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Bivariate Splines in Piecewise Constant Tension Kunimitsu Takahashi Masaru Kamada Graduate School of Science and Engineering Ibaraki University Hitachi, Ibaraki 316-8511, Japan Email: 14nm714l@vc.ibaraki.ac.jp Department of Computer and Information Sciences Ibaraki University Hitachi, Ibaraki 316-8511, Japan Email: masaru.kamada.snoopy@vc.ibaraki.ac.jp Abstract—An extension of the bivariate cubic spline on the uniform grid is derived in this paper to have different tensions in different square cells of the grid. The resulting function can be interpreted also as a bivariate extension of the univariate spline in piecewise constant tension which was applied to adaptive interpolation of digital images for their magnification and rotation. The bivariate function will hopefully make it possible to magnify and rotate images better and even to deform images into any shapes. A locally supported basis, which is crucial for the practical use of the bivariate functions, has not been constructed at the moment and its construction is left for the next step of study. I. I NTRODUCTION Interpolation of digital images for their magnification and rotation is a classical problem in image processing [1]. Once an image is represented by a bivariate function, we can resample it at any points in the domain to produce magnified and rotated images and even those deformed into arbitrary shapes [2], [3]. The bivariate cubic splines have been successfully applied [4] to image interpolation. The only drawback of the bivariate cubic splines is so-called “ringing” artifacts that may appear around “edges” where brightness of the image changes sharply. In order to suppress ringing around the edges adaptively, splines in piecewise constant tension have been derived [5] by extending the splines in tension [6], which used to have the tension fixed as a single constant over the entire domain [7], in the univariate setting, and applied to image interpolation [8], [9]. It was possible to magnify and rotate images by repeating the univariate spline interpolation in piecewise constant tension in the vertical and horizontal directions. But the quality of magnified and rotated images may have been degraded because the image was not represented by a bivariate function. Besides, it was not possible to deform images into arbitrary shapes for the same reason. In this paper, we aim at deriving splines in piecewise constant tension in the bivariate setting. The bivariate cubic spline is reviewed in the context of functional minimization in Section II. In Section III, the bivariate spline in piecewise constant tension is formulated and derived as a minimizer of a functional with piecewise constant tension incorporated. Its representation in terms of state transitions reveals how to c 978-1-4673-7353-1/15/$31.00 ⃝2015 IEEE evaluate the function for given tension and some Lagrange multipliers. A locally supported basis, which is crucial for the practical use, has not been obtained at the moment. A plan for its construction is mentioned in Section IV. II. B IVARIATE C UBIC S PLINES The bivariate cubic spline on the uniform grid in the plane with orthogonal x-y