Journal of Pattern Recognition Research 1 28 14-23 The Invariant Relations of 3D to 2D Projection of Point Sets Wang YuanBin wangyuanbin@ise.neu.edu.cn Zhang Bin zhangbin@ise.neu.edu.cn Yu Ge yuge@ise.neu.edu.cn Institute of Computer Application Technology, Northeastern University WenHua Road, ShenYang, 114, PRC Received November 23, 26. Accepted March 5, 28. Published online June 14, 28. Abstract The depth information is lost in the projection from a 3D object to a single 2D image and there are no geometric invariants of such a projection. However, there are invariant relations relating the projective invariants of a 3D point set and those of its 2D image under general projective projection. We present a general method to derive these relations in this paper. Keywords: Computer Vision, Invariants, Projective Invariants, Projection Invariants. 1. Introduction Geometric invariants are important for object recognition. A lot of work has been done to extract geometric invariants based on the assumption that the underlying transformations are 3D to 3D or 2D to 2D [1-5], which are usually inappropriate to general 3D object recognition using projected images. However, it is known that 3D to 2D projection invariants cannot be obtained for a set of 3D points in general configuration from a single view [6-8]. A method to overcome this barrier is to use multiple uncalibrated images of a 3D scene to derive invariants. Barrett et al. proposed a method to derive an invariant for six points by using stereo images [9]. Hartley has shown that the invariant could be computed from the images of the four lines in two distinct views with uncalibrated cameras once the epipolar geometry is known [1]. Quan has derived invariants for six 3D points from three images [11]. Another method to derive 3D to 2D projective invariants is to impose some restrictions on the structure of the geometric configurations. Rothwell et al. have shown that the projective invariants could be derived from a single view for two classes of structured objects [12]. Zhu et al. and Song et al. have derived an invariant for a structure with six points on two adjacent planes [7, 13]. Sugimoto has derived an invariant from six lines on three planes in a single view [14]. In this paper, we first derive a general invariant relation relating the 3D to 3D projective invariants of six 3D points to 2D to 2D projective invariants of the 2D images of the space points. Then, new invariant relations of six space points with different restrictions on the structure of the configurations are derived. 2. The Basic Invariant Relation In this section, we derive a basic invariant relation relating the 3D projective invariants of six space points and the 2D projective invariants of the planar projected points. Invariant c 28 JPRR. All rights reserved.
