Shape Representation Using Fourier Coefficients of the Sinusoidal
第2届丘成桐大学生数学竞赛试题
S.-T.Yau College Student Mathematics Contests 2011Analysis and Differential EquationsIndividual2:30–5:00pm,July 9,2011(Please select 5problems to solve)1.a)Compute the integral: ∞−∞x cos xdx (x 2+1)(x 2+2),b)Show that there is a continuous function f :[0,+∞)→(−∞,+∞)such that f ≡0and f (4x )=f (2x )+f (x ).2.Solve the following problem: d 2u dx 2−u (x )=4e −x ,x ∈(0,1),u (0)=0,dudx(0)=0.3.Find an explicit conformal transformation of an open set U ={|z |>1}\(−∞,−1]to the unit disc.4.Assume f ∈C 2[a,b ]satisfying |f (x )|≤A,|f(x )|≤B for each x ∈[a,b ]and there exists x 0∈[a,b ]such that |f (x 0)|≤D ,then |f (x )|≤2√AB +D,∀x ∈[a,b ].5.Let C ([0,1])denote the Banach space of real valued continuous functions on [0,1]with the sup norm,and suppose that X ⊂C ([0,1])is a dense linear subspace.Suppose l :X →R is a linear map (not assumed to be continuous in any sense)such that l (f )≥0if f ∈X and f ≥0.Show that there is a unique Borel measure µon [0,1]such that l (f )= fdµfor all f ∈X .6.For s ≥0,let H s (T )be the space of L 2functions f on the circle T =R /(2πZ )whose Fourier coefficients ˆf n = 2π0e−inx f (x )dx satisfy Σ(1+n 2)s ||ˆf n |2<∞,with norm ||f ||2s =(2π)−1Σ(1+n 2)s |ˆf n |2.a.Show that for r >s ≥0,the inclusion map i :H r (T )→H s (T )is compact.b.Show that if s >1/2,then H s (T )includes continuously into C (T ),the space of continuous functions on T ,and the inclusion map is compact.1S.-T.Yau College Student Mathematics Contests2011Geometry and TopologyIndividual9:30–12:00am,July10,2011(Please select5problems to solve)1.Suppose M is a closed smooth n-manifold.a)Does there always exist a smooth map f:M→S n from M into the n-sphere,such that f is essential(i.e.f is not homotopic to a constant map)?Justify your answer.b)Same question,replacing S n by the n-torus T n.2.Suppose(X,d)is a compact metric space and f:X→X is a map so that d(f(x),f(y))=d(x,y)for all x,y in X.Show that f is an onto map.3.Let C1,C2be two linked circles in R3.Show that C1cannot be homotopic to a point in R3\C2.4.Let M=R2/Z2be the two dimensional torus,L the line3x=7y in R2,and S=π(L)⊂M whereπ:R2→M is the projection map. Find a differential form on M which represents the Poincar´e dual of S.5.A regular curve C in R3is called a Bertrand Curve,if there existsa diffeomorphism f:C→D from C onto a different regular curve D in R3such that N x C=N f(x)D for any x∈C.Here N x C denotes the principal normal line of the curve C passing through x,and T x C will denote the tangent line of C at x.Prove that:a)The distance|x−f(x)|is constant for x∈C;and the angle made between the directions of the two tangent lines T x C and T f(x)D is also constant.b)If the curvature k and torsionτof C are nowhere zero,then there must be constantsλandµsuch thatλk+µτ=16.Let M be the closed surface generated by carrying a small circle with radius r around a closed curve C embedded in R3such that the center moves along C and the circle is in the normal plane to C at each point.Prove thatMH2dσ≥2π2,and the equality holds if and only if C is a circle with radius √2r.HereH is the mean curvature of M and dσis the area element of M.1S.-T.Yau College Student Mathematics Contests 2011Algebra,Number Theory andCombinatoricsIndividual2:30–5:00pm,July 10,2011(Please select 5problems to solve)For the following problems,every example and statement must be backed up by proof.Examples and statements without proof will re-ceive no-credit.1.Let K =Q (√−3),an imaginary quadratic field.(a)Does there exists a finite Galois extension L/Q which containsK such that Gal(L/Q )∼=S 3?(Here S 3is the symmetric group in 3letters.)(b)Does there exists a finite Galois extension L/Q which containsK such that Gal(L/Q )∼=Z /4Z ?(c)Does there exists a finite Galois extension L/Q which containsK such that Gal(L/Q )∼=Q ?Here Q is the quaternion group with 8elements {±1,±i,±j,±k },a finite subgroup of the group of units H ×of the ring H of all Hamiltonian quaternions.2.Let f be a two-dimensional (complex)representation of a finite group G such that 1is an eigenvalue of f (σ)for every σ∈G .Prove that f is a direct sum of two one-dimensional representations of G3.Let F ⊂R be the subset of all real numbers that are roots of monic polynomials f (X )∈Q [X ].(1)Show that F is a field.(2)Show that the only field automorphisms of F are the identityautomorphism α(x )=x for all x ∈F .4.Let V be a finite-dimensional vector space over R and T :V →V be a linear transformation such that(1)the minimal polynomial of T is irreducible;(2)there exists a vector v ∈V such that {T i v |i ≥0}spans V .Show that V contains no non-trivial proper T -invariant subspace.5.Given a commutative diagramA →B →C →D →E↓↓↓↓↓A →B →C →D →E1Algebra,Number Theory and Combinatorics,2011-Individual2 of Abelian groups,such that(i)both rows are exact sequences and(ii) every vertical map,except the middle one,is an isomorphism.Show that the middle map C→C is also an isomorphism.6.Prove that a group of order150is not simple.S.-T.Yau College Student Mathematics Contests 2011Applied Math.,Computational Math.,Probability and StatisticsIndividual6:30–9:00pm,July 9,2011(Please select 5problems to solve)1.Given a weight function ρ(x )>0,let the inner-product correspond-ing to ρ(x )be defined as follows:(f,g ):= baρ(x )f (x )g (x )d x,and let f :=(f,f ).(1)Define a sequence of polynomials as follows:p 0(x )=1,p 1(x )=x −a 1,p n (x )=(x −a n )p n −1(x )−b n p n −2(x ),n =2,3,···wherea n =(xp n −1,p n −1)(p n −1,p n −1),n =1,2,···b n =(xp n −1,p n −2)(p n −2,p n −2),n =2,3,···.Show that {p n (x )}is an orthogonal sequence of monic polyno-mials.(2)Let {q n (x )}be an orthogonal sequence of monic polynomialscorresponding to the ρinner product.(A polynomial is called monic if its leading coefficient is 1.)Show that {q n (x )}is unique and it minimizes q n amongst all monic polynomials of degree n .(3)Hence or otherwise,show that if ρ(x )=1/√1−x 2and [a,b ]=[−1,1],then the corresponding orthogonal sequence is the Cheby-shev polynomials:T n (x )=cos(n arccos x ),n =0,1,2,···.and the following recurrent formula holds:T n +1(x )=2xT n (x )−T n −1(x ),n =1,2,···.(4)Find the best quadratic approximation to f (x )=x 3on [−1,1]using ρ(x )=1/√1−x 2.1Applied Math.Prob.Stat.,2011-Individual 22.If two polynomials p (x )and q (x ),both of fifth degree,satisfyp (i )=q (i )=1i,i =2,3,4,5,6,andp (1)=1,q (1)=2,find p (0)−q (0)y aside m black balls and n red balls in a jug.Supposes 1≤r ≤k ≤n .Each time one draws a ball from the jug at random.1)If each time one draws a ball without return,what is the prob-ability that in the k -th time of drawing one obtains exactly the r -th red ball?2)If each time one draws a ball with return,what is the probability that in the first k times of drawings one obtained totally an odd number of red balls?4.Let X and Y be independent and identically distributed random variables.Show thatE [|X +Y |]≥E [|X |].Hint:Consider separately two cases:E [X +]≥E [X −]and E [X +]<E [X −].5.Suppose that X 1,···,X n are a random sample from the Bernoulli distribution with probability of success p 1and Y 1,···,Y n be an inde-pendent random sample from the Bernoulli distribution with probabil-ity of success p 2.(a)Give a minimum sufficient statistic and the UMVU (uniformlyminimum variance unbiased)estimator for θ=p 1−p 2.(b)Give the Cramer-Rao bound for the variance of the unbiasedestimators for v (p 1)=p 1(1−p 1)or the UMVU estimator for v (p 1).(c)Compute the asymptotic power of the test with critical region |√n (ˆp 1−ˆp 2)/ 2ˆp ˆq |≥z 1−αwhen p 1=p and p 2=p +n −1/2∆,where ˆp =0.5ˆp 1+0.5ˆp 2.6.Suppose that an experiment is conducted to measure a constant θ.Independent unbiased measurements y of θcan be made with either of two instruments,both of which measure with normal errors:fori =1,2,instrument i produces independent errors with a N (0,σ2i )distribution.The two error variances σ21and σ22are known.When ameasurement y is made,a record is kept of the instrument used so that after n measurements the data is (a 1,y 1),...,(a n ,y n ),where a m =i if y m is obtained using instrument i .The choice between instruments is made independently for each observation in such a way thatP (a m =1)=P (a m =2)=0.5,1≤m ≤n.Applied Math.Prob.Stat.,2011-Individual 3Let x denote the entire set of data available to the statistician,in this case (a 1,y 1),...,(a n ,y n ),and let l θ(x )denote the corresponding log likelihood function for θ.Let a =n m =1(2−a m ).(a)Show that the maximum likelihood estimate of θis given by ˆθ= n m =11/σ2a m −1 n m =1y m /σ2a m.(b)Express the expected Fisher information I θand the observedFisher information I x in terms of n ,σ21,σ22,and a .What hap-pens to the quantity I θ/I x as n →∞?(c)Show that a is an ancillary statistic,and that the conditional variance of ˆθgiven a equals 1/I x .Of the two approximations ˆθ·∼N (θ,1/I θ)and ˆθ·∼N (θ,1/I x ),which (if either)would you use for the purposes of inference,and why?S.-T.Yau College Student Mathematics Contests 2011Analysis and Differential EquationsTeam9:00–12:00am,July 9,2011(Please select 5problems to solve)1.Let H 2(∆)be the space of holomorphic functions in the unit disk ∆={|z |<1}such that ∆|f |2|dz |2<∞.Prove that H 2(∆)is a Hilbert space and that for any r <1,the map T :H 2(∆)→H 2(∆)given by T f (z ):=f (rz )is a compact operator.2.For any continuous function f (z )of period 1,show that the equation dϕdt=2πϕ+f (t )has a unique solution of period 1.3.Let h (x )be a C ∞function on the real line R .Find a C ∞function u (x,y )on an open subset of R containing the x -axis such that u x +2u y =u 2and u (x,0)=h (x ).4.Let S ={x ∈R ||x −p |≤c/q 3,for all p,q ∈Z ,q >0,c >0},show that S is uncountable and its measure is zero.5.Let sl (n )denote the set of all n ×n real matrices with trace equal to zero and let SL (n )be the set of all n ×n real matrices with deter-minant equal to one.Let ϕ(z )be a real analytic function defined in a neighborhood of z =0of the complex plane C satisfying the conditions ϕ(0)=1and ϕ (0)=1.(a)If ϕmaps any near zero matrix in sl (n )into SL (n )for some n ≥3,show that ϕ(z )=exp(z ).(b)Is the conclusion of (a)still true in the case n =2?If it is true,prove it.If not,give a counterexample.e mathematical analysis to show that:(a)e and πare irrational numbers;(b)e and πare also transcendental numbers.1S.-T.Yau College Student Mathematics Contests2011Applied Math.,Computational Math.,Probability and StatisticsTeam9:00–12:00am,July9,2011(Please select5problems to solve)1.Let A be an N-by-N symmetric positive definite matrix.The con-jugate gradient method can be described as follows:r0=b−A x0,p0=r0,x0=0FOR n=0,1,...αn= r n 22/(p TnA p n)x n+1=x n+αn p n r n+1=r n−αn A p nβn=−r Tk+1A p k/p TkA p kp n+1=r n+1+βn p nEND FORShow(a)αn minimizes f(x n+αp n)for allα∈R wheref(x)≡12x T A x−b T x.(b)p Ti r n=0for i<n and p TiA p j=0if i=j.(c)Span{p0,p1,...,p n−1}=Span{r0,r1,...,r n−1}≡K n.(d)r n is orthogonal to K n.2.We use the following scheme to solve the PDE u t+u x=0:u n+1 j =au nj−2+bu nj−1+cu njwhere a,b,c are constants which may depend on the CFL numberλ=∆t ∆x .Here x j=j∆x,t n=n∆t and u njis the numerical approximationto the exact solution u(x j,t n),with periodic boundary conditions.(i)Find a,b,c so that the scheme is second order accurate.