Feedback in Iterative Algorithms

Transcription

1 Feedback i Iterative Algorithms Charles Byre (Charles Byre@uml.edu), Departmet of Mathematical Scieces, Uiversity of Massachusetts Lowell, Lowell, MA October 17, 2005 Abstract Whe the oegative system of liear equatios y = Px has o oegative exact solutios, block-iterative methods fail to coverge, exhibitig, istead, subsequetial covergece leadig to a limit cycle of two or more distict vectors. From the vectors of the limit cycle we extract ew data, replace the data vector y with this ew data vector, ad restart the iterative process. Success with this approach as applied to the algebraic recostructio techique (ART) prompts us to apply feedback to other block-iterative methods. We cosider here several questios pertaiig to this feedback approach: Ca we show the existece of limit cycles for these other block-iterative algorithms? Does the sequece of successive limit cycles coverge to a sigle vector? If so, which vector is it? 1 Itroductio Whe the oegative system of liear equatios y = Px has o oegative solutios we say that we are i the icosistet case. I this case the SMART ad EMML algorithms still coverge, to a oegative miimizer of KL(Px, y) ad KL(y, Px), respectively. O the other had, the rescaled block-iterative versios of these algorithms, RBI-SMART ad RBI-EMML, do ot coverge. Istead they exhibit cyclic subsequetial covergece; for each fixed = 1,..., N, with N the umber of blocks, the subsequece {x mn+ } coverges to their ow limits. These limit vectors the costitute the limit cycle (LC). The LC for RBI-SMART is ot the same as for RBI- EMML, geerally, ad the LC varies with the choice of blocks. Our problem is to fid a way to calculate the SMART ad EMML limit vectors usig the RBI methods. More specifically, how ca we calculate the SMART ad EMML limit vectors from their associated RBI limit cycles? For details cocerig these algorithms see [3]. As is ofte the case with the algorithms based o the KL distace, we ca tur to the ART algorithm for guidace. What happes with the ART algorithm i the 1

2 icosistet case is ofte closely related to what happes with RBI-SMART ad RBI-EMML, although proofs for the latter methods are more difficult to obtai. For example, whe the system Ax = b has o solutio we ca prove that ART exhibits cyclic subsequetial covergece to a limit cycle. The same behavior is see with the RBI methods, but o oe kows how to prove this. Whe the system Ax = b has o solutio we usually wat to calculate the least squares (LS) approximate solutio. The problem the is to use the ART to fid the LS solutio. There are several ways to do this, as discussed i [2, 3]. We would like to be able to borrow some of these methods ad apply them to the RBI problem. I this article we focus o oe specific method that works for ART ad we try to make it work for RBI; it is the feedback approach. 2 Feedback i ART Suppose that the system Ax = b has o solutio. We apply the ART ad get the limit cycle {z 1, z 2,..., z M }, where M is the umber of equatios ad z 0 = z M. We assume that the rows of A have bee ormalized so that their legths are equal to oe. The the ART iterative step gives z m = z m 1 + A m (b m (Az m 1 ) m ) or z m z m 1 = A m (b m (Az m 1 ) m ). Summig over the idex m ad usig z 0 = z M we obtai zero o the left side, for each. Cosequetly A T b = A T c, where c is the vector with etries c m = (Az m 1 ) m. It follows that the systems Ax = b ad Ax = c have the same LS solutios ad that it may help to use both b ad c to fid the LS solutio from the limit cycle. The article [2] cotais several results alog these lies. Oe approach is to apply the ART agai to the system Ax = c, obtaiig a ew LC ad a ew cadidate for the right side of the system of equatios. If we repeat this feedback procedure, each time usig the LC to defie a ew right side vector, does it help us fid the LS solutio? Yes, as Theorem 4 of [2] shows. Our goal i this article is to explore the possibility of usig the same sort of feedback i the RBI methods. Some results i this directio are i [2]; we review those ow. 2

