Most of the mergesort implementations I see are similar to this. intro to algorithms book along with online implentations I search for. My recursion chops don't go much further than messing with Fibonacci generation (which was simple enough) so maybe it's the multiple recursions blowing my mind, but I can't even step through the code and understand whats going on even before I even hit the merge function.
How is it stepping through this? Is there some strategy or reading I should undergo to better understand the process here?
void mergesort(int *a, int*b, int low, int high)
{
int pivot;
if(low<high)
{
pivot=(low+high)/2;
mergesort(a,b,low,pivot);
mergesort(a,b,pivot+1,high);
merge(a,b,low,pivot,high);
}
}
and the merge(although frankly I'm mentally stuck before I even get to this part)
void merge(int *a, int *b, int low, int pivot, int high)
{
int h,i,j,k;
h=low;
i=low;
j=pivot+1;
while((h<=pivot)&&(j<=high))
{
if(a[h]<=a[j])
{
b[i]=a[h];
h++;
}
else
{
b[i]=a[j];
j++;
}
i++;
}
if(h>pivot)
{
for(k=j; k<=high; k++)
{
b[i]=a[k];
i++;
}
}
else
{
for(k=h; k<=pivot; k++)
{
b[i]=a[k];
i++;
}
}
for(k=low; k<=high; k++) a[k]=b[k];
}
MERGE SORT:
1) Split the array in half
2) Sort the left half
3) Sort the right half
4) Merge the two halves together
I think the "sort" function name in MergeSort is a bit of a misnomer, it should really be called "divide".
Here is a visualization of the algorithm in process.
Each time the function recurses, it's working on a smaller and smaller subdivision of the input array, starting with the left half of it. Each time the function returns from recursion, it will continue on and either start working on the right half, or recurse up again and work on a larger half.
Like this
[************************]mergesort
[************]mergesort(lo,mid)
[******]mergesort(lo,mid)
[***]mergesort(lo,mid)
[**]mergesort(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[***]mergesort(mid+1,hi)
[**]mergesort*(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[******]merge
[******]mergesort(mid+1,hi)
[***]mergesort(lo,mid)
[**]mergesort(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[***]mergesort(mid+1,hi)
[**]mergesort(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[******]merge
[************]merge
[************]mergesort(mid+1,hi)
[******]mergesort(lo,mid)
[***]mergesort(lo,mid)
[**]mergesort(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[***]mergesort(mid+1,hi)
[**]mergesort(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[******]merge
[******]mergesort(mid+1,hi)
[***]mergesort(lo,mid)
[**]mergesort*(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[***]mergesort(mid+1,hi)
[**]mergesort(lo,mid)
[**]mergesort(mid+1,hi)
[***]merge
[******]merge
[************]merge
[************************]merge
An obvious thing to do would be to try this merge sort on a small array, say size 8 (power of 2 is convenient here), on paper. Pretend you are a computer executing the code, and see if it starts to become a bit clearer.
Your question is a bit ambiguous because you don't explain what you find confusing, but it sounds like you are trying to unroll the recursive calls in your head. Which may or may not be a good thing, but I think it can easily lead to having too much in your head at once. Instead of trying to trace the code from start to end, see if you can understand the concept abstractly. Merge sort:
Splits the array in half
Sorts the left half
Sorts the right half
Merges the two halves together
(1) should be fairly obvious and intuitive to you. For step (2) the key insight is this, the left half of an array... is an array. Assuming your merge sort works, it should be able to sort the left half of the array. Right? Step (4) is actually a pretty intuitive part of the algorithm. An example should make it trivial:
at the start
left: [1, 3, 5], right: [2, 4, 6, 7], out: []
after step 1
left: [3, 5], right: [2, 4, 6, 7], out: [1]
after step 2
left: [3, 5], right: [4, 6, 7], out: [1, 2]
after step 3
left: [5], right: [4, 6, 7], out: [1, 2, 3]
after step 4
left: [5], right: [6, 7], out: [1, 2, 3, 4]
after step 5
left: [], right: [6, 7], out: [1, 2, 3, 4, 5]
after step 6
left: [], right: [7], out: [1, 2, 3, 4, 5, 6]
at the end
left: [], right: [], out: [1, 2, 3, 4, 5, 6, 7]
So assuming that you understand (1) and (4), another way to think of merge sort would be this. Imagine someone else wrote mergesort() and you're confident that it works. Then you could use that implementation of mergesort() to write:
sort(myArray)
{
leftHalf = myArray.subArray(0, myArray.Length/2);
rightHalf = myArray.subArray(myArray.Length/2 + 1, myArray.Length - 1);
sortedLeftHalf = mergesort(leftHalf);
sortedRightHalf = mergesort(rightHalf);
sortedArray = merge(sortedLeftHalf, sortedRightHalf);
}
Note that sort doesn't use recursion. It just says "sort both halves and then merge them". If you understood the merge example above then hopefully you see intuitively that this sort function seems to do what it says... sort.
