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15th May 2009, 23:54  #1  Link 
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on the theory of the Spline16/36/64 resizers
I contacted the author of Panorama Tools (this is where the Spline16/36/64 appear to come from) and asked him about the coefficient of the Spline polynomials. He sent me a short reply which i have worked out in detail:
Let me first say that the polynomials are not the interpolating polynomials, but they are certain blending polynomials as will be explained below. Let's consider Spline4 for simplicity. Consider one interpolating (cubic) spline for the interval [0,1]: Code:
S(x) = a[3]*x^3 + a[2]*x^2 + a[1]*x + a[0] for x in [0,1] Ok, let's set z0=z1=0. In other words, the second derivatives at the end points are zero. Under this condition (a[j])_j are fixed by (y0,y1). What happens if you replace y0 by y1 and y1 by y0? In that case the polynomial S will be mirrored at x=1/2, since (1) S is fixed by (y0,y1) and (2) z0=z1. A consequence of this is that: Code:
S[y0;y1](x) = S[y1;y0](1x) It turns out that S can be expressed as a linear combination of the control points (y0,y1): Code:
S(x) = w0(x) * y0 + w1(x) * y1 Code:
y0 = S(0) = w0(0) * y0 + w1(0) * y1 => w0(0)=1, w1(0)=0 y1 = S(1) = w0(1) * y0 + w1(1) * y1 => w0(1)=0, w1(1)=1 Code:
w0(x)+w1(x)=1 Because of the mirroring property we also have that Code:
S(x) = w0(1x) * y1 + w1(1x) * y0 Code:
w1(x) = w0(1x) w0(x) = w1(1x) Code:
w1(x) = w0(1x) > w0[x] Solving (w0,w1) using Maple: The boundary conditions result in: Code:
> eqn:={a0=y0, a3+a2+a1+a0=y1, 2*a2=0,6*a3+2*a2=0}; Code:
> solve(eqn,{a0,a1,a2,a3}); {a0 = y0, a2 = 0, a3 = 0, a1 = y0 + y1} Code:
> a0s:=y0; a1s:=y0+y1; a2s:=0; a3s:=0; > S:=a3s*x^3 + a2s*x^2 + a1s*x + a0s; S:= (y0 + y1) x + y0 > w0:=coeff(S,y0); w1:=coeff(S,y1); w0 := x + 1 w1 := x Two additional remarks: 1) In general for Spline k^2, we start with k1 interpolating polynomials. They are located at (k/2+1,k/2+2), ..., (k/21,k/2). The bending polynomials are calculated from the polynomial located at [0,1], which is expressed as a linear combination of (y0, ..., y(k1)). The boundary conditions are: Code:
y[1](k/2+1) = y0 y[1](k/2+2) = y1 y[2](k/2+2) = y1 ... y[k2](k/21) = y(k2) y[k2](k/21) = y(k2) y[k1](k/2) = y(k1) I have no idea how to derive a general formula. I can recalculate the known coefficients for Spline16 and Spline36. However for Spline64 I get slightly different ones. So, i assume that the known ones for Spline64 are simply wrong. edit Wilbert: I made a mistake with Spline64. Its coefficients are correct now. Last edited by Wilbert; 16th May 2009 at 22:52. 
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