If you know in advance that you’re going to want to multiple samples without overlaps, easiest is to do random.shuffle() on list(range(100)) (Python 3 – can skip the list() in Python 2), then peel off slices as needed. s = list(range(100)) random.shuffle(s) first_sample = s[-10:] del s[-10:] second_sample = s[-10:] del s[-10:] # etc Else … Read more
Poisson distribution Here’s how Wikipedia says Knuth says to do it: init: Let L ← e^(−λ), k ← 0 and p ← 1. do: k ← k + 1. Generate uniform random number u in [0,1] and let p ← p × u. while p > L. return k − 1. In Java, that would … Read more
The only deterministic, polynomial-time algorithm for primality testing I know of is the AKS primality test (http://en.wikipedia.org/wiki/AKS_primality_test). However, there are a lot of very good randomized primality tests that are fast and have extremely good probability of success. They usually work by finding whether the number is composite with exponentially good probability, so they’ll either … Read more
Equal hash means equal file, unless someone malicious is messing around with your files and injecting collisions. (this could be the case if they are downloading stuff from the internet) If that is the case go for a SHA2 based function. There are no accidental MD5 collisions, 1,47×10-29 is a really really really small number. … Read more
Duplicate of Expected collisions for perfect 32bit crc The answer referenced this article: http://arstechnica.com/civis/viewtopic.php?f=20&t=149670 Found the image below from: http://preshing.com/20110504/hash-collision-probabilities
Here’s a useful trick 🙂 function randomWithProbability() { var notRandomNumbers = [1, 1, 1, 1, 2, 2, 2, 3, 3, 4]; var idx = Math.floor(Math.random() * notRandomNumbers.length); return notRandomNumbers[idx]; }
Note you can generate the sequence by recursively generating all combinations with the first element, then all combinations without. In both recursive cases, you drop the first element to get all combinations from n-1 elements. In Python: def combination(l, r): if r == 0: yield [] elif len(l) == r: yield l else: for c … Read more
Yes, you can. Since GUIDs are 128 bits long, there is admittedly a minute possibility of a clash—but the word “minute” is nowhere near strong enough. There are so many GUIDs that if you generate several trillion of them randomly, you’re still more likely to get hit by a meteorite than to have even one … Read more
Generate a uniformly distributed random number. Iterate through your list until the cumulative probability of the visited elements is greater than the random number Sample code: double p = Math.random(); double cumulativeProbability = 0.0; for (Item item : items) { cumulativeProbability += item.probability(); if (p <= cumulativeProbability) { return item; } }
From Wikipedia: Studies by IBM in the 1990s suggest that computers typically experience about one cosmic-ray-induced error per 256 megabytes of RAM per month.[15] This means a probability of 3.7 × 10-9 per byte per month, or 1.4 × 10-15 per byte per second. If your program runs for 1 minute and occupies 20 MB … Read more