Now, if you look at it more carefully... sort() looks pretty much exactly like mergesort()! That's because it is mergesort() (except it doesn't have base cases because it's not recursive!).
But that's how I like thinking of recursive functions--assume that the function works when you call it. Treat it as a black box that does what you need it to. When you make that assumption, figuring out how to fill in that black box is often easy. For a given input, can you break it down into smaller inputs to feed to your black box? After you solve that, the only thing that's left is handling the base cases at the start of your function (which are the cases where you don't need to make any recursive calls. For example, mergesort([]) just returns an empty array; it doesn't make a recursive call to mergesort()).
Finally, this is a bit abstract, but a good way to understand recursion is actually to write mathematical proofs using induction. The same strategy used to write an proof by induction is used to write a recursive function:
Math proof:
Show the claim is true for the base cases
Assume it is true for inputs smaller than some n
Use that assumption to show that it is still true for an input of size n
Recursive function:
Handle the base cases
Assume that your recursive function works on inputs smaller than some n
Use that assumption to handle an input of size n
Concerning the recursion part of the merge sort, I've found this page to be very very helpful. You can follow the code as it's being executed. It shows you what gets executed first, and what follows next.
Tom
the mergesort() simply divides the array in two halves until the if condition fails that is low < high. As you are calling mergesort() twice : one with low to pivot and second with pivot+1 to high, this will divide the sub arrays even more further.
Lets take an example :
a[] = {9,7,2,5,6,3,4}
pivot = 0+6/2 (which will be 3)
=> first mergesort will recurse with array {9,7,2} : Left Array
=> second will pass the array {5,6,3,4} : Right Array
It will repeat until you have 1 element in each left as well as right array.
In the end you'll have something similar to this :
L : {9} {7} {2} R : {5} {6} {3} {4} (each L and R will have further sub L and R)
=> which on call to merge will become
L(L{7,9} R{2}) : R(L{5,6} R{3,4})
As you can see that each sub array are getting sorted in the merge function.
=> on next call to merge the next L and R sub arrays will get in order
L{2,7,9} : R{3,4,5,6}
Now both L and R sub array are sorted within
On last call to merge they'll be merged in order
Final Array would be sorted => {2,3,4,5,6,7,9}
See the merging steps in answer given by #roliu
My apologies if this has been answered this way. I acknowledge that this is just a sketch, rather than a deep explanation.
While it is not obvious to see how the actual code maps to the recursion, I was able to understand the recursion in a general sense this way.
Take a the example unsorted set {2,9,7,5} as input. The merge_sort algorithm is denoted by "ms" for brevity below. Then we can sketch the operation as:
step 1: ms( ms( ms(2),ms(9) ), ms( ms(7),ms(5) ) )
step 2: ms( ms({2},{9}), ms({7},{5}) )
step 3: ms( {2,9}, {5,7} )
step 4: {2,5,7,9}
It is important to note that merge_sort of a singlet (like {2}) is simply the singlet (ms(2) = {2}), so that at the deepest level of recursion we get our first answer. The remaining answers then tumble like dominoes as the interior recursions finish and are merged together.
Part of the genius of the algorithm is the way it builds the recursive formula of step 1 automatically through its construction. What helped me was the exercise of thinking how to turn step 1 above from a static formula to a general recursion.
