# More examples of recursion

## Merge sort

Divide and conquer sort:
```  /**
POST -- elementArray is sorted into non-decreasing order
**/
public void sort(Comparable[] elementArray)
{
recMergeSort(0,elementArray.length -1,elementArray);
}

/**
pre: first, last are legal indices of elementArray
post:  elementArray[firstIndex..last] is sorted in
non-decreasing order
**/
protected void recMergeSort(int first, int last,
Comparable[] elementArray)
{
int middle = (first+last)/2;    // middle index of array

if (last - first > 0)        // more than 1 elt
{
// Sort first half of list
recMergeSort(first,middle,elementArray);
// Sort second half of list
recMergeSort(middle+1,last,elementArray);
// Merge two halves
mergeRuns(first,middle,last,elementArray);
}
}

```
Easy to show recMergeSort is correct if mergeRuns is.

Method mergeRuns is where all the work takes place.

Notes.

• you can't merge the halves in place. Hence we need an auxiliary array to copy into.
• mergeRuns is not recursive!
```/**
PRE -- sortArray[first..middle] and
sortArray[middle+1..last] are sorted,
and each range is non-empty.
POST -- sortArray[first..last] is sorted
**/
protected void mergeRuns (int first, int middle, int last,
Comparable[] sortArray)
{
int elementCount = last-first+1;    // # elts in array
// temp array to hold elts to be merged
Comparable[] tempArray = new Comparable[elementCount];

// copy elts of sortArray into tempArray in preparation
// for merging
for (int index=0; index < elementCount; index++)
tempArray[index] = sortArray[first+index];

int outIndex = first;   // posn written to in sortArray
int run1 = 0;               // index of first, second runs
int run2 = middle-first+1;
int endRun1 = middle-first; // end of 1st, 2nd runs
int endRun2 = last-first;

// merge runs until one of them is exhausted
while (run1 <= endRun1 && run2 <= endRun2)
{
if (tempArray[run1].lessThan(tempArray[run2]))
{   // if elt from run1 is smaller add it to sortArray
sortArray[outIndex] = tempArray[run1];
run1++;
}
else
{               // add elt from run2 to sortArray
sortArray[outIndex] = tempArray[run2];
run2++;
}
outIndex++;
}  // while

// Out of elts from one run, but other may have elts
// add remaining elements from run1 if any left
while (run1 <= endRun1)
{
sortArray[outIndex] = tempArray[run1];
outIndex++;
run1++;
}

// add remaining elements from run2 if any left
while (run2 <= endRun2)
{
sortArray[outIndex] = tempArray[run2];
outIndex++;
run2++;
}
}

```
• It is easy to convince yourself that mergeRuns is correct. (A formal proof of correctness of iterative algorithms is actually harder than for recursive programs!)
• It is also easy to see that if the portion of the array under consideration has k elements (i.e., k = last-first+1), then the complexity of mergeRuns is O(k):
• If only look at comparisons then clear that every comparison (i.e., call to lessThan) in the if statement in the while loop results in an element being copied into sortArray.
• In the worst case, you run out of all elts in one run when there is only 1 element left in the other run: k-1 comparisons, giving O(k)
• If count copies of elements, then also O(k) since k copies in copying sortArray into tempArray, and then k more copies in putting elts back (in order) into sortArray.

### Determine the complexity of recMergeSort.

Claim: complexity is O(n log n) for sort of n elements.

Easiest to prove this if n = 2m for some m.

Prove by induction on m that sort of n = 2m elements takes <= n log n = 2m * m compares.

Base case: m=0, so n = 1. Don't do anything, so 0 compares <= 20 * 0.

Inductive Hypothesis: Suppose claim is true for m-1.

Inductive Step: Will show that claim is true for m.

recMergeSort of n = 2m elements proceeds by doing recMergeSort of two lists of size n / 2 = 2m-1, and then call of mergeRuns on list of size n = 2m.

Therefore,

```#(compares) <=  2m-1 * (m-1) + 2m-1 * (m-1) + 2m
= 2*(2m-1 * (m-1)) + 2m
= 2m * (m-1) + 2m
= 2m * ((m-1) + 1)
= 2m * m```
Therefore #(compares) <= 2m * m = n log n

### End of proof.

• Thus if n = 2m for some m, #(compares) <= n log n to do recursiveMergeSort
• It is not hard to show that a similar bound holds for n not a power of 2.
• Therefore O(n log n) compares. Same for number of copies.
• Can cut down number of copies significantly if merge back and forth between two arrays rather than copy and then merge.

## Quicksort

There is one last divide and conquer sorting algorithm: Quicksort.

While mergesort divided in half, sorted each half, and then merged (where all work is in the merge), Quicksort works in the opposite order.

That is, Quicksort splits the array (with lots of work), sorts each part, and then puts together (trivially).

