6 ms·
I am looking for a simple random number generator that fills a given amount of slots. Say it is initialized with "size=5" then it might output: 3,5,2,1,4 Is
by TekMol 6y ago
I am looking for a simple random number generator that fills a given amount of slots.
Say it is initialized with "size=5" then it might output:
3,5,2,1,4
Is there something like this?
It does not need much statistical resemblence to randomness. Just look kind of random to the eye. And the code should be short. A few lines of Javascript or so.
Maybe one approach might be to just loop through the sequence (1,2,3,4,5) and xor the number with some other number? Maybe with 0101010...?
- thewakalix 6y agoAre you talking about generating a random permutation?
- TekMol 6y agoI don't think so. Maybe a "random sequence" could be a name for it. But not sure.
- thewakalix 6y agoWhat I mean is, in your example, do you need to have precisely the numbers 1, 2, 3, 4, and 5, but in any order? Or is the constraint something else?
- TekMol 6y agoAh yes, you are right. In that sense, it is a random permutation of the numbers from 1 to max. Yes, when initialized with "size=5" the numbers in the output must be precisely 1,2,3,4,5 but in random looking order. And I don't want to store the whole sequence. I want to say gimmeTheNumberAtPosition(x) and it return just that number.
- FabHK 6y agoGenerating random permutations is quite a different problem from creating random numbers. The standard algo for random permutations is the Fisher-Yates Shuffle. https://en.wikipedia.org/wiki/Fisher–Yates_shuffle https://en.wikipedia.org/wiki/Fisher–Yates_shuffle ETA: As you don't want to store the permutation, you might want to pick a number randomly from 1 to n!, and then generate the permutation on the fly up to the desired element, using the techniques outlined here: https://stackoverflow.com/questions/29236556/how-can-i-calculate-n-th-permutation-or-tell-the-lexicographic-order-of-a-given https://stackoverflow.com/questions/29236556/how-can-i-calcu...
- SuchAnonMuchWow 6y agoStoring a number of the order of magnitude of n! will take the same amount of memory than storing a permutation of n elements, so what's the point ?
- FabHK 6y agoGood point. Storing a number of magnitude n! will take n log n bits, and storing n numbers up to n will also take n log n bit. In other words, good old Fisher Yates, storing the resulting permutation, and a simple lookup is probably the way to go.
- HelloNurse 6y agoIf one really wants to do it the slow and hard way, it's possible to compute a random permutation of n elements one element at a time with n bits of storage (less with fairly generic compression techniques): just remember the set of numbers that have been produced so far and sequentially pick one of the remaining ones. You'll waste some combination of additional memory (for arithmetic decoding state) and random bits (for rejected samples), but materializing a large randomly permuted vector should be slower.
- kinos 6y agofor JS you can probably just do `[...Array(n).keys()].sort(()=>Math.random() - 0.5)` ?
- matsemann 6y agoHaving an unstable comparator might wreak havoc and maybe not give a proper distributed space of possible results? Edit: yes, here is an example (in the comments) of how biased this is: https://stackoverflow.com/a/18650169/923847 https://stackoverflow.com/a/18650169/923847
- anderskaseorg 6y agoThat is exactly the algorithm that Microsoft used to bias its purportedly random “browser choice” screen. Don’t use it. https://www.robweir.com/blog/2010/02/microsoft-random-browser-ballot.html https://www.robweir.com/blog/2010/02/microsoft-random-browse... The Fisher–Yates shuffle is the right way to shuffle an array in an unbiased way. https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle https://en.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
- isaac21259 6y agoIf your value of size is a power of two you could just encrypt x with a block cipher where the block size is equal to size and the key is your seed.
- nightcracker 6y agoYou can construct arbitrarily-sized permutations with log(n) time random access using Sometimes-Recurse Shuffle: https://eprint.iacr.org/2013/560.pdf https://eprint.iacr.org/2013/560.pdf.
