Sorting an array diagonally - c++

Question

I've looked up some websites but I couldn't find an answer to my problem.
Here's my code:
#include "stdafx.h"
#include <iostream>
#include <math.h>
#include <time.h>
#include<iomanip>
#include<array>
#include <algorithm>
using namespace std;
const int AS = 6;
int filling(void);
void printing(int[AS][AS]);
int forsorting(int[][AS], int);
int main()
{
int funny = 0;
int timpa = 0;
int counter = 0;
int Array[AS][AS];
srand(time(0));
for (int i = 0; i<AS; i++)
{
for (int j = 0; j<AS; j++)
Array[i][j] = filling();
}
cout << "The unsorted array is" << endl << endl;
printing(Array);
cout << "The sorted array is" << endl << endl;
for (int il = 0; il<AS; il++)
{
for (int elle = 0; elle<AS; elle++)
Array[il][elle] =forsorting(Array, funny);
printing(Array);
}
system("PAUSE");
return 0;
}
int filling(void)
{
int kira;
kira = rand() % 87 + 12;
return kira;
}
void printing(int Array[AS][AS])
{
int counter = 0;
for (int i = 0; i<AS; i++)
{
for (int j = 0; j<AS; j++)
{
cout << setw(5) << Array[i][j];
counter++;
if (counter%AS == 0)
cout << endl << endl;
}
}
}
int forsorting(int Array[AS][AS], int funny)
{
int c, tmp, x;
int dice = 0;
int Brray[AS*AS];
int timpa = 0;
int super = 0;
//Transofrming Array[][] into Brray[]
for (int i = 0; i < AS; i++)
{
for (int k = 0; k < AS; k++)
{
Brray[timpa] = Array[i][k];
timpa++;
}
}
//Bubble sorting in Brray[]
for (int passer = 1; passer <= AS-1; passer++)
{
for (int timon = 1; timon <= AS-1; timon++)
{
if (Brray[timpa]>Brray[timpa + 1])
{
super = Brray[timpa];
Brray[timpa] = Brray[timpa + 1];
Brray[timpa + 1] = super;
}
}
}
//Transforming Brray[] into Array[][]
for (int e = 0; e<AS; e++)
{
for (int d = 0; d<AS; d++)
{
Brray[dice] = Array[e][d];
dice++;
}
}
***There's a part missing here***
}
What I have to do is, write a program using 3 functions.
The 1st function would fill my 2D array randomly (no problem with this part)
the 2nd function would print the unsorted array on the screen (no problem with this part)
and the 3rd function would sort my array diagonally as shown in this picture:
Then I need to call the 2nd function to print the sorted array. My problem is with the 3rd function I turned my 2D array into a 1D array and sorted it using Bubble sorting, but what I can't do is turn it back into a 2D array diagonaly sorted.

Answer 1

If you can convert from a 2D array to a 1D array, then converting back is the reverse process. Take the same loop and change around the assignment.
However in your case the conversion itself is wrong. It should take indexes in the order (0;0), (0;1), (1;0). But what it does is take indexes in the order (0;0), (0;1), (1;1).
My suggestion is to use the fact that the sum of the X and Y coordinates on each diagonal is the same and it goes from 0 to AS*2-2.
Then with another loop you can check for all possible valid x/y combinations. Something like this:
for ( int sum = 0; sum < AS*2-1; sum++ )
{
for ( int y = sum >= AS ? sum-AS+1 : 0; y < AS; y++ )
{
x = sum - y;
// Here assign either from Array to Brray or from Brray to Array
}
}
P.S. If you want to be really clever, I'm pretty sure that you can make a mathematical (non-iterative) function that converts from the index in Brray to an index-pair in Array, and vice-versa. Then you can apply the bubble-sort in place. But that's a bit more tricky than I'm willing to figure out right now. You might get extra credit for that though.
P.P.S. Realization next morning: you can use this approach to implement the bubble sort directly in the 2D array. No need for copying. Think of it this way: If you know a pair of (x;y) coordinates, you can easily figure out the next (x;y) coordinate on the list. So you can move forwards through the array from any point. That is all the the bubble sort needs anyway.

Answer 2

Suppose you have a 0-based 1-dimensional array A of n = m^2 elements. I'm going to tell you how to get an index into A, given and a pair of indices into a 2D array, according to your diagonalization method. I'll call i the (0-based) index in A, and x and y the (0-based) indices in the 2D array.
First, let's suppose we know x and y. All of the entries in the diagonal containing (x,y) have the same sum of their coordinates. Let sum = x + y. Before you got to the diagonal containing this entry, you iterated through sum earlier diagonals (check that this is right, due to zero-based indexing). The diagonal having sum k has a total of k + 1 entries. So, before getting to this diagonal, you iterated through 1 + 2 + ... + (sum - 1) entries. There is a formula for a sum of the form 1 + 2 + ... + N, namely N * (N + 1) / 2. So, before getting to this diagonal, you iterated through (sum - 1) * sum / 2 entries.
Now, before getting to the entry at (x,y), you went through a few entries in this very diagonal, didn't you? How many? Why, it's exactly y! You start at the top entry and go down one at a time. So, the entry at (x,y) is the ((sum - 1) * sum / 2 + y + 1)th entry, but the array is zero-based too, so we need to subtract one. So, we get the formula:
i = (sum - 1) * sum / 2 + y = (x + y - 1) * (x + y) / 2 + y
To go backward, we want to start with i, and figure out the (x,y) pair in the 2D array where the element A[i] goes. Because we are solving for two variables (x and y) starting with one (just i) and a constraint, it is trickier to write down a closed formula. In fact I'm not convinced that a closed form is possible, and certainly not without some floors, etc. I began trying to find one and gave up! Good luck!
It's probably correct and easier to just generate the (x,y) pairs iteratively as you increment i, keeping in mind that the sums of coordinate pairs are constant within one of your diagonals.

