> From: zfs-discuss-boun...@opensolaris.org [mailto:zfs-discuss- > boun...@opensolaris.org] On Behalf Of Peter Taps > > Thank you for sharing the calculations. In lay terms, for Sha256, how many > blocks of data would be needed to have one collision?
There is no point in making a generalization and a recommended "best practice." Because it's all just a calculation of probabilities and everyone will draw the line differently. My personal recommendation is to enable verification at work, just so you can never be challenged or have to try convincing your boss about something that is obvious to you. The main value of verification is that you don't need to explain anything to anyone. Two blocks would be needed to have one collision, at a probability of 2^-256 which is 10^-78. This is coincidentally very near the probability of randomly selecting the same atom in the observable universe twice consecutively. The more blocks, the higher the probability, and that's all there is to it (unless someone is intentionally trying to cause collisions with data generated specifically and knowledgeably for that express purpose). When you start reaching 2^128 (which is 10^38) blocks it becomes likely you have a collision. Every data pool in the world is someplace in between 2 and 2^128 blocks. Smack in the middle of the region where the probability is distinctly more likely than randomly selecting the same atom of the universe twice, and less likely than an armageddon caused by earth asteroid collision. _______________________________________________ zfs-discuss mailing list zfs-discuss@opensolaris.org http://mail.opensolaris.org/mailman/listinfo/zfs-discuss
