Merge Sort Time and Space Complexity

https://stackoverflow.com/questions/10342890

03-06-2021
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Pergunta

Let's take this implementation of Merge Sort as an example

void mergesort(Item a[], int l, int r) {
if (r <= l) return;
int m = (r+l)/2;
mergesort(a, l, m);  ------------ (1)
mergesort(a, m+1, r); ------------(2)
merge(a, l, m, r);

a) The time complexity of this Merge Sort is O(nlg(n)). Will parallelizing (1) and (2) give any practical gain ? Theorotically, it appears that after parallelizing them also you would end up in O(nlg(n). But practically can we get any gains ?

b) Space complexity of this Merge Sort here is O(n). However, if I choose to perform in-place merge sort using linked lists (not sure if it can be done with arrays reasonably) will the space complexity become O(lg(n)), since you have to account for recursion stack frame size ? Can we treat O(lg(n)) as constant since it cannot be more than 64 ? I may have misunderstood this at couple of places. What exactly is the significance of 64 ?

c) http://www.cprogramming.com/tutorial/computersciencetheory/sortcomp.html says merge sort requires constant space using linked lists. How ? Did they treat O(lg(n) constant ?

d) [Added to get more clarity] For space complexity calculation is it fair to assume the input array or list is already in memory ? When I do complexity calculations I always calculate the "Extra" space I will be needing besides the space already taken by input. Otherwise space complexity will always be O(n) or worse.

Solução

a) Yes - in a perfect world you'd have to do log n merges of size n, n/2, n/4 ... (or better said 1, 2, 3 ... n/4, n/2, n - they can't be parallelized), which gives O(n). It still is O(n log n). In not-so-perfect-world you don't have infinite number of processors and context-switching and synchronization offsets any potential gains.

b) Space complexity is always Ω(n) as you have to store the elements somewhere. Additional space complexity can be O(n) in an implementation using arrays and O(1) in linked list implementations. In practice implementations using lists need additional space for list pointers, so unless you already have the list in memory it shouldn't matter.

edit if you count stack frames, then it's O(n)+ O(log n) , so still O(n) in case of arrays. In case of lists it's O(log n) additional memory.

c) Lists only need some pointers changed during the merge process. That requires constant additional memory.

d) That's why in merge-sort complexity analysis people mention 'additional space requirement' or things like that. It's obvious that you have to store the elements somewhere, but it's always better to mention 'additional memory' to keep purists at bay.

Outras dicas

MergeSort time Complexity is O(nlgn) which is a fundamental knowledge. Merge Sort space complexity will always be O(n) including with arrays. If you draw the space tree out, it will seem as though the space complexity is O(nlgn). However, as the code is a Depth First code, you will always only be expanding along one branch of the tree, therefore, the total space usage required will always be bounded by O(3n) = O(n).

For example, if you draw the space tree out, it seems like it is O(nlgn)

                             16                                 | 16
                            /  \                              
                           /    \
                          /      \
                         /        \
                        8          8                            | 16
                       / \        / \
                      /   \      /   \
                     4     4    4     4                         | 16
                    / \   / \  / \   / \
                   2   2 2   2.....................             | 16
                  / \  /\ ........................
                 1  1  1 1 1 1 1 1 1 1 1 1 1 1 1 1              | 16

where height of tree is O(logn) => Space complexity is O(nlogn + n) = O(nlogn). However, this is not the case in the actual code as it does not execute in parallel. For example, in the case where N = 16, this is how the code for mergesort executes. N = 16.

notice how number of space used is 32 = 2n = 2*16 < 3n

Then it merge upwards

which is 34 < 3n. Then it merge upwards

36 < 16 * 3 = 48

then it merge upwards

16 + 16 + 14 = 46 < 3*n = 48

in a larger case, n = 64

which is 64*3 <= 3*n = 3*64

You can prove this by induction for the general case.

Therefore, space complexity is always bounded by O(3n) = O(n) even if you implement with arrays as long as you clean up used space after merging and not execute code in parallel but sequential.

Example of my implementation is given below:

templace<class X> 
void mergesort(X a[], int n) // X is a type using templates
{
    if (n==1)
    {
        return;
    }
    int q, p;
    q = n/2;
    p = n/2;
    //if(n % 2 == 1) p++; // increment by 1
    if(n & 0x1) p++; // increment by 1
        // note: doing and operator is much faster in hardware than calculating the mod (%)
    X b[q];

    int i = 0;
    for (i = 0; i < q; i++)
    {
        b[i] = a[i];
    }
    mergesort(b, i);
    // do mergesort here to save space
    // http://stackoverflow.com/questions/10342890/merge-sort-time-and-space-complexity/28641693#28641693
    // After returning from previous mergesort only do you create the next array.
    X c[p];
    int k = 0;
    for (int j = q; j < n; j++)
    {
        c[k] = a[j];
        k++;
    }
    mergesort(c, k);
    int r, s, t;
    t = 0; r = 0; s = 0;
    while( (r!= q) && (s != p))
    {
        if (b[r] <= c[s])
        {
            a[t] = b[r];
            r++;
        }
        else
        {
            a[t] = c[s];
            s++;
        }
        t++;
    }
    if (r==q)
    {
        while(s!=p)
        {
            a[t] = c[s];
            s++;
            t++;
        }
    }
    else
    {
        while(r != q)
        {
            a[t] = b[r];
            r++;
            t++;
        }
    }
    return;
}

Hope this helps!=)

Soon Chee Loong,

University of Toronto

a) Yes, of course, parallelizing merge sort can be very beneficial. It remains nlogn, but your constant should be significantly lower.

b) Space complexity with a linked list should be O(n), or more specifically O(n) + O(logn). Note that that's a +, not a *. Don't concern yourself with constants much when doing asymptotic analysis.

c) In asymptotic analysis, only the dominant term in the equation matters much, so the fact that we have a + and not a * makes it O(n). If we were duplicating the sublists all over, I believe that would be O(nlogn) space - but a smart linked-list-based merge sort can share regions of the lists.

Worst-case performance of merge sort : O(n log n), Best-case performance of merge sort : O(n log n) typicaly, O(n) natural variant, Average performance of merge sort : O(n log n), Worst-case space complexity of merge sort : О(n) total, O(n) auxiliary

Simple and smart thinking.

Total levels (L) = log2(N). At the last level number of nodes = N.

step 1 : let's assume for all levels (i) having nodes = x(i).

step 2 : so time complexity = x1 + x2 + x3 + x4 + .... + x(L-1) + N(for i = L);

step 3 : fact we know , x1,x2,x3,x4...,x(L-1) < N

step 4 : so let's consider x1=x2=x3=...=x(L-1)=N

step 5 : So time complexity = (N+N+N+..(L)times)

Time complexity = O(N*L); put L = log(N);

Time complexity = O(N*log(N))

We use the extra array while merging so,

Space complexity: O(N).

Hint: Big O(x) time means, x is the smallest time for which we can surely say with proof that it will never exceed x in average case

merge sort space complexity is O(nlogn), this is quite obvious considering that it can go to at maximum of O(logn) recursions and for each recursion there is additional space of O(n) for storing the merged array that needs to be reassigned. For those who are saying O(n) please don't forget that it is O(n) for reach stack frame depth.

for both best and worst case the complexity is O(nlog(n)) . though extra N size of array is needed in each step so space complexity is O(n+n) is O(2n) as we remove constant value for calculating complexity so it is O(n)

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