Algorithm Matrix


http://www.geeksforgeeks.org/print-matrix-alternate-manner-left-right-right-left/
Given a 2D array, the task is to print the 2D in alternate manner (First row from left to right, then from right to left, and so on).
void convert(int arr[R][C])
{
    bool leftToRight = true;
    for (int i=0; i<R; i++)
    {
        if (leftToRight)
        {
            for (int j=0; j<C; j++)
                printf("%d ", arr[i][j]);
        }
        else
        {
            for (int j=C-1; j>=0; j--)
                printf("%d ",arr[i][j]);
        }
 
        leftToRight = !leftToRight;
    }
}
http://www.geeksforgeeks.org/form-coils-matrix/
Given a positive integer n that represents dimensions of a 4n x 4n matrix with values from 1 to n filled from left to right and top to bottom. Form two coils from matrix and print the coils.
Examples:
Input  : n = 1;
Output : Coil 1 : 10 6 2 3 4 8 12 16 
         Coil 2 : 7 11 15 14 13 9 5 1
Explanation : Matrix is 
1  2  3  4 
5  6  7  8 
9  10 11 12 
13 14 15 16
Total elements in matrix are 16n2. All elements are divided in two coils. Every coil has 8n2 elements. We make two arrays of this size. We first fill elements in coil1 by traversing in given order. Once we have filled elements in coil1, we can get elements of other coil2 using formula coil2[i] = 16*n*n + 1 -coil1[i].
void printCoils(int n)
{
    // Number of elements in each coil
    int m = 8*n*n;
    // Let us fill elements in coil 1.
    int coil1[m];
    // First element of coil1
    // 4*n*2*n + 2*n;
    coil1[0] = 8*n*n + 2*n;
    int curr = coil1[0];
    int nflg = 1, step = 2;
    // Fill remaining m-1 elements in coil1[]
    int index = 1;
    while (index < m)
    {
        // Fill elements of current step from
        // down to up
        for (int i=0; i<step; i++)
        {
            // Next element from current element
            curr = coil1[index++] = (curr - 4*n*nflg);
            if (index >= m)
                break;
        }
        if (index >= m)
            break;
        // Fill elements of current step from
        // up to down.
        for (int i=0; i<step; i++)
        {
            curr = coil1[index++] = curr + nflg;
            if (index >= m)
                break;
        }
        nflg = nflg*(-1);
        step += 2;
    }
    /* get coil2 from coil1 */
    int coil2[m];
    for (int i=0; i<8*n*n; i++)
        coil2[i] = 16*n*n + 1 -coil1[i];
    // Print both coils
    cout << "Coil 1 : ";
    for(int i=0; i<8*n*n; i++)
        cout << coil1[i] << " ";
    cout << "\nCoil 2 : ";
    for (int i=0; i<8*n*n; i++)
        cout << coil2[i] << " ";
}

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