(ii)Verify that the scheme you derived in Part(i)is exact(i.e.u nj =u(x j,t n))ifλ=1orλ=2.Does this imply that the scheme is stable forλ≤2?If not,findλ0such that the scheme is stable forλ≤λ0. Recall that a scheme is stable if there exist constants M and C,which are independent of the mesh sizes∆x and∆t,such thatu n ≤Me CT u0for all∆x,∆t and n such that t n≤T.You can use either the L∞norm or the L2norm to prove stability.1Applied Math.Prob.Stat.,2011-Team2 3.Let X and Y be independent random variables,identically dis-tributed according to the Normal distribution with mean0and variance 1,N(0,1).(a)Find the joint probability density function of(R,),whereR=(X2+Y2)1/2andθ=arctan(Y/X).(b)Are R andθindependent?Why,or why not?(c)Find a function U of R which has the uniform distribution on(0,1),Unif(0,1).(d)Find a function V ofθwhich is distributed as Unif(0,1).(e)Show how to transform two independent observations U and Vfrom Unif(0,1)into two independent observations X,Y fromN(0,1).4.Let X be a random variable such that E[|X|]<∞.Show thatE[|X−a|]=infE[|X−x|],x∈Rif and only if a is a median of X.5.Let Y1,...,Y n be iid observations from the distribution f(x−θ), whereθis unknown and f()is probability density function symmetric about zero.Suppose a priori thatθhas the improper priorθ∼Lebesgue(flat) on(−∞,∞).Write down the posterior distribution ofθ.Provides some arguments to show that thisflat prior is noninforma-tive.Show that with the posterior distribution in(a),a95%probability interval is also a95%confidence interval.6.Suppose we have two independent random samples{Y1,i=1,...,n} from Poisson with(unknown)meanλ1and{Y i,i=n+1,...,2n}from Poisson with(unknown)meanλ2Letθ=λ1/(λ1+λ2).(a)Find an unbiased estimator ofθ(b)Does your estimator have the minimum variance among all un-biased estimators?If yes,prove it.If not,find one that has theminimum variance(and prove it).(c)Does the unbiased minimum variance estimator you found at-tain the Fisher information bound?If yes,show it.If no,whynot?S.-T.Yau College Student Mathematics Contests2011Geometry and TopologyTeam9:00–12:00am,July9,2011(Please select5problems to solve)1.Suppose K is afinite connected simplicial complex.True or false:a)Ifπ1(K)isfinite,then the universal cover of K is compact.b)If the universal cover of K is compact thenπ1(K)isfinite.pute all homology groups of the the m-skeleton of an n-simplex, 0≤m≤n.3.Let M be an n-dimensional compact oriented Riemannian manifold with boundary and X a smooth vectorfield on M.If n is the inward unit normal vector of the boundary,show thatM div(X)dV M=∂MX·n dV∂M.4.Let F k(M)be the space of all C∞k-forms on a differentiable man-ifold M.Suppose U and V are open subsets of M.a)Explain carefully how the usual exact sequence0−→F(U∪V)−→F(U)⊕F V)−→F(U∩V)−→0 arises.b)Write down the“long exact sequence”in de Rham cohomology as-sociated to the short exact sequence in part(a)and describe explicitly how the mapH kdeR (U∩V)−→H k+1deR(U∪V)arises.5.Let M be a Riemannian n-manifold.Show that the scalar curvature R(p)at p∈M is given byR(p)=1vol(S n−1)S n−1Ric p(x)dS n−1,where Ric p(x)is the Ricci curvature in direction x∈S n−1⊂T p M, vol(S n−1)is the volume of S n−1and dS n−1is the volume element of S n−1.1Geometry and Topology,2011-Team2 6.Prove the Schur’s Lemma:If on a Riemannian manifold of dimension at least three,the Ricci curvature depends only on the base point but not on the tangent direction,then the Ricci curvature must be constant everywhere,i.e.,the manifold is Einstein.S.-T.Yau College Student Mathematics Contests 2011Algebra,Number Theory andCombinatoricsTeam9:00–12:00pm,July 9,2011(Please select 5problems to solve)For the following problems,every example and statement must be backed up by proof.Examples and statements without proof will re-ceive no-credit.1.Let F be a field and ¯Fthe algebraic closure of F .Let f (x,y )and g (x,y )be polynomials in F [x,y ]such that g .c .d .(f,g )=1in F [x,y ].Show that there are only finitely many (a,b )∈¯F×2such that f (a,b )=g (a,b )=0.Can you generalize this to the cases of more than two-variables?2.Let D be a PID,and D n the free module of rank n over D .Then any submodule of D n is a free module of rank m ≤n .3.Identify pairs of integers n =m ∈Z +such that the quotient rings Z [x,y ]/(x 2−y n )∼=Z [x,y ]/(x 2−y m );and identify pairs of integers n =m ∈Z +such that Z [x,y ]/(x 2−y n )∼=Z [x,y ]/(x 2−y m ).4.Is it possible to find an integer n >1such that the sum1+12+13+14+ (1)is an integer?5.Recall that F 7is the finite field with 7elements,and GL 3(F 7)is the group of all invertible 3×3matrices with entries in F 7.(a)Find a 7-Sylow subgroup P 7of GL 3(F 7).(b)Determine the normalizer subgroup N of the 7-Sylow subgroupyou found in (a).(c)Find a 2-Sylow subgroup of GL 3(F 7).6.For a ring R ,let SL 2(R )denote the group of invertible 2×2matrices.Show that SL 2(Z )is generated by T = 1101 and S = 01−10 .What about SL 2(R )?1。
数字图像处理第二章课后习题及中文版解答
数字图像处理(冈萨雷斯版，第二版)课后习题及解答(部分)Ch 22.1使用2.1节提供的背景信息，并采用纯几何方法，如果纸上的打印点离眼睛0.2m 远，估计眼睛能辨别的最小打印点的直径。
为了简明起见，假定当在黄斑处的像点变得远比视网膜区域的接收器（锥状体）直径小的时候，视觉系统已经不能检测到该点。
进一步假定黄斑可用1.5mm × 1.5mm 的方阵模型化，并且杆状体和锥状体间的空间在该阵列上的均匀分布。
解：对应点的视网膜图像的直径x 可通过如下图题2.1所示的相似三角形几何关系得到，即()()220.20.014d x = 解得x =0.07d 。
根据2.1节内容，我们知道：如果把黄斑想象为一个有337000个成像单元的正方形传感器阵列，它转换成一个大小580×580成像单元的阵列。
假设成像单元之间的间距相等，这表明在总长为1.5 mm 的一条线上有580个成像单元和579个成像单元间隔。
则每个成像单元和成像单元间隔的大小为s =[(1.5 mm)/1159]=1.3×10-6 m 。
如果在黄斑上的成像点的大小是小于一个可分辨的成像单元，在我们可以认为改点对于眼睛来说不可见。
换句话说，眼睛不能检测到以下直径的点：x =0.07d<1.3×10-6m ，即d <18.6×10-6 m 。
下图附带解释：因为眼睛对近处的物体聚焦时，肌肉会使晶状体变得较厚，折射能力也相对提高，此时物体离眼睛距离0.2 m ，相对较近。
而当晶状体的折射能力由最小变到最大时，晶状体的聚焦中心与视网膜的距离由17 mm 缩小到14 mm ，所以此图中选取14mm(原书图2.3选取的是17 mm)。
图 题2.12.2 当在白天进入一个黑暗的剧场时，在能看清并找到空座位时要用一段时间适应，2.1节(视觉感知要素)描述的视觉过程在这种情况下起什么作用？解：根据人眼的亮度适应性，1）由于户外与剧场亮度差异很大，因此当人进入一个黑暗的剧场时，无法适应如此大的亮度差异，在剧场中什么也看不见；2）人眼不断调节亮度适应范围，逐渐的将视觉亮度中心调整到剧场的亮度范围，因此又可以看见、分清场景中的物体了。
TetGen用户手册中文版
一款优良的四面体格网生成和 3D Delaunay 三角剖分器
版本 1.5 用户手册 2012 年 11 月 5 日
Hang Si si@wias-berlin.de http://www.tetgen.osg c 2002 – 2012
周顺 陈强译 E_mail:shunzhou@ 2014 年 1 月
3 使用 TetGen.................................................................................................18
3.1 命令行语法..............................................................................................................18 3.2 命令行开关...............................................................................................................18 3.2.1 四面体格网化一个点集(-w, -v)...................................................................19 3.2.2 分段线性复合体(PLC)四面体格网化(-p, -Y).............................................21 3.2.3 生成优良格网 (-q).......................................................................................23 3.2.4 分配范围属性(-A)........................................................................................24 3.2.5 体积约束(-a).................................................................................................25 3.2.6 应用格网尺寸函数(-m)................................................................................25 3.2.7 重建四面体格网(-r) .....................................................................................25 3.2.8 插入附加点列表(-i)......................................................................................26 3.2.9 格网优化(-O)................................................................................................26 3.2.10 设置容差(-T) ..............................................................................................26 3.2.11 格网统计(-V)..............................................................................................27 3.2.12 其他选项.....................................................................................................27
数字图像处理-冈萨雷斯-课件(英文)Chapter11-表示与描述可编辑全文
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数字图像处理第四版拉斐尔课后答案
数字图像处理第四版拉斐尔课后答案数字图像处理（美）Rafael C. Gonzalez（拉斐尔·C. 冈萨雷斯），Richard E. Woods（理查德·E. 伍兹）课后习题答案1. 新增了关于精确直⽅图匹配、⼩波、图像变换、有限差分、k均值聚类、超像素、图割、斜率编码的内容。
2. 扩展了关于⾻架、中轴和距离变换的说明，增加了紧致度、圆度和偏⼼率等描述⼦。
3. 新增了哈⾥斯-斯蒂芬斯⾓点探测器及*稳定极值区域的内容。
扫⼀扫⽂末在⾥⾯回复答案+数字图像处理⽴即得到答案4. 重写了关于神经⽹络和深度学习的内容，全⾯介绍了全连接深度神经⽹络，新增了关于深度卷积神经⽹络的内容。
5. 为学⽣和教师提供⽀持包，⽀持包可从本书的配套⽹站下载。
6. 新增了⼏百幅图像、⼏⼗个新图表和上百道新习题。
在数字图像处理领域，本书作为主要教材已有40多年。
第四版是作者在前三版的基础上修订⽽成的，是前三版的发展与延续。
除保留前⼏版的⼤部分内容外，根据读者的反馈，作者对本书进⾏了全⾯修订，融⼊了近年来数字图像处理领域的重要进展，增加了⼏百幅新图像、⼏⼗个新图表和上百道新习题。
全书共12章，即绪论、数字图像基础、灰度变换与空间滤波、频率域滤波、图像复原与重构、⼩波变换和其他图像变换、彩⾊图像处理、图像压缩和⽔印、形态学图像处理、图像分割、特征提取、图像模式分类。
本书的读者对象主要是从事信号与信息处理、通信⼯程、电⼦科学与技术、信息⼯程、⾃动化、计数字图像处理课后答案（美）Rafael C.Gonzalez（拉斐尔·C. 冈萨雷斯），Richard E. Woods（理查德·E. 伍兹）算机科学与技术、地球物理、⽣物⼯程、⽣物医学⼯程、物理、化学、医学、遥感等领域的⼤学教师和科技⼯作者、研究⽣、⼤学本科⾼年级学⽣及⼯程技术⼈员。
Rafael C. Gonzalez: 1965于美国迈阿密⼤学获电⽓⼯程学⼠学位；1967年和1970年于美国佛罗⾥达⼤学盖恩斯维尔分校分别获电⽓⼯程硕⼠学位和博⼠学位。
冈萨雷斯_数字图像处理第3版第4章的习题集.doc
4.16 证明连续和离散二维傅里叶变换都是平移和旋转不变的。
首先列出平移和旋转性质：002(//)00(,)(,)j u x M v y N f x y e F u u v v π+⇔-- (4.6-3) 002(//)00(,)(,)j x r M y v N f x x y y F u v e π-+--⇔ (4.6-4)旋转性质：cos ,sin ,cos ,sin x r y r u v θθωϕωϕ====00(,)(,)f r F θθωϕϕ+⇔+ (4.6-5) 证明：由式(4.5-15)得：由式(4.5-16)得：依次类推证明其它项。
4.17 由习题4.3可以推出1(,)u v δ⇔和(,)1t z δ⇔。
使用前一个性质和表4.3中的平移性质证明连续函数00(,)cos(22)f t z A u t v z ππ=+的傅里叶变换是0000(,)[(,)(,)]2AF u v u u v v u u v v δδ=+++-- 证明：000000002()2()002()2()2()2()2()2()2((,)(,)cos(22)[]222j ut vz j ut vz j u t v z j u t v z j ut vz j u t v z j u t v z j ut vz j u F u v f t z e dtdzA u t v z e dtdzA e e e dtdzA A e e dtdz e e πππππππππππ∞∞-+-∞-∞∞∞-+-∞-∞∞∞+-+-+-∞-∞∞∞+-+-+--∞-∞==+=+=+⎰⎰⎰⎰⎰⎰⎰⎰)00000000(,)(,)22[(,)(,)]2t vz dtdz A Au u v v u u v v Au u v v u u v v δδδδ∞∞+-∞-∞=--+++=--+++⎰⎰ 4.18 证明离散函数(,)1f x y =的DFT 是1,0{1}(,)0,u v u v δ==⎧ℑ==⎨⎩其它证明：离散傅里叶变换112(//)00(,)(,)M N j ux M vy N x y F u v f x y e π---+===∑∑112(//)00112(//)00{1}M N j ux M vy N x y M N j ux M vy N x y e e ππ---+==---+==ℑ==∑∑∑∑如果0u v ==，{1}1ℑ=，否则：1100{1}{cos[2(//)]sin[2(//)]}M N x y ux M vy N j ux M vy N ππ--==ℑ=+-+∑∑考虑实部，1100{1}cos[2(//)]M N x y ux M vy N π--==ℑ=+∑∑，cos[2(//)]ux M vy N π+的值介于[-1, 1]，可以想象，1100{1}cos[2(//)]0M N x y ux M vy N π--==ℑ=+=∑∑，虚部相同，所以1,0{1}(,)0,u v u v δ==⎧ℑ==⎨⎩其它4.19 证明离散函数00cos(22)u x v y ππ+的DFT 是00001(,)[(,)(,)]2F u v u Mu v Nv u Mu v Nv δδ=+++--证明：000000112(//)00112(//)0000112()2()2(//)00112()2(//)00(,)(,)cos(22)1[]21{2M N j ux M vy N x y M N j ux M vy N x y M N j u x v y j u x v y j ux M vy N x y M N j u x v y j ux M vy N x y F u v f x y e u x v y e e e e e e πππππππππ---+==---+==--+-+-+==--+-+====+=+=∑∑∑∑∑∑∑∑000000112()2(//)0011112(//)2(//)2(//)2(//)00000000}1{}21[(,)(,)]2M N j u x v y j ux M vy N x y M N M N j Mu x M Nv y N j Mu x M Nv y N j ux M vy N j ux M vy N x y x y e e e e e e u Mu v Nv u Mu v Nv ππππππδδ---+-+==----+-+-+-+====+=+=+++--∑∑∑∑∑∑4.20 下列问题与表4.1中的性质有关。
pcl 法向夹角特征点提取
pcl 法向夹角特征点提取1. 什么是法向夹角特征点？法向夹角特征点是一种局部几何特征，它描述了表面法向向量之间的夹角。
法向夹角特征点可以用来检测表面上的突变、褶皱和边缘等特征。
2. PCL 中的法向夹角特征点提取PCL 中提供了多种法向夹角特征点提取算法，其中最常用的算法是曲率估计算法和主曲率算法。
曲率估计算法曲率估计算法通过计算表面曲率来检测法向夹角特征点。
曲率是曲面法向向量在曲线上变化的程度的度量。
曲率越大，曲面变化越快。
PCL 中提供了多种曲率估计算法，其中最常用的算法是法向向量法。
法向向量法通过计算曲面法向向量在曲线上变化的程度来估计曲率。
主曲率算法主曲率算法通过计算曲面的两个主曲率来检测法向夹角特征点。
主曲率是曲面法向向量在曲线上变化最快的两个方向上的曲率。
PCL 中提供了多种主曲率算法，其中最常用的算法是高斯曲率算法。
高斯曲率算法通过计算曲面高斯曲率来估计主曲率。
3. 法向夹角特征点提取的应用法向夹角特征点提取在计算机视觉和机器人领域有着广泛的应用，其中最常见的应用包括：表面重建法向夹角特征点可以用来重建曲面。
曲面重建是指从一组不规则的点云数据中恢复曲面的过程。
法向夹角特征点可以帮助确定曲面的边界和边缘，从而提高曲面重建的精度。
物体识别法向夹角特征点可以用来识别物体。
物体识别是指从一组图像或点云数据中识别物体的过程。
法向夹角特征点可以帮助确定物体的形状和轮廓，从而提高物体识别的准确率。
机器人导航法向夹角特征点可以用来帮助机器人导航。
机器人导航是指机器人自主地在环境中移动的过程。
法向夹角特征点可以帮助机器人检测障碍物和危险区域，从而提高机器人导航的安全性。
4. 总结法向夹角特征点提取是一种局部几何特征提取技术，它可以用来检测表面上的突变、褶皱和边缘等特征。
PCL 中提供了多种法向夹角特征点提取算法，其中最常用的算法是曲率估计算法和主曲率算法。
法向夹角特征点提取在计算机视觉和机器人领域有着广泛的应用，其中最常见的应用包括表面重建、物体识别和机器人导航。
Journal of Electronic Imaging 13(3), 411–417 (July 2004).
