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

Answer 1

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 2

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 3

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 4

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 5

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 6

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 :)