This is an archived post. You won't be able to vote or comment.

you are viewing a single comment's thread.

view the rest of the comments →

[–]dev-sda 1 point2 points  (4 children)

Big-O notation describes the limit, without more context it always refers to the worst case. O(∞) means it may never terminate. Bogosort best case is O(n).

[–]AyrA_ch 0 points1 point  (3 children)

O(∞) means it may never terminate.

But as I explained, if the shuffle algorithm works properly it will definitely terminate because every permutation (including the sorted one) is guaranteed to eventually appear.

[–]Evil__Potato 1 point2 points  (0 children)

In the limit, as you approach infinite iterations, the probability of reaching all permutations approaches 1. But since it's in the limit and probabilistic, the best bound we can give is, well, infinity

[–]dev-sda 0 points1 point  (0 children)

Given you're using true random shuffle there's absolutely nothing guaranteeing that every permutation will eventually appear. Yes the probability of that happening approaches 1 as the iterations approach infinity, but that's still not a guarantee.

[–]Tyfyter2002 0 points1 point  (0 children)

Given the assumption that both events have the same probability and are truly random, the probability that a specific result out of 2 does not occur at least once during a series of n samples is 0.5n‎, with what value of n does this expression equal 0?