Why should Insertion Sort be used after threshold crossover in Merge Sort

Question

I have read everywhere that for divide and conquer sorting algorithms like Merge-Sort and Quicksort, instead of recursing until only a single element is left, it is better to shift to Insertion-Sort when a certain threshold, say 30 elements, is reached. That is fine, but why only Insertion-Sort? Why not Bubble-Sort or Selection-Sort, both of which has similar O(N^2) performance? Insertion-Sort should come handy only when many elements are pre-sorted (although that advantage should also come with Bubble-Sort), but otherwise, why should it be more efficient than the other two?

And secondly, at this link, in the 2nd answer and its accompanying comments, it says that O(N log N) performs poorly compared to O(N^2) upto a certain N. How come? N^2 should always perform worse than N log N, since N > log N for all N >= 2, right?

Answer 1

If you bail out of each branch of your divide-and-conquer Quicksort when it hits the threshold, your data looks like this:

[the least 30-ish elements, not in order] [the next 30-ish ] ... [last 30-ish]

Insertion sort has the rather pleasing property that you can call it just once on that whole array, and it performs essentially the same as it does if you call it once for each block of 30. So instead of calling it in your loop, you have the option to call it last. This might not be faster, especially since it pulls the whole data through cache an extra time, but depending how the code is structured it might be convenient.

Neither bubble sort nor selection sort has this property, so I think the answer might quite simply be "convenience". If someone suspects selection sort might be better then the burden of proof lies on them to "prove" that it's faster.

Note that this use of insertion sort also has a drawback -- if you do it this way and there's a bug in your partition code then provided it doesn't lose any elements, just partition them incorrectly, you'll never notice.

Edit: apparently this modification is by Sedgewick, who wrote his PhD on QuickSort in 1975. It was analyzed more recently by Musser (the inventor of Introsort). Reference https://en.wikipedia.org/wiki/Introsort

Musser also considered the effect on caches of Sedgewick's delayed small sorting, where small ranges are sorted at the end in a single pass of insertion sort. He reported that it could double the number of cache misses, but that its performance with double-ended queues was significantly better and should be retained for template libraries, in part because the gain in other cases from doing the sorts immediately was not great.

In any case, I don't think the general advice is "whatever you do, don't use selection sort". The advice is, "insertion sort beats Quicksort for inputs up to a surprisingly non-tiny size", and this is pretty easy to prove to yourself when you're implementing a Quicksort. If you come up with another sort that demonstrably beats insertion sort on the same small arrays, none of those academic sources is telling you not to use it. I suppose the surprise is that the advice is consistently towards insertion sort, rather than each source choosing its own favorite (introductory teachers have a frankly astonishing fondness for bubble sort -- I wouldn't mind if I never hear of it again). Insertion sort is generally thought of as "the right answer" for small data. The issue isn't whether it "should be" fast, it's whether it actually is or not, and I've never particularly noticed any benchmarks dispelling this idea.

One place to look for such data would be in the development and adoption of Timsort. I'm pretty sure Tim Peters chose insertion for a reason: he wasn't offering general advice, he was optimizing a library for real use.

Answer 2

1. Insertion sort is faster in practice, than bubblesort at least. Their asympotic running time is the same, but insertion sort has better constants (fewer/cheaper operations per iteration). Most notably, it requires only a linear number of swaps of pairs of elements, and in each inner loop it performs comparisons between each of n/2 elements and a "fixed" element that can be stores in a register (while bubble sort has to read values from memory). I.e. insertion sort does less work in its inner loop than bubble sort.
2. The answer claims that 10000 n lg n > 10 n² for "reasonable" n. This is true up to about 14000 elements.

Answer 3

I am surprised no-one's mentioned the simple fact that insertion sort is simply much faster for "almost" sorted data. That's the reason it's used.

Answer 4

Here is an empirical proof the insertion sort is faster then bubble sort (for 30 elements, on my machine, the attached implementation, using java...).

I ran the attached code, and found out that the bubble sort ran on average of 6338.515 ns, while insertion took 3601.0

