I am having problems with bottoms-up mergesort. I have problems sorting/merging. Current code includes:
public void mergeSort(long[] a, int len) {
long[] temp = new long[a.length];
int length = 1;
while (length < len) {
mergepass(a, temp, length, len);
length *= 2;
}
}
public void mergepass(long[] a, long[] temp, int blocksize, int len) {
int k = 0;
int i = 1;
while(i <= (len/blocksize)){
if(blocksize == 1){break;}
int min = a.length;
for(int j = 0; j < blocksize; j++){
if(a[i*j] < min){
temp[k++] = a[i*j];
count++;
}
else{
temp[k++] = a[(i*j)+1];
count++;
}
}
for(int n = 0; n < this.a.length; n++){
a[n] = temp[n];
}
}
}
Obvious problems:
i is never incremented.
At no point do you compare two elements in the array. (Is that what if(a[i*j] < min) is supposed to be doing? I can't tell.)
Why are you multiplying i and j?
What's this.a.length?
Style problems:
mergeSort() takes len as an argument, even though arrays have an implicit length. To make matters worse, the function also uses a.length and length.
Generally poor variable names.
Nitpicks:
If you're going to make a second array of the same size, it is common to make one the "source" and the other the "destination" and swap them between passes, instead of sorting into a temporary array and copying them back again.
Related
Is there a better way to calculate a moving sum of a list?
List<double?> rollingSum({int window = 3, List data = const []}) {
List<double?> sum = [];
int i = 0;
int maxLength = data.length - window + 1;
while (i < maxLength) {
List tmpData = data.getRange(i, i + window).toList();
double tmpSum = tmpData.reduce((a, b) => a + b);
sum.add(tmpSum);
i++;
}
// filling the first n values with null
i = 0;
while (i < window - 1) {
sum.insert(0, null);
i++;
}
return sum;
}
Well, the code is already clean for what you need. Maybe just some improvements like:
Use a for loop
You can use the method sublist which creates a "view" of a list, which is more efficient
To insert some values in the left/right of a list, there is a specific Dart method called padLeft, where you specify the lenght of the list which you want it to become (first parameter), then the value you want to use to fill it (second parameter). For example, if you have an array of N elements, and you want to fill it with X "null"s to the left, use padLeft(N+X, null).
List<double?> rollingSum({int window = 3, List data = const []}) {
List<double?> sum = [];
for (int i = 0; i < data.length - window + 1; i++) {
List tmpData = data.sublist(i, i + window);
double tmpSum = tmpData.reduce((a, b) => a + b);
sum.add(tmpSum);
}
sum.padLeft(window - 1, null);
return sum;
}
if I understand your problem correctly you can just calculate the window one time and in one loop you can for each iteration you can add the current element to the sum and subtract i - (window - 1)
so for an input like this
data = [1,2,3,4,5,6]
window = 3
the below code will result in [6,9,12,15]
int sum = 0;
List<double> res = [];
for (int i = 0;i<data.length;i++) {
sum += data[i];
if (i < window - 1) {
continue;
}
res.add(sum);
sum -= data[i - (window - 1)]; // remove element that got out of the window size
}
this way you won't have to use getRange nor sublist nor reduce as all of those are expensive functions in terms of time and space complexity
I study the Quicksort algorithm and created a sample program like this :
// Java program for implementation of QuickSort
class QuickSort
{
/* This function takes last element as pivot,
places the pivot element at its correct
position in sorted array, and places all
smaller (smaller than pivot) to left of
pivot and all greater elements to right
of pivot */
int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
{
// If current element is smaller than or
// equal to pivot
if (arr[j] <= pivot)
{
i++;
// swap arr[i] and arr[j]
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
}
// swap arr[i+1] and arr[high] (or pivot)
int temp = arr[i+1];
arr[i+1] = arr[high];
arr[high] = temp;
return i+1;
}
/* The main function that implements QuickSort()
arr[] --> Array to be sorted,
low --> Starting index,
high --> Ending index */
void sort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
// Recursively sort elements before
// partition and after partition
sort(arr, low, pi-1);
sort(arr, pi+1, high);
}
}
/* A utility function to print array of size n */
static void printArray(int arr[])
{
int n = arr.length;
for (int i=0; i<n; ++i)
System.out.print(arr[i]+" ");
System.out.println();
}
// Driver program
public static void main(String args[])
{
int arr[] = {10, 7, 8, 9, 1, 5};
int n = arr.length;
QuickSort ob = new QuickSort();
ob.sort(arr, 0, n-1);
System.out.println("sorted array");
printArray(arr);
}
}
/*This code is contributed by Rajat Mishra */
Precisely in this part :
int partition(int arr[], int low, int high)
{
int pivot = arr[high];
int i = (low-1); // index of smaller element
for (int j=low; j<high; j++)
would you have an idea why the code says: int i = (low-1) ? The range of i will not become -1 with this instruction? I mean it was initialized previously to 0? Then do think is it possible such instruction? Or what as you it has to understand?
