Advanced SelectionSort - Search two elements in one iteration - c++

Question

I've got a homework where I have to "improve" SelectionSort with following parameters:
Sorting an given list with an "improved" SelectionSort
In one iteration, finding the smallest and 2nd smallest element
Bring the smallest and 2nd smallest element in the right position.
Ok so far I wrote this c++ code:
#include <iostream>
#include <array>
#include <string>
using namespace std;
int main()
{
int min, min2;
// doesn't work
array<int, 9> list = { 97,34,15,25,27,4,19,41,68 };
/* this one works:
array<int, 10> list = { 4,8,1,3,10,6,5,7,9,2 };
*/
// First loop
for (int i = 0; i < list.size(); i+=2) {
min = i;
min2 = i + 1;
// 2nd Loop for searching the smallest elements
for (int j = i + 2; j < list.size(); j++) {
if (list.at(j) < list.at(min)) {
min = j;
}
// outer if -> stop before out of array
if (j+1 < list.size()) {
if (list.at(j+1) < list.at(min2)) {
min2 = j+1;
}
}
}
swap(list.at(i), list.at(min));
// Prevent out of array error
if (i + 1 < list.size()) {
swap(list.at(i+1), list.at(min2));
}
}
cout << '\n' << '\n';
cout << "Sorted list: " << endl;
for (int elem : list) {
cout << elem << ", ";
}
}
Of course it's sorting and this is the result... but not the one I was hoping for:
4, 97, 15, 19, 25, 34, 27, 41, 68,
I'm out of ideas and the only hint I got was: "no third loop".
I would appreciate any help :-)
EDIT:
Due the voting to hold on I try to specify the problem.
When I'm using high int-values for example the ones in the code, the sorting-algorithm doesn't work properly
List: array<int, 9> list = { 97,34,15,25,27,4,19,41,68 };
Result: 4, 97, 15, 19, 25, 34, 27, 41, 68,
As you can see the values on position 0, 2, 4, 6, 8 from the first if-statement were sorted properly but the others form the 2nd if-statement not.
When I change the int values for example to values from 1-10 and mix them random, the algorithm seems to work properly (thanks for the comment!):
List: array<int, 10> list = { 4,8,1,3,10,6,5,7,9,2 };
Result: 1, 2, 4, 3, 5, 6, 8, 7, 9, 10,
I'm out of ideas - Is it a programming error or a bad algorithm?
EDIT 3:
Here is my (finally working) updated code:
//array<int, 10> list = { 4,8,1,3,10,6,5,7,9,2 };
//array<int, 4> list = { 97,15,25,18 };
//array<int, 2> list = { 97,18 };
array<int, 3> list = { 4,5,3 };
// First loop
for (int i = 0; i < list.size(); i+=2) {
if (i == list.size() - 1) {
break;
}
min = i;
min2 = i + 1;
// Enforce that list.at(min) <= list.at(min2) -> Sorting pivot (element) for the algorithm to smallest, 2nd smallest.
if (list.at(min) > list.at(min2)) {
swap(list.at(min), list.at(min2));
}
// Second Loop
for (int j = i + 2; j < list.size(); ++j) {
if (list.at(j) < list.at(min)) {
min2 = min; // min2 points now on the 2nd smallest element
min = j; // min points now on the smallest element
}
else if (list.at(j) < list.at(min2)) {
min2 = j;
}
}
// Swapping the elements in the right position.
swap(list.at(i + 1), list.at(min2));
swap(list.at(i), list.at(min));
}
Results:
{ 4,8,1,3,10,6,5,7,9,2 } -> 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
{ 97,15,25,18 } -> 15, 18, 25, 97,
{ 97,18 } -> 18, 97,
{ 4,5,3 } -> 3, 4, 5,

Answer 1

Try your program with the array [97,18]. I think you'll find that it doesn't work, because the 2nd loop is never entered, and the lines at the end of the first loop won't swap anything.
When I said in my comment that min must be less than or equal to min2, I meant to say that you must ensure list.at(min) <= list.at(min2). My example above of the 2-item array shows why that's important.
The way to enforce that is to modify your first loop:
// First loop
for (int i = 0; i < list.size(); i+=2) {
if (i == list.size()-1) {
// There is an odd number of items in the list.
// At this point, we know that the last item is in place.
// So just exit.
break;
}
min = i;
min2 = i + 1;
// enforce list(min) <= list(min2)
if (list.at(min) > list.at(min2)) {
swap(list.at(min), list.at(min2));
}
// second loop
And, yes, you must maintain that in the inner loop. If you try with the array [4,5,3], the result will be [3,5,4].
That's primarily because in your inner loop, when you find that list.at(j) < list.at(min), you swap the items. But when list.at(min2) > list.at(min), what you've done is swap things out of order.
You should single-step your code in the debugger using that 3-element list to understand what's happening. If you don't know how to use the debugger, stop right now and learn. This type of programming error is very easy to discover when you can watch a line-by-line execution of your code. The alternative is to do it by hand: with a piece of paper and pencil, walk through the code step by step, writing down the change to every variable.
The other problem with your inner loop is that you're checking list.at(j+1) with list.at(min2). But you're only incrementing j by 1 each time, so you end up doing extra work. There's no reason to do that extra check. It'll be handled the next time through the loop.
The fix in the inner loop, assuming that you maintain the proper relationship between min and min2, is easy enough. If list.at(j) < list.at(min), then set min2=min (because min is now pointing to the second-smallest item), and set min=j. If list.at(j) < list.at(min2), then just set min2=j. The code looks like this:
for (j = i+2; j < list.size(); ++j) {
if (list.at(j) < list.at(min)) {
min2 = min;
min = j;
} else if (list.at(j) < list.at(min2)) {
min2 = j;
}
}
Now your code at the end of the outer loop works correctly.
Debugging
Run your program in the debugger with the array [4,5,3]. Place a breakpoint on the line just after the inner loop, here:
swap(list.at(i), list.at(min));
If you examine the array, you'll see that it's still [4,5,3]. But look at min and min2. You'll see that min=2 and min2=0. i is equal to 0. Now, what happens when when you swap the items at list[i] and list[min]? You get [3,5,4], and min2 no longer points to the second smallest item.
There are several different conditions in which something like this can happen. You have to think of those conditions, and handle them. The code I provided lets you find the smallest and 2nd smallest items. But you have to figure out how to swap things into the correct places after you've found them.