In this work efficient geometric algorithms are provided for the linear approximation of digital signals under the uniform norm. Given a set of n points (xi, yi)i=1..n, with xi < xj if i < j, we give a new method to find the optimum linear approximation in O(n). Given also an error bound, we demonstrate how to construct in O(n) a non continuous piecewise solution such that the number k of segments is optimal. Furthermore we show that for such number of segments, the solution that is l∞ optimal can also be found in O(n) provided that n/k = O(1)

Dalai, Marco

Leonardi, Riccardo

DALAI, Marco

LEONARDI, Riccardo

Archivio istituzionale della ricerca - Università di Brescia

EFFICIENT (PIECEWISE) LINEAR MINMAX APPROXIMATION OF DIGITAL SIGNALSM. Dalai and R. LeonardiUniversita´ degli Studi di BresciaDipartimento di Elettronica per l’Automazionevia Branze, 38 - I-25123 Brescia, ItalyEmail: mdalai@inwind.it, Riccardo.Leonardi@ing.unibs.itABSTRACTIn this work efficient geometric algorithms are provided forthe linear approximation of digital signals under the uniformnorm.Given a set of n points (xi, yi)i=1..n, with xi < xj if i < j,we give a new method to find the optimum linear approxi-mation in O(n). Given also an error bound, we demonstratehow to construct in O(n) a non continuous piecewise solu-tion such that the number k of segments is optimal. Fur-thermore we show that for such number of segments, thesolution that is l∞ optimal can also be found in O(n) pro-vided that n/k = O(1).1. INTRODUCTIONThe problem of approximating a discrete time signal s in npoints xi, i = 1..n using the uniform norm basically con-sists on finding a function f pertaining to a class F of func-tions, such that||s− f ||∞ = max1≤i≤n|s(xi)− f(xi)|is smallest than a given value δ or, in other cases, it is thesmallest possible, that is||s− f ||∞ = ming∈F||s− g||∞In this paper we aim at finding linear (single segment(link))and noncontinuous piecewise linear (more segments (links))approximations of signals.In the first case, it is known that the problem is equiva-lent to a linear programming problem in three variables thatcan be solved in O(n) ([1], [2],[3]). We show instead a veryefficient geometric method based on the properties of theset of points S = {(xi, yi}i=1..n (where yi = s(xi)); wecompute the convex-hull Q of S and then we find a subsetof three points of Q that univocally determines the solution.This leads to an algorithm that can take up to 1/8 of the timenecessary for the linear programming solution.In the second case, when an error bound δ is given, weshow that it is possible to find, in O(n) time, a noncon-tinuous piecewise linear function with k segments that ap-proximate s with error  ≤ δ and such that the number ofsegments is optimal under this constraint; this means thatthere cannot exist a function with less than k segments ap-proximating s with error ′ ≤ δFinally, we show that given again an error bound δ, afterhaving found the minimum number of segments k and thek-link solution in the previous point, it is possible to find theoptimal k-link solution inO(n) if the length of the segmentsin the previous point is asymptotically bounded, which istrue for the majority of practical situations.2. SINGLE LINK SOLUTIONGiven a set S = {xi, yi} of n points that represents a dis-crete time signal, consider the convex hull Q of S.Suppose for simplicity that Q has no pair of parallel sides(the opposite case can be handled similarly with a few somemore considerations); then for every side XY of Q we canfind one vertex Z of Q which is the most distant one XY .We call Z opposite vertex to the side XY .Claim. There exists one side AB of Q such that its oppositevertex C satisfies xA < xC < xB . The best l∞ linear ap-proximation of S is the line r parallel to AB and equidistantfrom AB and C.Proof. Let pi, i = 1 . . . k, be the vertex of Q in counter-clockwise order, with p1 the left-most one. For clarity, letus add a point pk+1 = p1 and let m, m < k, be the integersuch that pm is the right-most vertex. For i = 1 . . . k, letli be the side pipi+1 and v(li) be the opposite vertex to theside li. We say that v(li) is x-internal to li if the verticalline through v(li) cuts li.We start by demonstrating that there exists at least one sidel whose opposite vertex v(l) is x-internal to it.Suppose that every side li has its opposite vertex v(li) thatis not x-internal; then, clearly, v(l1) must be on the right ofl1 and v(lm−1) must be on the left of lm−1. So, there mustexist an integer j < m such that v(lj−1) is on the right ofSignal samples Convex-hull found Solution foundFig. 