Trying to work out each and every step of a recursion is often not an ideal approach, but for beginners, it definitely helps to understand the basic idea behind recursion, and also to get better at writing recursive functions.
Here's a C solution to Merge Sort :-
#include <stdio.h>
#include <stdlib.h>
void merge_sort(int *, unsigned);
void merge(int *, int *, int *, unsigned, unsigned);
int main(void)
{
unsigned size;
printf("Enter the no. of integers to be sorted: ");
scanf("%u", &size);
int * arr = (int *) malloc(size * sizeof(int));
if (arr == NULL)
exit(EXIT_FAILURE);
printf("Enter %u integers: ", size);
for (unsigned i = 0; i < size; i++)
scanf("%d", &arr[i]);
merge_sort(arr, size);
printf("\nSorted array: ");
for (unsigned i = 0; i < size; i++)
printf("%d ", arr[i]);
printf("\n");
free(arr);
return EXIT_SUCCESS;
}
void merge_sort(int * arr, unsigned size)
{
if (size > 1)
{
unsigned left_size = size / 2;
int * left = (int *) malloc(left_size * sizeof(int));
if (left == NULL)
exit(EXIT_FAILURE);
for (unsigned i = 0; i < left_size; i++)
left[i] = arr[i];
unsigned right_size = size - left_size;
int * right = (int *) malloc(right_size * sizeof(int));
if (right == NULL)
exit(EXIT_FAILURE);
for (unsigned i = 0; i < right_size; i++)
right[i] = arr[i + left_size];
merge_sort(left, left_size);
merge_sort(right, right_size);
merge(arr, left, right, left_size, right_size);
free(left);
free(right);
}
}
/*
This merge() function takes a target array (arr) and two sorted arrays (left and right),
all three of them allocated beforehand in some other function(s).
It then merges the two sorted arrays (left and right) into a single sorted array (arr).
It should be ensured that the size of arr is equal to the size of left plus the size of right.
*/
void merge(int * arr, int * left, int * right, unsigned left_size, unsigned right_size)
{
unsigned i = 0, j = 0, k = 0;
while ((i < left_size) && (j < right_size))
{
if (left[i] <= right[j])
arr[k++] = left[i++];
else
arr[k++] = right[j++];
}
while (i < left_size)
arr[k++] = left[i++];
while (j < right_size)
arr[k++] = right[j++];
}
Here's the step-by-step explanation of the recursion :-
Let arr be [1,4,0,3,7,9,8], having the address 0x0000.
In main(), merge_sort(arr, 7) is called, which is the same as merge_sort(0x0000, 7).
After all of the recursions are completed, arr (0x0000) becomes [0,1,3,4,7,8,9].
| | |
| | |
| | |
| | |
| | |
arr - 0x0000 - [1,4,0,3,7,9,8] | | |
size - 7 | | |
| | |
left = malloc() - 0x1000a (say) - [1,4,0] | | |
left_size - 3 | | |
| | |
right = malloc() - 0x1000b (say) - [3,7,9,8] | | |
right_size - 4 | | |
| | |
merge_sort(left, left_size) -------------------> | arr - 0x1000a - [1,4,0] | |
| size - 3 | |
| | |
| left = malloc() - 0x2000a (say) - [1] | |
| left_size = 1 | |
| | |
| right = malloc() - 0x2000b (say) - [4,0] | |
| right_size = 2 | |
| | |
| merge_sort(left, left_size) -------------------> | arr - 0x2000a - [1] |
| | size - 1 |
| left - 0x2000a - [1] <-------------------------- | (0x2000a has only 1 element) |
| | |
| | |
| merge_sort(right, right_size) -----------------> | arr - 0x2000b - [4,0] |
| | size - 2 |
| | |
| | left = malloc() - 0x3000a (say) - [4] |
| | left_size = 1 |
| | |
| | right = malloc() - 0x3000b (say) - [0] |
| | right_size = 1 |
| | |
| | merge_sort(left, left_size) -------------------> | arr - 0x3000a - [4]
| | | size - 1
| | left - 0x3000a - [4] <-------------------------- | (0x3000a has only 1 element)
| | |
| | |
