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Comment by manimino | Hacker News Reader

The correct Fisher-Yates shuffle is this (from Wikipedia):

  -- To shuffle an array a of n elements (indices 0..n-1):
  for i from 0 to n−2 do
     j ← random integer such that i ≤ j < n
     exchange a[i] and a[j]

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For those interested, I wrote a small article on the various was of writing incorrect Fisher-Yates [0]. The post also covers generating single cycle permutations.

Fisher-Yates is pretty obvious but weirdly subtle in some ways. For someone not paying attention, it would be easy to get wrong, as the grandparent response points out.

[0] https://mechaelephant.com/dev/Fisher-Yates-Shuffle.html

Interestingly the method used to deal bridge hands for top tournaments doesn’t shuffle at all, it generates a random number between 1 and 52!/(13!^4) and maps that number to a corresponding deal of the cards.

Are you sure they do bijective mapping instead of simply running a RNG with the given seed?

Yes - see the “Making bridge deals” section of:

https://sater.home.xs4all.nl/doc.html

Right. The subtle difference being that because you only sample from position i to n you only ever exchange each card exactly once. In my algo you sample from 0 to n-1 so a card exchanged in a previous round can be exchanged again. This generalises into cards early in the shuffle being exchanged more than later cards, which changes the distribution of possible outcomes.

Nerd sidenote: Because generating random numbers tends to be expensive, I had an idea once where you generate the full permutation with only a single random sample. Basically you could use something which fully enumerates a space like Gray codes[1] (you want one with a closed form solution which can arbitrarily you the nth code) then you generate one random number, multiply by the number of combinations, look up that code and decode that into the sequence of cards. I never actually got around to implementing it, but I suspect if your code lookup was fast it would be much faster than Fisher-Yates.

The use case was if you wanted to make a Monte Carlo algorithm to evaluate strategies in a card game for instance.

[1] https://en.wikipedia.org/wiki/Gray_code

I've wondered about the same thing of generating a permutation with a minimum number of samples. I'm not sure how Gray codes helps you here, and the multiplication sounds wrong since many outputs aren't reachable from any input.

The idea is just a closed form solution that gets you from a number in 1...nPermutations to a single specific permutation. I was using Gray codes as an example of a sequence that exhausts all permutations with a closed form that gets you to a specific point in the sequence.

Might not work - as I say I never actually got around to implementing it.

I think you're looking for a factorial number system [1] and/or Lehmer code [2].

[1] https://en.wikipedia.org/wiki/Factorial_number_system [2] https://en.wikipedia.org/wiki/Lehmer_code

Thanks!

You’ll find that the number of ways to shuffle a card pack (52!) is about 8 × 10⁶⁷, so even if your RNG returns 64 bits, you can’t use a single call.

Also, you’ll have to make sure you pick each permutation is picked with equal probability. random() % factorial(52) likely won’t work because your RNG won’t return integers in a range k × 52! for some k > 1 (https://stackoverflow.com/questions/10984974/why-do-people-s...)

I wrote that code for fun 15-20 years ago (code definitely lost in the mists of time).

You can think of it as rolling one random number up to n!, then use x%n to pick the first element and then x/n to pick the next (n-1) elements (so the next step is then to pick the second element with (x/n)%(n-1) and use (x/n)/(n-1) for the remaining (n-2) elements).

Though as a sibling comment points out, since you need a random number up to n!, even 64 bits of randomness only get you so far (I think up to 20, but didn't check my math).

The n-2 bit is unnecessary and misleading because we've already established that an index can be swapped with itself.

for i in [0..n-1] swap(a[i], a[rand in [i..n-1])

where rand in [I..n-1] coud be implemented as i + (rand() % (n-i))

essentially start with your finger at 0, pick a card from the unpicked set which is everything on and after your finger, swap that card with your finger, move your finger up one card

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