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The second statement of the theorem differs from the first in the following respect: when K < M < 2 K , there will necessarily exist K -sparse signals x that cannot be uniquely recovered from the M -dimensional measurement vector y = Φ x . However, these signals form a set of measure zero within the set of all K -sparse signals and can safely be avoided if Φ is randomly generated independently of x .

Unfortunately, as discussed in Nonlinear Approximation from Approximation , solving this 0 optimization problem is prohibitively complex. Yet another challenge is robustness; in the setting ofTheorem "Recovery via ℓ 0 optimization" , the recovery may be very poorly conditioned. In fact, both of these considerations (computational complexity and robustness) can be addressed, but atthe expense of slightly more measurements.

Recovery via convex optimization

The practical revelation that supports the new CS theory is that it is not necessary to solve the 0 -minimization problem to recover α . In fact, a much easier problem yields an equivalent solution (thanks again to the incoherency of thebases); we need only solve for the 1 -sparsest coefficients α that agree with the measurements y [link] , [link] , [link] , [link] , [link] , [link] , [link] , [link]

α ^ = arg min α 1 s.t. y = Φ Ψ α .
As discussed in Nonlinear Approximation from Approximation , this optimization problem, also known as Basis Pursuit [link] , is significantly more approachable and can be solved with traditionallinear programming techniques whose computational complexities are polynomial in N .

There is no free lunch, however; according to the theory, more than K + 1 measurements are required in order to recover sparse signals via Basis Pursuit. Instead, one typically requires M c K measurements, where c > 1 is an oversampling factor . As an example, we quote a result asymptotic in N . For simplicity, we assume that the sparsity scales linearly with N ; that is, K = S N , where we call S the sparsity rate .

Theorem

[link] , [link] , [link] Set K = S N with 0 < S 1 . Then there exists an oversampling factor c ( S ) = O ( log ( 1 / S ) ) , c ( S ) > 1 , such that, for a K -sparse signal x in the basis Ψ , the following statements hold:

  1. The probability of recovering x via Basis Pursuit from ( c ( S ) + ϵ ) K random projections, ϵ > 0 , converges to one as N .
  2. The probability of recovering x via Basis Pursuit from ( c ( S ) - ϵ ) K random projections, ϵ > 0 , converges to zero as N .

In an illuminating series of recent papers, Donoho and Tanner [link] , [link] , [link] have characterized the oversampling factor c ( S ) precisely (see also "The geometry of Compressed Sensing" ). With appropriate oversampling, reconstruction via Basis Pursuit is also provably robust tomeasurement noise and quantization error [link] .

We often use the abbreviated notation c to describe the oversampling factor required in various settings even though c ( S ) depends on the sparsity K and signal length N .

A CS recovery example on the Cameraman test image is shown in [link] . In this case, with M = 4 K we achieve near-perfect recovery of the sparse measured image.

Compressive sensing reconstruction of the nonlinear approximation Cameraman image from [link] (b). Using M = 16384 random measurements of the K -term nonlinear approximation image (where K = 4096 ), we solve an 1 -minimization problem to obtain the reconstruction shown above. The MSE with respect to the measured image is 0.08 , so the reconstruction is virtually perfect.

Questions & Answers

show that the set of all natural number form semi group under the composition of addition
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Dominic
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_3_2_1
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The denominator of a certain fraction is 9 more than the numerator. If 6 is added to both terms of the fraction, the value of the fraction becomes 2/3. Find the original fraction. 2. The sum of the least and greatest of 3 consecutive integers is 60. What are the valu
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1. x + 6 2 -------------- = _ x + 9 + 6 3 x + 6 3 ----------- x -- (cross multiply) x + 15 2 3(x + 6) = 2(x + 15) 3x + 18 = 2x + 30 (-2x from both) x + 18 = 30 (-18 from both) x = 12 Test: 12 + 6 18 2 -------------- = --- = --- 12 + 9 + 6 27 3
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Mark and Don are planning to sell each of their marble collections at a garage sale. If Don has 1 more than 3 times the number of marbles Mark has, how many does each boy have to sell if the total number of marbles is 113?
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Mark = x,. Don = 3x + 1 x + 3x + 1 = 113 4x = 112, x = 28 Mark = 28, Don = 85, 28 + 85 = 113
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Solve for the first variable in one of the equations, then substitute the result into the other equation. Point For: (6111,4111,−411)(6111,4111,-411) Equation Form: x=6111,y=4111,z=−411x=6111,y=4111,z=-411
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Need help solving this problem (2/7)^-2
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x+2y-z=7
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-1
Shedrak
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
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Jeannette has $5 and $10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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Source:  OpenStax, Concise signal models. OpenStax CNX. Sep 14, 2009 Download for free at http://cnx.org/content/col10635/1.4
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