| | merge_sort(right, right_size) -----------------> | arr - 0x3000b - [0]
| | | size - 1
| | right - 0x3000b - [0] <------------------------- | (0x3000b has only 1 element)
| | |
| | |
| | merge(arr, left, right, left_size, right_size) |
| | i.e. merge(0x2000b, 0x3000a, 0x3000b, 1, 1) |
| right - 0x2000b - [0,4] <----------------------- | (0x2000b is now sorted) |
| | |
| | free(left) (0x3000a is now freed) |
| | free(right) (0x3000b is now freed) |
| | |
| | |
| merge(arr, left, right, left_size, right_size) | |
| i.e. merge(0x1000a, 0x2000a, 0x2000b, 1, 2) | |
left - 0x1000a - [0,1,4] <---------------------- | (0x1000a is now sorted) | |
| | |
| free(left) (0x2000a is now freed) | |
| free(right) (0x2000b is now freed) | |
| | |
| | |
merge_sort(right, right_size) -----------------> | arr - 0x1000b - [3,7,9,8] | |
| size - 4 | |
| | |
| left = malloc() - 0x2000c (say) - [3,7] | |
| left_size = 2 | |
| | |
| right = malloc() - 0x2000d (say) - [9,8] | |
| right_size = 2 | |
| | |
| merge_sort(left, left_size) -------------------> | arr - 0x2000c - [3,7] |
| | size - 2 |
| | |
| | left = malloc() - 0x3000c (say) - [3] |
| | left_size = 1 |
| | |
| | right = malloc() - 0x3000d (say) - [7] |
| | right_size = 1 |
| | |
| | merge_sort(left, left_size) -------------------> | arr - 0x3000c - [3]
| left - [3,7] was already sorted, but | | size - 1
| that doesn't matter to this program. | left - 0x3000c - [3] <-------------------------- | (0x3000c has only 1 element)
| | |
| | |
| | merge_sort(right, right_size) -----------------> | arr - 0x3000d - [7]
| | | size - 1
| | right - 0x3000d - [7] <------------------------- | (0x3000d has only 1 element)
| | |
| | |
| | merge(arr, left, right, left_size, right_size) |
| | i.e. merge(0x2000c, 0x3000c, 0x3000d, 1, 1) |
| left - 0x2000c - [3,7] <------------------------ | (0x2000c is now sorted) |
| | |
| | free(left) (0x3000c is now freed) |
| | free(right) (0x3000d is now freed) |
| | |
| | |
| merge_sort(right, right_size) -----------------> | arr - 0x2000d - [9,8] |
| | size - 2 |
| | |
| | left = malloc() - 0x3000e (say) - [9] |
| | left_size = 1 |
| | |
| | right = malloc() - 0x3000f (say) - [8] |
| | right_size = 1 |
| | |
| | merge_sort(left, left_size) -------------------> | arr - 0x3000e - [9]
| | | size - 1
| | left - 0x3000e - [9] <-------------------------- | (0x3000e has only 1 element)
| | |
| | |
| | merge_sort(right, right_size) -----------------> | arr - 0x3000f - [8]
| | | size - 1
| | right - 0x3000f - [8] <------------------------- | (0x3000f has only 1 element)
| | |
| | |
| | merge(arr, left, right, left_size, right_size) |
| | i.e. merge(0x2000d, 0x3000e, 0x3000f, 1, 1) |
| right - 0x2000d - [8,9] <----------------------- | (0x2000d is now sorted) |
| | |
| | free(left) (0x3000e is now freed) |
| | free(right) (0x3000f is now freed) |
| | |
| | |
| merge(arr, left, right, left_size, right_size) | |
| i.e. merge(0x1000b, 0x2000c, 0x2000d, 2, 2) | |
right - 0x1000b - [3,7,8,9] <------------------- | (0x1000b is now sorted) | |
| | |
| free(left) (0x2000c is now freed) | |
| free(right) (0x2000d is now freed) | |
| | |
| | |
merge(arr, left, right, left_size, right_size) | | |
i.e. merge(0x0000, 0x1000a, 0x1000b, 3, 4) | | |
(0x0000 is now sorted) | | |
| | |
free(left) (0x1000a is now freed) | | |
free(right) (0x1000b is now freed) | | |
| | |
| | |
| | |
I know this is an old question but wanted to throw my thoughts of what helped me understand merge sort.
