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Related question: What are the practically relevant differences between various image resampling methods?

Bilinear and bicubic interpolation for image resampling seem to be fairly common, but biquadratic is in my experience rarely heard of. To be sure, it's available in some programs and libraries, but generally it doesn't seem to be popular. This is further evidenced by there being Wikipedia articles for bilinear and bicubic interpolation, but none for biquadratic. Why is this the case?

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3 Answers 3

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Bilinear and biquadratic interpolation gives you a $C^0$ interpolating function. That is, a function that is continuous but has a discontinuous first derivative. On the other hand, Bicubic interpolation gives you a $C^1$ interpolating function. That is, a function that is continuous and has a continuous first derivative (the second derivative is discontinuous though). So you are not gaining anything in terms of "smoothness" (continuity of higher derivatives) by using biquadratic over bilinear, just more complexity. To get a smoother interpolation, you have to step up from bilinear to bicubic.

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  • $\begingroup$ Are you sure about this? If I understand correctly, quadratic interpolation (or at least one case of it) is equivalent to convolving the original image point samples (spaced 1 unit apart) by a box filter (of width and height 1) three times, or equivalently, convolving with a kernel constructed as the convolution of three such box filters. That kernel is the piecewise quadratic function q(x) = abs(x)<=.5 ? .75 - x**2 : abs(x)<=1.5 ? (abs(x)-1.5)**2/2 : 0 which can be plotted in gnuplot or other program: i.sstatic.net/9eMko.png ; it is evidently $C^1$. $\endgroup$
    – Don Hatch
    Commented Dec 19, 2023 at 10:54
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You want to have an interpolation with a equal number of weights on both sides of the points you want to interpolate in between. So you choose either one or two weights on each side, resulting in an interpolation of two (linear) or four (cubic) points.

An quadratic interpolation would need three points, which would only make sense at the border of a grid, where you can choose only one point on one side and two on the other.

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More expensive computationally with marginal results. Most images don't require such interpolation technique. You can also look at this question: Higher order spline interpolation

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