axes is usually represented in the literature in terms of its basis given by the Cartesian product of the cardinal cubic B-splines of x and those of y. In this section, we review its formulation in terms of a minimum principle as a preparation for its extension to allow for different tensions in different square cells of the grid. Let the bivariate cubic spline s(x, y) be a square-integrable function having square-integrable first and second derivatives sxy (x, y) and sxxyy (x, y). It should minimize ∫ ∫ ∞ −∞ ∞ −∞ 2 (sxxyy (x, y)) dxdy (1) subject to the interpolation constraints s(k, l) = skl , k, l = 0, ±1, ±2, · · · (2) for some fixed values skl at the grid points. Incorporating these constraints by the Lagrange multipliers λkl and the identity ∫ ∫ ∞ ∞ s(k, l) = −∞ −∞ δ(x − k)δ(y − l)s(x, y)dxdy (3) valid for the Dirac delta function δ, we have the unconstrained objective functional ∫ ∞ ∫ ∞ 2 Q[s] = (sxxyy (x, y)) dxdy ∫ ∞∫ ∞ ∑∑ − λkl δ(x − k)δ(y − l)s(x, y)dxdy + skl .(4) −∞ k l −∞ −∞ −∞ Denoted by ∆ the variational operator. Then the first variation of Q[s] is reduced to ∫ ∞∫ ∞ ∆Q[s] = 2 sxxyy (x, y)∆sxxyy (x, y) −∞ −∞ ∑∑ − λkl δ(x − k)δ(y − l)∆s(x, y)dxdy k l x→∞ [[2 sxxyy (x, y)∆sxy (x, y)]x→−∞ ]y→∞ y→−∞ = x→∞ −[[2 sxxxyyy (x, y)∆s(x, y)]x→−∞ ]y→∞ y→−∞ ∫ ∞∫ ∞ + 2 sxxxxyyyy (x, y)∆s(x, y) −∞ −∞ ∑∑ − λkl δ(x − k)δ(y − l)∆s(x, y)dxdy ∫ = ∞ ∫k ∞ l ( 2 sxxxxyyyy (x, y) −∞ ) ∑∑ − λkl δ(x − k)δ(y − l) ∆s(x, y)dxdy (5) l by means of integration by parts and because ∆s(x, y) → 0 and ∆sxy (x, y) → 0 at x → ±∞ and y → ±∞ for the square-integrable s(x, y) and sxy (x, y). Equating (5) to zero, we have the Euler-Lagrange equation ∑∑ 0 = 2 sxxxxyyyy (x, y) − λkl δ(x − k)δ(y − l). (6) k l Its four-fold integration by each of x and y shows that s(x, y) should be in the form s(x, y) = ∑∑ λkl k l 2 We apply the piecewise constant tension p(x, y) = Pkl , (x, y) ∈ [k, k + 1) × [l, l + 1), This functional falls back to the original (1) in the case p(x, y) ≡ 0. On the other hand, as p(x, y) gets large, the first term dominates in (10) and s(x, y) approaches the simple polyhedral interpolation that is a minimizer of the integral of squared first derivative under the interpolation constraints (2). We seek such s(x, y) that minimizes (10) subject to (2) among the square-integrable functions with square-integrable first and second derivatives. B. Euler-Lagrange equation In the same way as Section II, we can incorporate the constraints (2) into (10) to have the unconstrained objective functional ∫ ∞∫ ∞ 2 2 (p(x, y))2 (sxy (x, y)) +(sxxyy (x, y)) dxdy −∞ −∞ ∑∑ ∫ ∞ ∫ ∞ − λkl δ(x − k)δ(y − l)s(x, y)dxdy + skl . (11) k (◦ − •)+ = ◦−• 0 if ◦ > • otherwise −∞ −∞ l −∞ +g3 (x)y 3 + g2 (x)y 2 + g1 (x)y + g0 (x), (7) { −∞ Its first variation is reduced to ∫ ∞∫ ∞ 2(p(x, y))2sxy (x, y)∆sxy (x, y) (x − k)3+ (y − l)3+ +f3 (y)x3 + f2 (y)x2 + f1 (y)x + f0 (y) where (9) which can be a different non-negative constant in each square cell of the grid, to s(x, y) by extending the functional (1) to ∫ ∞∫ ∞ 2 2 (p(x, y))2 (sxy (x, y)) + (sxxyy (x, y)) dxdy. (10) −∞ −∞ k A. Functional to be minimized −∞ +2sxxyy (x, y)∆sxxyy (x, y) ∑∑ λkl δ(x − k)δ(y − l)∆s(x, y)dxdy − k (8) and fi (x) and gi (y), (i = 0, 