Superquadric representation of automotive parts applying part decompositionYan ZhangAndreas KoschanMongi A.AbidiUniversity of TennesseeDepartment of Electrical and Computer Engineering334Ferris HallKnoxville,Tennessee37996-2100E-mail:yzhang@Abstract.Superquadrics are able to represent a large variety of objects with only a few parameters and a single equation.We present a superquadric representation strategy for automotive parts composed of3-D triangle meshes.Our strategy consists of two ma-jor steps of part decomposition and superquadricfitting.The origi-nalities of this approach include the following two features.First,our approach can represent multipart objects with superquadrics suc-cessfully by applying part decomposition.Second,superquadrics re-covered from our approach have the highest confidence and accu-racy due to the3-D watertight surfaces utilized.A novel,generic3-D part decomposition algorithm based on curvature analysis is also proposed.Experimental results demonstrate that the proposed part decomposition algorithm is able to segment multipart objects into meaningful single parts efficiently.The proposed superquadric rep-resentation strategy can then represent each individual part of the original objects with a superquadric model successfully.©2004 SPIE and IS&T.[DOI:10.1117/1.1762516]1IntroductionObject representation denotes representing real-world ob-jects with known graphic or mathematical primitives that can be recognized by computers.This research has numer-ous applications in areas including computer vision,com-puter graphics,and reverse engineering.An object can be represented by three levels of primitives in terms of the dimensional complexity:volumetric primitives,surface el-ements,and contours.The primitive selected to describe the object depends on the complexity of the object and the tasks involved.As the highest level primitives,volumetric primitives can better represent global features of an object with a significantly reduced amount of information com-pared with surface elements and contours.In addition, volumetric primitives have the ability to achieve the highest data compression ratio without losing the accuracy of the original data.The primarily used volumetric primitives in-clude generalized cylinders,geons,and superquadrics.1Su-perquadrics are a generalization of basic quadric surfaces and they can represent a large variety of shapes with only a few parameters and a single equation.An object initially represented by thousands of triangle meshes can be repre-sented by only a small set of superquadrics.This compact representation can be applied to object recognition to aid, for example,automated depalletizing of industrial parts or robot-guided bin picking of mixed nuclear waste in a haz-ardous environment.The quality control of both tasks can be enhanced by employing superquadrics.Furthermore,the registration of multiview data is indispensable to measure the size of partially occluded objects or their distances from each other in several image-based quality control tasks.Su-perquadrics can be used to efficiently register multiview range data of scenes with small overlap.2Most early research on superquadric representation con-centrated on representing single-part objects from single-view intensity or range images by assuming that the objects have been appropriately segmented.3–13This category of research focused on the data-fitting process,including ob-jective function selection,fitting criteria measurements,andPaper ORNL-007received Jul.30,2003;accepted for publication Feb.23,2004. 1017-9909/2004/$15.00©2004SPIE andIS&T.Fig.1Real range image of a multipart object obtained from Ref.18. Journal of Electronic Imaging13(3),411–417(July2004).Journal of Electronic Imaging/July2004/Vol.13(3)/411convergence analysis.For complex,multipart objects or scenes,there are two major types of approaches in the re-search literature.The first type of method incorporates an image segmentation step prior to the superquadric fitting.11–15The other type of method directly recovers su-perquadrics from a range image withoutpresegmentation.16–19Compared with superquadric repre-sentation of single-part object,these two types of methods can represent more complex objects and have wider appli-cations in related tasks including robotic navigation,object recognition,and virtual reality.However,existing super-quadric representation methods have several weaknesses.First,existing methods cannot handle arbitrary shapes or significant occlusions in the scene.Figure 1shows an ex-ample of the most complicated object that can be repre-sented by superquadrics appeared in the research literature.18We observe that the range image shown in Fig.1con-tains very few occlusions due to the simplicity of the ob-ject.In this case,an optimal viewpoint can easily be found from which each part of the object is visible.When an automotive part,i.e.,a complex,multipart object such as shown in Fig.2,is of interest,no existing methods can represent this object correctly because heavy occlusions are inevitable from any single viewpoint due to the complexity of the object.The second weakness of existing methods is that they utilize only single-view images.Again,for the automotive part shown in Fig.2͑a ͒,it is too difficult to find an optimal viewpoint from which all the parts are visible due to self-occlusions and occlusions,as shown in Fig.2͑b ͒.In addi-tion,the confidence of recovered superquadrics is low due to incomplete single-view data utilized and the accuracy of the recovered models highly depends on the viewpoint used to acquire the data.How complicated,multipart objects canbe represented by superquadrics with high confidence and accuracy remains unknown from the literature.In this paper,we propose an efficient strategy to repre-sent multipart objects with superquadrics.We also present a novel 3-D part decomposition algorithm based on curvature analysis to facilitate our superquadric representation strat-egy.Experiments are shown for automotive parts composed of 3-D triangulated surfaces.The remainder of this paper proceeds as follows.Section 2presents a superquadric representation approach for mul-tipart objects.Section 3proposes the 3-D part decomposi-tion algorithm for triangle meshes.The experimental results are presented in Sec.4and Sec.5concludes the paper.2Superquadric Representation of Multipart Objects Utilizing Part DecompositionA diagram for the proposed superquadric representation al-gorithm is illustrated in Fig.3.Beginning with a multipart object composed of triangle meshes,we propose a part de-composition algorithm to segment the meshes into single parts.Next,each single part is fitted with a superquadric model.Utilizing part decomposition,the difficult represen-tation problem of complicated objects is solved.We use 3-D triangulated surfaces reconstructed from multiview range images as input so that the recovered superquadrics have significantly higher confidence than those recovered from single-view images.In addition,our proposed algo-rithms are generic and flexible in the sense of triangle mesh handling ability since triangle meshes have been the stan-dard surface representation elements in many computer-related areas.A triangulation step is required only if un-structured 3-D point clouds areprovided.Fig.2Distributor cap:(a)photograph of the object,(b)rendering of 3-D triangulated surfaces scanned from view 1,and (c)rendering of 3-D triangulated surfaces scanned from view2.Fig.3Diagram of the proposed superquadric representation strategy utilizing part decomposition.Zhang,Koschan,and Abidi412/Journal of Electronic Imaging /July 2004/Vol.13(3)2.1Introduction to SuperquadricsA set of superquadrics with various shape factors is shown in Fig.4.The implicit definition of superquadrics is ex-pressed as 18F ͑x ,y ,z ͒ϭͫͩx a 1ͪ2/2ϩͩy a 2ͪ2/2ͬ2/1ϩͩz a 3ͪ2/1ϭ1,1,2͑0,2͒,͑1͒where (x ,y ,z )represents a surface point of the superquad-ric,(a 1,a 2,a 3)represent sizes in the (x ,y ,z )directions,and (1,2)represent shape factors.To represent a super-quadric model with global deformations in the world coor-dinate system,15parameters are needed.They are summa-rized as 18∧ϭ͑a 1,a 2,a 3,1,2,,,,p x ,p y ,p z ,k x ,k y ,␣,k ͒,͑2͒where the first 11parameters define a regular superquadric.Parameters k x and k y define the tapering deformations and ␣and k define the bending deformations.Most approaches define an objective function and find the superquadric pa-rameters through minimizing this objective function.The objective function used in this paper is 1G ͑∧͒ϭa 1a 2a 3͚i ϭ1N͓F 1͑x c ,y c ,z c ͒Ϫ1͔2.͑3͒The Levenberg-Marquardt method 20was implemented tominimize the objective function due to its stability and ef-ficiency.In addition,our superquadric fitting algorithm is able to recover superquadrics with global deformations from unstructured 3-D data points.3Curvature-Based 3-D Part DecompositionMany tasks in computer vision,computer graphics,and re-verse engineering involve objects or models.These tasks become extremely difficult when the object of interest is complicated,e.g.,it contains multiple parts.Part decompo-sition can simplify the original task performed on multipart objects into several subtasks each dealing with their con-stituent single,much simpler parts.21,22While a significant amount of research for part decomposition of 2-D intensity or 2.5-D range images has been conducted over the last 2decades,23–25little effort has been made on part segmenta-tion of 3-D data.26,27Therefore,a novel 3-D part decompo-sition algorithm is proposed in this paper.Figure 5illus-trates the difference between region segmentation and part decomposition.A scene consisting of a barrel on the floor is segmented into three surfaces by a region segmentation al-gorithm and two single-part objects by a part decomposi-tion algorithm.We can observe that the scene can be rep-resented by two superquadrics,which is consistent with the part decomposition result.Therefore,we conclude that part decomposition is more appropriate for high-level tasks such as volumetric primitives-based object representation and recognition.A diagram of the proposed part decomposition algorithm is shown in Fig.6.The proposed part decomposition consists of four major steps:Gaussian curvature estimation,boundary detection,region growing,and postprocessing.Boundaries between two articulated parts are composed of points with highly negative curvature according to the transversality regularity.21,22These boundaries are therefore detected by thresholding estimated curvatures for each vertex.A component-labeling operation is then performed to grow nonboundary vertices into parts.Finally,a postprocessing step is performed to assign nonlabeled vertices,including boundary vertices,to one of the parts and merge parts con-taining fewer vertices than a prespecified threshold into their neighbor parts.This part decomposition algorithm is summarized as follows.ˆAlgorithm 1…3-D part decomposition of triangle meshes …‰ˆInput:‰Triangulated surfaces.ˆStep 1.‰Compute Gaussian curvature for each vertex on the surface.ˆStep 2.‰Label vertices of highly negative curvatureasFig.4Superquadrics with various shapeparameters.Fig.6Diagram of the proposed 3-D part decomposition algorithm.Superquadric representation of automotive parts ...Journal of Electronic Imaging /July 2004/Vol.13(3)/413boundaries and the remaining vertices as seeds.ˆStep 3.‰Perform iterative region growing on each seed vertex.ˆStep 4.‰Assign nonlabeled vertices to parts and merge parts having less than a prespecified number of vertices into their neighboring parts.