Im working on a program that utilizes many of the sorts and allows me to compare them. However I ran into a situation I am curious about.

For selection sort, for both string and integer sorting, why would the number of comparisons be the same even though the sorting order is different? Then this also poses another question, why would string sorting take more time?

I also just wanted to confirm some of my findings.
Either merge-sort or quick sort is good for large input correct? Is there a best?
Selection / Insertion sort for small input?
Also why does the pivot point matter on quicksort? So less items have to be compared?

1 - selection sort is order (n^2)
this because it work as follow
1 it select the min from 1 -> n and put it in the 1st place (n compareson)
2 it select the min from 2 ->n and put it in the 2nd place (n-1 compareson)
.....
n it select the min from n ->n and put it in the nth place (n-n compareson)
so always the iteration is n+(n-1)+(n-2)+...+0 = n/2*(n-1) ie n^2

mege sort and quik sort is order n*lg(n) and this is best order for sorting but complectaed so when we user small numbers we user Selection / Insertion since in this case n^2 and n*lg(n) will be very near

the most problem in quik sort is the pivot because the the chossing of the pivot may change the order to o(n^2)
for example if you use sorted array and you choose the pivot as 1st element
you then divide array to 2 parts 1 greater than pivot 2 less than or equal pivot (if the pivot divide the array to 2 equal size array you reach optimal) but since array sorted the 1st array will contain 1 element and the second contain (n-1) so it as you do nothing

How does selection sort work? Think about that for just a second.

Selection sort runs two count controlled loops. One indexes from the first to the last element in the array. The second, inner loop, indexed from the first index to the end of the loop looking for the index of the smallest element in that range. After the inner loop finishes, the smallest element is swapped with the element indexed by the first index.

Do you see anything interesting in that description of selection sort? No matter which element is the smallest in the remaining area to check, one item is compared with the current smallest in each iteration of the inner loop. So the number of times keys are compared is actually CONSTANT for any given input size regardless of the type or ordering of the input.

Your code counts KEY COMPARISONS. That is, comparisons between whole values stored inside the array to be sorted. Thus comparisons conunts comparing two strings exactly the same as it counts comparing two integers. It is true that the assembly code for comparing two integers is one (or two if you count the branch) instructions and comparing strings may be many instructions (it is a loop even if it looks like "=="). So the TIME for comparing two keys depends on the type and (perhaps) value of the key but the number of key comparisons does not.

-bcl

0

Featured Post

We value your feedback.

Take our survey and automatically be enter to win anyone of the following:
Yeti Cooler, Amazon eGift Card, and Movie eGift Card!

this because it work as follow

1 it select the min from 1 -> n and put it in the 1st place (n compareson)

2 it select the min from 2 ->n and put it in the 2nd place (n-1 compareson)

.....

n it select the min from n ->n and put it in the nth place (n-n compareson)

so always the iteration is n+(n-1)+(n-2)+...+0 = n/2*(n-1) ie n^2

mege sort and quik sort is order n*lg(n) and this is best order for sorting but complectaed so when we user small numbers we user Selection / Insertion since in this case n^2 and n*lg(n) will be very near

the most problem in quik sort is the pivot because the the chossing of the pivot may change the order to o(n^2)

for example if you use sorted array and you choose the pivot as 1st element

you then divide array to 2 parts 1 greater than pivot 2 less than or equal pivot (if the pivot divide the array to 2 equal size array you reach optimal) but since array sorted the 1st array will contain 1 element and the second contain (n-1) so it as you do nothing