1, 2, 3) are arbitrary four-times differentiable functions of x and y, respectively. Choosing cubic polynomials for fi (x) and gi (y), we have s(x, y) being a cubic spline in x and also in y as one of the minimizers. The bivariate cubic spline represented in (7) in terms of the truncated power functions can also be represented in terms of the basis constructed as the Cartesian product of the cubic cardinal B-splines of x and those of y. III. B IVARIATE S PLINES IN P IECEWISE T ENSION We shall extend the functional (1) for the bivariate cubic splines to include a term that incorporates different tensions in different square cells of the grid. Then we will derive its minimizer which is to be called bivariate splines in piecewise constant tension. l 2 y→∞ = [[2 (p(x, y)) sxy (x, y)∆s(x, y)]x→∞ x→−∞ ]y→−∞ ∫ ∞∫ ∞ ( ) ∂2 − (p(x, y))2 sxy (x, y) ∆s(x, y)dxdy 2 ∂x∂y −∞ −∞ y→∞ +[[2 sxxyy (x, y)∆sxy (x, y)]x→∞ x→−∞ ]y→−∞ y→∞ −[[2 sxxxyyy (x, y)∆s(x, y)]x→∞ x→−∞ ]y→−∞ ∫ ∞∫ ∞ + 2 sxxxxyyyy (x, y)∆s(x, y) −∞ −∞ ∑∑ − λkl δ(x − k)δ(y − l)∆s(x, y)dxdy ∫ = k ∞ ∫ ∞( −∞ −∞ l ( ) ∂2 −2 ∂x∂y (p(x, y))2 sxy (x, y) ∆s(x, y) +2 sxxxxyyyy (x, y)∆s(x, y) ) ∑∑ − λkl δ(x − k)δ(y − l) ∆s(x, y)dxdy(12) k l by means of integration by parts and because ∆s(x, y) → 0 and ∆sxy (x, y) → 0 at x → ±∞ and y → ±∞ for the square-integrable s(x, y) and sxy (x, y). That yields the EulerLagrange equation ( ) ∂2 2sxxxxyyyy (x, y)−2 ∂x∂y (p(x, y))2sxy (x, y) ∑∑ = λkl δ(x − k)δ(y − l). (13) k l This differential equation includes p(x, y) being constant inside each cell of the grid and discontinuous at the cell borders. We shall handle this equation in the two kinds of domains: the open domain within each cell and the infinitesimal domains across the borders of adjacent cells, in the following subsections. We take a state-space representation of the solution inside the cells because of its compatibility with boundary conditions written in terms of derivatives at the borders. C. State transition inside each cell In each open square (k + x, l + y) for 0 < x, y < 1, the differential equation (13) falls back to the homogeneous one 2 sxxxxyyyy (k+x, l+y)−Pkl sxxyy (k+x, l+y) = 0 (14) since p(k + x, l + y) = Pkl according to (9) and because the delta functions in (13) are identically zero anywhere other than the grid lines. Denote w(k+x, l+y) = sxxyy (k+x, l+y). (15) Then (14) becomes 2 wxxyy (k+x, l+y)−Pkl w(k+x, l+y) = 0 (16) of which the general solution is represented in the form w(k + x, l + y) (√ ) (√ ) = akl cosh Pkl (x + y) +bkl cosh Pkl (x − y) ) (√ ) (√ + ckl sinh Pkl (x + y) +dkl sinh Pkl (x − y) . (17) Integrating (17) twice by each of x and y with the simplest choice of integral constants, we have a solution for (14) as s(k + x, l + y) (√ ) (√ ) 1 { = 2 akl cosh Pkl (x + y) + bkl cosh Pkl (x − y) Pkl (√ ) (√ )} + ckl sinh Pkl (x + y) + dkl sinh Pkl (x − y) +αkl xy + βkl x + γkl y + θkl . (18) The eight coefficients akl , bkl , ckl , dkl , αkl , βkl , γkl , and θkl determine s(k + x, l + y) within the open cell and also on its boundary since s(x, y) is assumed to be twice differentiable over the entire domain. Across the cell borders, the function should be connected to the next ones via the boundary