Are there any worse sorting algorithms than Bogosort (a.k.a Monkey Sort)?

Question

My co-workers took me back in time to my University days with a discussion of sorting algorithms this morning. We reminisced about our favorites like StupidSort, and one of us was sure we had seen a sort algorithm that was O(n!). That got me started looking around for the "worst" sorting algorithms I could find.

We postulated that a completely random sort would be pretty bad (i.e. randomize the elements - is it in order? no? randomize again), and I looked around and found out that it's apparently called BogoSort, or Monkey Sort, or sometimes just Random Sort.

Monkey Sort appears to have a worst case performance of O(∞), a best case performance of O(n), and an average performance of O(n·n!).

What is the currently official accepted sorting algorithm with the worst average sorting performance (and there fore beeing worse than O(n·n!))?

Answer 1

From David Morgan-Mar's Esoteric Algorithms page: Intelligent Design Sort

Introduction

Intelligent design sort is a sorting algorithm based on the theory of intelligent design.

Algorithm Description

The probability of the original input list being in the exact order it's in is 1/(n!). There is such a small likelihood of this that it's clearly absurd to say that this happened by chance, so it must have been consciously put in that order by an intelligent Sorter. Therefore it's safe to assume that it's already optimally Sorted in some way that transcends our naïve mortal understanding of "ascending order". Any attempt to change that order to conform to our own preconceptions would actually make it less sorted.

Analysis

This algorithm is constant in time, and sorts the list in-place, requiring no additional memory at all. In fact, it doesn't even require any of that suspicious technological computer stuff. Praise the Sorter!

Feedback

Gary Rogers writes:

Making the sort constant in time denies the power of The Sorter. The Sorter exists outside of time, thus the sort is timeless. To require time to validate the sort diminishes the role of the Sorter. Thus... this particular sort is flawed, and can not be attributed to 'The Sorter'.

Heresy!

Answer 2

Many years ago, I invented (but never actually implemented) MiracleSort.

Start with an array in memory.
loop:
    Check to see whether it's sorted.
    Yes? We're done.
    No? Wait a while and check again.
end loop

Eventually, alpha particles flipping bits in the memory chips should result in a successful sort.

For greater reliability, copy the array to a shielded location, and check potentially sorted arrays against the original.

So how do you check the potentially sorted array against the original? You just sort each array and check whether they match. MiracleSort is the obvious algorithm to use for this step.

EDIT: Strictly speaking, this is not an algorithm, since it's not guaranteed to terminate. Does "not an algorithm" qualify as "a worse algorithm"?

Answer 3

Quantum Bogosort

A sorting algorithm that assumes that the many-worlds interpretation of quantum mechanics is correct:

1. Check that the list is sorted. If not, destroy the universe.

At the conclusion of the algorithm, the list will be sorted in the only universe left standing. This algorithm takes worst-case Θ(N) and average-case θ(1) time. In fact, the average number of comparisons performed is 2: there's a 50% chance that the universe will be destroyed on the second element, a 25% chance that it'll be destroyed on the third, and so on.

Answer 4

Jingle Sort, as described here.

You give each value in your list to a different child on Christmas. Children, being awful human beings, will compare the value of their gifts and sort themselves accordingly.

Answer 5

I'm surprised no one has mentioned sleepsort yet... Or haven't I noticed it? Anyway:

#!/bin/bash
function f() {
    sleep "$1"
    echo "$1"
}
while [ -n "$1" ]
do
    f "$1" &
    shift
done
wait

example usage:

./sleepsort.sh 5 3 6 3 6 3 1 4 7
./sleepsort.sh 8864569 7

In terms of performance it is terrible (especially the second example). Waiting almost 3.5 months to sort 2 numbers is kinda bad.

Answer 6

I had a lecturer who once suggested generating a random array, checking if it was sorted and then checking if the data was the same as the array to be sorted.

Best case O(N) (first time baby!) Worst case O(Never)

Answer 7

There is a sort that's called bogobogosort. First, it checks the first 2 elements, and bogosorts them. Next it checks the first 3, bogosorts them, and so on.

Should the list be out of order at any time, it restarts by bogosorting the first 2 again. Regular bogosort has a average complexity of O(N!), this algorithm has a average complexity of O(N!1!2!3!...N!)