ˆOutput:‰Decomposed single parts.The major steps of this part decomposition algorithm are described respectively in the following sections.3.1Gaussian Curvature Estimation and BoundaryDetectionThe Gaussian curvature for each vertex on a triangulated surface is estimated by K ͑p ͒ϭ3͑2Ϫ͚i ϭ1N i ͒i ϭ1NA i␦2͑p Ϫp i ͒,͑4͒using the method proposed in Ref.28.Variable p representsthe point of interest,p i represents a neighboring vertex of the point p ,and A i represents the area of the triangle con-taining the point p .Variable i represents the interior angle of the triangle at p ,and ␦is the Dirac delta function.The triangles sharing the vertex p are illustrated in Fig.7.After Gaussian curvature is obtained for each vertex on the surface,a prespecified threshold is applied to label ver-tices as boundary or seed.Vertices of highly negative cur-vature are labeled as boundaries between two parts accord-ing to the transversality regularity,21while the rest are labeled as seeds.The threshold is critical and affects the performance of region growing.This threshold is deter-mined in a heuristic manner and depends on mesh resolu-tion.Two types of isolated vertices defined in this work according to their labels include:͑1͒a point that is labeled as boundary while all of its neighbors are labeled as seeds and ͑2͒a point that is labeled as a seed while all of its neighbors are labeled as boundary.The isolated vertices are removed by changing their labels to be the same as those of their neighbors.3.2Region Growing and PostprocessingAfter the vertices are labeled,a region-growing step is per-formed on each vertex labeled as seed.Figure 8showstriangle meshes around the point p .To illustrate the region growing process,a set of two-ring neighbor meshes around point p is shown in this figure.Region growing is performed as follows.Starting from a seed vertex p ,the region number 1is first assigned to the vertex.Second,all the neighbors p i initially labeled as seeds are then labeled with the same region number as the point p .The same labeling process is performed for each neighbor p i to label vertices p i j .This process terminates when the grown region is surrounded by boundary vertices,i.e.,the neighbors of the edge vertices of the region are all labeled as boundaries (Ϫ1).This process is repeated for every other vertex labeled as seed ͑0͒,but not for a vertex that has been grown and labeled with one of the region numbers (1,2,...,N ).After all the seed vertices are assigned new labels,a postprocessing is performed for each bound-ary vertex.Given a seed point x ,all its neighbors x i are first sorted in an ascending order based on their Euclidean distance to the point x .Next,a neighboring vertex x i ,which is the first point labeled with a region number ͑Ͼ0͒,is picked up.The boundary vertex x is then labeled the same as the vertex x i ,i.e.,the label of x is changed from Ϫ1to a region number (Ͼ0).Finally,with the exception of a few missing vertices,each vertex is labeled as 1,2,3,...,N ,the number of the parts.Missing vertices are usually located around boundaries between two articulated parts,and they are further assigned to parts during the post-processing step.Finally,a postprocessing step is performed to assign the nonlabeled vertices to parts.For example,the vertex p is an unlabeled vertex and needs further postprocessing.Assum-ing that p i (i ϭ1,2,...,N )represents a neighboring vertex of the point p ,the neighboring vertices are first selected if they have the same sign of curvature as that of the vertex p and belong to one of the segmented parts.Next,among those neighbor vertices,the vertex that has the smallest Euclidean distance to the vertex p is selected as a target vertex.For example,the vertex p 1is assumed to be the target vertex of the vertex p .Vertex p is assigned thesameFig.7Curvature estimation for the vertex p utilizing triangle meshinformation.Fig.8Region growing process for the vertex p .Zhang,Koschan,and Abidi414/Journal of Electronic Imaging /July 2004/Vol.13(3)Superquadric representation of automotive parts...Fig.5Region and part segmentation of a synthetic scene:(a)rendering of a synthetic scene consist-ing of a barrel on thefloor,(b)three segmented regions rendered in different colors,and(c)twocolors.decomposed parts rendered in differentview range images from the IVP Ranger system29and consists of37,171vertices and73,394tri-angles.The part decomposition results consist of two parts:(a)photograph of the original object,(b)rendering of the reconstructed mesh,(c)decomposed parts rendered in different colors,and(d)twocolors.recovered superquadrics rendered in differentmultiview range images from the IVP Ranger system29and consists of58,975vertices and117,036triangles.The part decomposition results consist of13parts:(a)photograph of the original object,(b)rendering of the reconstructed mesh,(c)decomposed parts rendered in different colors,and(d)colors.recovered superquadrics rendered in differentmultiview range images from the IVP Ranger system29and consists of58,784vertices and117,564triangles.The part decomposition results consist of nine parts:(a)photograph of the original object,(b)rendering of the reconstructed mesh,(c)decomposed parts labeled in different colors,and(d)recov-ered superquadrics rendered in different colors.Journal of Electronic Imaging/July2004/Vol.13(3)/415label as vertex p1,i.e.,the same segmented part.Further-more,parts composed of fewer vertices than a specified threshold are merged with adjacent regions.4Experimental ResultsExperimental results on superquadric representation for multipart,automotive objects including a disk brake,a dis-tributor cap,and a water neck are shown in this section. The meshes were reconstructed from multiview range im-ages scanned from the IVP Ranger System.29The recovered superquadrics were rendered in three dimensions using quad meshes.30Figure9shows the disk brake and its part decomposition and superquadric representation results.The reconstructed3-D triangulated surface shown in Fig.9͑b͒consists of37,171vertices and73,394triangles.Starting from this reconstructed mesh,our part decomposition algo-rithmfirst decomposed the disk brake into two single parts, as shown in Fig.9͑c͒.Each decomposed part was nextfit-ted to a superquadric model,as shown in Fig.9͑d͒.The decomposed parts and recovered superquadrics are ren-dered in different colors.We observe that our part decom-position algorithm successfully decomposed the disk brake into its constituent parts and the superquadric representa-tion strategy recovered correct superquadrics in terms of their size,shape,and pared to the original triangle mesh representation consisting of37,171vertices and73,394triangles,the recovered superquadrics describe the disk brake with only22parameters͑11parameters for each superquadric without global deformations͒.This low representation cost of superquadric representation can sig-nificantly benefit tasks including virtual reality,object rec-ognition,and robotic navigation.However,the hole at the center of the disk brake failed to be represented since su-perquadrics can only represent objects with genus of zero.19 Figure10shows the distributor cap and its part decom-position and superquadric representation results.The recon-structed mesh shown in Fig.10͑b͒consists of58,975ver-tices and117,036triangles and was decomposed into13 single parts,as shown in Fig.10͑c͒.We observe that this decomposition result is consistent with human perception. The recovered superquadrics shown in Fig.10͑d͒correctly represent the distributor cap.The recovered superquadric parameters and the ground truths for one of the small cyl-inders on top of the distributor cap are shown in Table1. We can observe that the recovered superquadric parameters for this cylinder have the correct size and shape informa-tion when compared with the ground truth parameters of the object.In addition,superquadrics represent the distribu-tor cap with only143floating numbers,while the original triangle mesh consists of58,975vertices and117,036tri-angles.Figure11shows the water neck and its part decomposi-tion and superquadric representation results.The recon-structed mesh shown in Fig.11͑b͒consists of58,784ver-tices and117,564triangles and was decomposed into nine single parts,as shown in Fig.11͑c͒.We observe that the decomposed parts are consistent with human perception. The recovered superquadrics shown in Fig.11͑d͒correctly represent the water neck.The recovered superquadric pa-rameters and the ground truths for the handle,the ball,and the small cylinder next to the handle of the water neck are shown in Table2.From this table,we observe that the recovered superquadric parameters have the correct size and shape information when compared with the ground truth parameters of the objects.Again,superquadrics repre-sent the water neck in a desirable accuracy with only99 parameters while the original triangle mesh consists of 58,784vertices and117,564triangles.5ConclusionsThis paper proposed a superquadric representation ap-proach for multipart objects.Superquadrics can represent objects in an acceptable accuracy with only a few param-eters,while other surface primitives and contours usually require thousands of representation elements.Such a com-pactness and low representation cost can significantly ben-efit tasks including virtual reality,object recognition,and robot navigation,e.g.,it enables these tasks to run in a real-time manner.The advantages of the proposed super-quadric representation approach include:͑1͒it can success-fully represent complicated,multipart objects byfirst de-composing them into single-part objects,and͑2͒the recovered superquadrics have the highest confidence and accuracy since the input we use are3-D triangulated sur-faces reconstructed from multiview range images.The in-completeness and ambiguities contained in single-view im-ages were eliminated during the multiview surface reconstruction process.We also proposed a3-D part de-composition algorithm to decompose compound objects represented by triangle meshes into their constituent single parts based on curvature analysis.Considering the fact that the triangle mesh has been a standard surface representation element in computer vision and computer graphics,the pro-posed part decomposition algorithm is generic,flexible,and can facilitate computer vision tasks such as shape descrip-tion and object recognition.Furthermore,the part decom-position algorithm can segment a large number of triangle meshes͑over100,000͒in only seconds on an SGI Octane workstation.Table1Recovered superquadric parameters and ground truths for one of the small cylinders shown in Fig.10(d)where the unit is millimeters.Parameters a1a2a312 Ground truths15.215.620.10.1 1.0 Superquadric parameters16.4515.6720.420.120.96Table2Recovered superquadric parameters and ground truths for the water neck shown in Fig.11(d)where the unit is millimeters. Object Parameters a1a2a312 Handle Ground truths39.739.417.60.1 1.0 Superquadric parameters40.2340.5866.830.130.98 Ball Ground Truths50.047.656.0 1.0 1.0 Superquadric parameters51.6247.5654.28 1.020.95 Cylinder Ground truths16.517.844.20.1 1.0 Superquadric parameters17.5617.9443.380.110.95Zhang,Koschan,and Abidi 416/Journal of Electronic Imaging/July2004/Vol.13(3)AcknowledgmentsThis work was supported by the University Research Pro-gram in Robotics under Grant No.DOE-DE-FG02-86NE37968,by the Department of Defense/U.S.Army Tank-automotive and Armaments Command/National Au-tomotive Center/Automotive Research