ≥ (cid:107)vSc(cid:107)1, but since |S| (cid:28) N it is unlikely for A...
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6.4 Partial Fourier matrices satisfying the Restricted Isometry Property While the results above are encouraging, rarely one has the capability of designing random gaussian measurements. A more realistic measurement design is to use rows of the Discrete Fourier Transform: Consider the random M × N matrix obtained by dr...
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8In these references the sets considered are slightly different than the one described here, as the goal is to ensure recovery of just one sparse vector, and not all of them simultaneously. 107 hard [BDMS13, TP13]). Despite suboptimal, coherence based methods are still among the most popular ways of building RIP matrice...
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th (cid:111) (cid:110) . If B has zero diagonal, then this reads: B (cid:107) (cid:107) ≤ maxi |Bij|. corresponds to λ : |λ − Bii| ≤ j=i |Bij| 1, . . . , aN ∈ R we define its worst-case coherence µ as M (cid:12) (cid:12) Given a set of N vectors a of B are contained T (cid:12)ai aj(cid:12) (cid:80) (cid:12) (cid:12) i a...
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example [BBRV01]) This motivates the question of how many orthonormal basis can be made simultaneously (or mutually) unbiased in Cd, such sets of bases are called mutually unbiased bases. Let M(d) denote the maximum number of such bases. It is known that M(d) ≤ d + 1 and that this upper bound is achievable when d is a ...
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(cid:113) (ETFs), these There are constructions that achieve the Welch bound, known as Equiangular Tight Frames are tight frames for which all inner products between pairs of vectors have the same N −M , meaning that they are “equiangular”. It is known that for there to exist an modulus µ = M (N −1) ETF in CM one needs...
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14] no unconditional result is known for s ≈ p but conjectured [BFMW13] to be RIP for s ≈ (cid:28) p. This motivates the following Open Problem. p polylogp √ √ Open Problem 6.4 Does the Paley Equiangular tight frame satisfy the Restricted Isometry Property pass the square root bottleneck? (even by logarithmic factors?)...
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to reduce the randomness needed in certain RIP constructions [BFMM14] 110 7 Group Testing and Error-Correcting Codes 7.1 Group Testing During the Second World War the United States was interested in weeding out all syphilitic soldiers called up for the army. However, syphilis testing back then was expensive and testing...
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in (Combinatorial) Group testing, introduced by Robert Dorfman [Dor43] with essentially the motivation described above.31 Let Ai be a subset of [T ] = {1, . . . , T } that indicates the tests for which soldier i participates. Consider A the family of n such sets A = {A1, . . . , An}. We say that A satisfies the k-disjun...
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A0 ⊆ (A1 ∪ · · · ∪ Ak)] = (cid:16) 1 − p(1 − p)k (cid:17) T . This is minimized for p = 1 k+1 . For this choice of p, we have 1 − p(1 − p)k = 1 − 1 k + 1 (cid:18) 1 − (cid:19) k 1 k + 1 Given that there are n such sets, there are (k + 1) n k+1 different ways of picking a set and k others to test whether the first is cont...
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)(cid:17) (cid:16) − log 1 − 1 k+1 e−1 k+1 k (cid:17) = (cid:16) (cid:1)(cid:17) log k(cid:0) n k+1 − log (1 − (ek)−1) = O(k2 log(n/k)), where the last inequality uses the fact that log the Taylor expansion − log(1 − x−1)−1 = O(x) (cid:16)(cid:0) n k+1 (cid:1)(cid:17) = O (cid:0)k log (cid:0) n k (cid:1)(cid:1) due to ...
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48) ∈ A, F (cid:54)⊆ A(cid:48) (cid:9) . We will procede by giving an upper bound to A0 ∪ A1. For that, we will need a coup family of sets. Let F denote the family of sets F in the definition of A1. More precisely, le of auxiliary F := {F ∈ [T ] : |F | = u and ∃!A ∈ A : F ⊆ A} . By construction |A1| ≤ |F| Also, let B be...
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cid:12) (cid:12)A0 ∪ B(cid:48)(cid:12) (cid:12) ≤ (cid:19) (cid:18) T u (cid:12)B B ∪ (cid:48)(cid:12) (cid:12) = |B| + (cid:12) (cid:12)B(cid:48) = (cid:12) (cid:12) . (cid:12) This implies that A | 0| ≤ |B| . Because A satisfies the k-disjunct property, no two sets in A can contain eachother. This implies that the fam...
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which would contradict the k-disjunct property. (cid:83) k If |A2| > k then we can take A0, A1, . . . , Ak distinct elements of A2. For this choice and any j = 0, . . . , k This means that (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) Aj \ (cid:91) Ai 0 ≤i<j (cid:12) (cid:12) (cid:12) (cid:12) (cid:12) (cid:12)...
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