3 ms·
That doesn’t address the issue at all. The issue is that discarding small changes in any color space is fundamentally lossy, so the algorithm isn’t lossless.
by oivey 1y ago
That doesn’t address the issue at all. The issue is that discarding small changes in any color space is fundamentally lossy, so the algorithm isn’t lossless.
- arcastroe 1y agoWell yes.. I agree. But my comment was addressing parent's point that pure blue and pure red are currently considered close together since they share the same greyscale values.