There are two big parts to merge sort
Splitting of the array into smaller chunks (dividing)
Merging the array together (conquering)
The role of the recurison is simply the dividing portion.
I think what confuses most people is that they think there is a lot of logic in the splitting and determining what to split, but most of the actual logic of sorting happens on the merge. The recursion is simply there to divide and do the first half and then the second half is really just looping, copying things over.
I see some answers that mention pivots but I would recommend not associating the word "pivot" with merge sort because that's an easy way to confuse merge sort with quicksort (which is heavily reliant on choosing a "pivot"). They are both "divide and conquer" algorithms. For merge sort the division always happens in the middle whereas for quicksort you can be clever with the division when choosing an optimal pivot.
process to divide the problem into subproblems
Given example will help you understand recursion. int A[]={number of element to be shorted.}, int p=0; (lover index). int r= A.length - 1;(Higher index).
class DivideConqure1 {
void devide(int A[], int p, int r) {
if (p < r) {
int q = (p + r) / 2; // divide problem into sub problems.
devide(A, p, q); //divide left problem into sub problems
devide(A, q + 1, r); //divide right problem into sub problems
merger(A, p, q, r); //merger the sub problem
}
}
void merger(int A[], int p, int q, int r) {
int L[] = new int[q - p + 1];
int R[] = new int[r - q + 0];
int a1 = 0;
int b1 = 0;
for (int i = p; i <= q; i++) { //store left sub problem in Left temp
L[a1] = A[i];
a1++;
}
for (int i = q + 1; i <= r; i++) { //store left sub problem in right temp
R[b1] = A[i];
b1++;
}
int a = 0;
int b = 0;
int c = 0;
for (int i = p; i < r; i++) {
if (a < L.length && b < R.length) {
c = i + 1;
if (L[a] <= R[b]) { //compare left element<= right element
A[i] = L[a];
a++;
} else {
A[i] = R[b];
b++;
}
}
}
if (a < L.length)
for (int i = a; i < L.length; i++) {
A[c] = L[i]; //store remaining element in Left temp into main problem
c++;
}
if (b < R.length)
for (int i = b; i < R.length; i++) {
A[c] = R[i]; //store remaining element in right temp into main problem
c++;
}
}
When you call the recursive method it does not execute the real function at the same time it's stack into stack memory. And when condition not satisfied then it's going to next line.
Consider that this is your array:
int a[] = {10,12,9,13,8,7,11,5};
So your method merge sort will work like below:
mergeSort(arr a, arr empty, 0 , 7);
mergeSort(arr a, arr empty, 0, 3);
mergeSort(arr a, arr empty,2,3);
mergeSort(arr a, arr empty, 0, 1);
after this `(low + high) / 2 == 0` so it will come out of first calling and going to next:
mergeSort(arr a, arr empty, 0+1,1);
for this also `(low + high) / 2 == 0` so it will come out of 2nd calling also and call:
merger(arr a, arr empty,0,0,1);
merger(arr a, arr empty,0,3,1);
.
.
So on
So all sorting values store in empty arr.
It might help to understand the how recursive function works
This is a hard algorithms problem that :
Divide the list in 2 parts (sum) that their sum closest to (most) each other
list length is 1 <= n <= 100 and their(numbers) weights 1<=w<=250 given in the question.
For example : 23 65 134 32 95 123 34
1.sum = 256
2.sum = 250
1.list = 1 2 3 7
2.list = 4 5 6
I have an algorithm but it didn't work for all inputs.
init. lists list1 = [], list2 = []
Sort elements (given list) [23 32 34 65 95 123 134]
pop last one (max one)
insert to the list which differs less
Implementation :
list1 = [], list2 = []
select 134 insert list1. list1 = [134]
select 123 insert list2. because if you insert to the list1 the difference getting bigger 3. select 95 and insert list2 . because sum(list2) + 95 - sum(list1) is less.
and so on...