3 3 Feedback i RBI methods Oe issue that makes the KL methods more complicated tha the ART is the support of the limit vectors, meaig the set of idices for which the etries of the vector are positive. I [1] it was show that whe the system y = Px has o oegative solutios ad P has the full rak property there is a subset S of { = 1,..., J} with cardiality at most I 1, such that every oegative miimizer of KL(Px, y) has zero for its -th etry wheever is ot i S. It follows that the miimizer is uique. The same result holds for the EMML, although it has ot bee prove that the set S is the same as i the SMART case. The same result holds for the vectors of the LC for both RBI-SMART ad RBI-EMML. A simple, yet helpful, example to refer to as we proceed is the followig. P = [ ] [ ] 1.5.5, y = There is o oegative solutio to this system of equatios ad the support set S for SMART, EMML ad the RBI methods is S = { = 2}. 3.1 The RBI-SMART Our aalysis of the SMART ad EMML methods has show that the theory for SMART is somewhat icer tha that for EMML ad the resultig theorems for SMART are a bit stroger. The same is true for RBI-SMART, compared to RBI- EMML. For that reaso we begi with RBI-SMART. Recall that the iterative step for RBI-SMART is x k+1 = x k exp( s 1 P i log(y i /(Px k ) i )), where = k(modn) + 1, s = I i=1 P i, s = P i ad m = max{s /s, = 1,..., J}. For each let G (x, z) = J =1 s KL(x, z ) Exercise 1: Show that J =1 KL((Px) i, (Pz) i ) + s KL(x, z ) KL((Px) i, (Pz) i ) 0, 3 KL((Px) i, y i ).

4 so that G (x, z) 0. Exercise 2: Show that where J G (x, z) = G (z, z) + s KL(x, z ), =1 z = z exp( s 1 P i log(y i /(Pz) i ). We assume that there are o oegative solutios to the oegative system y = Px. We apply the RBI-SMART ad get the limit cycle {z 1,..., z N }, where N is the umber of blocks. We also let z 0 = z N ad for each i let w i = (Pz 1 ) i where i B, the -th block. Prompted by what we leared cocerig the ART, we ask if the oegative miimizers of KL(Px, y) ad KL(Px, w) are the same. This would be the correct questio to ask if we were usig the slower urescaled block-iterative SMART, i which the m are replaced by oe. For the rescaled case it turs out that the proper questio to ask is: Are the oegative miimizers of the fuctios ad N =1 N =1 KL((Px) i, y i ) KL((Px) i, w i ) the same? The aswer is Yes, probably. The difficulty has to do with the support of these miimizers; specifically: Are the supports of both miimizers the same as the support of the LC vectors? If so, the we ca prove that the two miimizers are idetical. This is our motivatio for the feedback approach. The feedback approach is the followig: begiig with y 0 = y we apply the RBI- SMART ad obtai the LC, from which we extract the vector w, which we also call w 0. We the let y 1 = w 0 ad apply the RBI-SMART to the system y 1 = Px. From the resultig LC we extract w 1 = y 2, ad so o. I this way we obtai a ifiite sequece of data vectors {y k }. We deote by {z k,1,..., z k,n } the LC we obtai from the system y k = Px, so that y k+1 i = (Pz k, ) i, fori B. Oe issue we must cofrot is how we use the support sets. At the first step of feedback we apply RBI-SMART to the system y = y 0 = Px, begiig with a positive vector 4

5 x 0. The resultig limit cycle vectors are supported o a set S 0 with cardiality less tha I. At the ext step we apply the RBI-SMART to the system y 1 = Px. Should we begi with a positive vector (ot ecessarily the same x 0 as before) or should our startig vector be supported o S 0? Exercise 3: Show that the RBI-SMART sequece {x k } is bouded. Hits: For each let M = max{y i /P i, P i > 0} ad let C = max{x 0, M }. Show that x k C for all k. Exercise 4: Let S be the support of the LC vectors. Show that N =1 P i log(y i /w i ) 0 (3.1) for all, with equality for those S. Coclude from this that N =1 KL((Px) i, y i ) N =1 N =1 (y i w i ), KL((Px) i, w i ) with equality if the support of the vector x lies withi the set S. Hits: For S cosider log(z /z 1 ) ad sum over the idex, usig the fact that z N = z 0. For geeral assume there is a for which the iequality does ot hold. Show that there is M ad ɛ > 0 such that for m M log(x (m+1)n /x mn ) ɛ. Coclude that the sequece {x mn } is ubouded. Exercise 5: Show that N N G (z k,, z k, 1 ) = =1 =1 (yi k yi k+1 ), ad coclude that the sequece { N =1 ( y k i )} is decreasig ad that the sequece { N =1 G (z k,, z k, 1 )} 0 as k. Hits: Calculate G (z k,, z k, 1 ) usig Exercise 2. 5