- resolvent 6y agoSo you want a mapping of [1, n] to a permutation of [1, n] without having to generate the list [1, n] and shuffling it. An affine transformation modular n should work. Instead of [1, n], look at [0, n). Find a value `a` in [0, n) such that gcd(a, n) = 1 and pick a random integer b in [0, n). Then the `random` number at each position is `a * x + b mod n`. This is simple and fast, but is not secure at all. You can solve for a, b by solving the linear congruence. It also does not generate every permutation of `n`. For n = 5, only 20 sequences can be found out of 5! = 120.
- johndough 6y agoThe value `a` has to fulfill some more properties: https://en.wikipedia.org/wiki/Linear_congruential_generator#c_%E2%89%A0_0 https://en.wikipedia.org/wiki/Linear_congruential_generator#... Since the "randomness" of the permutation is not that great, I was looking for something better, but could not find anything. The closest I got was https://en.wikipedia.org/wiki/Xorshift#xoshiro_and_xoroshiro https://en.wikipedia.org/wiki/Xorshift#xoshiro_and_xoroshiro which only works for powers of two. A workaround would be to choose the next larger power of two and reject all random values which are smaller than `n`, but that introduces unpredictable latency and destroys the cool jump-ahead feature.
- resolvent 6y agoThe LCG is the improved extension which causes you to have to compute values x_0 = seed to x_n in order to calculate x_{n+1}. What I am talking about is just a mapping of index x -> f(x) which is what the OP seemed to have wanted. In that case, `a` only needs to be relatively prime. This cannot account for every permutation of n since the number of values relatively prime to n, the totient, has an upper bound of n, which is the possible values for a. The number of values for b is also n, so at most n^2 possible sequences are generated, which is less than n! for n > 3. For example, with n = 5: Let a = 3 and b = 2. x = [0, 1, 2, 3, 4], a * x = [0, 3, 6, 9, 12], a * x + b = [2, 5, 8, 11, 14], a * x + b mod n = [2, 0, 3, 1, 4]
- e12e 6y agoI guess you could predictably enumerate the permutations, so that you could lookup with parameters size/n, position/i, seed/r - where r is the permutation. For size=1, all parameters are 1, and only result is 1. For size 2, r can be 1 or 2, naming the permutations [1,2];[2,1] - and eg: rand_at(n=2,r=2,i=2) would return 2, but i=1, would return 2. I'm not sure how I'd implement this - I suppose it might be possible to generate a predictable "walk" based on n/r/i? But for sizes less than, say, a million i would think that a shuffled array would be easier? r would need to be in the range 1..n!
- CipherThrowaway 6y agoWhat you want is a cipher from 1..N to 1..N. This problem is often approached as an encryption problem known as Format Preserving Encryption. The Sometimes Recurse Shuffle linked elsewhere in this thread is one solution but the most foundational and accessible paper on this is Black and Rogaway's "Ciphers with Arbitrary Domains" The Generalized Feistel Network (Method 3) is fast and easy to implement. https://web.cs.ucdavis.edu/~rogaway/papers/subset.pdf https://web.cs.ucdavis.edu/~rogaway/papers/subset.pdf
- wizeman 6y agoLook into linear feedback shift registers, they are very simple and can produce random-looking permutations.
- vince14 6y agohttps://news.ycombinator.com/item?id=25497357 https://news.ycombinator.com/item?id=25497357
- Ayesh 6y agoProbably looking for this: https://en.m.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle https://en.m.wikipedia.org/wiki/Fisher%E2%80%93Yates_shuffle
- TekMol 6y agoThis seems to use an external random number generator. I want the code to be complete and reproducible. So gimmeTheNumberAtPosition(x) will always return the same for the same x. Also, FYShuffle is much more complex than I would like the algo to be.
- jimktrains2 6y agoIf you seed your random number generator, then it is reproducible.
- jhgb 6y agoFisher-Yates is one of the simplest things out there IMO. Isn't the Lehmer code the thing that you want? Then you can specify your permutation with an integer and extract the individual digits from it repeatedly.