values of s(k + x, l + y) and seven of its derivatives constituting a state vector. Let such a state vector and the coefficient vector be composed by sxxxyyy (x+k, y+l) akl bkl sxyy (x+k, y+l) ckl sxxy (x+k, y+l) sxxyy (x+k, y+l) and ckl = dkl , (19) s(k+x, l+y)= sxy (x+k, y+l) αkl sx (x+k, y+l) βkl sy (x+k, y+l) γkl s(x+k, y+l) θkl respectively. Then they are linearly related to each other by s(k+x, l+y) = Ψkl (x, y)ckl , (20) where the matrix Ψkl (x, y) and its inverse Ψ−1 kl (x, y) are calculated with labor as follows: √ Pkl cosh(√Pkl (x + y)) √ 1 sinh( Pkl (x + y)) Pkl √ √ 1 sinh( Pkl (x + y)) Pkl √ cosh(√Pkl (x + y)) 1 Ψkl (x, y) = Pkl cosh(√Pkl (x + y)) 1 √ 3 sinh( Pkl (x + y)) ( Pkl ) √ √1 ( Pkl )3 sinh(√Pkl (x + y)) 1 Pkl (x + y)) 2 cosh( Pkl √ √ −Pkl cosh(√Pkl (x − y)) Pkl sinh(√Pkl (x + y)) √ 1 sinh( Pkl (x − y)) √ 1 cosh( Pkl (x + y)) Pkl Pkl √ √ 1 √ 1 cosh( Pkl (x + y)) − √P sinh( Pkl (x − y)) P kl kl √ √ sinh(√Pkl (x + y)) cosh(√Pkl (x − y)) 1 − P1kl cosh( Pkl (x − y)) Pkl sinh(√Pkl (x + y)) √ 1 1 √ sinh( Pkl (x − y)) (√P )3 cosh( Pkl (x + y)) ( Pkl )3 kl √ √ 1 √ − ( P )3 sinh( Pkl (x − y)) (√P1 )3 cosh( Pkl (x + y)) kl kl √ √ 1 1 Pkl (x − y)) Pkl (x + y)) 2 cosh( 2 sinh( Pkl Pkl √ −Pkl sinh(√Pkl (x − y)) 0 0 0 0 √ 1 cosh( Pkl (x − y)) 0 0 0 0 Pkl √ 1 − √P cosh( Pkl (x − y)) 0 0 0 0 kl √ sinh(√Pkl (x − y)) 0 0 0 0 (21) − P1kl sinh( Pkl (x − y)) 1 0 0 0 √ 1 √ cosh( Pkl (x − y)) y 1 0 0 ( Pkl )3 √ − (√P1 )3 cosh( Pkl (x − y)) x 0 1 0 kl √ 1 sinh( P (x − y)) xy x y 1 2 kl P kl −1 Ψkl (x, y) = √ 1 Pkl (x + y)) 2Pkl cosh( √ 1 − 2Pkl cosh( Pkl (x − y)) √ − 2P1kl sinh( Pkl (x + y)) √ 1 2Pkl sinh( Pkl (x − y)) − P12 kl y 2 Pkl x 2 Pkl − Pxy2 kl √ √ − √P2 kl sinh( Pkl (x + y)) √ −√ P2 kl sinh( Pkl (x − y)) √ Pkl cosh( Pkl (x + y)) √2 √ Pkl 2 cosh( Pkl (x − y)) 0 − P1kl 0 √ Pkl Pkl (x + y)) 2 sinh( √ Pkl sinh( P (x − y)) √2 √ kl Pkl cosh( Pkl (x + y)) √2 √ − P2 kl cosh( Pkl (x − y)) x Pkl y Pkl √ √ −√ 0 0 0 0 0 − P12 kl 0 0 0 0 1 −y −x xy 0 0 0 0 0 1 0 −x 0 0 0 0 0 0 1 −y 0 0 0 0 0 0 0 1 s(k+x, l+y)=Ψkl (x, y)Ψ−1 kl (x0 , y0 )s(k+x0 , l+y0 ), 1 ∑ ∑ λij+f k+g(l+y)+h(26) 2 sxxxyyy (k + 0, l + y) 1∑ ∑ 2 = Pkl sxy (k, l + y) + λij+f k+g(l+y)+h. (27) 2 i≤k j≤l (22) s(k+x, l+y)=Ψkl (x, y)Ψ−1 kl (0, 0)s(k+0, l+0) Subtracting (26) from (27), we have sxxxyyy (k + 0, l + y) − sxxxyyy (k − 0, l + y) ( ) 1∑ 2 2 λkj = Pkl − P(k−1)l sxy (k, l + y) + 2 (28) j≤l (23) where 0 < x, y, x0 , y0 < 1. Given the state vector s(k + 0, l + 0) = limx0 →+0,y0 →+0 s(k + x0 , l + y0 ) at the left bottom corner (k + 0, l + 0) of the cell, for example, we can evaluate the state at anywhere inside the same cell by (24) for 0 < x, y < 1. Here we used the fact that every entry of the matrix Ψ−1 kl (x, y) is a continuous function of x and y so that Ψ−1 (+0, +0) = Ψ−1 kl kl (0, 0). The same is true of Ψkl (x, y) so that we can evaluate the state even at x = 1 − 0 and y = 1 − 0 by using Ψkl (1, y) and Ψkl (x, 1) instead of Ψkl (1 − 0, y) and Ψkl (x, 1 − 0), respectively. D. State transition across the cell borders Integrating the differential equation (13) once by each of x and y, we have 2 = P⌊x⌋⌊y⌋ sxy (x, y) + 2 = P(k−1)l sxy (k, l + y) + and Using the above state transition matrices, we can