Edit: To give you an idea of how large this number is, for 20 elements, this algorithm takes an average of 3.930093*10^158 years,well above the proposed heat death of the universe(if it happens) of 10^100 years,

whereas merge sort takes around .0000004 seconds, bubble sort .0000016 seconds, and bogosort takes 308 years, 139 days, 19 hours, 35 minutes, 22.306 seconds, assuming a year is 365.242 days and a computer does 250,000,000 32 bit integer operations per second.

Edit2: This algorithm is not as slow as the "algorithm" miracle sort, which probably, like this sort, will get the computer sucked in the black hole before it successfully sorts 20 elemtnts, but if it did, I would estimate an average complexity of 2^(32(the number of bits in a 32 bit integer)*N)(the number of elements)*(a number <=10^40) years,

since gravity speeds up the chips alpha moving, and there are 2^N states, which is 2^640*10^40, or about 5.783*10^216.762162762 years, though if the list started out sorted, its complexity would only be O(N), faster than merge sort, which is only N log N even at the worst case.

Edit3: This algorithm is actually slower than miracle sort as the size gets very big, say 1000, since my algorithm would have a run time of 2.83*10^1175546 years, while the miracle sort algorithm would have a run time of 1.156*10^9657 years.

Answer 8

If you keep the algorithm meaningful in any way, O(n!) is the worst upper bound you can achieve.

Since checking each possibility for a permutations of a set to be sorted will take n! steps, you can't get any worse than that.

If you're doing more steps than that then the algorithm has no real useful purpose. Not to mention the following simple sorting algorithm with O(infinity):

list = someList
while (list not sorted):
    doNothing

Answer 9

Bogobogosort. Yes, it's a thing. to Bogobogosort, you Bogosort the first element. Check to see if that one element is sorted. Being one element, it will be. Then you add the second element, and Bogosort those two until it's sorted. Then you add one more element, then Bogosort. Continue adding elements and Bogosorting until you have finally done every element. This was designed never to succeed with any sizable list before the heat death of the universe.

Answer 10

You should do some research into the exciting field of Pessimal Algorithms and Simplexity Analysis. These authors work on the problem of developing a sort with a pessimal best-case (your bogosort's best case is Omega(n), while slowsort (see paper) has a non-polynomial best-case time complexity).

Answer 11

Here's 2 sorts I came up with my roommate in college

1) Check the order 2) Maybe a miracle happened, go to 1

and

1) check if it is in order, if not 2) put each element into a packet and bounce it off a distant server back to yourself. Some of those packets will return in a different order, so go to 1

Answer 12

There's always the Bogobogosort (Bogoception!). It performs Bogosort on increasingly large subsets of the list, and then starts all over again if the list is ever not sorted.

for (int n=1; n<sizeof(list); ++n) {
  while (!isInOrder(list, 0, n)) {
    shuffle(list, 0, n);
  }
  if (!isInOrder(list, 0, n+1)) { n=0; }
}

Answer 13

1 Put your items to be sorted on index cards
2 Throw them into the air on a windy day, a mile from your house.
2 Throw them into a bonfire and confirm they are completely destroyed.
3 Check your kitchen floor for the correct ordering.
4 Repeat if it's not the correct order.

Best case scenerio is O(∞)

Edit above based on astute observation by KennyTM.

Answer 14

The "what would you like it to be?" sort

1. Note the system time.
2. Sort using Quicksort (or anything else reasonably sensible), omitting the very last swap.
3. Note the system time.
4. Calculate the required time. Extended precision arithmetic is a requirement.
5. Wait the required time.
6. Perform the last swap.

Not only can it implement any conceivable O(x) value short of infinity, the time taken is provably correct (if you can wait that long).

Answer 15

Nothing can be worse than infinity.

Answer 16

Segments of π

Assume π contains all possible finite number combinations. See math.stackexchange question

1. Determine the number of digits needed from the size of the array.
2. Use segments of π places as indexes to determine how to re-order the array. If a segment exceeds the size boundaries for this array, adjust the π decimal offset and start over.
3. Check if the re-ordered array is sorted. If it is woot, else adjust the offset and start over.

Answer 17

Bozo sort is a related algorithm that checks if the list is sorted and, if not, swaps two items at random. It has the same best and worst case performances, but I would intuitively expect the average case to be longer than Bogosort. It's hard to find (or produce) any data on performance of this algorithm.