```/**
POST -- "elementArray" sorted into non-decreasing order
**/
public void quicksort(Comparable[] elementArray)
{
Q_sort(0, elementArray.length - 1, elementArray);
}

/**
PRE -- left <= right are legal indices of table.
POST -- table[left..right] sorted in non-decreasing order
**/
protected void Q_sort (int left, int right, Comparable[] table)
{
if (right > left)   // More than 1 elt in table
{
int pivotIndex = partition(left,right,table);
// table[Left..pivotIndex] <= table[pivotIndex+1..right]
Q_sort(left, pivotIndex-1, table);      // Quicksort small elts
Q_sort(pivotIndex+1, right, table);     // Quicksort large elts
}
}
```
If partition works then Q_sort (and hence quicksort) clearly works.

Note: it always makes a recursive call on a smaller array (easy to blow so it doesn't and then never terminates).

Partition: Algorithm below starts out by ensuring the elt at the left edge of the table is <= the one at the right. This allows guards on the while loops to be simpler and speeds up the algorithm by about 20% or more. Other optimizations can make it even faster.

```/**
post: table[left..pivotIndex-1] <= pivot
and pivot <= table[pivotIndex+1..right]
**/
protected int partition (int left, int right, Comparable[] table)
{
Comparable tempElt;         // used for swaps
int smallIndex = left;      // index of current posn in left (small elt) partition
int bigIndex = right;       // index of current posn in right (big elt) partition

if (table[bigIndex].lessThan(table[smallIndex]))
{   // put sentinel at table[bigIndex] so don't
// walk off right edge of table in loop below
tempElt = table[bigIndex];
table[bigIndex] = table[smallIndex];
table[smallIndex] = tempElt;
}

Comparable pivot = table[left]; // pivot is fst elt
// Now table[smallIndex] = pivot <= table[bigIndex]
do
{
do                          // scan right from smallIndex
smallIndex++;
while (table[smallIndex].lessThan(pivot));

do                          // scan left from bigIndex
bigIndex--;
while (pivot.lessThan(table[bigIndex]));

// Now table[smallIndex] >= pivot >= table[bigIndex]

if (smallIndex < bigIndex)
{   // if big elt to left of small element, swap them
tempElt = table[smallIndex];
table[smallIndex] = table[bigIndex];
table[bigIndex] = tempElt;
} // if
} while (smallIndex < bigIndex);
// Move pivot into correct pos'n bet'n small & big elts

int pivotIndex = bigIndex;      // pivot goes where bigIndex got stuck

// swap pivot elt w/small elt at pivotIndex
tempElt = table[pivotIndex];
table[pivotIndex] = table[left];
table[left] = tempElt;

return pivotIndex;
}
```
The basic idea of the algorithm:
• Start with smallIndex and bigIndex at the left and right edges of the array.
• Move each of them toward the middle until smallIndex is on a "big" element (one >= than pivot) and bigIndex is on a small one.
• As long as the indices haven't crossed (i.e. as long as smallIndex < bigIndex) swap them so that the small elt goes on the small side and the big elt on the big side.
• When they cross, swap the rightmost small elt (pointed to by bigIndex) with the pivot element and return its new index to Q_sort. Clearly at the end,
• table[left..pivotIndex-1] <= pivot <= table[pivotIndex+1..right]
• Note that the algorithm given above will not work when the first and last elements have the same value and that value happens to be either the smallest or largest in the array. Class discussion will reveal ways to handle that tricky case.

The complexity of QuickSort is harder to evaluate than MergeSort because the pivotIndex need not always be in the middle of the array (in the worst case pivotIndex = left or right).

Partition is clearly O(n) because every comparison results in smallIndex or bigIndex moving toward the other and quit when they cross.

In the best case the pivot element is always in the middle and the analysis results in
O(n log n), exactly like MergeSort.

In the worst case the pivot is at the ends and QuickSort behaves like SelectionSort, giving O(n2).

Careful analysis shows that QuickSort is O(n log n) in the average case (under reasonable assumptions on distribution of elements of array).

Compare the algorithms with real data:
```Complxity   100 elts    100 elts    500 elts    500 elts    1000 elts   1000 elts
unordered   ordered     unordered   ordered     unordered   ordered
Insertion   0.033       0.002       0.75        0.008       3.2         .017
Selection   0.051       0.051       1.27        1.31        5.2         5.3
Merge       0.016       0.015       0.108       0.093       0.24        0.20
Quick       0.009       0.044       0.058       1.12        0.13        4.5

```
Notice that for Insertion or Selection sorts, doubling size of list increases time by 4 times (for unordered case), whereas for Merge and Quick sorts bit more than doubles time. Calculate (1000 log 1000) / (500 log 500) = 2 * (log 1000 / log 500) ~ 2 * (10/9) ~ 2.2