Optimized Selection Sort? - c

Question

I have read sources that say that the time complexities for Selection sort are:
Best-case: O(n^2)
Average-case: O(n^2)
Worst-case: O(n^2)
I was wondering if it is worth it to "optimize" the algorithm by adding a certain line of code to make the algorithm "short-circuit" itself if the remaining part is already sorted.
Here's the code written in C:
I have also added a comment which indicates which lines are part of the "optimization" part.
void printList(int* num, int numElements) {
int i;
for (i = 0; i < numElements; i ++) {
printf("%d ", *(num + i));
}
printf("\n");
}
int main() {
int numElements = 0, i = 0, j = 0, min = 0, swap = 0, numSorted = 0;
printf("Enter number of elements: ");
scanf("%d", &numElements);
int* num = malloc(sizeof(int) * numElements);
for (i = 0; i < numElements; i ++) {
printf("Enter number = ");
scanf(" %d", num + i);
}
for (i = 0; i < numElements-1; i++) {
numSorted = i + 1; // "optimized"
min = i;
for (j = i + 1; j < numElements; j++) {
numSorted += *(num + j - 1) <= *(num + j); // "optimized"
if (*(num + min) > *(num + j))
min = j;
}
if (numSorted == numElements) // "optimized"
break;
if (min != i) {
swap = *(num + i);
*(num + i) = *(num + min);
*(num + min) = swap;
}
printList(num, numElements);
}
printf("Sorted list:\n");
printList(num, numElements);
free(num);
getch();
return 0;
}

Answer 1

Optimizing selection sort is a little silly. It has awful best-case, average, and worst-case time complexity, so if you want a remotely optimized sort you would (almost?) always pick another sort. Even insertion sort tends to be faster and it's hardly much more complicated to implement.
More to the point, checking if the list is sorted increases the time the algorithm takes in the worst case scenarios (the average case too I'm inclined to think). And even a mostly sorted list will not necessarily go any faster this way: consider 1,2,3,4,5,6,7,9,8. Even though the list only needs two elements swapped at the end, the algorithm will not short-circuit as it is not ever sorted until the end.

Answer 2

Just because something can be optimized, doesn't necessarily mean it should. Assuming profiling or "boss-says-so" indicates optimization is warranted there are a few things you can do.
As with any algorithm involving iteration over memory, anything that reduces the number of iterations can help.
keep track of the min AND max values - cut number of iterations in half
keep track of multiple min/max values (4 each will be 1/8th the iterations)
at some point temp values will not fit in registers
the code will get more complex
It can also help to maximize cache locality.
do a backward iteration after the forward iteration
the recently accessed memory should still be cached
going straight to another forward iteration would cause a cache miss
since you are moving backward, the cache predictor may prefetch the rest
this could actually be worse on some architectures (RISC-V)
operate on a cache line at a time where possible
this can allow the next cache line to be prefetched in the mean time
you may need to align the data or specially handle the first and last data
even with increased alignment, the last few elements may need "padding"
Use SIMD instructions and registers where useful and practical
useful for non-branching rank order sort of temps
can hold many data points simultaneously (AVX-512 can do a cache line)
avoids memory access (thus less cache misses)
If you use multiple max/min values, optimize sorting the n values of max and min
see here for techniques to sort a small fixed number of values.
save memory swaps until the end of each iteration and do them once
keep temporaries (or pointers) in registers in the mean time
There are quite a few more optimization methods available, but eventually the resemblance to selection sort starts to get foggy. Each of these is going to increase complexity and therefore maintenance cost to the point where a simpler implementation of a more appropriate algorithm may be a better choice.

Answer 3

The only way I see how this can be answered is if you define the purpose of why you are optimizing it.
Is it worth it in a professional setting: on the job, for code running "in production" - most likely (even almost certainly) not.
Is it worth it as a teaching/learning tool - sometimes yes.
I teach programming to individuals and sometimes I teach them algorithms and datastructures. I consider selection sort to be one of the easiest to explain and teach - it flows so naturally after explaining the algorithm for finding the minimum and swapping two values (swap()). Then, at the end I introduce the concept of optimization where we can implement this counter "already sorted" detection.
Admittedly bubble sort is even better to introduce optimization, because it has at least 3 easy to explain and substantial optimizations.

Answer 4

I was wondering if it is worth it to "optimize" the algorithm by adding a certain line of code to make the algorithm "short-circuit" itself if the remaining part is already sorted.
Clearly this change reduces the best-case complexity from O(n2) to O(n). This will be observed for inputs that are already sorted except for O(1) leading elements. If such inputs are a likely case, then the suggested code change might indeed yield an observable and worthwhile performance improvement.
Note, however, that your change more than doubles the work performed in the innermost loop, and consider that for uniform random inputs, the expected number of outer-loop iterations saved is 1. Consider also that any outer-loop iterations you do manage to trim off will be the ones that otherwise would do the least work. Overall, then, although you do not change the asymptotic complexity, the actual performance in the average and worst cases will be noticeably worse -- runtimes on the order of twice as long.
If you're after better speed then your best bet is to choose a different sorting algorithm. Among the comparison sorts, Insertion Sort will perform about the same as your optimized Selection Sort on the latter's best case, but it has a wider range of best-case scenarios, and will usually outperform (regular) Selection Sort in the average case. How the two compare in the worst case depends on implementation.
If you want better performance still then consider Merge Sort or Quick Sort, both of which are pretty simple to implement. Or if your data are suited to it then Counting Sort is pretty hard to beat.

Answer 5

we can optimize selection sort in best case which will be O(n) instead of O(n^2).
here is my optimization code.
public class SelectionSort {
static void selectionSort(int[]arr){
for(int i=0; i< arr.length; i++){
int maxValue=arr[0];
int maxIndex=0;
int cnt=1;
for (int j=1; j< arr.length-i; j++){
if(maxValue<=arr[j]){
maxValue=arr[j];
maxIndex=j;
cnt++;
}
}
if(cnt==arr.length)break;
arr[maxIndex]=arr[arr.length-1-i];
arr[arr.length-1-i]=maxValue;
}
}
public static void main(String[] args) {
int[]arr={1,-3, 0, 8, -45};
selectionSort(arr);
System.out.println(Arrays.toString(arr));
}
}