Answer 18

Recursive Bogosort (probably still O(n!){
if (list not sorted)
list1 = first half of list.
list 2 = second half of list.
Recursive bogosort (list1);
Recursive bogosort (list2);
list = list1 + list2
while(list not sorted)
    shuffle(list);
}

Answer 19

A worst case performance of O(∞) might not even make it an algorithm according to some.

An algorithm is just a series of steps and you can always do worse by tweaking it a little bit to get the desired output in more steps than it was previously taking. One could purposely put the knowledge of the number of steps taken into the algorithm and make it terminate and produce the correct output only after X number of steps have been done. That X could very well be of the order of O(n²) or O(n^n!) or whatever the algorithm desired to do. That would effectively increase its best-case as well as average case bounds.

But your worst-case scenario cannot be topped :)

Answer 20

My favorite slow sorting algorithm is the stooge sort:

void stooges(long *begin, long *end) {
   if( (end-begin) <= 1 ) return;
   if( begin[0] < end[-1] ) swap(begin, end-1);
   if( (end-begin) > 1 ) {
      int one_third = (end-begin)/3;
      stooges(begin, end-one_third);
      stooges(begin+one_third, end);
      stooges(begin, end-one_third);
   }
}

The worst case complexity is O(n^(log(3) / log(1.5))) = O(n^2.7095...).

Another slow sorting algorithm is actually named slowsort!

void slow(long *start, long *end) {
   if( (end-start) <= 1 ) return;
   long *middle = start + (end-start)/2;
   slow(start, middle);
   slow(middle, end);
   if( middle[-1] > end[-1] ) swap(middle-1, end-1);
   slow(start, end-1);
}

This one takes O(n ^ (log n)) in the best case... even slower than stoogesort.

Answer 21

Double bogosort

Bogosort twice and compare results (just to be sure it is sorted) if not do it again

Answer 22

This page is a interesting read on the topic: http://home.tiac.net/~cri_d/cri/2001/badsort.html

My personal favorite is Tom Duff's sillysort:

/*
 * The time complexity of this thing is O(n^(a log n))
 * for some constant a. This is a multiply and surrender
 * algorithm: one that continues multiplying subproblems
 * as long as possible until their solution can no longer
 * be postponed.
 */
void sillysort(int a[], int i, int j){
        int t, m;
        for(;i!=j;--j){
                m=(i+j)/2;
                sillysort(a, i, m);
                sillysort(a, m+1, j);
                if(a[m]>a[j]){ t=a[m]; a[m]=a[j]; a[j]=t; }
        }
}

Answer 23

One I was just working on involves picking two random points, and if they are in the wrong order, reversing the entire subrange between them. I found the algorithm on http://richardhartersworld.com/cri_d/cri/2001/badsort.html, which says that the average case is is probably somewhere around O(n^3) or O(n^2 log n) (he's not really sure).

I think it might be possible to do it more efficiently, because I think it might be possible to do the reversal operation in O(1) time.

Actually, I just realized that doing that would make the whole thing I say maybe because I just realized that the data structure I had in mind would put accessing the random elements at O(log n) and determining if it needs reversing at O(n).

Answer 24

Randomsubsetsort.

Given an array of n elements, choose each element with probability 1/n, randomize these elements, and check if the array is sorted. Repeat until sorted.

Expected time is left as an exercise for the reader.

Answer 25

You could make any sort algorithm slower by running your "is it sorted" step randomly. Something like:

1. Create an array of booleans the same size as the array you're sorting. Set them all to false.
2. Run an iteration of bogosort
3. Pick two random elements.
4. If the two elements are sorted in relation to eachother (i < j && array[i] < array[j]), mark the indexes of both on the boolean array to true. Overwise, start over.
5. Check if all of the booleans in the array are true. If not, go back to 3.
6. Done.

Answer 26

Yes, SimpleSort, in theory it runs in O(-1) however this is equivalent to O(...9999) which is in turn equivalent to O(∞ - 1), which as it happens is also equivalent to O(∞). Here is my sample implementation:

/* element sizes are uneeded, they are assumed */
void
simplesort (const void* begin, const void* end)
{
  for (;;);
}