The Invariant Relations of 3D to 2D Projection of Point Sets relations can be used for model-based recognition of 3D objects [8,15-17]. The derivation method is similar to [8, 17]. It is much simpler and easier to understand. The form of the relation is slightly different also. Let P 1, P 2, P 3, P 4, P 5, and P 6 be six distinct points in 3D space. We suppose that P 1, P 2, P 3, and P 4 are not coplanar. Let p 1, p 2, p 3, p 4, p 5, and p 6 be the planar image of the given six 3D points under a general projective transformation. We suppose that p 1, p 2, and p 3 are not collinear. The homogeneous representation of the 3D points and the 2D points are x i, y i, z i, 1 T and u i, v i, 1 T, respectively. The relations between the 3D points and their 2D projection points are k i u i v i 1 = M x i y i z i 1, 1 where i = 1, 2, 3, 4, 5, and 6, M is a 3 4 projective camera matrix of rank 3, and k i are non zero scalar factors. The homogeneous points of a 3D projective space are also vectors of a four dimension vector space. From the principles of elementary linear algebra, no five homogeneous points in 3D projective space can be linearly independent. We can represent P 5 and P 6 as the linear combinations of other four points P 5 = α 1 P 1 + α 2 P 2 + α 3 P 3 + α 4 P 4, 2 P 6 = β 1 P 1 + β 2 P 2 + β 3 P 3 + β 4 P 4. 3 Since P 1, P 2, P 3, and P 4 are not coplanar, they are linearly independent. This means that the coefficients in the above representations are unique. Since the last coordinates of these points all equal 1, it is obvious that α 1 + α 2 + α 3 + α 4 = 1 and β 1 + β 2 + β 3 + β 4 = 1. Now we begin the derivation of the basic invariant relation. Multiply each side of 2 and 3 by a 3D to 2D projection matrix M, we have MP 5 = α 1 MP 1 + α 2 MP 2 + α 3 MP 3 + α 4 MP 4, 4 MP 6 = β 1 MP 1 + β 2 MP 2 + β 3 MP 3 + β 4 MP 4. 5 From the relations defined in equations 1, 4 and 5, we can derive the following relations k 5 p 5 = k 1 α 1 p 1 + k 2 α 2 p 2 + k 3 α 3 p 3 + k 4 α 4 p 4, 6 k 6 p 6 = k 1 β 1 p 1 + k 2 β 2 p 2 + k 3 β 3 p 3 + k 4 β 4 p 4. 7 Similarly, since p 1, p 2, and p 3 are not collinear, we can uniquely represent p 4, p 5, and p 6 as the linear combinations of them p 4 = λ 1 p 1 + λ 2 p 2 + λ 3 p 3 p 5 = ρ 1 p 1 + ρ 2 p 2 + ρ 3 p 3. 8 p 6 = τ 1 p 1 + τ 2 p 2 + τ 3 p 3 15