1. Steps of the geometric method for single link optimal solutionlj−1 and v(lj) is on the left of lj . It is easy to see that pj isthe opposite vertex to every side between v(lj−1) and v(lj);the vertical line through pj must cut one of these sides andso there exists a side whose opposite vertex, pj , is x-internalto it, so that the initial hypothesis was inconsistent.Now, suppose we have three points A, B and C of Q suchthat C is the x-internal opposite vertex to the side AB. Forthese three points the optimal linear approximation is easilyproved to be the line r parallel to AB and equidistant fromAB and C. The error committed by this line in approximat-ing s at every x coordinate xi is proportional to the distanceof the point si = (xi, yi) from the line; the way r has beenselected 1 assures thatA, B andC are the points of S mostlydistante from r and so, the l∞ approximation error of r isdue to A, B and C. But for this three points r is optimal andso it is for the whole set S, as A, B, C are peculiar verticesof the convex hull.Finally we have to show that there cannot exist another tripletof points A′, B′ and C ′ such that C ′ is x-internal to the sideA′B′. Supposing these three points exist, they should leadto an optimal solution r′. Calling e(t; q1, q2, q3) the errorcommitted by the line t over the points q1, q2 and q3 weshould havee(r′;A′, B′, C ′) ≥ e(r′;A,B,C) ≥ e(r;A,B,C) (1)because r′ lead to its maximum error on A′, B′ and C ′, andr is optimum for A, B and C. But symmetrically we havee(r,A,B,C) ≥ e(r;A′, B′, C ′) ≥ e(r′;A′, B′, C ′) (2)So the only possibility is that all these ≥ must be replacedby = and, consequently, r = r′, which means that AB isparallel to A′B′, contrarily to the hypothesis on Q. These considerations lead to the following efficient al-gorithm for finding the optimal line r (compare with fig. 1):1. compute the convex-hull of the set S; since the pointsin S can be ranked by increasing x coordinate, suchcomputation can be performed in O(n) with Graham1Consider that all the points of S lie in the strip of plane between thethe line t passing through A and B and its parallel t′ passing through C.The line r is exactly in the medium of this strip and the most distant pointsare the ones lying on t and t′.method ([4]), by incrementally adding points fromleft to right.2. scan the M sides of the convex-hull computing theiropposite vertex until the three points A, B and C areidentified; with some tricks this can be performed inO(M).3. compute the solution of r in O(1).Compared to the linear programming solution this algo-rithm has many advantages. It is very easy to implementand generates a very compact code; all the computationsin the construction of the convex-hull and the scanning ofits vertices can be reduced to a scalar product operation ofthe type e=ac+cd. Moreover the only memory requirement(apart from the input sequence) is a vector containing theindices of the convex hull vertices which represent at mostn × q bits, given a q bit digital sequence. One importantthing to note is that, if we are working with q bits samples,all the computations in the first two steps of the method canbe performed using only fixed point arithmetic. Thus thealgorithm can be implemented without requiring a floatingpoint unit.Furthermore, since the optimal solution is based on select-ing three particular points of S the solution for the wholeset provides a mean to irregularly decimate the original se-quence, with a controlled mean for reconstruction of theoriginal sequence, as it is often required for example in nearlossless data compression. So it provides a strategy for ir-regular sampling or a new mean for detection of relevantsamples in a digital sequence.3. PIECEWISE SOLUTION WITH ERROR BOUNDIn this section we want to study a different problem and,in particular, we are interested to non-continuous piecewiselinear approximations 2 .For the sake of clarity, we suppose that the set S is char-acterized by xi = i, i = 1 . . . n (even though the presented2The proposed strategy for approximation shown in this section andthe next one can be easily extended to any piecewise optimally minmaxfunctional approximation. The difference simply lies in the procedure andcomputational complexity individually attached to each segment.results hold in the case of non uniform samples). Given allthe possible partitions of the domain, the problem of min-max approximating the sequence within each segment of thepartition is a mere application of what explained in the pre-vious section.Given any piecewise minmax approximation g of the signal,we characterize it with an error e(g), a number of segmentsν(g) and a partition set P (g) = {pi(g)}i=1..