- 52 Views
- Uploaded on
- Presentation posted in: General

Inventing A Really Bad Sort: It Seemed Like A Good Idea At The Time

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Inventing A Really Bad Sort:It Seemed Like A Good Idea At The Time

Jim Huggins

Kettering University

http://www.kettering.edu/~jhuggins

Bless me, Father Knuth, for I have sinned …

Writing questions for take-home exam for Advanced Algorithms course (sophomore)

Desired: an algorithm which:

- does something useful
- is simple to analyze
- hasn’t been done before on the web

public void StoogeSort(int[] arr, int start, int stop) {

if (arr[start] > arr[stop]) {

int swap = arr[start];

arr[start] = arr[stop];

arr[stop] = swap;

}

if (start+1 >= stop)

return;

int third = (stop - start + 1) / 3;

StoogeSort(arr, start, stop-third); // First two-thirds

StoogeSort(arr, start+third, stop); // Last two-thirds

StoogeSort(arr, start, stop-third); // First two-thirds

}

Comparison count:

T(n) = 3T(⅔n) + 1; T(0)=0, T(1)=0, T(2) = 1

T(n) = Θ(nlog3/2 3) ≈ Θ(n2.7)

public void goofySort (int[] array, int start, int stop) {

if (start>=stop) return;

goofySort(array, start, stop-1); // first n-1 items

if (array[stop-1] > array[stop]) {

int swap = array[stop-1]; // swap last item

array[stop-1] = array[stop];

array[stop] = swap;

goofySort(array, start, stop-1); // first n-1 items

}// again

}

An added bonus: different behavior in best case, worst case. (More questions!)

public void goofySort (int[] array, int start, int stop) {

if (start>=stop) return;

goofySort(array, start, stop-1);

if (array[stop-1] > array[stop]) { // best case: false

int swap = array[stop-1];

array[stop-1] = array[stop];

array[stop] = swap;

goofySort(array, start, stop-1);

}

}

Comparison count:

T(n) = T(n-1) + 1, T(1) = 0

T(n) = O(n)

public void goofySort (int[] array, int start, int stop) {

if (start>=stop) return;

goofySort(array, start, stop-1);

if (array[stop-1] > array[stop]) { // worst case: true

int swap = array[stop-1];

array[stop-1] = array[stop];

array[stop] = swap;

goofySort(array, start, stop-1);

}

}

Comparison count:

T(n) = 2T(n-1) + 1, T(1) = 0

T(n) = O(2n)

“Beware of bugs in the above code; I have only proved it correct, not tried it.”

- Coded the algorithm in Java
- Tested with a variety of random inputs
- Tested with a variety of list sizes20, 30, 40, …
- And it all works! Great!(What could possibly go wrong?)

- T(n) = O(n3)
- T(n) = T(n-1) + O(n2) = O(n3)
- T(n) = T(n-1) + Σ1n i = ?
- T(n) = O(2n)
- One bright student, at least!

- Preparing my rant …
- “you completely missed the point”
- “we did this in class … ”
- “I even tested this on lots of inputs ...”

- And then I remember:
- If this is really exponential time, how did I run it on an input of size 40?
- @#@!. What if I’m wrong and they’re right?

Paraphrasing:

“If you double the input size, and the instruction count quadruples, you’ve got a quadratic algorithm.”

(Watch the Google TechTalk … it’s neat.)

Take the average over 100 random runs …

@#$!. It looks like it’s cubic!

public void goofySort (int[] array, int start, int stop) {

if (start>=stop) return;

goofySort(array, start, stop-1);

if (array[stop-1] > array[stop]) {

int swap = array[stop-1];

array[stop-1] = array[stop];

array[stop] = swap;

goofySort(array, start, stop-1);

}

}

The first recursive call is always bad;the array could be completely unordered

The second recursive call is always good;all but the last item are ordered

- What’s the worst case input?
- Reverse sorted, right?

- At this point, I don’t trust myself, so …
- Generate all permutations on a list of size n
- We just covered this in class! (Lucky this is an algorithms course!)

- Verified: worst case happens in reverse order

- Generate all permutations on a list of size n

- Do a bunch of input sizes …
- … and putz around with a calculator …
- The closed-form formula appears to be:T(n) = n(n-1)(n-2)/6 + (n-1)
- So this does appear to be cubic after all.(Now, how do I prove it?)

- T(n) = 1 + T(n-1) + GT(n-1); T(1) = 0
- GT(n) = (n-1) + GT(n-1); GT(1) = 0…GT(n) = n(n-1)/2
- T(n) = 1 + T(n-1) + (n-1)(n-2)/2; T(1) = 0…T(n) = (n-1) + n(n-1)(n-2)/6
(see me for the full proof … it’s not that bad)

- Deep apologies to the students
- They were gracious

- Q: “How did y’all know it was cubic?”A: “We ran it and it looked cubic.”
- They had no idea how to proceed …so they did the empirical analysis first to find the “right” answer!

“Beware of bugs in the above code; I have only proved it correct, not tried it.”

“Beware of bugs in the above analysis; I have only proved it correct, not verified it empirically.”