evaluate the state vector at any point (k + x, l + y) inside the cell from the state vector at any other point (k + x0 , l + y0 ) inside the same cell by sxxxyyy (x, y) sxxxyyy (k − 0, l + y) i≤k−1 j≤l − P1kl 1 2 cosh(√Pkl (x + y)) 1 2 cosh(√Pkl (x − y)) − 12 sinh(√Pkl (x + y)) − 12 sinh( Pkl (x − y)) The two ways of crossing the borders are investigated in the following. For 0 < y < 1, (25) gives 1 ∑ ∑ λkl + f x + gy + h,(25) 2 k≤⌊x⌋ l≤⌊y⌋ where f , g and h are integral constants and ⌊•⌋ denotes the largest integer not exceeding •. We recall sxy (x, y) is continuous over the entire domain because s(x, y) is assumed to be twice differentiable with respect to x and y. Then (25) means that sxxxyyy (x, y) is a continuous function over the entire domain except at the cell borders where it is 2 discontinuous with a finite height of jump caused by P⌊x⌋⌊y⌋ ∑ ∑ and k≤⌊x⌋ l≤⌊y⌋ λkl . That also means that all the other state variables in s(x, y) being integrals of sxxxyyy (x, y) are continuous over the entire domain. So we have only to take care of the change of sxxxyyy (x, y) in the state transitions across the cell borders. which tells how to update the state variable sxxxyyy (k−0, l+y) to sxxxyyy (k + 0, l + y) when crossing the border x = k from left to right. In the same way, for 0 < x < 1, subtracting sxxxyyy (k + x, l − 0) 2 = Pk(l−1) sxy (k + x, l) + 1∑ ∑ λij+f (k+x)+gl+h(29) 2 i≤k j≤l−1 from sxxxyyy (k + x, l + 0) 1∑ ∑ 2 λij+f (k+x)+gl+h, (30) = Pkl sxy (k + x, l) + 2 i≤k j≤l we have sxxxyyy (k + x, l + 0) − sxxxyyy (k + x, l − 0) ( ) 1∑ 2 2 = Pkl − Pk(l−1) sxy (k + x, l) + λil 2 (31) i≤k which tells how to update the state variable sxxxyyy (k + x, l − 0) to sxxxyyy (k + x, l + 0) when crossing the border y = l upward. By the notation 1 1 0 0 0 P 0 0 0 2 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 Φ(P ) = and u = 0 , (32) 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 we can write the state transitions instructed by (28) and (31) as follows: ∑ 2 2 s(k+0, l+y) = Φ(Pkl −P(k−1)l )s(k−0, l+y)+u λkj ,(33) j≤l ∑ 2 2 s(k+x, l+0) = Φ(Pkl −Pk(l−1) )s(k+x, l−0)+u i≤k λil . (34) E. Evaluation of bivariate splines in piecewise tension Making use of the state transitions inside the cell (24) and those across the cell borders (33) and (34), we can evaluate the state s(k+x, l+y) in the given piecewise constant tension Pij (i, j = 0, ±1, ±2, · · · ) for a given series of parameters λij (i, j = 0, ±1, ±2, · · · ). The function value s(k +x, l +y) is included in the state s(k+x, l+y) at its bottom row. Suppose that we know the state s(k +0, l+0) at the left bottom corner of the cell (k, k + 1) × (l, l + 1). Then the state at (k + x, l + y) for 0 < x, y < 1 inside the cell can be evaluated by (24), i.e., s(k+x, l+y)=Ψkl (x, y)Ψ−1 kl (0, 0)s(k+0, l+0). As a special case, the state at the right bottom corner is given by s(k+1−0, l+0)=Ψkl (1, 0)Ψ−1 kl (0, 0)s(k+0, l+0). In order to cross the border x = k + 1 to the right, we use the state transition (33) to evaluate ∑ 2 2 s(k+1+0, l+0)= Φ(P(k+1)l −Pkl )s(k+1−0, l+0)+u λ(k+1)j j≤l to arrive at the left bottom corner of the next cell (k + 1, k + 2) × (l, l + 1) on the right. In the same way, the state at the left top