- deleted 6y ago[deleted]
- miked85 6y agoAre you saying you want a reproducible random number generator?
- jamespwilliams 6y agoIf you’re using Go, you can use rand.Perm (https://golang.org/pkg/math/rand/#Perm https://golang.org/pkg/math/rand/#Perm), which uses a Fisher-Yates shuffle under the hood.
- NotCamelCase 6y agoHave you looked up LFSRs yet? https://en.wikipedia.org/wiki/Linear-feedback_shift_register https://en.wikipedia.org/wiki/Linear-feedback_shift_register They are quite efficient for both HW and SW implementations.
- jchook 6y agoThe equivalence classes modulo some n form a partition of the integers so you can use that to create a very efficient solution with very little code. Here is a neat explanation: https://preshing.com/20121224/how-to-generate-a-sequence-of-unique-random-integers https://preshing.com/20121224/how-to-generate-a-sequence-of-... If you need even better properties (eg cryptographically secure) you can also look at PCG with k-dimensional equidistribution: http://www.pcg-random.org/index.html http://www.pcg-random.org/index.html
- jlouis 6y agoIf size is a prime, repeated multiplication on a generator element (any element) will work if you compute `mod p`. But it's not going to have nice random properties for the most part. It's just a "permutation". The reason it works is because the subgroup order generated by the element has to be divisible in the group order. There's only 1 and p. So unless it's trivial (i.e., the element 1), it's going to be p and thus it generates the whole group in p steps.
- nestorD 6y agoFrom memory you don't need size to be a prime, what you need is that it is coprime with your step. Thus, the easiest solution is to take a step-size that is prime and different from you size (that works for any size). Given the index of the current element, the next index would be `(current_index + step) % size`.
- Someone 6y agoXoring won’t work: - xoring will only produce permutations that have period two, and every element will be part of a 2-cycle. - if your n isn’t a power of two, it may produce numbers larger than n The set of 2-cycles will have a lot of structure, too (for example, if your magic constant is even, all cycles will have either two even numbers or two odd numbers; if it is odd, all cycles will have an even and an odd number), but the above should already be bad enough to drop that idea.
- TekMol 6y agoif your n isn’t a power of two, it may produce numbers larger than n This is the biggest problem. The others are not such big problems as I don't need strong resemblence to real randomness.
- Someone 6y agoI can’t look into your use case, but I’m not sure you realize how spectacularly bad xoring with a fixed value is as a way to generate a ‘random’ permutation. For example, it will still alternate odd and even numbers. Also, an adversary can derive the xor key from a single sample, and predict the entire sequence from it.
- TekMol 6y agoAlterning odd and even numbers is fine. There is no adversary. The use case is not about security.
- ascar 6y agoIf you want a simple pseudo random sequence of N that just looks random (pseudocode): int start = getRandomInt(seed) % N; double k = N / 1.618; k = k % 2 == 0 ? k - 1 : k; // k and N must be coprime int nextElement = k * start % N; start++; This will jump wildly across your sequence and visit all elements (as k and N are chosen coprime). No need to store a full array. Is it statistically random? Of course not, e.g. you will never see two values close to each other right after another.
- anderskaseorg 6y agoYour algorithm fails to choose k and N coprime for about 19% of N, such as N = 6, 10, 14, 15, 25, 27, 35, 36, 45, 54, 55, 65, 66, 75, 84, 85, 90, 93, 95 ….
- amitp 6y agoYes, xor can be used as part of this — see the paper[1] and code[2]. The idea is that you're generating a random shuffle of a sequence, but you can do it in a way that doesn't require storing the entire shuffled sequence. You can ask "what's in position 4" and it will return "item 1". [1]: http://pcgworkshop.com/archive/mawhorter2019anarchy.pdf http://pcgworkshop.com/archive/mawhorter2019anarchy.pdf [2]: https://github.com/solsword/anarchy https://github.com/solsword/anarchy