Center Program R01-1344-18,and by the Federal Aviation Administration National Safe Skies Alliance Program R01-1344-48/49. References1.F.Solina and R.Bajcsy,‘‘Recovery of parametric models from rangeimages:the case for superquadrics with global deformations,’’IEEE Trans.Pattern Anal.Mach.Intell.12͑2͒,131–147͑1990͒.2.Y.Zhang,J.Paik,A.Koschan,and M.Abidi,‘‘3-D object represen-tation from multi-view range data applying deformable superquad-rics,’’in Proc.Int.Conf.Pattern Recognition,V ol.3,pp.273–276͑2002͒.3.A.Gross and T.Boult,‘‘Error offit measures for recovering paramet-ric solids,’’in Proc.2nd puter Vision,pp.690–694͑1988͒.4.A.P.Pentland and S.Sclaroff,‘‘Closed-form solutions for physicallybased shape modeling and recognition,’’IEEE Trans.Pattern Anal.Mach.Intell.13͑7͒,715–729͑1991͒.5.A.P.Pentland,‘‘Automatic extraction of deformable part models,’’put.Vis.4,107–126͑1990͒.6.M.Leyton,‘‘A process-grammar for shape,’’Artif.Intell.34͑99͒,213–247͑1988͒.7.D.Terzopoulos,A.Witkin,and M.Kass,‘‘Constraints on deformablemodels:recovering3D shape and nongrid motion,’’Artif.Intell.36, 91–123͑1988͒.8.D.Metaxas and D.Terzopoulos,‘‘Constrained deformable superquad-rics and nongrid motion tracking,’’in puter Vi-sion and Pattern Recognition,pp.337–343͑1991͒.9.D.Terzopoulos and D.Metaxas,‘‘Dynamic3-D models with localand global deformations:deformable superquadrics,’’IEEE Trans.Pattern Anal.Mach.Intell.13͑7͒,703–714͑1991͒.10.E.Bardinet,L.D.Cohen,and N.Ayache,‘‘Fitting3D data usingsuperquadrics and free-form deformations,’’in Proc.12th IAPR Int.Conf.on Pattern Recognition,V ol.1,pp.79–83͑1994͒.11.D.Metaxas,E.Koh,and N.I.Badler,‘‘Multi-level shape representa-tion using global deformations and locally adaptivefinite elements,’’put.Vis.25͑1͒,49–61͑1997͒.12.D.Metaxas,Physics-Based Deformable Models:Applications toComputer Vision,Graphics and Medical Imaging,Kluwer Academic Press,͑1997͒.13.M.Pilu,A.W.Fitzgibbon,and R.B.Fisher,‘‘Training PDMs onmodels:the case of deformable superellipses,’’in Proc.British Ma-chine Vision Conf.,pp.373–382͑1996͒.14.K.Wu and M.D.Levine,‘‘Recovering parametric geons from 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Documentation:MAPP2500Ranger PCI System,Version1.6,Integrated Vision Products,Sweden͑2000͒.30.J.Wernecke,The Inventor Mentor:Programming Object-oriented3DGraphics with Open Inventor,Addison-Wesley,Reading,MA͑1994͒.Yan Zhang received her BS and MS de-grees in electrical engineering from Hua-zhong University of Science and Technol-ogy,China,in1994and1997,respectively,and her PhD degree in electrical engineer-ing from the University of Tennessee,Knoxville,in2003.Her research interestsinclude3-D image processing,computervision,and patternrecognition.Andreas Koschan received his MS de-gree in computer science and his PhD incomputer engineering from the TechnicalUniversity Berlin,Germany,in1985and1991,respectively.He is currently a re-search associate professor with the De-partment of Electrical and Computer Engi-neering,the University of Tennessee,Knoxville.His work has primarily focusedon color image processing and3-D com-puter vision including stereo vision and la-ser rangefinding techniques.He is a coauthor of two textbooks on 3-D image processing and a member of IS&T andIEEE.Mongi A.Abidi is a W.Fulton Professorwith the Department of Electrical and Com-puter Engineering,the University of Ten-nessee,Knoxville,which he joined in1986.Dr.Abidi received his MS and PhD degreesin electrical engineering in1985and1987,both from the University of Tennessee,Knoxville.His interests include image pro-cessing,multisensor processing,3-D imag-ing,and robotics.He has published over120papers in these areas and coedited the book Data Fusion in Robotics and Machine Intelligence(Academic Press,1992).He is the recipient of the1994to1995Chancellor’s Award for Excellence in Research and Creative Achievement and the2001Brooks Distinguished Professor Award.He is a member of the IEEE,the Computer Society,the Pattern Recognition Society, SPIE and the Tau Beta Pi,Phi Kappa Phi,and Eta Kappa Nu honor societies.He also received the First Presidential Principal Engineer Award prior to joining the University of Tennessee.Superquadric representation of automotive parts...Journal of Electronic Imaging/July2004/Vol.13(3)/417。
基于球形坐标映射和傅立叶描述符的三维人脸特征提取
Ｘ ｓ 进行离散傅立叶变换得到傅 立叶变换系数ａ ， Ｃ（ ０
对这些系数进行归一化 ，得 到的就是描述 目标边 界的傅 立叶描述符。 的计算公 式为：
信息系统工程 Ｉ ０ １ ０ ０ １ １ １ ． ２ ３ ２ １
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Ａ Ａ Ｅ ＣＲ Ｓ Ａ Ｃ 学术 研 究 Ｃ Ｄ ＭＩ Ｅ Ｅ Ｒ Ｈ
物 体 边 界 是 一 条 封 闭 的 曲线 ，设 其 边 界 坐 标 为
｛ （，（ ） ＝０１．Ｎ一 ｝。可用复数 ｆ＋ （ 的形式 （ ｆ ｙｆ， ， ． ） ）ｆ ，． １ ， （ ｆ ） ） 来表示给定边界上的每个点 ，则该封闭曲线可描述为 ：
ｓ ） （ ＋ （ ， ＝ ，… Ｎ一 （） （ ＝ ｆ ｆ ｔ ０１ ， １ ２ ｔ ） ） ，
＜＜ ¨ ＡＡ Ｅ ＩＲＳＡ Ｃ 学 研究 ＿ ＣＤＭＣ ＥＥＲＨ 术
基于 球形 坐标映射和
傅立 叶描述符 的三维 人脸特征提取
◆王 垫 郭 哲
摘要 ：将傅 立叶描 述符和 算法扩展 应 用 于三 维人脸 特征提 取的研
究 ，在 对 三 维人 脸 模 型 进行 预 处理 及 二 维 球 形 映射 的基 础 上 ，建 立 了具
其 次 ，对厶 进行二维傅立 叶变换 ，得 到初始变换 系 数数列Ｆ Ｄ：
－。 ） Ｈ， １ ， 厂 ： ｉ ＝， ． （， ｉ ２ Ｎ，这可以被近似看作二维图
像 。其 中，连续均值 曲率Ｈ可由文献 计算得到 。
Ｆｐ Ｚ ， ｃ＋ ｛） Ｄ （￣ｐ 云 叫（ （ Ｅ ｒｘ ｐ ｆ ０ｘ ） ｅ ４
描述符 （ Ｄ） 。 Ｆ 】
化的方法为将第一个系数 除以包含图像的区域 ，其他全 部余下的系数除以第一个系数来得到。具体公式如下：
三维网格模型的分割及应用技术综述
第17卷第8期2005年8月计算机辅助设计与图形学学报JOURNAL OF COMPU TER 2AIDED DESIGN &COMPU TER GRAPHICSVol 117,No 18Aug 1,2005 收稿日期:2004-03-09;修回日期:2004-07-08 基金项目:国家“八六三”高技术研究发展计划重点项目(2001AA231031,2002AA231021);国家重点基础研究发展规划项目(G1998030608);国家科技攻关计划课题(2001BA904B08);中国科学院知识创新工程前沿研究项目(20006160,20016190(C ))三维网格模型的分割及应用技术综述孙晓鹏1,2) 李 华1)1(中国科学院计算技术研究所智能信息处理重点实验室 北京 100080)2(中国科学院研究生院 北京 100039)(xpsun @ict 1ac 1cn )摘要 对三维网格模型分割的定义、分类和应用情况做了简要回顾,介绍并评价了几种典型的网格模型分割算法,如分水岭算法、基于拓扑和几何信息的分割算法等;同时,对网格分割在几种典型应用中的研究工作进行了分类介绍和评价1最后对三维分割技术今后的发展方向做出展望1关键词 分割Π分解;三维分割;形状特征;网格模型中图法分类号 TP391A Survey of 3D Mesh Model Segmentation and ApplicationSun Xiaopeng 1,2) Li Hua 1)1(Key L aboratory of Intelligent Inf ormation Processi ng ,Instit ute of Com puti ng Technology ,Chi nese Academy of Sciences ,Beiji ng 100080)2(Graduate School of the Chi nese Academy of Sciences ,Beiji ng 100039)Abstract In this paper ,we present a brief summary to 3D mesh model segmentation techniques ,includ 2ing definition ,latest achievements ,classification and application in this field 1Then evaluations on some of typical methods ,such as Watershed ,topological and geometrical !method ,are introduced 1After some ap 2plications are presented ,problems and prospect of the techniques are also discussed 1K ey w ords segmentation Πdecomposition ;3D segmentation ;shape features ;mesh model1 引 言基于三维激光扫描建模方法的数字几何处理技术,继数字声音、数字图像、数字视频之后,已经成为数字媒体技术的第四个浪潮,它需要几何空间内新的数学和算法,如多分辨率问题、子分问题、第二代小波等,而不仅仅是欧氏空间信号处理技术的直接延伸[1]1在三维网格模型已成为建模工作重要方式的今天,如何重用现有网格模型、如何根据新的设计目标修改现有模型,已成为一个重要问题1网格分割问题由此提出,并成为近年的热点研究课题[223]12 网格分割概述三维网格模型分割(简称网格分割),是指根据一定的几何及拓扑特征,将封闭的网格多面体或者可定向的二维流形,依据其表面几何、拓扑特征,分解为一组数目有限、各自具有简单形状意义的、且各自连通的子网格片的工作1该工作被广泛应用于由点云重建网格、网格简化、层次细节模型、几何压缩与传输、交互编辑、纹理映射、网格细分、几何变形、动画对应关系建立、局部区域参数化以及逆向工程中的样条曲面重建等数字几何处理研究工作中[223]1同时,三维网格模型的局部几何拓扑显著性也是对三维网格模型进行检索的一种有效的索引[4]1与网格曲面分割有关、并对其影响巨大的一个早期背景工作是计算几何的凸分割,其目的是把非凸的多面体分解为较小的凸多面体,以促进图形学的绘制和渲染效率1该工作已经有了广泛的研究,但多数算法难以实现和调试,实际应用往往不去分割多面体,而是分割它的边界———多边形网格1多面体网格边界的分割算法有容易实现、复杂形体输出的计算量往往是线性的等优势[5]1另外一个早期背景工作是计算机视觉中的深度图像分割,其处理的深度图像往往具有很简单的行列拓扑结构,而不是任意的,故其分割算法相对简单[6]1三维网格模型的分割算法一般是从上述两类算法推广而来1心理物理学认为:人类对形状进行识别时,部分地基于分割,复杂物体往往被看作简单的基本元素或组件的组合[728]1基于这个原理,Hoffman 等[9]于1984年提出人类对物体的认知过程中,倾向于把最小的负曲率线定义为组成要素的边界线,并据此将物体分割为几个组成要素,即视觉理论的“最小值规则”1由此得到的分割结果称为“有意义的”分割,它是指分割得到的子网格必须具有和其所在应用相关的相对尺寸和组织结构1由于曲率计算方法不同,很多算法给出的有意义的分割结果也存在差异1诸多应用研究[10214]证明,网格模型基于显著性特征的形状分割,是物体识别、分类、匹配和跟踪的基本问题1而有意义的分割对于网格模型显著占优特征的表示和提取、多尺度的存储和传输以及分布式局部处理都是十分有意义的1211 网格分割的发展较早的三维网格分割工作可以追溯到1991年,Vincent 等[15]将图像处理中的分水岭算法推广到任意拓扑连接的3D 曲面网格的分割问题上11992年,Falcidieno 等[16]按照曲率相近的原则,把网格曲面分割为凹面片、凸面片、马鞍面片和平面片11993年,Maillot 等[17]将三角片按法向分组,实现了自动分割;1995年,Hebert 等[18]给出了基于二次拟合曲面片的曲率估计方法,并把区域增长法修改推广应用到任意拓扑连接的网格曲面分割问题中;1995年,Pedersen [19]和1996年Krishnamurthy 等[20]在他们的动画的变形制作过程中,给出了用户交互的分割的方法11997年,Wu 等[3]模拟电场在曲面网格上的分布,给出了基于物理的分割方法;1998年Lee 等[21]和2000年Guskov 等[22]给出了几个对应于简化模型的多分辨率方法;1999年Mangan 等[2]使用分水岭算法实现网格分割,并较好地解决了过分割问题;2001年,Pulla 等[23224]改进了Mangan 的曲率估计工作;1999年,Gregory 等[25]提出一个动画设计中的交互应用,根据用户选择的特征点将网格曲面分割为变形对应片;1999年,Tan 等[26]基于顶点的简化模型建立了用于碰撞检测的、更紧致于网格曲面分割片的层次体包围盒12000年,Rossl 等[27]在逆向工程应用中,在网格曲面上定义了面向曲率信号的数学形态学开闭操作,从而得到去噪后的特征区域骨架,并实现了网格分割;2001年,Yu 等[28]的视觉系统自动将几何场景点云分割为独特的、用于纹理映射和绘制的网格曲面片二叉树;Li 等[29]为了碰撞检测,给出了基于边收缩得到描述几何和拓扑特征的骨架树,然后进行空间扫描自动分割;Sander 等[30]使用区域增长法,按照分割结果趋平、紧凑的原则分割、合并分割片1所有这些方法都是为了使分割的结果便于参数化,即只能产生凸的分割片1由此产生边界不连续的效果12002年,Werghi 等[31]识别三维人体扫描模型的姿态,根据人体局部形状索引进行网格模型的分割;Bischoff 等[32]和Alface 等[33]分别给出了网格分割片光谱在几何压缩和传输中的应用;Levy 等[34]在纹理生成工作中,以指定的法向量的夹角阈值对尖锐边滤波,对保留下来的边应用特征增长算法,最后使用多源Dijkstra 算法扩张分割片实现了网格模型的分割;2003年,Praun 等[35]将零亏格网格曲面投影到球面上,然后把球面投影到正多面体上得到与多面体各面对应的网格模型分割,最后将多面体平展为平面区域以进行参数化,但其结果不是有意义的分割1212 网格分割的分类早期的网格分割算法多为手工分割或者半自动分割,近两年出现了基于自动分割的应用工作1从网格模型的规则性来看,可将分割算法分为规则网格分割、半规则网格分割和任意结构的网格分割算法,根据分割结果可以分为有意义的分割和非有意义的分割1同时,面向不同的应用目标出现了不同的分割策略(见第4节)1目前,网格分割的质量指标主要有三个方面:边界光顺程度、是否有意义、过分割处理效果1多数分8461计算机辅助设计与图形学学报2005年割算法以边界光顺为目标,采用的方法有在三角网格上拟合B样条曲面然后采样[20],逼近边界角点(两个以上分割片的公共顶点)间的直线段[30]等1近年来多数分割算法都追求产生有意义的分割结果1对于过分割的处理方法目前主要有忽略、合并和删除三种方式1多数三维网格分割算法是从二维图像分割的思想出发,对图像分割算法作三维推广得到其三维网格空间的应用1如分水岭算法[2,15,23224,36239]、K2 means算法[40]、Mean2shift算法[41]以及区域增长算法[18,30]等1同样,与图像处理问题类似,光谱压缩[33,42243]、小波变换[31]等频谱信息处理方法在三维网格分割中也有算法1除此之外,同时考虑几何与拓扑信息的分割会产生较好的结果1这方面的工作主要有基于特征角和测地距离度量[44]、基于高斯曲率平均曲率[45247]、基于基本体元[32]、基于Reeb图[48250]、基于骨架提取和拓扑结构扫描[27,29,51252]等使用三维网格曲面形状特征的算法1作为网格模型的基础几何信息,曲率估计方法目前主要为曲面拟合、曲线拟合以及离散曲率等三种1其中曲面拟合法较为健壮,但是计算量大;离散曲率法计算量小,但是除个别算法外都不是很健壮,且无主方向主曲率信息;曲线拟合的曲率估计方法则集中了上述两种方法的优势[3],实际研究中使用较多13 典型三维网格分割算法311 分水岭算法1999年,Mangan等[2]的工作要求输入的是三角网格曲面,以及任何一种可以用来计算每个顶点曲率的附加信息(如曲面法向量等),并针对体数据和网格数据给出了两种曲率计算方法;但是分水岭算法本身和曲率的类型无关1首先,计算每个顶点的曲率(或者其他高度函数),寻找每个局部最小值,并赋予标志,每一个最小值都作为网格曲面的初始分割;然后,开始自下而上或者自上而下地合并分水岭高度低于指定阈值的区域,有时平坦的部分也会得到错误的分割,后处理解决过分割问题1分割为若干简单的、无明确意义的平面或柱面,属于非有意义的分割1Rettmann等[36237]结合测地距离,并针对分水岭算法的过分割给出一个后处理,实现了MRI脑皮层网格曲面的分割12002年,Marty[38]以曲率作为分水岭算法的高度函数,给出了有意义的分割结果1 2003年,Page等[39]的算法同样只分割三角网格,依据最小值规则,他们试图得到网格模型高层描述1其主要贡献为:创建了一个健壮的、对三角网格模型进行分割的贪婪分水岭法;使用局部主曲率定义了一个方向性的、遵循最小值规则的高度图;应用形态学操作,改进了分水岭算法的初始标识集1文献[39]在网格的每一个顶点计算主方向和主曲率,根据曲率阈值,使用贪婪的分水岭算法分割出由最小曲率等高线确定的区域1形态学的开闭操作应用于网格模型每个顶点的k2ring碟状邻域,闭操作会连接空洞,而开操作会消除峡部1创建了标识集后,依据某顶点与其邻接顶点之间的方向,由欧拉公式和已知主曲率计算该顶点在该方向上的法曲率从而得到在该方向上、该顶点与邻接顶点之间的方向曲率高度图,并将其作为方向梯度1对该顶点所在的标识区域使用分水岭算法得到分割片1上述工作表明,分水岭算法在改进高度函数的定义后,可以得到有意义的分割效果1312 基于拓扑信息的网格分割基于几何以及拓扑信息的形状分割方法可以归结为Reeb图[50]、中轴线[52]和Shock图[53254]等1基于拓扑信息的形状特征描述主要有水平集法[55]和基于拓扑持久性的方法[56]11999年,Lazarus等[51]提出从多面体顶点数据集提取轴线结构,在关键点处分割网格的水平集方法,如图1所示1这种轴线结构与定义在网格模型顶点集上的纯量函数关联,称之为水平集图,它能够为变形和动画制作提供整体外形和拓扑信息1图1 人体网格模型及其水平集图文献[51]针对三角剖分的多面体,使用与源点之间的最短路径距离作为水平集函数,基于Dijkstra 算法构造记录水平集图的结构树,其根结点、内部结点和叶子分别表示源点、水平集函数的鞍点和局部最大值点1该工作可以推广到非三角网格模型1 2001年,Li等[29]基于PM算法[57]的边收缩和94618期孙晓鹏等:三维网格模型的分割及应用技术综述空间扫掠,给出了一个有效的、自动的多边形网格分割框架1该工作基于视觉原理,试图将三维物体分割为有视觉意义和物理意义的组件1他们认为三维物体最显著的特征是几何特征和拓扑特征,由此,定义几何函数为扫掠面周长在扫掠结点之间的积分为骨架树中分支的面积;定义拓扑函数为相邻两个扫掠面拓扑差异的符号函数,并定义了基于微分几何和拓扑函数的关键点1文献[29]首先基于PM算法将每条边按照其删除误差函数排序,具有最小函数值的边收缩到边中点,删除其关联的三角形面片;如果某边没有关联任何三角片则指定为骨架边,保持其顶点不变;循环上述过程,得到一个新的、通过抽取给定多边形网格曲面骨架的方法1其次,加入虚拟边连接那些脱节的骨架边,称这些虚拟边以及原有的骨架边组成的树为骨架树,即为扫掠路径1扫掠路径为分段线条1然后,定义骨架树中分支面积(扫掠面周长函数在扫掠结点之间的积分),分支面积较小的首先扫掠,以保证小的、但是重要的分割片被首先抽取出来,以免被其他较大的分割片合并1最后,沿扫掠路径计算网格的几何、拓扑函数的函数值1一旦发现几何函数、拓扑函数的关键点,抽取两个关键点之间的网格曲面得到一个新的分割片1整个过程无需用户干涉12003年,Xiao等[48249]的工作基于人体三维扫描点云的离散Reeb图,给出了三维人体扫描模型的一个拓扑分割方法:通过探测离散Reeb图的关键点,抽取表示身体各部分的拓扑分支,进而进行分割1水平集法具有较高的计算速度和健壮的计算精度1基于拓扑持久性的方法结合代数学,能更准确地计算形状特征,但是没有解决分割问题[55256]1 313 基于实体表示的网格分割2002年,Bischoff等[32]把几何形状分割为表示其粗糙外形的若干椭球的集合,并附加一个独立的网格顶点的采样集合来表示物体的细节1生成的椭球完全填充了物体的内部,采样点就是原始的网格顶点1该方法的步骤如下:Step11首先,在物体原始网格的每一顶点上生成一个椭球,或者随机在物体原始网格上采样选择种子点;每个种子点作为球面上的一个顶点,沿该点的网格法向做球面扩展,直至与网格上另外一个顶点相交;然后沿此两点的垂直方向将球面扩张为最大椭球,直至与第三个网格顶点相交;最后沿此三点平面的法向(即该三点所在平面的柱向)扩张,直至与第四个网格顶点相近,由此得到一个椭球1Step21对生成的椭球进行优化选择,体积最大的椭球首先被选中,以后每一次都将选出对累计体积贡献最大的椭球1如果有若干体积累计贡献相近的椭球同时出现的情况发生,则最小半径最短的椭球被选出1为了简化体积累计贡献的计算,对椭球体素化后计算完全包含在椭球内的体素的数目进行堆排序1发送方传送选出的椭球集合;接收方得到包含基本几何和拓扑信息的椭球集合后,使用Marching