One of mergesort’s most attractive properties is that it guarantees to sort any array of N items in time proportional to N * log N. Its prime disadvantage is that it uses extra space proportional to N.
Top-down mergesort
It is one of the best-known examples of the utility of the divide-and-conquer paradigm for efficient algorithm design. This recursive code is the basis for an inductive proof that the algorithm sorts the array: if it sorts the two subarrays, it sorts the whole array, by merging together the subarrays.
To understand mergesort, it is worthwhile to consider carefully the dynamics of the method calls, shown in the trace at right. To sort a[0..15], the sort() method calls itself to sort a[0..7] then calls itself to sort a[0..3] and a[0..1] before finally doing the first merge of a[0] with a[1] after calling itself to sort a[0] and then a[1] (for brevity, we omit the calls for the base-case 1-entry sorts in the trace). Then the next merge is a[2] with a[3] and then a[0..1] with a[2..3] and so forth. From this trace, we see that the sort code simply provides an organized way to sequence the calls to the merge() method.
public class Merge {
// stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
public static void merge(Comparable[] a, Comparable[] aux, int lo, int mid, int hi) {
// precondition: a[lo .. mid] and a[mid+1 .. hi] are sorted subarrays
assert isSorted(a, lo, mid);
assert isSorted(a, mid + 1, hi);
// copy to aux[]
for (int k = lo; k <= hi; k++) {
aux[k] = a[k];
}
// merge back to a[]
int i = lo, j = mid + 1;
for (int k = lo; k <= hi; k++) {
if (i > mid) a[k] = aux[j++];
else if (j > hi) a[k] = aux[i++];
else if (less(aux[j], aux[i])) a[k] = aux[j++];
else a[k] = aux[i++];
}
// postcondition: a[lo .. hi] is sorted
assert isSorted(a, lo, hi);
}
// mergesort a[lo..hi] using auxiliary array aux[lo..hi]
private static void sort(Comparable[] a, Comparable[] aux, int lo, int hi) {
if (hi <= lo) return;
int mid = lo + (hi - lo) / 2;
sort(a, aux, lo, mid);
sort(a, aux, mid + 1, hi);
merge(a, aux, lo, mid, hi);
}
public static void sort(Comparable[] a) {
Comparable[] aux = new Comparable[a.length];
sort(a, aux, 0, a.length - 1);
assert isSorted(a);
}
/**
* ********************************************************************
* Helper sorting functions
* *********************************************************************
*/
// is v < w ?
private static boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
// exchange a[i] and a[j]
private static void exch(Object[] a, int i, int j) {
Object swap = a[i];
a[i] = a[j];
a[j] = swap;
}
/**
* ********************************************************************
* Check if array is sorted - useful for debugging
* *********************************************************************
*/
private static boolean isSorted(Comparable[] a) {
return isSorted(a, 0, a.length - 1);
}
private static boolean isSorted(Comparable[] a, int lo, int hi) {
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i - 1])) return false;
return true;
}
}
Another approach is described as Indirect sort. This sort algorithm does not rearrange the array, but returns an int[] array, say perm, such that perm[i] is the index of the ith smallest entry in the array:
public class Merge {
/**
* ********************************************************************
* Helper sorting functions
* *********************************************************************
*/
// is v < w ?