Best regard,
Thanks,
Intelego.
Yes of course,
First, "i" is initialized to 0
QuickSort ob = new QuickSort();
ob.sort(arr, 0, n-1);
And then,
high --> Ending index */
void sort(int arr[], int low, int high)
{
if (low < high)
{
/* pi is partitioning index, arr[pi] is
now at right place */
int pi = partition(arr, low, high);
Low = 0 until here,
and the code gives "i" the value of -1.
from where "i" wanted to know if it is possible "i" could be -1 ?
In this case what could represent the index -1 ?
Thanks very much,
Intelegoit
Today, when trying quicksort, instead of taking last element as pivot and partitioning,i took the first element as pivot, But it is not producing the correct partitioned output.
int pivot = ar[0];
int pindex = 0;
for(int i = 0;i < ar.size();i++)
{
if(ar[i] <= pivot)
{
swap(ar[i],ar[pindex]);
pindex++;
}
}
swap(ar[pindex],ar[ar.size()-1]);
I could not understand why, i always use this for partition, but this is not working when i take first element as partition.
But this worked even if i took first element as partition
int i, j, pivot, temp;
pivot = ar[0];
i = 0;
j = ar.size()-1;
while(1)
{
while(ar[i] < pivot && ar[i] != pivot)
i++;
while(ar[j] > pivot && ar[j] != pivot)
j--;
if(i < j)
{
temp = ar[i];
ar[i] = ar[j];
ar[j] = temp;
}
else
{
break;
}
}
What are the differences between them.
At last found that, this method is Hoare's partition method, where as the typical quick sort method we all follow is lomuto's partition.
See this wiki page, it has all details https://en.wikipedia.org/wiki/Quicksort
I am taking about 50 times as long as expected to loop through a simple assignment. My first reaction was that I had disordered my memory access in the arrays, resulting in cache misses. This doesn't seem the case, however.
The pixel value assignment and updating the arrays takes a dogs age. Do any one of you folks have an inclining as to why this is happening? (I am compiling for an iPod with an A4)
memset(columnSumsCurrentFrameA, 0, sizeof(unsigned int) * (_validImageWidth/numSubdivisions) );
memset(rowSumsCurrentFrameA, 0, sizeof(unsigned int) * (_validImageHeight/numSubdivisions) );
int pixelValue = 0;
int startingRow = 0;
int startingColumn = 0;
for (int i = 0; i < _validImageHeight/numSubdivisions; i++)
{
int index = (i + startingRow) * _imageWidth;
for( int j = 0; j < (_validImageWidth/numSubdivisions); j++)
{
pixelValue = imageData[index + startingColumn + j];
columnSumsCurrentFrameA[j] += pixelValue;
rowSumsCurrentFrameA[i] += pixelValue;
}
}
The result of _validImageWidth/numSubdivisions must be an integer, are you sure that is always the case?
Also, you should calculate _validImageWidth/numSubdivisions before entering the double loops, it's not safe to assume your compiler takes care of it.
int limit = _validImageHeight/numSubdivisions;
for (int i = 0; i < limit; i++)
{
int index = (i + startingRow) * _imageWidth;
for( int j = 0; j < limit; j++)
{
pixelValue = imageData[index + startingColumn + j];
columnSumsCurrentFrameA[j] += pixelValue;
rowSumsCurrentFrameA[i] += pixelValue;
}
}
Imagine a std:vector, say, with 100 things on it (0 to 99) currently. You are treating it as a loop. So the 105th item is index 4; forward 7 from index 98 is 5.
You want to delete N items after index position P.
So, delete 5 items after index 50; easy.
Or 5 items after index 99: as you delete 0 five times, or 4 through 0, noting that position at 99 will be erased from existence.
Worst, 5 items after index 97 - you have to deal with both modes of deletion.
What's the elegant and solid approach?