Cubes算法或者Shrink2wrapping算法抽取0等值面1显然即使部分椭球丢失,工作依然可以继续:因为椭球是互相重叠的,抽取等值面不影响它们的拓扑关系,而且如果重叠充分,丢失少部分椭球不会影响重要形状信息的重构1如图2所示1图2 以不同数目椭球表示的网格分割Step31在生成很好地逼近原始物体的初始网格后,开始将采样点(即原始网格顶点)插入网格[58]1为了提高最终重构结果的质量,由Marching Cubes算法生成的临时网格顶点在网格原始顶点陆续到来后,最终被删除,因为它们不是物体的原始顶点1314 基于模糊聚类的层次分解2003年,Katz等[44]提出了模糊聚类的层次分解算法,算法处理由粗到精,得到分割片层次树1层次树的根表示整个网格模型S1在每个结点,首先确定需要进一步分割为更精细分割片的数目,然后执行一个k2way分割1如果输入的网格模型S由多个独立网格构成,则分别对每个网格进行同样的操作1分割过程中,算法不强调每个面片必须始终属于特定的分割片1大规模网格模型的分割在其简化模型上进行,然后将分割片投影到原始网格模型上,在不同的尺度下计算分割片之间的精确边界1文献[44]算法优点是:可以对任意拓扑连接的或无拓扑连接的、可定向的网格进行处理;避免了过分割和边界锯齿;考虑测地距离和凸性,使分割边界通过凹度最深的区域,从而得到有意义的分割结果1分割结果适用于压缩和纹理映射14 三维网格分割应用411 三维检索中的网格分割算法在三维VRML数据库中寻找一个与给定物体0561计算机辅助设计与图形学学报2005年相似的模型的应用需求,随着WWW的发展正变得越来越广泛,如计算生物学、CAD、电子商务等1形状描述子和基于特征的表示是实体造型领域中基本的研究问题,它们使对物体的识别和其他处理变得容易1因为相似的物体有着相似的分割,所以分割结构形状描述子可以用于匹配算法1中轴线、骨架等网格模型拓扑结构的形状描述子在三维模型检索中也得到研究,它可以从离散的体数据以及边界表示数据(网格模型)中抽取出来1对于后者,目前还没有精确、有效的结果[39]1但我们相信,依据拓扑信息进行分割得到的分布式形状描述子也是一种值得尝试的三维模型检索思路1 2002年,Bischoff等[32]提出从椭球集合中得到某种统计信息,如椭球半径的平均方差或者标准方差,以及它们的比率,由于这些统计信息在不同的形状修改中都保持不变,作为一种检索鉴别的标识的想法1但是没有严格的理论或者实验结果证明1 2002年,Zuckerberger等[59]在一个拥有388个VRML三维网格模型的数据库上,进行基于分割的变形、简化、检索等三个应用1首先将三维网格模型分割为数目不多的有意义的分割片,然后评价每一个分割片形状,确定它们之间的关系1为每个分割建立属性图,看作是与原模型关联的索引,当数据库中检索到与给定网格模型相似的物体时,只是去比较属性图相似的程度1属性图与其三维模型的关联过程分为三步:(1)分割网格曲面为有限数目的分割片;(2)每一个分割片拟合为基本二次曲面形状;(3)依据邻接分割片的相对尺寸关系进行过分割处理,最后构造网格曲面模型的属性图1对分割片作二次拟合,由此产生检索精确性较差的问题;分割片属性图的比较采用图同构的匹配方法,计算量较大,且是一个很困难的问题;从其实验结果看,有意义的分割显然还不够,出现飞机、灯座等模型被检索为与猫相似的结构;区分坐、立不同的人体模型效果显然也很差等12003年,Dey等[4]基于网格模型的拓扑信息,给出了名为“动力学系统”的形状特征描述方法,并模拟连续形状定义离散网格形状特征1实验表明该算法十分有效地分割二维及三维形状特征1他们还给出了基于此健壮特征分割方法的形状匹配算法1 412 几何压缩传输中的网格分割健壮的网格模型压缩传输方法必须保证即使部分几何信息丢失,剩下的部分至少能够得到一个逼近原始物体的重构,即逼近的质量下降梯度,要大大滞后于信息丢失梯度1无论是层次结构的还是过程表示的多边形网格模型,它们的缺陷是:严格的拓扑信息一致性要求1顶点和面片之间的交叉引用导致即使在传输中丢失了1%的网格数据,也将导致无法从99%的剩余信息里重建网格曲面的任何一部分1对此可以考虑引入高度的冗余信息,即使传输中丢失一定额度的数据,接收方依然可以重构大部分的几何信息1问题的关键是将几何体分割为相互独立的大块信息,如单个点,这样接收方可以在不依赖相关索引信息的情况下,重构流形的邻域关系1为了避免接收方从点云重构曲面的算法变得复杂,早期的健壮传输方法总假设至少整体拓扑信息可以无损地传送1一旦知道了粗糙的形状信息,接收方可以插入一些附加点生成逼近网格12002年,Bischoff等[32,58]在网格分割工作中将每个椭球互相独立地定义自己的几何信息1由于椭球的互相重叠,冗余信息由此产生,因此如果只有很少的椭球丢失,网格曲面的拓扑信息和整体形状不会产生变化1冗余信息不会使存储需求增加,因为每个椭球和三角网格中每个顶点一样,只需要9个存储纯量1其传送过程如下:种子点采样生成椭球集合;传送优化选择的椭球子集;接收方抽取等值面重构逼近网格;以陆续到来的原始网格顶点替换临时网格顶点11996年,Taubin等[42]首先在几何压缩处理中提出光谱压缩,其工作在三维网格模型按如下方式应用傅里叶变换:由任意拓扑结构的网络顶点邻接矩阵及其顶点价数,得到网格Laplacian矩阵的定义及由其特征向量构成的R n空间的正交基底,相对应的特征值即为频率1三维网格顶点的坐标向量在该空间的投影即为该网格模型的几何光谱1网格表面较为光顺的区域即为低频信号12000年,Karni等[43]将几何网格分割片光谱推广到传输问题上1光谱直接应用于定义几何网格的拓扑信息时,会产生伪频率信息1对于大规模的网格,由于在网格顶点数目多于1000时,Laplacian矩阵特征向量的计算几乎难以进行,因此该工作在最小交互前提下,将网格模型分割为有限数目的分割片1该方法有微小的压缩损失,且在分割片边界出现人工算法痕迹12003年,Alface等[33]提出了光谱表示交叠方法:扩张分割片,使分割片之间产生交叠1具体方法是把被分割在其他邻接分割片中的、但与该分割片15618期孙晓鹏等:三维网格模型的分割及应用技术综述邻接的三角片的顶点,按旋转方向加入到该分割片中,从而由于分割片重叠搭接产生冗余信息,并称这种分割片扩展冗余处理的光谱变换为交叠的正交变换1该工作在几何网格压缩和过程传输的应用中明显地改进了Karni等的工作1显然上述工作的基础是良好的网格分割1建立分片独立的基函数将使得分割效果更为理想1413 纹理贴图中的网格分割如果曲面网格的离散化是足够精细的,如细分网格,那么直接对顶点进行纹理绘制就足够了;否则就要把网格模型分割为一组与圆盘同胚的、便于进行参数化的分割片,再对每片非折叠的分割片参数化,最后分割片在纹理空间里拼接起来1网格模型的分割显然会因其局部性而降低纹理映射纹理贴图、网格参数化的扭曲效果1面向纹理的分割算法一般要求满足两个条件:(1)分割片的不连续边界不能出现人工算法痕迹;(2)分割片与圆盘同胚,而且不引入太大的变形就可以参数化1不要求有意义的分割结果12001年,Sander等[30]基于半边折叠的PM算法,使用贪婪的分割片合并方法(区域增长法)对网格模型进行分割1首先将网格模型的每一个面片都看作是独立的分割片,然后每个分割片与其邻接分割片组对、合并1在最小合并计算量的前提下,循环执行分割片对的合并操作,并更新其他待合并分割片的计算量1当计算量超出用户指定的阈值时,停止合并操作1分割片之间的边界为逼近角点间直线段的最短路径,从而减轻了锯齿情况12002年,Levy等[34]将网格模型分割为具有自然形状的分割片,但仍然没有得到有意义的分割结果1为了与圆盘同胚,该算法自动寻找位于网格模型高曲率区域的特征曲线,避免了在平展区域内产生分割片边界,并增长分割片使他们在特征曲线上相交,尽量获得尺寸较大的分割片1414 动画与几何变形中的网格分割影视动画制作中,多个对象间的几何变形特技使用基于网格分割的局部区域预处理1如建立动画区域对应关系,对多个模型进行一致分割,然后在多个模型的对应分割片之间做变形,将提高动画制作的精度和真实性;且每个“Polygon Soup”模型都可用来建立分割片对应;模型间的相似分割有利于保持模型的总体特征1目前,多数的自动对应算法精度较低,手工交互指定对应关系的效率又太低1 1996年,Krishnamurthy等[20]从高密度、非规则、任意拓扑结构的多边形网格出发,手工指定分割边界,构造张量积样条曲面片的动画模型1文献[20]首先在多边形网格的二维投影空间交互选择一个顶点序列,然后自动地将顶点序列关联到网格上最近的顶点上;对于序列中前后两个顶点计算在网格曲面上连接它们的最短路径;对该路径在面片内部进行双三次B样条曲面拟合、光顺、重新采样,得到分割片在两个顶点之间的边界曲线1但计算量的付出依然是非常昂贵的11999年,Gregory等[25]在两个输入的多面体曲面上交互选择多面体顶点,作为一个对应链的端点,对应链上其他顶点通过计算曲面上端点对之间的最短路径上的顶点确定,由此得到这些顶点和边构成的多面体表面网格的连通子图;然后将每一个多面体分割为相同数目的分割片,每个分割片都与圆盘同胚;在分割片之间建立映射、重构、局部加细,完成对应关系的建立;最后插值实现两个多面体之间的变形12002年,Shlafman等[40]的工作不再限制输入网格必须是零亏格或者是二维流形1该算法通过迭代,局部优化面片的归属来改进某些全局函数,因此与图像分割K2means方法相近,属于非层次聚类算法1最终分割片的数目可以由用户预先指定,从而避免了过分割,且适用于动画制作的需求1分割过程的关键在于确认给定的两个面片是否属于同一个分割片1其分割工作是非层次的,因为面片可能会在优化迭代中被调整到另外一个分割片去1该工作表明,基于分割的变形对于保持模型的特征有着重要的意义1局部投影算法能够产生精细的对应区域,且能自动产生有意义的分割片1415 模型简化中的网格分割网格简化是指把给定的一个有n个面片的网格模型处理为另一个保持原始模型特征的、具有较少面片、较大简化Π变形比的新模型1三维网格分割显然可以被看作是一种网格简化,其基本思想是在简化中增加一个预处理过程,先按模型显著特征将其分割为若干分割片,然后在每个分割片内应用简化算法,由此保持了模型的显著特征,如特征边、特征尖锐以及其他精细的细节1例如,把曲率变化剧烈的区域作为分割边界,将曲率变化平缓的区域各自分割开来,就是基于曲率阈值的网格简化方法1网格曲面分割结果的分割片数目在去除过分割后被限制在指定的范围内12561计算机辅助设计与图形学学报2005年。
平面几何研究（叶中豪）
高中平面几何(上海教育出版社叶中豪)知识要点三角形的特殊点重心，外心，垂心，内心，旁心，类似重心，九点圆心，Spieker点，Gergonne点，Nagel点，等力点，Fermat 点，Napoleon点，Brocard点，垂聚点，切聚点，X点，Tarry点，Steiner点，Soddy点，Kiepert双曲线特殊直线、圆Euler线，Lemoine线，极轴，Brocard轴，九点圆，Spieker圆，Brocard圆，Neuberg圆，McCay圆，Apollonius圆，Schoute圆系，第一Lemoine圆，第二Lemoine圆，Taylor圆，Fuhrmann圆特殊三角形中点三角形，垂三角形，切点三角形，切线三角形，旁心三角形，弧中点三角形，反弧中点三角形，第一Brocard三角形，第二Brocard三角形，D-三角形，协共轭中线三角形相关直线及相关三角形Simson线，垂足三角形，Ceva三角形，反垂足三角形，反Ceva三角形重心坐标和三线坐标四边形和四点形质点重心，边框重心，面积重心，Newton线，四点形的核心，四点形的九点曲线完全四边形Miquel点，Newton线，垂心线，外心圆，Gauss-Bodenmiller定理重要轨迹平方差，平方和，Apollonius圆三角形和四边形中的共轭关系等角共轭点，等角共轭线，等截共轭点，等截共轭线几何变换及相似理论平移，旋转（中心对称），对称，相似和位似，相似不动点，逆相似轴，两圆外位似中心及内位似中心Miquel定理内接三角形，外接三角形，Miquel点根轴圆幂，根轴，共轴圆系，极限点反演反演，分式线性变换（正定向和反定向）配极极点与极线，共轭点对，三线极线及三线极点，垂极点射影几何点列的交比，线束的交比，射影几何基本定理，调和点列与调和线束，完全四边形及完全四点形的调和性，Pappus 定理，Desargues定理，Pascal定理，Brianchon定理著名定理三大作图问题，勾股定理，黄金分割，鞋匠的刀，P’tolemy定理，Menelaus定理，Ceva定理，Stewart定理，Euler线，Fermat-Torricelli问题，Fagnano-Schwarz问题，Newton线，Miquel定理，Simson线，Steiner 定理，九点圆，Feuerbach定理，Napoleon定理，蝴蝶定理，Morley定理，Mannheim定理例题和习题1．已知：ABCD是圆外切四边形，内切圆心O在对角线BD上射影为M。
再看看美国的这本代数教材
作者： 刘俊杰
作者机构： 上海师范大学
出版物刊名： 上海中学数学
页码： F002-F002页
主题词： 教材 代数 师生 几何 编者 出版 课题 彩色
摘要：上一期我们简单地介绍了Geometry,这一次我们聊一聊Algebra.它也是美国Prentice Hall 2001年出版的教材,风格和几何教材一样,不过这一次我们集中在一个方面来介绍,比如说,介绍project(课题)吧.其实,书中引人思考的地方不少,浏览过它的精装、彩色页面、照片和图画的外貌之后,我们看到的是编者们为师生们所做的细考虑,有些项目是我们编教材时不太注意的.。
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第五届MathorCup全球大学生数学建模挑战赛题目D题图像去噪中几类稀疏变换的矩阵表示
m ⨯ m D V m ⨯1 MathorCup 全球大学生数学建模挑战赛
暨 CAA 世界大学生数学建模竞赛
D 题 图像去噪中几类稀疏变换的矩阵表示
假设一幅二维灰度图像 X 受到加性噪声的干扰：Y = X + N ， Y 为观察到的噪声图像， N 为噪声。
通过对于图像Y 进行稀疏表示可以达到去除噪声的目的。
任务：
1. 将图像Y 分割为相互重叠的小块{Y ij }
，对于讨论Y ij ( m ⨯ m) 四类稀疏变换的矩阵表 示：离散余弦变换（DCT ），离散小波变换（DWT ，用 DB4 小波），主成分分析（PCA ）和奇异值分解（SVD ）。
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2. 利用Cameraman 图像中的一个小图像块（见图 1）进行验证。
3. 分析稀疏系数矩阵，比较四种方法的硬阈值稀疏去噪性能，并提出可能的新的稀疏去噪方法。
图1 实验图像，第一行表示Cameraman 图像及其噪声干扰图像（高斯噪声，标准偏差为10）；第二行表示上述两幅图像相同位置的一个图像小块（行数：144-151，列数：167-174），数字为对应位置像素的灰度值。
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Shape Representation using Fourier Coefficients of theSinusoidal TransformIan PrattDepartment of Computer Science,University of Manchester,Manchester M139PL,U.K.ipratt@AbstractThis paper investigates the use of the sinusoidal transform to represent convex regions of the plane.Formulae are derived for the determination of various geometrical features and relations in terms of sinu-soidal transforms and their Fourier descriptors.Necessary and sufficient conditions for a periodic functionto be the sinusoidal transform of a convex region are derived.Sinusoidal Fourier descriptors are shown tocompare favourably with other shape-representation schemes based on Fourier descriptors.Keywords:Shape representation,Fourier descriptors,convex,plane,Hough transform.1BackgroundThe sinusoidal transform is a member of the general class of geometrical representation schemes based on boundary functions.A boundary function is a periodic function of one variable which picks out a bounded region of the plane in terms of the contour that encloses it.Various types of boundary function have been explored in the literature,but one feature they all share is that,being periodic,they can be approximated bya sequence of Fourier coefficients,which therefore serve to represent the original region.For example,Zahn and Roskies[11]investigate the boundary function which gives(a normalized version of)the tangent angle at any point of the contour as a function of the normalized arc-length to that point from afixed starting point.Granlund[2],Persoon and Fu[6]and Lin and Hwang[4]consider instead the boundary functions and giving the Cartesian coordinates of any point of the contour as a function of the normalized arc-length to that point from afixed starting point.And Kiryati and Maydan[3]analyse the boundary function such that are the polar coordinates of the contour points relative to an origin within the region.Exactly which geometric features of regions can be read off from their representations in terms of Fourier coefficients depends on the type of boundary function used.In this paper,we investigate what happens when we take the boundary function to be the sinusoidal transform(Pratt[7]).This method of representing convex regions is essentially equivalent to the‘rotating callipers’method(Toussaint[9])and to the well-studied and widely-applied Hough-and Radon transforms(see,for example,Leavers[5]).According to this method, a convex region of the plane can be represented by rotating it around the origin and plotting the height(the y-coordinate)of its highest and lowest points over time.Figure1shows some convex regions of the plane and their corresponding representations.Spatial representation schemes based on similar geometrical ideas have also been proposed as models of spatial representation in animal brains(Touretzky,Redish and Wan[10]). However,our concern in this paper is not to apply,or even to recommend,the sinusoidal transform,but rather, to establish its properties.The plan of the paper is as follows.We begin by introducing the sinusoidal transform,explaining its most basic properties.We then investigate the transform from a more analytical perspective,deriving further properties on our way.Armed with this analysis,we establish an inverse of the sinusoidal transform,and discuss the representation of convex regions based on sinusoidal Fourier descriptors.2The sinusoidal transformGiven a vector whose polar coordinates are,we can associate with the sine wave(1)2a)b)c)d)Figure 1:Some convex regions and their representations over one revolution.The symbol marks the origin.3with amplitude and phase-lead.We shall say that is the sine wave corresponding to the vector. Geometrically,if we imagine the vector rotating about the origin with constant angular velocity1radian per unit time,then the sine wave corresponding to is the graph obtained by plotting the height(i.e.the -coordinate)of against time.The following well-known fact states that we can add vectors by superposing their corresponding sine waves:Observation1Let,be vectors having polar coordinates,respectively,and let ,be the corresponding sine waves of frequency.If is the point-wise sum,and if is the vector sum,then is also a sine wave of frequency,,where are the polar coordinates of the vector.Henceforth,we take the term region to denote a compact,nonempty set of points(or vectors)in the plane. We now show how a pair of periodic functions of one variable can represent a convex region.Definition1If is a convex region,we define the sinusoidal hull corresponding to to be the pair of functions:(By the compactness of,these limits will always exist.)In addition,we call the function the sinusoidal transform of.Geometrically,if we imagine the region rotating anti-clockwise about the origin with constant an-gular velocity of1radian per until time,then is the graph obtained by plotting the height(i.e.the -coordinate)of the highest point on,and is the graph obtained by plotting the height of the lowest point on.That is,and give the projection of on the-axis as is rotated about the origin(see figure1).The ability to represent a convex region of the plane using the functions and depends on the following observation(which is easy to verify from the geometrical interpretation of sinusoidal hulls):Observation2Let be a convex region,and its corresponding sinusoidal hull.Let be a vector,and its corresponding sine wave.Then the following are equivalent:(i);(ii)for all, ;(iii)for all,.Thus,the function(or,alternatively,)represents uniquely,because it suffices to determine of any given vector,represented by a sine wave,whether lies in.Although each of the functions4and by itself contains all the information needed to represent S,we shall sometimesfind it convenient to represent a region using both functions—that is,using its sinusoidal hull.If we represent convex regions by their sinusoidal hulls,a range of geometrical tests and operations becomes available.Suppose we want to translate a convex region by a vector,where S is represented by the functions and by the function.The following theorem follows simply from observations1and2:Theorem1Let be a region with sinusoidal hull,a vector corresponding to sine wave ,and a real number.Let denote the region obtained by translating by,and the region obtained by rotating anti-clockwise about the origin through radians.Then has sinusoidal hull,and has sinusoidal hull.Thus:to translate a convex region represented by its sinusoidal hull,add an appropriate sine wave to the functions and;to rotate about the origin,shift the functions and by an appropriate amount along the-axis.In the remainder of this section,we show how various relationships between a pair of convex regions can be read off from the sinusoidal hulls of those regions.Theorem2Let and be sinusoidal hulls corresponding to convex regions and ,respectively.1.The following are equivalent:(i);(ii)for all,;(iii)for all,.2.The following are equivalent:(i);(ii)for all,;(iii)for all,.Figure2illustrates.If and are non-overlapping convex regions,then we take the separation between them to be the length of the shortest line segment with endpoints in and.The following theorem is not trivial,but still straightforward.Theorem3If and are sinusoidal hulls corresponding to convex regions and ,respectively,such that,then the distance between and is given by5a T (t)a (t)S T (t)b (t)b S TS a)a T (t)T (t)b a (t)S (t)S b T S b)Figure 2:Convex regions and and their sinusoidal hulls,showing the relations:a);b).A more general measure of the distance between two convex regions—overlapping or not—is as follows.(See Eggleston [1]ch.4for a full analysis).Letbe the smallest non-negative number such that every point inis no more than distance from ,and let be the smallest non-negative number such that every point in is no more than distance from .Then we define .Theorem 4Let ,be the sinusoidal transforms of convex regionsand ,respectively.Then This result establishes the stability of sinusoidal representations,because ‘similar’convex regions (according to the standard -metric)yield ‘similar’sinusoidal representations (according to an appropriate metric on bounded periodic functions).Finally,we note in passing that,if and are non-overlapping convex regions,there exist four lines of support—and hence eight directed lines of support—betweenand ,as shown in figure 3.The sinusoidal hulls and will have eight crossing points,corresponding to these eight directed lines of support.Geometrically,if we imagine and rotating together about the origin,the crossing points occur at times when their corresponding lines of support are horizontal.This correspondence is the basis for the use of sinusoidal hulls for route-planning,as described in Pratt [8].6a (t)S (t)S b a (t)T (t)T b TSFigure 3:The eight crossing-points of two sinusoidal hulls3Analysing the sinusoidal transformIn this section,we analyse the sinusoidal transform in more detail and lay the groundwork for section 4,where we investigate its inverse.Along the way,we also establish some surprising properties of the sinusoidal transform.Let be a convex region of the plane.Each point on the boundary of lies on one or more tangents to .(We take a tangent to to be a line intersecting the boundary of but not its interior).Some tangents to may lie on a straight-line segment of ’s boundary;all other tangents to will intersect in just one point.Likewise,some points on the boundary of may lie on more than one tangent (we refer to these points as kinks in the boundary of );all other points will lie on just one tangent.If we give the boundary of an orientation—say,clockwise—any tangent to picks up a direction in the obvious way,and can be uniquely identified by the angle it makes with the horizontal (measured clockwise from the positive x-axis)in the range .Thus,every angle in the range corresponds to either a unique point ,on the boundary of ,or to a straight line segment of the boundary of .Figure 4illustrates.It follows that,if the boundary of contains no straight line segments,as in fig 4a),it can be described in parametric form by the pair of functions:.Geometrically:tours the boundary of clockwise in such a way that,at any time ,we are at a point where there is a tangent to making angle with the x-axis.Kinks in the boundary of correspond to intervals over which is constant,as in fig 4b).7a)b)S(t)Saw(t)t0Figure5:A boundary point to a region and its corresponding sine wave.gives the x-coordinate of as is rotated about the origin along with.Thus,just as gives the -coordinate of the highest point on as is rotated about the origin,so gives the-coordinate of that same point.A rigorous proof of the following theorem can be given using a limit-argument,but the details are routine.Theorem5Let be a well-behaved,convex region of the plane,and suppose to rotate anticlockwise about the origin with unit angular velocity.Let be a boundary point of,and suppose to rotate with.If the boundary of contains no straight line segments in the region of,and is a time when is uppermost,then exists,and are the coordinates of at that time.If lies at the start of a straight line segment of the boundary of(proceeding clockwise around),and that line segment is uppermost at time,then exists,and are the coordinates of at that time.If lies at the end of a straight line segment of the boundary of(proceeding clockwise around),and that line segment is uppermost at time,then exists,and are the coordinates of at that time.As an immediate corollary of theorem5,if’s boundary contains a straight-line segment,which becomes9uppermost on at time,then is discontinuous at.Moreover,the length of this line segment is given by the size of the discontinuity:(2)Note that this quantity must,therefore,be positive.Finally,having dealt with straight-line segments in’s boundary,what can we say of kinks?If lies at a kink in’s boundary,then,assuming that has some interior points,there exist,with, such that a tangent to passes through just in case the angle of that tangent lies in the interval (modulo).The quantity is the external angle of the kink.Clearly,will be a sine wave over the interval.To summarize:(i)a point on a smoothly curving portion of’s boundary corresponds to a single time (in each revolution)lying within some non-sinusoidal region of;(ii)all points on a straight-line segment of’s boundary correspond to a single time(in each revolution)at which the derivative is discontinuous,with the length of the line segment equal to the size of the discontinuity;(iii)a point at a kink on’s boundary corresponds to all points within a time interval(in each revolution)over whichis a pure sine wave,with the external angle of the kink equal to the duration of the interval.Having established the geometrical interpretation of,we can now relate to the curve. Let be a well-behaved,convex region of the plane.By simple rotation of axes,we have:(3)(4)Thefirst of these equations holds for all,the second,for all such that is not on a straight line segment of’s boundary.We note in passing that,in image processing,edge-detection byfiltering standardly delivers,for each boundary-point detected,the slope of the tangent to the boundary at that point.That is,in terms of our parametric equations,boundary points are naturally detected along with their tangent slope. This means that equation(3)can be applied directly tofiltered images so as to generate(noisy)sinusoidal transforms from images.However,in this paper,we shall not concern ourselves with the practical difficulties of extracting the sinusoidal transform(or its relatives)from an image,since that would involve us in issues10of image processing on which we have nothing to add.Let us now turn our attention to the significance of the second derivative of the sinusoidal transform, .Again,let be a well-behaved,convex region of the plane.Suppose that is a point on the boundary of lying neither on a straight-line segment nor at a kink.By hypothesis,’s boundary has afinite, positive radius of curvature at,which we may denote.We shall now show:(5)We supposefirst that is the origin,.(This simplifies the mathematics.)Figure6shows a nearby point with coordinates,together with normals to the boundary of subtended from and. Let be the point of intersection of these normals.We may take the radius of curvature to be the limit of the length of as.By theorem5,exists in some region of.Moreover,and ,where is the length of.Hence,(6) with this second derivative guaranteed to exist if there is a radius of curvature.To generalize to the case where is other than the origin,let be the sine wave corresponding to this boundary point.We have.Moreover,by shifting so that is moved to the origin,we obtain,by theorem1,the region with sinusoidal transform.Therefore the radius of curvature we are seeking is given by,which immediately yields equation(5).Kinks in the boundary of—where the radius of curvature is zero—correspond to intervals over which is constant.But we have established that will be a sine wave over this interval,and so that ,again,giving the correct radius of curvature.Straight line segments in the boundary of —where the radius of curvature is infinite,correspond to values of where—and hence—does not exist.But we have established that,at such points,so that we might(with a little notational licence)write.The following theorem is a somewhat imprecise summary of the foregoing observations:11and(8),guarantees the existence of and.Routine calculation gives(9)(10)Hence(11)Finally,by the assumption that is well-behaved,theorem(6)guarantees that,if does not lie on a straight line segment or at a kink in’s boundary,(12)By equation(11),for such values of,and cannot both be zero.This observation will be useful presently.We can summarize the properties of sinusoidal transforms of well-behaved regions in the following theo-rem.Definition3Let be a periodic function with period.We say that is well-behaved if is continuous and,over every period:(i)and both exist except possibly at afinite number of points such that and both exist with;(ii)whenever exists,except possibly for afinite number of closed intervals such that and throughout.Theorem7If is a well-behaved,convex region of the plane,then is a well-behaved periodic function with period.We shall see in section4that a converse of theorem7obtains:every well-behaved periodic function with period is the sinusoidal transform of a well-behaved,convex region of the plane.Now let us return to extracting geometrical information