corner is given by s(k+0, l+1−0)=Ψkl (0, 1)Ψ−1 kl (0, 0)s(k+0, l+0). In order to cross the border y = l + 1 upward, we use the state transition (34) to evaluate ∑ 2 2 λi(l+1) s(k+0, l+1+0)= Φ(Pk(l+1) −Pkl )s(k+0, l+1−0)+u i≤k to arrive at the left bottom corner of the next upper cell (k, k+ 1) × (l + 1, l + 2). The state at the right top corner given by −1 s(k+1−0, l+1−0)=Ψkl (1, 1)Ψkl (0, 0)s(k+0, l+0) is moved to s(k+1+0, l+1−0) 2 2 = Φ(P(k+1)l −Pkl )s(k+1−0, l+1−0)+u ∑ λ(k+1)j j≤l+1 by (33) to cross the border x = k + 1, and then to s(k+1+0, l+1+0) 2 2 = Φ(P(k+1)(l+1) −P(k+1)l )s(k+1+0, l+1−0)+u ∑ λi(l+1) i≤k+1 by (34) to cross the border y = l + 1 to reach the state at the left bottom corner of the next upper right cell (k + 1, k + 2) × (l + 1, l + 2). Applying the above transitions successively for larger k and l, we can evaluate s(k +x, l+y) for any k, l, x and y if we are given the series of parameters λkl . However, λkl will never be given in practice but must be determined so that s(k, l) = skl for the given data skl . IV. T OWARD LOCALLY SUPPORTED BASIS It is crucial for practical use of the splines to have a locally supported basis like the Cartesian product of the cubic cardinal B-splines of x and those of y for the bivariate cubic splines. It was possible in the univariate case to construct a locally supported basis for the splines in piecewise tension by making use of the state transitions [9]. Construction of a locally supported basis in the bivariate case has not yet been made but only planned as follows: Consider the problem of constructing a basis function with the support [0, 4] × [0, 4] by choosing a set of 25 possibly non-zero λkl for (k, l) ∈ [0, 4] × [0, 4]. We start from the zero state s(−0, −0) = 0 just outside the support. Then the 2 state transitions (33) and (34) give s(+0, +0) = Φ(P00 − 2 2 2 P0 −1 )Φ(P0 −1 − P−1 −1 )s(−0, −0) + uλ00 = uλ00 . Successive application of the state transitions described in the last subsection for four times will give the terminal states s(4 + 0, +0), s(+0, 4 + 0) and s(4 + 0, 4 + 0) represented as linear combinations of some eight-dimensional vectors with the 25 coefficients λkl . Fixing λ00 = 1 and requesting s(4 + 0, +0) = 0, s(+0, 4 + 0) = 0 and s(4 + 0, 4 + 0) = 0, we may determine the other 24 coefficients λkl so that the states be identically zero outside the support [0, 4] × [0, 4]. V. C ONCLUSION We derived the expression of bivariate splines in piecewise constant tension in terms of state transitions of the characterizing Euler-Lagrange equation. The derived function is an extension of the bivariate cubic splines to piecewise constant tensions and is also an extension of the univariate spline in piecewise constant tension to the bivariate case. A plan for constructing a locally supported basis was briefly described. ACKNOWLEDGMENT This work was partially supported by the JSPS grant-in-aid #26420409. R EFERENCES [1] H. S. Hou and H. C. Andrews, Cubic splines for image interpolation and digital filtering, IEEE Trans. Acoustic, Speech, Signal Process., 26, 508–517, 1978. [2] P. Thévenaz,T. Blu and M. Unser, Interpolation revisited, IEEE Trans. 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