Here's a boring routine I wrote
-(void)knotRemovalHelper:(NSMutableArray*)original
after:(NSInteger)nn howManyToDelete:(NSInteger)desired
{
#define ORCO ((NSInteger)[original count])
static NSInteger kount, howManyUntilLoop, howManyExtraAferLoop;
if ( ... our array is NOT a loop ... )
// trivial, if messy...
{
for ( kount = 1; kount<=desired; ++kount )
{
if ( (nn+1) >= ORCO )
return;
[original removeObjectAtIndex:( nn+1 )];
}
return;
}
else // our array is a loop
// messy, confusing and inelegant. how to improve?
// here we go...
{
howManyUntilLoop = (ORCO-1) - nn;
if ( howManyUntilLoop > desired )
{
for ( kount = 1; kount<=desired; ++kount )
[original removeObjectAtIndex:( nn+1 )];
return;
}
howManyExtraAferLoop = desired - howManyUntilLoop;
for ( kount = 1; kount<=howManyUntilLoop; ++kount )
[original removeObjectAtIndex:( nn+1 )];
for ( kount = 1; kount<=howManyExtraAferLoop; ++kount )
[original removeObjectAtIndex:0];
return;
}
#undef ORCO
}
Update!
InVariant's second answer leads to the following excellent solution. "starting with" is much better than "starting after". So the routine now uses "start with". Invariant's second answer leads to this very simple solution...
N times do if P < currentsize remove P else remove 0
-(void)removeLoopilyFrom:(NSMutableArray*)ra
startingWithThisOne:(NSInteger)removeThisOneFirst
howManyToDelete:(NSInteger)countToDelete
{
// exception if removeThisOneFirst > ra highestIndex
// exception if countToDelete is > ra size
// so easy thanks to Invariant:
for ( do this countToDelete times )
{
if ( removeThisOneFirst < [ra count] )
[ra removeObjectAtIndex:removeThisOneFirst];
else
[ra removeObjectAtIndex:0];
}
}
Update!
Toolbox has pointed out the excellent idea of working to a new array - super KISS.
Here's an idea off the top of my head.
First, generate an array of integers representing the indices to remove. So "remove 5 from index 97" would generate [97,98,99,0,1]. This can be done with the application of a simple modulus operator.
Then, sort this array descending giving [99,98,97,1,0] and then remove the entries in that order.
Should work in all cases.
This solution seems to work, and it copies all remaining elements in the vector only once (to their final destination).
Assume kNumElements, kStartIndex, and kNumToRemove are defined as const size_t values.
vector<int> my_vec(kNumElements);
for (size_t i = 0; i < my_vec.size(); ++i) {
my_vec[i] = i;
}
for (size_t i = 0, cur = 0; i < my_vec.size(); ++i) {
// What is the "distance" from the current index to the start, taking
// into account the wrapping behavior?
size_t distance = (i + kNumElements - kStartIndex) % kNumElements;
// If it's not one of the ones to remove, then we keep it by copying it
// into its proper place.
if (distance >= kNumToRemove) {
my_vec[cur++] = my_vec[i];
}
}
my_vec.resize(kNumElements - kNumToRemove);
There's nothing wrong with two loop solutions as long as they're readable and don't do anything redundant. I don't know Objective-C syntax, but here's the pseudocode approach I'd take:
endIdx = after + howManyToDelete
if (Len <= after + howManyToDelete) //will have a second loop
firstloop = Len - after; //handle end in the first loop, beginning in second
else
firstpass = howManyToDelete; //the first loop will get them all
for (kount = 0; kount < firstpass; kount++)
remove after+1
for ( ; kount < howManyToDelete; kount++) //if firstpass < howManyToDelete, clean up leftovers
remove 0
This solution doesn't use mod, does the limit calculation outside the loop, and touches the relevant samples once each. The second for loop won't execute if all the samples were handled in the first loop.
