4 ms·
With k = 1/x, the inverse weight interpolation is equivalent to bilinear. Perhaps it's better to show the result for k = 1/x^2. Neat aside, k = 1/x^∞ seems to
by GrantMoyer 2y ago
With k = 1/x, the inverse weight interpolation is equivalent to bilinear. Perhaps it's better to show the result for k = 1/x^2.
Neat aside, k = 1/x^∞ seems to be equivalent to nearest neighbor. (please excuse the abuse of notation)
- mark-r 2y agoBilinear isn't 1/x, it's 1-x.
- GrantMoyer 2y agoRight, but the more complicated expression ends up simplifying to the bilinear expression.
- MereInterest 2y agoThe abuse of notation is a useful one, as it also captures the sense that as the exponent approaches infinity, the interpolation approaches nearest neighbor interpolation. This seems similar to the “max norm”, described in the wiki link below, which also uses infinity to describe the relationship with other distance formulas. https://en.wikipedia.org/wiki/Uniform_norm https://en.wikipedia.org/wiki/Uniform_norm
- lloeki 2y agoCould be nice if there were a log-scale slider to interactively vary the exponent from 1 to ∞-ish and see how that affects the frame-cropped preview.