ν(g)−1 of val-ues such that pi = m+ 1/2 if m is the last point of the i-thsegment and m+1 is the first point of the (i+1)-th segment.Moreover, we set p0(g) = 1/2 and pν(g)(g) = n+1/2, andit is implicitly considered that the partition points pi canonly take values of the type m + 1/2 with m ∈ N. Allconsidered intervals are measured on a discrete half integervalued lattice. Thus we will identify the ”approximation onthe interval [3/2,7/2]” as”the one of locations 2 and 3”; sim-ilarly, by stating ”the partition point pi is in ]3/2, 7/2]” isequivalent to say ”pi ∈ {5/2, 7/2}”.The problem to be solved is the following: given the setS of n samples of s and an error bound δ, we want to find a(non continuous) piecewise linear approximation g of switherror e(g) ≤ δ such that the number ν(g) of segments usedis the smallest possible. In general there are more solutionsto the problem but, interestingly enough, two of them (notnecessarily distinct) can be found in O(n) in a simple way.We call them −→g (left-to-right solution) and←−g (right-to-leftsolution), the motivation will become obvious. It is implicitthat −→g and ←−g depend on the value δ. The algorithm thatdetermines −→g is the following:• Starting from the first point on the left, incrementallyadd points until the error bound of the associated 1-link solution δ is reached. This requires the minmaxerror to be computed at every step. However, givenvertices Am, Bm, Cm at step m and the convex-hullat stepm+1, verticesAm+1, Bm+1 andCm+1 at stepm+1 in O(1) are straightforward to derive. Thus thetotal computational time remains simply proportionalto the number sample points.• When the error bound is reached, this determines thevalue p1(−→g ). The process can now be started for anew segment, until p2(−→g ) is found, i.e. when theerror bound is reached a second time. The procedureis similarly carried out till all points in S have beendealt with.Using the above construction process the number of seg-ments used by −→g is optimal, which means that every piece-wise function h approximating S with error e(h) ≤ δ has anumber of segments ν(h) ≥ ν(−→g ).The way p1(−→g ) was obtained when constructing the so-lution −→g implicitly says that it is not possible to approxi-mate the interval [1, p1(−→g )+1] with a single segment (with-out exceeding the value of δ); h cannot be an exceptionand thus P (h) must have a partition point p1(h) in the in-terval [1, p1(−→g )]. Similarly it is not possible to approxi-mate with a single segment the interval [p1(−→g ), p2(−→g ) +1], so that P (h) must have at least another point p2(h) in]p1(−→g ), p2(−→g )] as p1(h) ≤ p1(−→g ). By iterating the argu-ment, this proves by induction, that for 1 ≤ i ≤ ν(−→g ) − 2there must exist a point ofP (h) in the interval ]pi(−→g ), pi+1(−→g )]and thus ν(h) ≥ ν(−→g ).By symmetry of the construction process, the same proofcan be applied by replacing −→g with ←−g while proceedingfrom right to left. This will ensure that ν(h) ≥ ν(←−g ). Now,setting h = ←−g has lead to ν(←−g ) ≥ ν(−→g ) while settingh = −→g has lead to ν(−→g ) ≥ ν(←−g ).This proves that ν(−→g ) = ν(←−g ) = k and that this number kof segments defines the minimum of the piecewise minmaxoptimal solution.4. OPTIMAL K-LINK SOLUTIONWe have shown in the preceding section how to find a piece-wise solution with optimal number of segments in O(n)when an error bound is given. Once we know the mini-mum number k of segments, we can ask for the solution ink segments with minimum error. The general problem offinding the optimal solution given the number of segmentsis solvable in O(n log n) as shown by Goodrich in [5]. Inour case, however, we can take advantage of the fact that wealready know two solutions (−→g and←−g ) with k segments. Infact we will show that these two solutions provide two parti-tion sets P (−→g ) and P (←−g ) that determine a sort of extremesof the partition set P (h) of any possible k-link solution h).More precisely, if h satisfies ν(h) = k and e(h) ≤ δ thenpi(←−g ) ≤ pi(h) ≤ pi(−→g ) for every 1 ≤ i ≤ k − 1.If e(h) ≤ δ we have already proven that there exists apoint of P (h) in ]pi(−→g ), pi+1(−→g )] for 0 ≤ i ≤ k − 2.If ν(h) = k then in each interval there is exactly one point,which has to be pi+1(h). This holds for h = ←−g so thatpi(−→g ) < pi+1(←−g ) ≤ pi+1(−→g ). By symmetry we can saythat if e(h) ≤ δ and ν(h) = k there is exactly one pointpi(h) in [pi(←−g ), pi+1(←−g )[ for 1 ≤ i ≤ k − 1 and, forh = −→g we obtain pi(←−g ) ≤ pi(−→g ) < pi+1(←−g ). By com-bining these inequalities we reach the result that if h is aoptimal k-link solution then pi(←−g ) ≤ pi(h) ≤ pi(−→g ) forevery 1 ≤ i ≤ k − 1.The above consideration provides a very important prop-erty of the possible partitions of the domain to obtain a k-link solution.We now show how to find the solution f that minimizes themaximum error in O(n). The idea is to proceed from leftto right and find at every location of the candidate i-th par-tition point the partition point pi−1 that minimizes the errorfor all x < pi (which will be indicated with e(pi)).Let us call wi = pi(−→g ) − pi(←−g ) + 1; we know thatPartition of the domain for −→g (up) and←−g (down) Possible positions for p1 and p2, with relative link errorsFig. 2. Finding the optimum partition set P . By computing−→g and←−g we first find the possible ranges for the values pi. Then,for every possible value of pi+1 we compute the value of pi that minimizes the maximum error on the corresponding link; atthe last step pk−1 will be determined and consequently, every pi.for the first point, p1, we can choose between w1 differentpositions, that is p1(←−g ), p1(←−g ) + 1, p1(←−g ) + 2, ... p1(−→g ).For every candidate pl1 = p1(←−g ) + l − 1, l = 1 . . . w1, wecompute the approximation error of the first segment, as wedid when finding −→g , and we denote the corresponding val-ues e(p11), e(p21) . . . e(pw11 ), respectively.Now, for every possible position pr2, r = 1 . . . w2 of p2 andfor every possible position pl1 of p1, we compute the er-ror ε(pl1, pr2) of approximating the interval [pl1, pr2] with onesegment.For every point pr2 we compute the value e(pr2) given bye(pr2) = min1≤l≤w1(max(e(pl1), ε(pl1, pr2))) (3)that is the minimum error that we can induce in the domainx < pr2 if we choose the second partition point p2 = pr2.The value of l for which the minimum is reached in (3)determines the best choice for p1 given that p2 = pr2. Byiterating the process (as e(p12), e(p22), ... are known) we candetermine the best choice for p2 for all possible candidatepositions of p3 and so on. At the last stage we will findthe best position of pk−1 given that pk = n + 1/2. Nowby backpropagation, we can determine the optimum pk−2,pk−3, ..., p1.Let us now assess the computational complexity of thedescribed procedure. At step i, wi · wi−1 errors need tobe estimated. If we move pi in the inner cycle and pi−1 inthe outer one we can assume the number of computationsbe less than wi−1 · (pi(−→g ) − pi(←−g )). By calling li thelength pi(−→g )−pi−1(−→g ) of the i-th segment of−→g , we havewi−1 · (pi(−→g )− pi(←−g )) < li−1 · (li−1 + li).If we consider that the maximum length of the segmentsis asymptotically constant in the number of points, whichactually makes sense in many practical situations (considerfor example electrocardiographic signals), then the numberof operations for every pi is bounded by a constant and sothe complexity of the algorithm is O(n).5. CONCLUSIONWe have presented a new geometric algorithm for the linearapproximation of signals in the l∞ norm. The algorithm isvery efficient both in terms of its computational efficiency,memory usage and ease of implementation. Given an errorbound δ, we have shown how to find a piecewise linear ap-proximation of a signal with minimum number of segmentsk in O(n). Finally, we have shown how to minimize theerror for the same number k of segments in O(n).6. REFERENCES[1] N Megiddo, “Linear-time algorithms for linear pro-gramming in R3 and related problems,” SIAM J. Com-put., vol. 12, pp. 12:759–776, 1983.[2] N. Megiddo, “Linear programming in linear time whenthe dimension is fixed,” J. ACM, pp. 12:114–127, 1984.[3] R. Seidel, “Small-dimensional linear programming andconvex hulls made easy,” Discrete Comput. Geom., vol.6, no. 593-613, 1991.[4] R. Graham, “An efficient algorithm for determinimgthe convex hull of a finite point set,” Info. Proc. Letters,vol. 1, pp. 132–133, 1972.[5] M. T. Goodrich, “Efficient piece-wise linear functionapproximation using the uniform metric,” DiscreteComput. Geom., vol. 14, pp. 445–462, 1995.[6] H. Imai and M. Iri, “Computational gemoetric methodsfor polygonal approximations of a curve,” Computervision, Graphics and Image Processing, vol. 36, pp. 31–41, 1986.[7] B. K. Natarajan, “On piece-wise linear approxima-tions to curves,” SIAM Conference on geometric design,1991.

Efficient (Piecewise) Linear Minmax Approximation of Digital Signals

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