6 Exercise 6: Show that for all vectors x 0 the sequece N { =1 is decreasig ad the sequece as k. Hits: Calculate N { =1 N =1 ad use the previous exercise. KL((Px) i, y k i )} (yi k yi k+1 ) 0, N KL((Px) i, yi k )} { =1 KL((Px) i, yi k+1 )} Exercise 7: Exted the boudedess result obtaied earlier to coclude that for each fixed the sequece {z k, } is bouded. Sice the sequece {z k,0 } is bouded there is a subsequece {z kt,0 } covergig to a limit vector z,0. Sice the sequece {z kt,1 } is bouded there is subsequece covergig to some vector z,1. Proceedig i this way we fid subsequeces {z km, } covergig to z, for each fixed. Our goal is to show that, with certai restrictios o P, z, = z for each. We the show that the sequece {y k } coverges to Pz ad that z miimizes N KL((Px) i, y i ). =1 It follows from Exercise 5 that N { G (z,, z, 1 )} = 0. =1 Exercise 8: Fid suitable restrictios o the matrix P to permit us to coclude from above that z, = z, 1 = z for each. Exercise 9: Show that the sequece {y k } coverges to Pz. Hits: Sice the sequece { N =1 m 1 KL((Pz ) i, yi k )} is decreasig ad a subsequece coverges to zero, it follows that the whole sequece coverges to zero. Exercise 10: Use Exercise 4 to obtai coditios that permit us to coclude that the vector z is a oegative miimizer of the fuctio N =1 KL((Px) i, y i ). 6

7 3.2 The RBI-EMML We tur ow to the RBI-EMML method, havig the iterative step x k+1 = (1 s 1 s )x k + s 1 x k P i y i /(Px k ) i, with = k(modn)+1. As we wared earlier, developig the theory for feedback with respect to the RBI-EMML algorithm appears more difficult tha i the RBI-SMART case. Applyig the RBI-EMML algorithm to the system of equatios y = Px havig o oegative solutio, we obtai the LC {z 1,..., z N }. As before, for each i we let w i = (Pz 1 ) i where i B. There is a subset S of { = 1,..., J} with cardiality less tha I such that for all we have z = 0 if is ot i S. ad The first questio that we ask is: Are the oegative miimizers of the fuctios the same? N =1 N =1 KL(y i, (Px) i ) KL(w i, (Px) i ) As before, the feedback approach ivolves settig y 0 = y, w 0 = w = y 1 ad for each k defiig y k+1 = w k, where w k is extracted from the limit cycle LC(k) = {z k,1,..., z k,n = z k,0 } obtaied from the system y k = Px as w k i = (Pz k, 1 ) i where is such that i B. Agai, we must cofrot the issue of how we use the support sets. At the first step of feedback we apply RBI-EMML to the system y = y 0 = Px, begiig with a positive vector x 0. The resultig limit cycle vectors are supported o a set S 0 with cardiality less tha I. At the ext step we apply the RBI-EMML to the system y 1 = Px. Should we begi with a positive vector (ot ecessarily the same x 0 as before) or should our startig vector be supported o S 0? Oe approach could be to assume first that J < I ad that S = { = 1,..., J} always ad the see what ca be discovered. Our coectures, subect to restrictios ivolvig the support sets, are as follows: 1: The sequece {y k } coverges to a limit vector y ; 2: The system y = Px has a oegative solutio, say x ; 7

8 3: The LC obtaied for each k coverge to the sigleto x ; 4: The vector x miimizes the fuctio over oegative x. N =1 KL(y i, (Px) i ) Some results cocerig feedback for RBI-EMML were preseted i [2]. We sketch those results ow. Exercise 11: Show that the quatity is the same for k = 0, 1,... N =1 y k i Hits: Show that J N s (z k, =1 =1 ad rewrite it i terms of y k ad y k+1. z k, 1 ) = 0 Exercise 12: Show that there is a costat B > 0 such that z k, ad. B for all k, Exercise 13: Show that s log(z k, 1 /z k, ) m 1 P i log(yi k+1 /yi k ). Hits: Use the covexity of the log fuctio ad the fact that the terms 1 s ad P i, i B sum to oe. Exercise 14: Use the previous exercise to prove that the sequece N { =1 KL((Px) i, y k i )} is decreasig for each oegative vector x ad the sequece is icreasig. N { =1 P i log(y k i )} 8

9 Refereces [1] Byre, C. (1993) Iterative image recostructio algorithms based o crossetropy miimizatio. IEEE Trasactios o Image Processig IP-2, pp [2] Byre, C. (1997) Coverget block-iterative algorithms for image recostructio from icosistet data. IEEE Trasactios o Image Processig IP-6, pp [3] Byre, C. (2005) Sigal Processig: A Mathematical Approach, AK Peters, Publ., Wellesley, MA. 9