compare elements with each other

[glao]

Hello , I want to compare each element of an array with every other.
For example, array[0] with array[1] , with array[2] ...
Then , array[1] with array[2],with array[3] ...

So , I thought something like this :

for ( int i = 0; i < N; i ++ )
{
    for ( int k = i + 1; k < N; k++ )
    {
        temp[ i * N + k ] = (array[ i ] == array[ k ]);
      }
 }

But , for example , for i = 0 , temp indexing will be : temp[ 0 + 1] ,hence temp[ 1] .
There will not be temp[ 0 ] as first index!

I can't think how to solve this and also what the size of temp will be.If I leave it as N * N ,it will cover all cases but it is more than it is needed.

Thanks!

[glao]

Hmm, I figured something!

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <vector>

using namespace std;

int main( int argc, char** argv )
{
    int N = 16;

    int numCompares = (N / 2) * (N -1);

    //vector<int> test;
    //test.resize(numCompares*N);

    int test[ numCompares*N ];
    int array[ N ];

    array[0] = 0;
    array[1] = 1;
    array[2] = 1;
    array[3] = 6;
    array[4] = 8;
    array[5] = 4;
    array[6] = 0;
    array[7] = 11;
    array[8] = 4;
    array[9] = 23;
    array[10] = 3;
    array[11] = 15;
    array[12] = 2;
    array[13] = 12;
    array[14] = 19;
    array[15] = 12;

    int count = 0;
    //compare 
    for ( int p = 0; p < numCompares; p++ )
    {
        for ( int idx = 0; idx < N; idx++ )
        {
            for ( int k = idx + 1; k < N; k ++ )
            {
                if ( array[ idx ] == array[ k ] ) 
                {  
                       //test.push_back( 1 );
                      //test[ p ] = 1;
                       test[ p * N + idx * N + k ] = 1;
                  }
                  else
                  {

                        test[ p * N + idx * N + k ] = 0;
                        //test[ p ] = 0;
                        //test.push_back(0);
                  }

            }
        }

    }




    return 0;
}

I figured that the number of comparisons would be the sum of an arithmetic progression.
As for the size of test , since every comparison ( array[0] with array[1]) will result in a N dimension array, so N * numCompares will be the size of test.

Now , I am not sure how to handle the filling ( and index ) and size of test data.
That's why I have different version in the if statement.I also tried to use vector for test.

Any ideas?

Thanks!

[tinstaafl]

One of the simplest ways to get the index of test is to use a counter:

    int index = 0;
    for ( int idx = 0; idx < N; idx++ )
    {
        for ( int k = idx + 1; k < N; k ++ )
        {
            if ( array[ idx ] == array[ k ] ) 
            {  
                   test[ index++ ] = 1;
              }
              else
              {
                    test[ index++ ] = 0;
              }
        }

[glao]

Hello and thanks for helping.

Yes, I agree with the index.As you can see I already have defines numCompares which is 120 ( and using your code index = 120 ).

But,I think my problem is logical?
I think test should hold 120 * N elements because N is the size of the array and so 120*N for all the calculations.( and every comparison gives a N size array filled with 0 or 1 )

I can't find how to index test according to my above statement.

[tinstaafl]

I'm not sure I understand what you're getting at. With the above code, test will contain the result of every comparison for that array. What else do you need?

[glao]

If I compare array[0] with array[1] , then with array[2] , array[3] ...so on , the result is a N dimension array ( flled with 0 or 1 ).

Then , I am comparing array[1] with array[2] , array[3] ..so on , the result again is a N dimension array.

So , I will have 120 such comparisons , so 120 * N will be the elements of test.

So,I can't figure how to store test.
I hope this is clear.

Thanks!

[tinstaafl]

I think you may be confused. Comparing each element of the array with each other element of the array is 120 comparisons. If test is set to a size of 120 then test is storing all the comparisons you need. Why do you need more storage? Why do you feel that test will be 120 * N when there are only 120 elements in total?

[glao]

Yes , you are right!I was confused.
But then , the problem how to index(fill) test array still remains..
If you look to my code , let's correct the
int test[ numCompares*N ]; to int test[ numCompares ];

Inside the loop ,how can I handle test?
Right now it is test[ p * N + idx * N + k ] = 1; which is wrong.
using test[ p ] = 1; is also wrong since it will write all comparisons at the same p.

How can I overcome this?
Thank you.

[tinstaafl]

The code I showed you before will do that:

        int index = 0;
        for ( int idx = 0; idx < N; idx++ )
        {
            for ( int k = idx + 1; k < N; k ++ )
            {
                if ( array[ idx ] == array[ k ] ) 
                {  
                       test[ index++ ] = 1;
                }
                else
                {
                    test[ index++ ] = 0;
                }
            }
        }

As long as test has sufficient size it will contain all the comparisons.

[glao]

Yes, you are right!I can see that.

But , if I want to follow an indexing like I am trying?Is there a way?

Thank you.