The common way to do this in DSP is with a circular buffer. This is just a fixed length buffer with two associated counters:
//make sure BUFSIZE is a power of 2 for quick mod trick
#define BUFSIZE 1024
int CircBuf[BUFSIZE];
int InCtr, OutCtr;
void PutData(int *Buf, int count) {
int srcCtr;
int destCtr = InCtr & (BUFSIZE - 1); // if BUFSIZE is a power of 2, equivalent to and faster than destCtr = InCtr % BUFSIZE
for (srcCtr = 0; (srcCtr < count) && (destCtr < BUFSIZE); srcCtr++, destCtr++)
CircBuf[destCtr] = Buf[srcCtr];
for (destCtr = 0; srcCtr < count; srcCtr++, destCtr++)
CircBuf[destCtr] = Buf[srcCtr];
InCtr += count;
}
void GetData(int *Buf, int count) {
int srcCtr = OutCtr & (BUFSIZE - 1);
int destCtr = 0;
for (destCtr = 0; (srcCtr < BUFSIZE) && (destCtr < count); srcCtr++, destCtr++)
Buf[destCtr] = CircBuf[srcCtr];
for (srcCtr = 0; srcCtr < count; srcCtr++, destCtr++)
Buf[destCtr] = CircBuf[srcCtr];
OutCtr += count;
}
int BufferOverflow() {
return ((InCtr - OutCtr) > BUFSIZE);
}
This is pretty lightweight, but effective. And aside from the ctr = BigCtr & (SIZE-1) stuff, I'd argue it's highly readable. The only reason for the & trick is in old DSP environments, mod was an expensive operation so for something that ran often, like every time a buffer was ready for processing, you'd find ways to remove stuff like that. And if you were doing FFT's, your buffers were probably a power of 2 anyway.
These days, of course, you have 1 GHz processors and magically resizing arrays. You kids get off my lawn.
Another method:
N times do {remove entry at index P mod max(ArraySize, P)}
Example:
N=5, P=97, ArraySize=100
1: max(100, 97)=100 so remove at 97%100 = 97
2: max(99, 97)=99 so remove at 97%99 = 97 // array size is now 99
3: max(98, 97)=98 so remove at 97%98 = 97
4: max(97, 97)=97 so remove at 97%97 = 0
5: max(96, 97)=97 so remove at 97%97 = 0
I don't program iphone for know, so I image std::vector, it's quite easy, simple and elegant enough:
#include <iostream>
using std::cout;
#include <vector>
using std::vector;
#include <cassert> //no need for using, assert is macro
template<typename T>
void eraseCircularVector(vector<T> & vec, size_t position, size_t count)
{
assert(count <= vec.size());
if (count > 0)
{
position %= vec.size(); //normalize position
size_t positionEnd = (position + count) % vec.size();
if (positionEnd < position)
{
vec.erase(vec.begin() + position, vec.end());
vec.erase(vec.begin(), vec.begin() + positionEnd);
}
else
vec.erase(vec.begin() + position, vec.begin() + positionEnd);
}
}
int main()
{
vector<int> values;
for (int i = 0; i < 10; ++i)
values.push_back(i);
cout << "Values: ";
for (vector<int>::const_iterator cit = values.begin(); cit != values.end(); cit++)
cout << *cit << ' ';
cout << '\n';
eraseCircularVector(values, 5, 1); //remains 9: 0,1,2,3,4,6,7,8,9
eraseCircularVector(values, 16, 5); //remains 4: 3,4,6,7
cout << "Values: ";
for (vector<int>::const_iterator cit = values.begin(); cit != values.end(); cit++)
cout << *cit << ' ';
cout << '\n';
return 0;
}
However, you might consider:
creating new loop_vector class, if you use this kind of functionality enough
using list if you perform many deletions (or few deletions (not from end, that's simple pop_back) but large array)
If your container (NSMutableArray or whatever) is not list, but vector (i.e. resizable array), you most definitely don't want to delete items one by one, but whole range (e.g. std::vector's erase(begin, end)!
Edit: reacting to comment, to fully realize what must be done by vector, if you erase element other than the last one: it must copy all values after that element (e.g. 1000 items in array, you erase first, 999x copying (moving) of item, that is very costly).
Example:
#include <iostream>
#include <vector>
#include <ctime>
using namespace std;
int main()
{
clock_t start, end;
vector<int> vec;
const int items = 64 * 1024;
cout << "using " << items << " items in vector\n";
for (size_t i = 0; i < items; ++i) vec.push_back(i);
start = clock();
while (!vec.empty()) vec.erase(vec.begin());
end = clock();
cout << "Inefficient method took: "
<< (end - start) * 1.0 / CLOCKS_PER_SEC << " ms\n";
for (size_t i = 0; i < items; ++i) vec.push_back(i);
start = clock();
vec.erase(vec.begin(), vec.end());
end = clock();
cout << "Efficient method took: "
<< (end - start) * 1.0 / CLOCKS_PER_SEC << " ms\n";
return 0;
}
Produces output:
using 65536 items in vector
Inefficient method took: 1.705 ms
Efficient method took: 0 ms
Note it's very easy to get inefficient, look e.g. have at http://www.cplusplus.com/reference/stl/vector/erase/