You can reformulate this as the knapsack problem.
You have a list of items with total weight M that should be fitted into a bin that can hold maximum weight M/2. The items packed in the bin should weigh as much as possible, but not more than the bin holds.
For the case where all weights are non-negative, this problem is only weakly NP-complete and has polynomial time solutions.
A description of dynamic programming solutions for this problem can be found on Wikipedia.
The problem is NPC, but there is a pseudo polynomial algorithm for it, this is a 2-Partition problem, you can follow the way of pseudo polynomial time algorithm for sub set sum problem to solve this. If the input size is related polynomially to input values, then this can be done in polynomial time.
In your case (weights < 250) it's polynomial (because weight <= 250 n => sums <= 250 n^2).
Let Sum = sum of weights, we have to create two dimensional array A, then construct A, Column by Column
A[i,j] = true if (j == weight[i] or j - weight[i] = weight[k] (k is in list)).
The creation of array with this algorithm takes O(n^2 * sum/2).
At last we should find most valuable column which has true value.
Here is an example:
items:{0,1,2,3}
weights:{4,7,2,8} => sum = 21 sum/2 = 10
items/weights 0| 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10
---------------------------------------------------------
|0 | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | 0
|1 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1 | 0
|2 | 0 | 1 | 0 | 0 | 0 | 1 | 0 | 0 | 1 | 1
|3 | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 1
So because a[10, 2] == true the partition is 10, 11
This is an algorithm I found here and edited a little bit to solve your problem:
bool partition( vector< int > C ) {
// compute the total sum
int n = C.size();
int N = 0;
for( int i = 0; i < n; i++ ) N += C[i];
// initialize the table
T[0] = true;
for( int i = 1; i <= N; i++ ) T[i] = false;
// process the numbers one by one
for( int i = 0; i < n; i++ )
for( int j = N - C[i]; j >= 0; j--)
if( T[j] ) T[j + C[i]] = true;
for(int i = N/2;i>=0;i--)
if (T[i])
return i;
return 0;
}
I just returned first T[i] which is true instead of returning T[N/2] (in max to min order).
Finding the path which gives this value is not hard.
This problem is at least as hard as the NP-complete problem subset sum. Your algorithm is a greedy algorithm. This type of algorithm is fast, and can generate an approximate solution quickly but it cannot find the exact solution to an NP-complete problem.
A brute force approach is probably the simplest way to solve your problem, although it is will be to slow if there are too many elements.
Try every possible way of partitioning the elements into two sets and calculate the absolute difference in the sums.
Choose the partition for which the absolute difference is minimal.
Generating all the partitions can be done by considering the binary representation of each integer from 0 to 2^n, where each binary digit determines whether the correspending element is in the left or right partition.
Trying to resolve the same problem I have faced into the following idea which seems too much a solution, but it works in a linear time. Could one provide an example which would show that it does not work or explain why it is not a solution?
arr = [20,10,15,6,1,17,3,9,10,2,19] # a list of numbers
g1 = []
g2 = []
for el in reversed(sorted(arr)):
if sum(g1) > sum(g2):
g2.append(el)
else:
g1.append(el)
print(f"{sum(g1)}: {g1}")
print(f"{sum(g2)}: {g2}")
Typescript code:
import * as _ from 'lodash'
function partitionArray(numbers: number[]): {
arr1: number[]
arr2: number[]
difference: number
} {
let sortedArr: number[] = _.chain(numbers).without(0).sortBy((x) => x).value().reverse()
let arr1: number[] = []
let arr2: number[] = []
let median = _.sum(sortedArr) / 2
let sum = 0
_.each(sortedArr, (n) => {
let ns = sum + n
if (ns > median) {
arr1.push(n)
} else {
sum += n
arr2.push(n)
}
})
return {
arr1: arr1,
arr2: arr2,
difference: Math.abs(_.sum(arr1) - _.sum(arr2))
}
}