from sinusoidal transforms.The arc-length be-tween any two points on’s boundary can also be simply given in terms of.Suppose that’s boundary contains no kinks or straight line-segments at or between the points and.Then,as we have13observed,and cannot both be zero for in this interval.By a standard result,’s boundary between and will then have a well-defined arc-length arc,given byTheorem9If is a well-behaved,convex region,its total perimeter is given by:Given theorem1,and noting that,for a sine wave,we see instantly that the measurement of arc-length as given by theorem8is invariant under translation,as of course it should be.Invariance under rotation is even more obvious.The area of can also be simply expressed in terms of.Suppose for the moment that the interior of contains the origin.And let us assume that the boundary of contains no straight line segments,so that and exist everywhere.Then’s boundary can be described by the periodic function,where give the polar coordinates of points on that boundary.When using polar coordinates,the standard convention is that is measured in an anticlockwise direction.In that case,the area of is given by the standard formula:(16)with the quadrant of being defined by the signs of and in the obvious way.Ifthen by elementary calculus:(17)A simple rotation of shows that this formula is also correct when.In order for to move from to(measured anticlockwise),must move backwards from some point to(since is measured in the clockwise direction).Hence(18) Moreover,using the integration-by-parts formula,we can show that,if is a sine wave of frequency1,(19) whence(20) so that equation(18)holds even when does not contain the origin.Finally,we note that,again using integration by parts,we haveso that equation(18)becomesThe extension of this theorem to cases where’s boundary contains straight line segments is completely routine.164Inverting the sinusoidal transformThe job of this section is to answer the following questions:1.Given a periodic function with period,under what conditions is the sinusoidal transformof some well-behaved,convex region?2.For those functions satisfying these conditions,how can we characterize the mapping from to(i.e.the inverse of the sinusoidal transform)?3.Can this inverse mapping be extended to functions which are not the sinusoidal transforms of anywell-behaved convex region?If so,what regions of the plane are obtained?Let be any twice differentiable periodic function with period,such that.(We will relax the differentiability requirement and the strict inequality later.)Now define:(22)(23)By the periodicity of,equations(22)and(23)are the parametric equations of a closed curve in the plane as moves from to.We will now show that is simple(does not cross over itself),that the region it encloses is convex and that the enclosing contour is toured in a clockwise manner. Differentiating equations(22)and(23)gives(24)(25)Moreover,the tangent angle to any parametrically specified curve at is given byThus,the directed tangent to at any point makes angle with the positive-axis.By equation(25),crosses zero at;moreover,since,is positive just before and negative just after.Thus,attains a local maximum at,which means that the tangent to at must lie to the left of the curve.By considering shifted versions of the function ,we can generalize this to the observation that the tangent to at any point must lie to the left of the curve.In other words,as moves from to,and we travel around,we are always turning to the right.Since we can only turn through a total of radians,it follows that is simple and encloses a convex region about which it winds clockwise.Note that,if satisfied the reverse inequality, then parallel reasoning would show that as moves from to,and we travel around,we would always turn to the left.Thus,in this case too would enclose a convex region,but would travel anti-clockwise around it.So:given a twice differentiable periodic function with period such that, equations(22)and(23)generate a convex ing these equations together with equation(3)gives:(28)In other words,the sinusoidal transform of a region generated from by equations(22)and(23)is just .Moreover,given that,we can reverse the derivation(6)in section3to conclude that must have a well-defined,positive radius of curvature at every point of its boundary.Hence,is a well-behaved region(in fact,one with no kinks or straight line segments in its boundary).Conversely,we know from theorems6and7that,given a well-behaved,convex region with no kinks or straight line segments in its boundary,its sinusoidal transform is a twice-differentiable periodic function with period such that.Moreover,we have.In other words,the mappings andare inverses for the case where is a well-behaved,convex region with no kinks or straight line segments in its boundary,and satisfies the conditions above.Having sorted out the correspondence between convex regions and periodic functions where always exists and does not vanish,we now relax these conditions on slightly.Suppose that over an interval with.By equations(24)and(25),we have,so that and are constant over.This in turn implies the existence of a kink in the curve traced out by,with external angle.Thus,traces out a simple closed curve18enclosing a convex region as before.Hence,we can weaken the inequality satisfied by to allow that over afinite collection of intervals in a period:equation(28)will still hold,of course.Finally,we consider the case where fails to exist at a point,but is continuous,and the limits and both exist with.Thus,traces out a discontinuous curve with the limits and both existing.But we have already agreed that,in such cases,we should take the line from to as part of the boundary of.The continuity of and the condition guarantee that the directed line segment from to will make an angle with the positive x-axis.Thus,in jumping from to,we continue to turn clockwise as we move around so that is still simple and encloses a convex region.This means that we can weaken the differentiability requirement on to allow that fail to exist at afinite number of points,as long as and both exist and.Now we can summarize the observations of this section in the following two theorems.Theorem11If is a well-behaved,periodic function with period,then is a well-behaved,convex region of the plane.Theorem11is the promised converse to theorem7of section3.Moreover:Theorem12For any well-behaved,periodic function with period,we have;and for any well-behaved,convex region of the plane,we have.Theorems7,11and12effectively wrap up thefirst two questions set at the beginning of this section.We now know exactly which periodic functions are sinusoidal transforms of well-behaved,convex regions,and,given any such function,equations(22)and(23)enable us to reconstruct the region whose sinusoidal transform it is.But of course,these equations can be used to generate curves from any differentiable periodic function, well-behaved or not.And the question arises as to what happens in the non-well-behaved case.Functions which fail the double-differentiability requirement,or for which has a minimum at zero are of little computational interest.Moreover,reversing the inequalities and insisting that and that generates the same set of regions,but circumnavigated in an anticlockwise direction.The only interesting cases left to understand are(i)when crosses or(ii)wheneven though is not uniformly less than.19c)Figure7:Schematic illustration of various types of cusp in:the arrows on the curves indicate the motion of the point as increases.Let us assume then,that is a well-behaved periodic function with period,and suppose crosses at,where exists.It helps to assume that does not take one of the values,,,. Then,from equations(24)and(25),we know that both and reverse signs simultaneously,at a point where the tangent to the curve is neither horizontal nor vertical.Clearly,this behaviour means that has a cusp at.This situation is illustrated schematically infigure7a).Routine checking shows that this result is also true where is one of,,,.Note that might linger on the t-axis rather than crossing it instantaneously.In that case,the internal angle of the cusp is equal to the amount of time spends at0.This situation is illustrated schematically infigure7b).Supposefinally that we have at one of thefinitely many points for which does not exist.Since is continuous and differentiable in regions either side of,we have. (Otherwise,by a standard result of analysis,would exist.)Therefore.In other words, we must consider the possibility of an instantaneous decrease in thefirst derivative.As might be expected, under the convention of joining up discontinuities in with straight line segments,this case corresponds to a cusp with a straight line segment inserted in it.Exactly what happens at the cusp depends on the sign ofeither side;the details are unsurprising.One such situation—where changes sign across—is illustrated schematically infigure7c).Thus,non-well-behavedness in leads invariably to cusps in.Since the curve lies to the right or to the left of the directed tangent according to whether is positive or negative,there is no20a)b)c)d)Figure8:Some functions for which crosses zero,and their associated curves.Note that the‘lower’function,is also shown.reason,in this case,why need trace out a simple curve as moves though.That can fail to be simple when crosses zero is shown byfigure8where we have four closed contours,a)–d), corresponding to four arbitrarily chosen,doubly differentiable periodic functions.The preceding discussion effectively wraps up the third question set at the beginning of this section,since we now know what happens when we extend the inverse transform to functions which are not the sinusoidal transforms of well-behaved,convex regions.The only geometrically interesting sort of misbehaviour of arises when crosses zero or.Such a case represents a cusp,and frequently, though not always,leads to curves which are not simply closed.This(negative)result is of practical interest,because it means that there is no natural extension of the sinusoidal transform which represents any particularly useful class of regions.215Fourier series representationsLet us round off with some results of a more positive character.We shall exploit the the sinusoidal transform to represent regions using sinusoidal Fourier descriptors.Let be a convex region of the plane.Sinceis a periodic function with period,it can be expanded as a Fourier series:where()Thus,instead of thinking of as being the maximum of a collection of sine waves of the same frequency, we now think of it as being a sum of sine waves of different frequencies.Moreover thefinite initial segment of the sequence of Fourier coefficientsconstitutes an approximate(and easily stored)representation of a region.In practice,a handful of these Fourier coefficients give a convincing representation of most regions.The most noticeable deficits for small values of are that straight lines tend to be rounded and corners are sometimes replaced by thefish-tail patters characteristic of the failure of the condition.Figure9shows two regions followed by(an initial segment of)their Fourier descriptors and the reconstructions of these regions from the given Fourier descriptors.Of course,since and can be recovered from their Fourier descriptors,all of the results derived above can be applied to extract geometric information about the represented regions. Sometimes,however,there are short cuts,as we shall now show.Thefirst three Fourier descriptors have simple geometric interpretations.From theorem9,we know that。