4 ms·
That's exactly what I assume. If I know 90% of the image, but need to brute force 10%, I just reduce the possible field of values by an order of magnitude. The
by IgorPartola 9y ago
That's exactly what I assume. If I know 90% of the image, but need to brute force 10%, I just reduce the possible field of values by an order of magnitude. The better I know what the image was to begin with, the easier it gets. That's exactly why poor RNGs are bad: I can brute force the value by knowing your method + approximately your input.
And if I can affect the image in a way that makes it even more predictable (black sheet over the camera, bright white light into the camera), it gets even easier.
- vokep 9y agobut unless the image is very low resolution, chances are sha256 is a smaller space no?
- silotis 9y agoDoing a little back-of-the-envelope math, a 1080p image has a total of 1920x1080x3 = 6,220,800 subpixels. That means each subpixel needs to have less than 0.00004 bits of entropy for there to be less than 256 bits in a single image. Surely a typical consumer CCD will introduce more noise than that. A potential catch is that the noise is likely correlated to a degree. Still, it's not clear that even a webcam pointed at a black sheet of paper is insufficient as a source of entropy.
- tjoff 9y ago10% of 1 megapixels divided by 24 is 4166. So, even if you only consider the least significant bit of one of the three subpixels in 10% of the image you still have 16 times more information than what you can represent in a SHA256 hash from a measly 1 megapixel image. I think it is valid to question how you could reduce the search space enough to be more beneficial than brute forcing SHA256.