Answer 3

Store the "diagonally sorted" numbers into an array and use this to display your sorted array. For ease, assume 0-based indexing:
char order[] = { 0, 1, 3, 6, 10, 2, 4, 7, 11, 15, .. (etc)
Then loop over this array and display as
printf ("%d", Array[order[x]]);
Note that it is easier if your sorted Array is still one-dimensional at this step. You'd add the second dimension only when printing.

Answer 4

Following may help you:
#include <algorithm>
#include <iomanip>
#include <iostream>
#include <vector>
template<typename T>
class DiagArray
{
public:
DiagArray(int size) : width(size), data(size * size), orders(size * size)
{
buildTableOrder(size);
}
const T& operator() (int x, int y) const { return data[orders[width * y + x]]; }
T& operator() (int x, int y) { return data[orders[width * y + x]]; }
void sort() { std::sort(data.begin(), data.end()); }
void display() const {
int counter = 0;
for (auto index : orders) {
std::cout << std::setw(5) << data[index];
counter++;
if (counter % width == 0) {
std::cout << std::endl;
}
}
}
private:
void buildTableOrder(int size)
{
int diag = 0;
int x = 0;
int y = 0;
for (int i = 0; i != size * size; ++i) {
orders[y * size + x] = i;
++y;
--x;
if (x < 0 || y >= size) {
++diag;
x = std::min(diag, size - 1);
y = diag - x;
}
}
}
private:
int width;
std::vector<T> data;
std::vector<int> orders;
};
int main(int argc, char *argv[])
{
const int size = 5;
DiagArray<int> da(size);
for (int y = 0; y != size; ++y) {
for (int x = 0; x != size; ++x) {
da(x, y) = size * y + x;
}
}
da.display();
std::cout << std::endl;
da.sort();
da.display();
return 0;
}

Answer 5

Thank you for your assistance everyone, what you said was very useful to me. I actually was able to think about clearly and came up with a way to start filling the array based on your recommendation, but one problem now, Im pretty sure that my logic is 99% right but there's a flaw somewhere. After I run my code the 2nd array isnt printed on the screen. Any help with this?
#include "stdafx.h"
#include <iostream>
#include <math.h>
#include <time.h>
#include<iomanip>
#include<array>
#include <algorithm>
using namespace std;
const int AS = 5;
int filling(void);
void printing(int[AS][AS]);
int forsorting(int[][AS], int);
int main()
{
int funny = 0;
int timpa = 0;
int counter = 0;
int Array[AS][AS];
srand(time(0));
for (int i = 0; i<AS; i++)
{
for (int j = 0; j<AS; j++)
Array[i][j] = filling();
}
cout << "The unsorted array is" << endl << endl;
printing(Array);
cout << "The sorted array is" << endl << endl;
for (int il = 0; il<AS; il++)
{
for (int elle = 0; elle<AS; elle++)
Array[il][elle] =forsorting(Array, funny);
}
printing(Array);
system("PAUSE");
return 0;
}
int filling(void)
{
int kira;
kira = rand() % 87 + 12;
return kira;
}
void printing(int Array[AS][AS])
{
int counter = 0;
for (int i = 0; i<AS; i++)
{
for (int j = 0; j<AS; j++)
{
cout << setw(5) << Array[i][j];
counter++;
if (counter%AS == 0)
cout << endl << endl;
}
}
}
int forsorting(int Array[AS][AS], int funny)
{int n;
int real;
int dice = 0;
int Brray[AS*AS];
int timpa = 0;
int super = 0;
int median;
int row=0;
int col=AS-1;
//Transofrming Array[][] into Brray[]
for (int i = 0; i < AS; i++)
{
for (int k = 0; k < AS; k++)
{
Brray[timpa] = Array[i][k];
timpa++;
}
}
//Bubble sorting in Brray[]
for (int passer = 1; passer <= AS-1; passer++)
{
for (int timon = 1; timon <= AS-1; timon++)
{
if (Brray[timpa]>Brray[timpa + 1])
{
super = Brray[timpa];
Brray[timpa] = Brray[timpa + 1];
Brray[timpa + 1] = super;
}
}
}
//Transforming Brray[] into sorted Array[][]
for(int e=4;e>=0;e--)//e is the index of the diagonal we're working in
{
if(AS%2==0)
{median=0.5*(Brray[AS*AS/2]+Brray[AS*AS/2-1]);
//We start filling at median - Brray[AS*AS/2-1]
while(row<5 && col>=0)
{real=median-Brray[AS*AS/2-1];
Array[row][col]=Brray[real];
real++;
col--;
row++;}
}
else {
median=Brray[AS*AS/2];
//We start filling at Brray[AS*AS/2-AS/2]
while(row<5 && col>=0)
{real=Brray[AS*AS/2-AS/2];
n=Array[row][col]=Brray[real];
real++;
col--;
row++;}
}
}
return n;
}
Thanks again for your assistance