private static boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
// exchange a[i] and a[j]
private static void exch(Object[] a, int i, int j) {
Object swap = a[i];
a[i] = a[j];
a[j] = swap;
}
/**
* ********************************************************************
* Check if array is sorted - useful for debugging
* *********************************************************************
*/
private static boolean isSorted(Comparable[] a) {
return isSorted(a, 0, a.length - 1);
}
private static boolean isSorted(Comparable[] a, int lo, int hi) {
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i - 1])) return false;
return true;
}
/**
* ********************************************************************
* Index mergesort
* *********************************************************************
*/
// stably merge a[lo .. mid] with a[mid+1 .. hi] using aux[lo .. hi]
private static void merge(Comparable[] a, int[] index, int[] aux, int lo, int mid, int hi) {
// copy to aux[]
for (int k = lo; k <= hi; k++) {
aux[k] = index[k];
}
// merge back to a[]
int i = lo, j = mid + 1;
for (int k = lo; k <= hi; k++) {
if (i > mid) index[k] = aux[j++];
else if (j > hi) index[k] = aux[i++];
else if (less(a[aux[j]], a[aux[i]])) index[k] = aux[j++];
else index[k] = aux[i++];
}
}
// return a permutation that gives the elements in a[] in ascending order
// do not change the original array a[]
public static int[] indexSort(Comparable[] a) {
int N = a.length;
int[] index = new int[N];
for (int i = 0; i < N; i++)
index[i] = i;
int[] aux = new int[N];
sort(a, index, aux, 0, N - 1);
return index;
}
// mergesort a[lo..hi] using auxiliary array aux[lo..hi]
private static void sort(Comparable[] a, int[] index, int[] aux, int lo, int hi) {
if (hi <= lo) return;
int mid = lo + (hi - lo) / 2;
sort(a, index, aux, lo, mid);
sort(a, index, aux, mid + 1, hi);
merge(a, index, aux, lo, mid, hi);
}
}
Top-down merge sort uses between 1⁄2*N*lgN and N*lgN compares to sort any array of length N, also it uses at most 6 N lgN array accesses to sort an array of length N.
We can cut the running time of mergesort substantially with some carefully considered modifications to the implementation.
Use insertion sort for small subarrays. We can improve most recursive algorithms by handling small cases differently, because the recursion guarantees that the method will be used often for small cases, so improvements in handling them lead to improvements in the whole algorithm. In the case of sorting, we know that insertion sort (or selection sort) is simple and therefore likely to be faster than mergesort for tiny subarrays. As usual, a visual trace provides insight into the operation of mergesort. The visual trace on the facing page shows the operation of a mergesort implementation with a cutoff for small subarrays. Switching to insertion sort for small subarrays (length 15 or less, say) will improve the running time of a typical mergesort implementation by 10 to 15 percent.
Test whether the array is already in order. We can reduce the running time to be linear for arrays that are already in order by adding a test to skip the call to merge() if a[mid] is less than or equal to a[mid+1]. With this change, we still do all the recursive calls, but the running time for any sorted subarray is linear.
Eliminate the copy to the auxiliary array. It is possible to eliminate the time(but not the space) taken to copy to the auxiliary array used for merging. To do so, we use two invocations of the sort method: one takes its input from the given array and puts the sorted output in the auxiliary array; the other takes its input from the auxiliary array and puts the sorted output in the given array. With this approach, in a bit of recursive trickery, we can arrange the recursive calls such that the computation switches the roles of the input array and the auxiliary array at each level.
Below is the code to implemente all of the improvments:
public class MergeX {
private static final int CUTOFF = 7; // cutoff to insertion sort
// This class should not be instantiated.
private MergeX() { }
private static void merge(Comparable[] src, Comparable[] dst, int lo, int mid, int hi) {
// precondition: src[lo .. mid] and src[mid+1 .. hi] are sorted subarrays
assert isSorted(src, lo, mid);
assert isSorted(src, mid+1, hi);
int i = lo, j = mid+1;
for (int k = lo; k <= hi; k++) {
if (i > mid) dst[k] = src[j++];
else if (j > hi) dst[k] = src[i++];
else if (less(src[j], src[i])) dst[k] = src[j++]; // to ensure stability
else dst[k] = src[i++];
}
// postcondition: dst[lo .. hi] is sorted subarray
assert isSorted(dst, lo, hi);
}
private static void sort(Comparable[] src, Comparable[] dst, int lo, int hi) {
// if (hi <= lo) return;
if (hi <= lo + CUTOFF) {
insertionSort(dst, lo, hi);
return;
}
int mid = lo + (hi - lo) / 2;
sort(dst, src, lo, mid);
sort(dst, src, mid+1, hi);
// if (!less(src[mid+1], src[mid])) {
// for (int i = lo; i <= hi; i++) dst[i] = src[i];
// return;
// }
// using System.arraycopy() is a bit faster than the above loop
if (!less(src[mid+1], src[mid])) {
System.arraycopy(src, lo, dst, lo, hi - lo + 1);
return;
}
merge(src, dst, lo, mid, hi);
}
/**
* Rearranges the array in ascending order, using the natural order.
