Merge Sort Time and Space Complexity
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.
Answer
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.
16
/
8
/
4
/
2
/ \
1 1
notice how number of space used is 32 = 2n = 2*16 < 3n
Then it merge upwards
16
/
8
/
4
/ \
2 2
/ \
1 1
which is 34 < 3n. Then it merge upwards
16
/
8
/ \
4 4
/
2
/ \
1 1
36 < 16 * 3 = 48
then it merge upwards
16
/ \
8 8
/ \
4 4
/ \
2 2
/\
1 1
16 + 16 + 14 = 46 < 3*n = 48
in a larger case, n = 64
64
/ \
32 32
/ \
16 16
/ \
8 8
/ \
4 4
/ \
2 2
/\
1 1
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