I used wilcoxon signed test to check the probability that this is a mistake and they should actually be the same - but the result is below the range of the numerical error (and effectively P_VALUE ~= 0)

private static void swap(int[] arr, int i, int j) { 
    int temp = arr[i];
    arr[i] = arr[j];
    arr[j] = temp;
}

public static void insertionSort(int[] arr) { 
    for (int i = 1; i < arr.length; i++) {
        int j = i;
        while (j > 0 && arr[j-1] > arr[j]) { 
            swap(arr, j, j-1);
            j--;
        }
    }
}
public static void bubbleSort(int[] arr) { 
    for (int i = 0 ; i < arr.length; i++) { 
        boolean bool = false;
        for (int j = 0; j < arr.length - i ; j++) { 
            if (j + 1 < arr.length && arr[j] > arr[j+1]) {
                bool = true;
                swap(arr,j,j+1);
            }
        }
        if (!bool) break;
    }
}

public static void main(String... args) throws Exception {
    Random r = new Random(1);
    int SIZE = 30;
    int N = 1000;
    int[] arr = new int[SIZE];
    int[] millisBubble = new int[N];
    int[] millisInsertion = new int[N];
    System.out.println("start");
    //warm up:
    for (int t = 0; t < 100; t++) { 
        insertionSort(arr);
    }
    for (int t = 0; t < N; t++) { 
        arr = generateRandom(r, SIZE);
        int[] tempArr = Arrays.copyOf(arr, arr.length);

        long start = System.nanoTime();
        insertionSort(tempArr);
        millisInsertion[t] = (int)(System.nanoTime()-start);

        tempArr = Arrays.copyOf(arr, arr.length);

        start = System.nanoTime();
        bubbleSort(tempArr);
        millisBubble[t] = (int)(System.nanoTime()-start);
    }
    int sum1 = 0;
    for (int x : millisBubble) {
        System.out.println(x);
        sum1 += x;
    }
    System.out.println("end of bubble. AVG = " + ((double)sum1)/millisBubble.length);
    int sum2 = 0;
    for (int x : millisInsertion) {
        System.out.println(x);
        sum2 += x;
    }
    System.out.println("end of insertion. AVG = " + ((double)sum2)/millisInsertion.length);
    System.out.println("bubble took " + ((double)sum1)/millisBubble.length + " while insertion took " + ((double)sum2)/millisBubble.length);
}

private static int[] generateRandom(Random r, int size) {
    int[] arr = new int[size];
    for (int i = 0 ; i < size; i++) 
        arr[i] = r.nextInt(size);
    return arr;
}

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EDIT:
(1) optimizing the bubble sort (updated above) reduced the total time taking to bubble sort to: 6043.806 not enough to make a significant change. Wilcoxon test is still conclusive: Insertion sort is faster.

(2) I also added a selection sort test (code attached) and compared it against insertion. The results are: selection took 4748.35 while insertion took 3540.114.
P_VALUE for wilcoxon is still below the range of numerical error (effectively ~=0)

code for selection sort used:

public static void selectionSort(int[] arr) {
    for (int i = 0; i < arr.length ; i++) { 
        int min = arr[i];
        int minElm = i;
        for (int j = i+1; j < arr.length ; j++) { 
            if (arr[j] < min) { 
                min = arr[j];
                minElm = j;
            }
        }
        swap(arr,i,minElm);
    }
}

Answer 5

The easier one first: why insertion sort over selection sort? Because insertion sort is in O(n) for optimal input sequences, i.e. if the sequence is already sorted. Selection sort is always in O(n^2).

Why insertion sort over bubble sort? Both need only a single pass for already sorted input sequences, but insertion sort degrades better. To be more specific, insertion sort usually performs better with a small number of inversion than bubble sort does. Source This can be explained because bubble sort always iterates over N-i elements in pass i while insertion sort works more like "find" and only needs to iterate over (N-i)/2 elements in average (in pass N-i-1) to find the insertion position. So, insertion sort is expected to be about two times faster than insertion sort on average.

Answer 6

EDIT: As IVlad points out in a comment, selection sort does only n swaps (and therefore only 3n writes) for any dataset, so insertion sort is very unlikely to beat it on account of doing fewer swaps -- but it will likely do substantially fewer comparisons. The reasoning below better fits a comparison with bubble sort, which will do a similar number of comparisons but many more swaps (and thus many more writes) on average.

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One reason why insertion sort tends to be faster than the other O(n^2) algorithms like bubble sort and selection sort is because in the latter algorithms, every single data movement requires a swap, which can be up to 3 times as many memory copies as are necessary if the other end of the swap needs to be swapped again later.

With insertion sort OTOH, if the next element to be inserted isn't already the largest element, it can be saved into a temporary location, and all lower elements shunted forward by starting from the right and using single data copies (i.e. without swaps). This opens up a gap to put the original element.

C code for insertion-sorting integers without using swaps:

void insertion_sort(int *v, int n) {
    int i = 1;
    while (i < n) {
        int temp = v[i];         // Save the current element here
        int j = i;

        // Shunt everything forwards
        while (j > 0 && v[j - 1] > temp) {
            v[j] = v[j - 1];     // Look ma, no swaps!  :)
            --j;
        }

        v[j] = temp;
        ++i;
    }
}