YuanBin et al. Again, it is clear that λ 1 + λ 2 + λ 3 = 1, ρ 1 + ρ 2 + ρ 3 = 1 and τ 1 + τ 2 + τ 3 = 1 because the last coordinates of these points are all 1. Substitute 8 into equations 6 and 7, we have k 5 ρ 1 p 1 + ρ 2 p 2 + ρ 3 p 3 = k 1 α 1 p 1 + k 2 α 2 p 2 + k 3 α 3 p 3 + k 4 α 4 λ 1 p 1 + λ 2 p 2 + λ 3 p 3, 9 k 6 τ 1 p 1 + τ 2 p 2 + τ 3 p 3 = k 1 β 1 p 1 + k 2 β 2 p 2 + k 3 β 3 p 3 + k 4 β 4 λ 1 p 1 + λ 2 p 2 + λ 3 p 3. 1 Rewrite 9 and 1 k 1 α 1 + k 4 α 4 λ 1 k 5 ρ 1 p 1 + k 2 α 2 + k 4 α 4 λ 2 k 5 ρ 2 p 2 + k 3 α 3 + k 4 α 4 λ 3 k 5 ρ 3 p 3 =, 11 k 1 β 1 + k 4 β 4 λ 1 k 6 τ 1 p 1 + k 1 β 1 + k 4 β 4 λ 2 k 6 τ 2 p 2 + k 1 β 1 + k 4 β 4 λ 3 k 6 τ 3 p 3 =. 12 Since p 1, p 2, and p 3 are not collinear, they are independent. So the coefficients in the above equations must be all zero. We have a set of linear equations then. That is k 1 α 1 + k 4 α 4 λ 1 k 5 ρ 1 = k 2 α 2 + k 4 α 4 λ 2 k 5 ρ 2 = k 3 α 3 + k 4 α 4 λ 3 k 5 ρ 3 = k 1 β 1 + k 4 β 4 λ 1 k 6 τ 1 = k 2 β 2 + k 4 β 4 λ 2 k 6 τ 2 = k 3 β 3 + k 4 β 4 λ 3 k 6 τ 3 = Rewrite the system of linear equations 13 in the matrix form α 1 λ 1 α 4 ρ 1 α 2 λ 2 α 4 ρ 2 α 3 λ 3 α 4 ρ 3 β 1 λ 1 β 4 τ 1 β 2 λ 2 β 4 τ 2 β 3 λ 3 β 4 τ 3 k 1 k 2 k 3 k 4 k 5 k 6. 13 =. 14 Because we have known in advance that all k i are non zero, this means that the set of homogeneous linear equations 14 should have a non trivial solution. From basic theorems of linear algebra, the determinant of the matrix must be zero in order for the set of homogeneous linear equations to have a non trivial solution. This is the main trick of our derivation. That is α 1 λ 1 α 4 ρ 1 α 2 λ 2 α 4 ρ 2 α 3 λ 3 α 4 ρ 3 β 1 λ 1 β 4 τ 1 =. 15 β 2 λ 2 β 4 τ 2 β 3 λ 3 β 4 τ 3 Calculate the determinant in 15, we obtain the basic invariant relation ρ 3 λ 1 τ 2 λ 2 τ 1 α 1 α 2 β 3 β 4 ρ 2 λ 1 τ 3 λ 3 τ 1 α 1 α 3 β 2 β 4 + τ 1 λ 2 ρ 3 λ 3 ρ 2 α 1 α 4 β 2 β 3 + 16
The Invariant Relations of 3D to 2D Projection of Point Sets ρ 1 λ 2 τ 3 λ 3 τ 2 α 2 α 3 β 1 β 4 τ 2 λ 1 ρ 3 λ 3 ρ 1 α 2 α 4 β 1 β 3 + τ 3 λ 1 ρ 2 λ 2 ρ 1 α 3 α 4 β 1 β 2 =. 16 Divide each side of 16 by α 2α 3 α 4 β 1 β 1 α 1, we have ρ 3 λ 1 τ 2 λ 2 τ 1 α 1β 3 α 1 β 4 α 3 β 1 α 4 β 1 ρ 3 λ 1 τ 3 λ 3 τ 1 α 1β 2 α 1 β 4 α 2 β 1 α 4 β 1 + τ 1 λ 2 ρ 3 λ 3 ρ 2 α 1β 2 α 1 β 3 α 2 β 1 α 3 β 1 +ρ 1 λ 2 τ 3 λ 3 τ 2 α 1β 4 α 4 β 1 τ 2 λ 1 ρ 3 λ 3 ρ 1 α 1β 3 α 3 β 1 + τ 3 λ 1 ρ 2 λ 2 ρ 1 α 1β 2 α 2 β 1 =. 17 It is well known that 6 distinct points in 3D projective space produce 3 6 15 = 3 independent 3D to 3D projective transformation invariants [3]. Let us define I 12 = α 1β 2 α 2 β 1, I 13 = α 1β 3, and I 14 = α 1β 4. They are cross ratios of coefficients. It is easy to check that α 3 β 1 α 4 β 1 I 12, I 13, and I 14 are invariant under 3D to 3D projective transformations [8, 17]. Rewrite 17 ρ 3 λ 1 τ 2 λ 2 τ 1 I 13 I 14 ρ 2 λ 1 τ 3 λ 3 τ 1 I 12 I 14 + τ 1 λ 2 ρ 3 λ 3 ρ 2 I 12 I 13 +ρ 1 λ 2 τ 3 λ 3 τ 2 I 14 τ 2 λ 1 ρ 3 λ 3 ρ 1 I 13 + τ 3 λ 1 ρ 2 λ 2 ρ 1 I 12 =. 