* @param a the array to be sorted
*/
public static void sort(Comparable[] a) {
Comparable[] aux = a.clone();
sort(aux, a, 0, a.length-1);
assert isSorted(a);
}
// sort from a[lo] to a[hi] using insertion sort
private static void insertionSort(Comparable[] a, int lo, int hi) {
for (int i = lo; i <= hi; i++)
for (int j = i; j > lo && less(a[j], a[j-1]); j--)
exch(a, j, j-1);
}
// exchange a[i] and a[j]
private static void exch(Comparable[] a, int i, int j) {
Comparable swap = a[i];
a[i] = a[j];
a[j] = swap;
}
// is a[i] < a[j]?
private static boolean less(Comparable a, Comparable b) {
return (a.compareTo(b) < 0);
}
/***********************************************************************
* Check if array is sorted - useful for debugging
***********************************************************************/
private static boolean isSorted(Comparable[] a) {
return isSorted(a, 0, a.length - 1);
}
private static boolean isSorted(Comparable[] a, int lo, int hi) {
for (int i = lo + 1; i <= hi; i++)
if (less(a[i], a[i-1])) return false;
return true;
}
}
Bottom-up mergesort The recursive implementation of mergesort is prototypical of the divide-and-conquer algorithm design paradigm, where we solve a large problem by dividing it into pieces, solving the subproblems, then using the solutions for the pieces to solve the whole problem. Even though we are thinking in terms of merging together two large subarrays, the fact is that most merges are merging together tiny subarrays. Another way to implement mergesort is to organize the merges so that we do all the merges of tiny subarrays on one pass, then do a second pass to merge those subarrays in pairs, and so forth, continuing until we do a merge that encompasses the whole array. This method requires even less code than the standard recursive implementation. We start by doing a pass of 1-by-1 merges (considering individual items as subarrays of size 1), then a pass of 2-by-2 merges (merge subarrays of size 2 to make subarrays of size 4), then 4-by-4 merges, and so forth. The sec- ond subarray may be smaller than the first in the last merge on each pass (which is no problem for merge()), but otherwise all merges involve subar- rays of equal size, doubling the sorted subarray size for the next pass.
public class MergeBU {
// stably merge a[lo..m] with a[m+1..hi] using aux[lo..hi]
private static void merge(Comparable[] a, Comparable[] aux, int lo, int m, int hi) {
// copy to aux[]
System.arraycopy(a, lo, aux, lo, hi + 1 - lo);
// merge back to a[]
int i = lo, j = m + 1;
for (int k = lo; k <= hi; k++) {
if (i > m) a[k] = aux[j++];
else if (j > hi) a[k] = aux[i++];
else if (less(aux[j], aux[i])) a[k] = aux[j++];
else a[k] = aux[i++];
}
}
// bottom-up mergesort
public static void sort(Comparable[] a) {
int N = a.length;
Comparable[] aux = new Comparable[N];
for (int n = 1; n < N; n = n + n) {
for (int i = 0; i < N - n; i += n + n) {
int m = i + n - 1;
int hi = Math.min(i + n + n - 1, N - 1);
merge(a, aux, i, m, hi);
}
}
assert isSorted(a);
}
/**
* ********************************************************************
* Helper sorting functions
* *********************************************************************
*/
// is v < w ?
private static boolean less(Comparable v, Comparable w) {
return (v.compareTo(w) < 0);
}
// exchange a[i] and a[j]
private static void exch(Object[] a, int i, int j) {
Object swap = a[i];
a[i] = a[j];
a[j] = swap;
}
/**
* ********************************************************************
* Check if array is sorted - useful for debugging
* *********************************************************************
*/
private static boolean isSorted(Comparable[] a) {
for (int i = 1; i < a.length; i++)
if (less(a[i], a[i - 1])) return false;
return true;
}
}
Bottom-up merge sort uses between 1⁄2 * N* lgN and N* lgN compares and at most 6 * N * lgN array accesses to sort an array of length N. When the array length is a power of 2, top-down and bottom-up mergesort perform precisely the same compares and array accesses, just in a different order. When the array length is not a power of 2, the sequence of compares and array accesses for the two algorithms will be different.