# User talk:MichaelChrisco

Thanks for squelching that spam. (Google Translate says it was some kind of advert for a Russian factory. Inappropriate for here for sure.) –Donal Fellows 08:41, 27 July 2010 (UTC)

ya no kidding. Your welcome.

### Bead sort: a Unique Solution

Ive been working on an elegant solution to a paper I found on the internet. Its called Bead-sort http://en.wikipedia.org/wiki/Bead_sort. I am not sure that this is an original idea but I seem to have implemented it in C++. Why C++.....because I couldn't do it in Newlisp (yet!). I think that the elegance of this algorithm is cool (as it was an A HA!! moment for me). It uses a distribution algorithm to distribute the "beads" one at a time though "buckets". These buckets, in turn, are the accumulators (or the after affects of gravity on bead sort).

I was trying to figure out a solution into turning them back into a list/array sorted format when it hit me! Use the same algorithm twice! So i did. And it worked! It works because gravity works both ways.

The following code is open source. Do what you want with it (just gimme a little credit and the original authors of the paper):

<lang cpp>

//this algorithm only works with positive, whole numbers. It can be made to work with other numbers but the performance would be horrific. //its a proof of concept. For actual sorting, it probably has worse time and space complexity than some algorithms. //O(2n) time complexity where n is the summation of the whole list to be sorted. //O(3n) space complexity. There are three lists that I created. 2 on the fly.

//Michael Chrisco Aug. 22, 2010 //michaelachrisco@gmail.com

//Based off of paper here: http://www.cs.auckland.ac.nz/~jaru003/research/publications/journals/beadsort.pdf

1. include<iostream>
2. include<vector>

using namespace std;

void distribute( int dist, vector<int> &List)//in theory makes *beads* go down into different buckets using gravity. {

if (dist > List.size() ) { List.resize(dist,0);//can be done differently but *meh* }

for (int i=0; i < dist; i++) { List[i]=List[i]+1;

} } //end of distribute

int main(){ vector<int> list; vector<int> list2; int myints[] = {5,3,1,7,4,1,1};

``` vector<int> fifth (myints, myints + sizeof(myints) / sizeof(int) );
```
``` cout << "#1 Beads falling down: ";
for (int i=0; i < fifth.size(); i++)
```

distribute (fifth[i], list);

``` cout << endl;
```

cout <<endl<< "Beads on their sides: "; for (int i=0; i < list.size(); i++) cout << " " << list[i]; cout << endl;

//second part

cout << "#2 Beads right side up: "; for (int i=0; i < list.size(); i++) distribute (list[i], list2);

cout << endl;

cout <<endl<< "Sorted list/array"; for (int i=0; i < list2.size(); i++) cout << " " << list2[i]; cout << endl;

return 0;

}

</lang>

### Output:

Beads falling down:

Beads on their sides: 7 4 4 3 2 1 1 Beads right side up:

Sorted list/array 7 5 4 3 1 1 1

### Time trials

• small dataset:

time ./test

Sorted list/array 7 5 4 3 1 1 1

real 0m0.011s

user 0m0.001s

sys 0m0.002s

• larger dataset

Sorted list/array 1000 7 5 4 3 1 1

real 0m0.009s

user 0m0.001s

sys 0m0.002s

• even larger

Sorted list/array 9999 1000 7 7 6 6 5 5 5 4 4 4 4 3 3 3 3 2 2 1 1 1 1 1 1 1 1 1 1 1 1 1 1

real 0m0.019s

user 0m0.003s

sys 0m0.002s

### For those who what a challenge

I dont feel like doing the newLISP implementation at the moment but here is what I was working on:

<lang newLISP> (define (distribute X L List2) (while (< (length L) X) (push 0 L));;make sure that you can add the X to the list

(let (List2 '())) for some reason this list does not work.

(for (y 0 (- X 1) 1) (let (POS (+ (L y) 1))(println "P: "POS) ;;POS->position of pointer (push POS List2 )))List2);;create new list adding 1 to each element in the origional.

(define (over L)
(let (SORTED_LIST '()))
(for (z 0 (- (length L) 1) 1) (SORTED_LIST (distribute (L z) '()))))

(define Y '(1 2)) (define T '())

(distribute 10 Y T)

(define Sorted '())

now we try it over a list

(define (over L);;do recursion here (if (empty? L) (print "a:") (over (pop L))))

(over '(4 4 5)) (over '()) </lang> Its a work in progress. So if your lisp savvy, try it out and tell me what you get.....