18 Divide each side of 18 by λ 2 ρ 3 τ 1 λ1 τ 2 1 I 13 I 14 λ 3ρ 2 λ 2 τ 1 λ 2 ρ 3 ρ1 τ 3 ρ 3 τ 1 λ 3ρ 2 ρ 1 τ 2 λ 2 ρ 3 ρ 2 τ 1 I 14 λ1 τ 2 λ 2 τ 1 λ 3ρ 2 ρ 1 τ 2 λ 2 ρ 3 ρ 2 τ 1 λ1 τ 3 1 I 12 I 14 + 1 λ 3ρ 2 λ 3 τ 1 λ 2 ρ 3 I 13 + λ3 ρ 2 λ 1 τ 3 λ 2 ρ 3 λ 3 τ 1 ρ 1τ 3 ρ 3 τ 1 I 12 I 13 + I 12 =. 19 It is well known that 6 points in the plane produce 2 6 8 = 4 independent 2D to 2D projective transformation invariants [3]. Let us define K λτ 12 = λ 1τ 2 λ 2 τ 1, K λτ 13 = λ 1τ 3 λ 3 τ 1, K ρτ 12 = ρ 1τ 2 ρ 2 τ 1, K ρτ 13 = ρ 1τ 3 ρ 3 τ 1, and K λρ 32 = λ 3ρ 2 λ 2 ρ 3. It is easy to check that K λτ 12, Kλτ 13, Kρτ 12, Kρτ 13, and Kλρ 32 2D projective transformations [8, 17]. It is also clear that K λρ are invariant under 2D to 32 = Kλτ 12 Kρτ 13 K13 λτ Kρτ 12 At this point, we have derived the basic invariant relation relating the 3D projective invariants of the six space points and the 2D projective invariants of their projected planar points. Theorem 1 The invariant relation relating the absolute 3D to 3D projective transformation invariants of six 3D points and the 2D to 2D projective transformation invariants of their 2D images is K ρτ K12 λτ 1 I 13 I 14 Kλτ 12 Kρτ 13 K λτ 13 Kλτ 12 Kρτ 13 K13 λτ I 14 K λτ K13 λτ Kρτ 12 13 1 12 Kλτ 12 Kρτ 13 K13 λτ I 12 I 14 + 1 Kλτ 12 Kρτ 13 K λτ 12 I 13 + Kρτ 13 K ρτ 12 K13 λτ Kρτ 12. I 12 I 13 + K ρτ 13 I 12 =. 2 17
YuanBin et al. We can easily extend the above method to derive invariant relations of point sets with more than six points. Suppose P 7 are the seventh point of the point set and p 7 is the projective image of P 7. Let us also represent P 7 as the linear combination of P 1, P 2, P 3, and P 4 and represent p 7 as the linear combination of p 1, p 2, and p 3, respectively P 7 = δ 1 P 1 + δ 2 P 2 + δ 3 P 3 + δ 4 P 4, 21 p 7 = ε 1 p 1 + ε 2 p 2 + ε 3 p 3. 22 Using similar derivation method, we can derive another invariant relation ρ 3 λ 1 ε 2 λ 2 ε 1 α 1 α 2 δ 3 δ 4 ρ 2 λ 1 ε 3 λ 3 ε 1 α 1 α 3 δ 2 δ 4 + ε 1 λ 2 ρ 3 λ 3 ρ 2 α 1 α 4 δ 2 δ 3 + ρ 1 λ 2 ε 3 λ 3 ε 2 α 2 α 3 δ 1 δ 4 ε 2 λ 1 ρ 3 λ 3 ρ 1 α 2 α 4 δ 1 δ 3 + ε 3 λ 1 ρ 2 λ 2 ρ 1 α 3 α 4 δ 1 δ 2 =. 23 Generally speaking, if the 3D point set contains n points in general position, then the number of independent 3D to 2D projective invariant relations determined by this set is n 5 provided that no three of the projected points are collinear. 3. Invariant Relations of Restricted Configurations 3.1 The Invariant Relation when Four of the Six 3D Points Are Coplanar Now we will derive the 3D to 2D projection invariant relation when four of the six 3D points are coplanar. Let us still suppose that P 1, P 2, P 3, and P 4 are not coplanar and that p 1, p 2, and p 3 are not collinear. These are the prerequisites for the equation 16 to be sound. Let us suppose that P 3, P 4, P 5, and P 6 are coplanar. So we can represent P 6 as the linear combination of P 3, P 4, and P 5 From 3 and 24, we have another representation of P 5 P 6 = η 3 P 3 + η 4 P 4 + η 5 P 5. 