4 ms·
Not really, because in each image parts of the watermark will be in different places. You don't even need to warp the watermark to achieve that, it's enough to
by pps43 9y ago
Not really, because in each image parts of the watermark will be in different places. You don't even need to warp the watermark to achieve that, it's enough to randomly reposition it.
- Retric 9y agoSupose you have 4 points on a box. If you randomly increase or decrease the location of each point by some random value say +-3% then averaged 100,000 of those boxes, you get a smaller box(1) surrounded by a larger increasingly blurry box. This is true even if no single picture shows that single smaller box, but it will show up on the average. (1) Rather than a true box the midpoints are going to be slightly buldging, but with a large sample set it's very close to a box. Sure, you can get into cat and mouse games ever stranger geometry. But, the water mark is limited by how much it distracts from the image.
- pps43 9y agoThe article shows the result of this approach, and it clearly does not work well.
- Retric 9y agoAs I said you need a new step: "Estimating that transformation for a specific image should be fairly straightforward". It does show a good average and simply moving the watermark was easily reversible suggesting detection of a watermark in an image was easy.
- Dylan16807 9y agoIt has crisp edges. If the same watermark is repositioned it's trivial to figure out the adjustment and remove it. Warping is harder but still largely doable by finding each part of the watermark's exact location.