[tinstaafl]

The problem with calculating the index is there is an offset that gets progressively bigger and relies on the previous offset. One way around this is to use a helper function that recaclulates the offset:

int FactorialOffset(int value)
{
    int outval = 0;
    for (int i = 1; i <= value + 1; i++)
    {
        outval += i;
    }
    return outval;
}

Using this function with this formula ((N*i) + k) - FactorialOffset(i), you can calculate the index for each comparison. In your code it would look like this:

    for ( int idx = 0; idx < N; idx++ )
    {
        //Calculate the offset here, since it won't change until i does
        int fo = FactorialOffset(i);
        for ( int k = idx + 1; k < N; k ++ )
        {
            if ( array[ idx ] == array[ k ] ) 
            {  
                   test[ ((N*i) + k) - fo ] = 1;
              }
              else
              {
                    test[ ((N*i) + k) - fo ] = 0;
              }
        }
    }

[glao]

Excellent!I could never thought of this..

Now, I have one more thing to solve.
Right now, the test array holds indices from 0-120 ( numCompares = 120 ).

We are comparing element 0 with element 1 , then with element 2..element 15
Comparison of array[0] with array[1]will be written to test[0] , comparison array[0] with array[2]will be writte to test[1]and so on.

I want a new vector to hold N ( N =16 right now ) labels (indices) :

testVector[ 0 ]will contain all the comparison of array[0] with every other element ( until N )

0->1
0->2
0->3
0->4
...
0->15   ( we said until N =16 )

Count sum = 15 times ( N-1)
So testVector[ 0 ] will hold 15 values

testVector[ 1 ]will contain all the comparison of array[1] with every other element

1->2
1->3
1->4
..
1->15    

Count sum = 14 times
SotestVector[ 1 ] will hold 14 values

and so on.

So , code will be like this:

vector< vector<int> > testVector( numCompares );
int jCount = N - 1; // count sum for every 'label' 

for ( int i = 0; i < N; i++ ) // I want vector to have N labels
{
    while ( jCount >= 1 )
    {
        for ( int j = 0; j < numCompares; j++ )
        {                   
            testVector[ i ].push_back( test[ j ] );
            continue; // must jump to next i
        }
        jCount--;
    }
}

The code is not right.
I can't figure a way to manipulate the 'j' index related to i and jCount.
Remember, that, for testVector[ 0], jCount = 15 and we must read the test array from j indices (0-14).
Then, we must jump to testVector[1], jCount =14 and we must read test array from j = 15-28 and so on.

I hope it is clear!

[tinstaafl]

Basically you use the same nested loop as you've already used. Before the start of the k loop, instantiate a temporary vector to hold the results of that set of comparisons. Add the results of each comparison to that vector inside the k loop. After the k loop is finished add that vector to the 2D vector that will hold all the results:

vector< vector<int> > testVector( numCompares );
    for ( int idx = 0; idx < N; idx++ )
    {
        vector<int> temp;
        //Calculate the offset here, since it won't change until i does
        int fo = FactorialOffset(i);
        for ( int k = idx + 1; k < N; k ++ )
        {

            if ( array[ idx ] == array[ k ] ) 
            { 
                    temp.push_back(1);
                   test[ ((N*i) + k) - fo ] = 1;
              }
              else
              {
                    temp.push_back(0);
                    test[ ((N*i) + k) - fo ] = 0;
              }
        }
        testVector.push_back(temp);
    }

[glao]

Hmm. Nice!

I am little confused how to access the testVector now..
How can I print the contents?

Because if I try something like this:

 for (int i = 0; i< testVector.size(); i++) 
 {
        for (int j = 0; j< 120; j++)
        {
            cout << testVector[ i ][ j ] << " ";
        }
        cout << endl;
    }

gives seg fault.
My vector now is 3D vector?

Also, what about the way I tried to solve this problem? (my previous post) Is there a solution according that?

Thank you very much!

[glao]

Ok, I figured!

    for (int i = 0; i< testVector.size(); i++) {
        for (int j = 0; j< testVector[i].size(); j++) {
            cout << testVector[i][j] << " ";
        }
        cout << endl;
    }

And it works fine!
So , it is still a 2D vector.
As far as I understand , testVector is just declared as a 2D vector ( we don't fill it as a 2D vector yet) , and then we fill it with a 1D vector.So , we have a vector (testVector) of vectors(temp).

Also , we don't need the test array anymore like that.

If you can answer me about my last comment ( if it is possible to solve it using my procedure)

[tinstaafl]

You will have to be more specific. You have code to:

• use a counter as the index to fill an array
• calculate the index and use that when filling the array
• fill a 2D vector

What other ways of storing/accessing this data do you need?

[glao]

Ok, your solution is very clear.
Maybe my thought is too complicated.
I was just wondering if there was a solution based on the flow I provided.

Anyway, thank you very much for your help.

[tinstaafl]

If your question is answered please mark this solved. Thank You