24 P 5 = β 1 η 5 P 1 + β 2 η 5 P 2 + β 3 η 3 η 5 P 3 + β 4 η 4 η 5 P 4. 25 Since P 1, P 2, P 3, and P 4 are linearly independent, the representation of P 5 is unique. So the coefficients in 2 and 25 satisfy β 1 = η 5 α 1, β 2 = η 5 α 2, β 3 = η 5 α 3 + η 3, β 4 = η 5 α 4 + η 4. 26 Substitute the equations in 26 into 16 and cancel α 1 α 2 ρ 3 λ 1 τ 2 λ 2 τ 1 α 3 α 4 η 5 η 5 + α 4 η 3 η 5 + α 3 η 4 η 5 + η 3 η 4 ρ 2 λ 1 τ 3 λ 3 τ 1 α 3 α 4 η 5 η 5 + α 3 η 4 η 5 +τ 1 λ 2 ρ 3 λ 3 ρ 2 α 3 α 4 η 5 η 5 + α 4 η 3 η 5 + ρ 1 λ 2 τ 3 λ 3 τ 2 α 3 α 4 η 5 η 5 + α 3 η 4 η 5 τ 2 λ 1 ρ 3 λ 3 ρ 1 α 3 α 4 η 5 η 5 + α 4 η 3 η 5 + τ 3 λ 1 ρ 2 λ 2 ρ 1 α 3 α 4 η 5 η 5 =. 27 Rewrite the above equation ρ 3 λ 1 τ 2 λ 2 τ 1 η 3 η 4 + ρ 1 λ 2 τ 3 λ 3 τ 2 ρ 2 λ 1 τ 3 λ 3 τ 1 + ρ 3 λ 1 τ 2 λ 2 τ 1 α 3 η 4 η 5 +λ 3 ρ 1 τ 2 ρ 2 τ 1 α 4 η 3 η 5 =. 28 18
The Invariant Relations of 3D to 2D Projection of Point Sets Divide each side of 28 by α 4 η 3 η 5 =. From 26, it is obvious So 29 can be written as That is ρ 3 λ 1 τ 2 λ 2 τ 1 η 4 α 4 η 5 + ρ 1 λ 2 τ 3 λ 3 τ 2 ρ 2 λ 1 τ 3 λ 3 τ 1 +ρ 3 λ 1 τ 2 λ 2 τ 1 α 3η 4 α 4 η 3 + λ 3 ρ 1 τ 2 ρ 2 τ 1 =. 29 β η 3 β 1 α 3 3 α = 1 α 1 β 3 α 3 β 1 = = I α 3 η 5 β 13 1, 3 1 α α 3 β 1 3 α 1 β η 4 β 1 α 4 4 α = 1 α 1 β 4 α 4 β 1 = = I α 4 η 5 β 14 1. 31 1 α α 4 β 1 4 α 1 ρ 3 λ 1 τ 2 λ 2 τ 1 I 14 1 + ρ 1 λ 2 τ 3 λ 3 τ 2 ρ 2 λ 1 τ 3 λ 3 τ 1 +ρ 3 λ 1 τ 2 λ 2 τ 1 = I 14 1 I 13 1 + λ 3ρ 1 τ 2 ρ 2 τ 1 =. 32 ρ 3 λ 1 τ 2 λ 2 τ 1 I 13 I 14 + λ 3 ρ 1 τ 2 ρ 2 τ 1 ρ 3 λ 1 τ 2 λ 2 τ 1 I 13 + ρ 1 λ 2 τ 3 λ 3 τ 2 ρ 2 λ 1 τ 3 λ 3 τ 1 I 14 + τ 3 λ 1 ρ 2 λ 2 ρ 1 =. 33 This is a new invariant relation when four of the six 3D points are coplanar. 3.2 The Invariant Relation when the Six 3D Points Are on Two Adjacent Planes Now we will derive the 3D to 2D projection invariant relation when the six 3D points are on two adjacent planes [7, 13]. Let us again suppose that P 1, P 2, P 3, and P 4 are not coplanar and that p 1, p 2, and p 3 are not collinear. These are the prerequisites for the equation 15 to be sound. Let us suppose that P 1, P 2, P 5, and P 6 are coplanar and that P 3, P 4, P 5, and P 6 are coplanar. It is obvious that P 5 and P 6 are on the intersection line of these two planes. We can represent P 6 as the linear combinations of P 1, P 2, and P 5 and P 3, P 4, and P 5, respectively P 6 = η 1 P 1 + η 2 P 2 + η 3 P 5, 34 From 34 and 35 P 6 = µ 1 P 3 + µ 2 P 4 + µ 3 P 5. 35 P 5 = η 1 P 1 + η 2 P 2 µ 1 P 3 µ 2 P 4, 36 P 6 = η 1µ 3 P 1 + η 2µ 3 P 2 η 3µ 1 P 3 η 3µ 2 P 4. 37 19
YuanBin et al. Table 1: Test results of invariant 42 on the six views of a 3D scene. Group No. Projection 1 Projection 2 Projection 3 Projection 4 Projection 5 Projection 6 1.8.83.79.81.8.8 2.5.48.48.48.48.48 3.89.89.88.9.89.89 4 -.3 -.2 -.3 -.2 -.4 -.3 So we have α 1 = η 1, α 2 = η 2, α 3 = µ 1, α 4 = µ 2, 38 β 1 = η 1µ 3, β 2 = η 2µ 3, β 3 = η 3µ 1, β 4 = η 3µ 2. 38 Substitute 38 and 39 into 16 and cancel common factors We finally obtain ρ 3 λ 1 τ 2 λ 2 τ 1 ρ 3 λ 1 τ 2 λ 2 τ 1 η 3 η 3 ρ 2 λ 1 τ 3 λ 3 τ 1 η 3 µ 3 + τ 1 λ 2 ρ 3 λ 3 ρ 2 η 3 µ 3 + ρ 1 λ 2 τ 3 λ 3 τ 2 η 3 µ 3 τ 2 λ 1 ρ 3 λ 3 ρ 1 η 3 µ 3 + τ 3 λ 1 ρ 2 λ 2 ρ 1 µ 3 µ 3 =. 4 η3 µ 3 2 ρ 3 λ 1 τ 2 λ 2 τ 1 +τ 3 λ 1 ρ 2 λ 2 ρ 1 η 3 µ 3 +τ 3 λ 1 ρ 2 λ 2 ρ 1 =. 41 Solve this equation, we obtain η 3 µ 3 = τ 3λ 1 ρ 2 λ 2 ρ 1 ρ 3 λ 1 τ 2 λ 2 τ 1 or η 3 µ 3 = 1. The solution η 3 µ 3 = 1 corresponds to the coplanar information, so we simply drop it. Since η 3 = α 1 β 4 = I 14, µ 3 β 1 α 4 η 3 = α 1 β 3 = I 13, we have µ 3 β 1 α 3 I 13 = I 14 = τ 3λ 1 ρ 2 λ 2 ρ 1 ρ 3 λ 1 τ 2 λ 2 τ 1. 42 This is the invariant relation when the six 3D points are on two adjacent planes. Compared with the work in [7] and [13], we derived the invariant using only elementary method from linear algebra which is much simpler. 4. Test Results We present test results for invariant 42 in this section. Six different views of a 3D scene were taken as depicted in Fig. 1 to Fig. 6. Nine critical points were marked as in Fig. 1. From these 9 points, 4 groups were selected to compute their invariant values. The test results are presented in Table 1. The small differences among the same invariants may be caused by computation precisions or camera distortions. They are quite satisfactory though. 5. Conclusion We have presented a simple derivation of a basic invariant relation relating the projective invariants of six 3D points and those of their 2D projection image points. Based on this 2
The Invariant Relations of 3D to 2D Projection of Point Sets Fig. 1: The first view of the 3D scene. Fig. 2: The second view of the 3D scene. Fig. 3: The third view of the 3D scene. 21
YuanBin et al. Fig. 4: The fourth view of the 3D scene. Fig. 5: The fifth view of the 3D scene. Fig. 6: The sixth view of the 3D scene. 22
The Invariant Relations of 3D to 2D Projection of Point Sets basic relation, new invariant relations with assumptions on the structure of the point configurations can be derived. We have given two such examples. It can be seen that they are quite clear and simple. This is the main attribute of the paper. Acknowledgments The author wishes to thank the editors and reviewers for their kindly help and useful suggestions. References [1] S. Carlsson, The double algebra: An effective tool forcomputing invariants in computer vision, In Applications of Invariance in Computer Vision, J.L. Mundy et al. Eds., Springer Verlag, pp. 145 164, 1994. [2] D. A. Forsyth, J. L. Mundy, A. P. Zisserman, C. Coelho, A. Heller, and C. A. Rothwell, Invariant descriptors for 3-D object recognition and pose, IEEE TPAMI, vol. 13, no. 1, 1991, pp. 971 991. [3] J. L. Mundy and A. Zisserman, Geometric Invariance in Computer Vision, MIT Press, Cambridge, 1992. [4] I. Weiss, Noise-resistant invariants of curves, IEEE TPAMI, vol. 15, no. 9, 1993, pp. 943 948. [5] I. Weiss, Geometric invariants and object recognition, International Journal Computer Vision, vol. 1, no. 3, 1993, pp. 27 231. [6] J. B. Burns, R. S. Weiss, and E. M. Riseman, View variation of point-set and line-segment features IEEE TPAMI, vol. 15, no. 1, 1993, pp. 51 68. [7] B. S. Song, K. M. Lee, and S. U. Lee, Model-based object recognition using geometric invariants of points and lines, Computer Vision and Image Understanding, vol. 84, no. 3, pp. 361 383, 21. [8] I. Weiss and M. Ray, Model-based recognition of 3D objects from single images, IEEE TPAMI, vol. 23, no. 2, pp. 116 128, 21. [9] E. B. Barrett, P. M. Payton, N. N. Haag, and M. H. Brill, General methods for determining projective invariants in imagery, CVGIP: Image Understanding, vol. 53, 1991, pp. 46 65. [1] R. Hartley, Invariants of lines in space, in Proceedings of Image Understanding Workshop, 1993, pp. 737 744. [11] L. Quan, Invariants of six points and projective reconstruction from three uncalibrated images, IEEE TPAMI, vol. 17, no. 1, 1995, pp. 34 46. [12] C. A. Rothwell, D. A. Forsyth, A. Zisserman, and J. L. Mundy, Extracting projective structure from single perspective views of 3D point sets, in Proceedings of IEEE International Conference on Computer Vision, Berlin, Germany, 1993, pp. 573 582. [13] Y. Zhu, L. D. Seneviratne, and S. W. E. Earles, New algorithm for calculating an invariant of 3D point sets from a single view, Image and Vision Computing 14, 1996, pp. 179 188. [14] Akihiro Sugimoto, Geometric invariant of non coplanar lines in a single view, in Proceedings of IEEE International Conference on Pattern Recognition, 1994, pp. 19 195. [15] P. C. Wayner, Efficiently using invariant theory for model-based matching, in Proceedings of IEEE Conference on Computer Vision and Pattern Recognition, 1991, pp. 473 478. [16] D. Weinshall, Model-based invariants for 3D vision, International Journal Computer Vision, vol. 1, no. 1, 1993, pp. 27 42. [17] I. Weiss and M. Ray, Model-Based Recognition of 3D Objects from One View, in Lecture Notes in Computer Science 146, vol. I, pp. 716 732, Springer, 1998. 23