[HELP] Eigenvalues in C

This is a discussion on [HELP] Eigenvalues in C within the C Programming forums, part of the General Programming Boards category; Guys I have an assignment on programming that needs to implement the eigenvalues algorithm in C. The maximum MATRIX is ...

  1. #1
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    [HELP] Eigenvalues in C

    Guys I have an assignment on programming that needs to implement the eigenvalues algorithm in C. The maximum MATRIX is 5X5 and the minimum is 2X2.


    My code already checks the upper half of the matrix (X) and the Diagonalization Values (O). I left with the lower right of the matrix (Y).


    O X X X
    X O X Y
    X X O Y
    X Y Y O

    Here is what I got so far.
    Code:
    #include <math.h>
    #include <stdio.h>
    #include <stdlib.h>
    
    
    int
    main(void)
    {
        int num, col, row, i, j, sum = 0, p = 1;
        char stream[256];
    
    
        // correct input from the user
        printf("Enter the dimension of your matrix [2-5]: ");
        do
        {
            char *line = fgets(stream, 256, stdin);
    
    
            char c;
            if (sscanf(line, " %d %c", &num, &c) == 1)
            {
                system("cls");
                if (num > 5)
                    printf("Dimension too big!\nRetry [2-5]: ");
                else if (num < 2)
                    printf("Dimension too small!\nRetry [2-5]: ");
            }
            else
            {
                system("cls");
                printf("Invalid input!\nEnter the dimension of your matrix [2-5]: ");
            }
        }
        while (num > 5 || num < 2);
    
    
        // create the matrix based on the size of input
        int matrix[num][num];
    
    
        system("cls");
        printf("Enter the matrix:\n\n");
        for (i = 0; i < num; i++)
            for (col = 0; col < num; col++)
                scanf("%d", &matrix[i][col]);
    
    
        // print out the diagonalized values
        printf("\nDiagonalization Values: ");
        for (i = 0; i < num; i++)
            printf("%d ", matrix[i][i]);
    
    
        // check the upper-left half of the matrix
        printf("\nThe constant is: ");
        for (i = 0; i < num; i++)
        {
            int cnt = 0, p = 1;
            if (i < num)
                for (row = 0; row < num; row++)
                {
                    col = (num - 1) - row;
                    if (row != col - i && col - i >= 0)
                    {
                        p *= matrix[row][col - i];
                        cnt++;
                    }
                }
            if (cnt != 0)
                sum += p;
        }
    
    
        // check the lower-right half of the matrix
        for (j = 1; j < num - 1; j++)
        {
            int cnt = 0, p = 1;
            if (j < num - 1)
                for (row = 1, col = (num - 1); row < num; row++, col--)
                {
                    if (row != col)
                    {
                        p *= matrix[row][col];
                        cnt++;
                    }
                }
            if (cnt != 0)
                sum += p;
        }
        printf("\n\n%d", sum);
    
    
        return 0;
    }

  2. #2
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    Unfortunately "help" isn't a question, and leaves no clue what you are having a problem with. We're not going to extend your program to include another part of it that you haven't finished. That's doing your work for you.

    If you have a specific problem, ask away. "Help" won't do, however.

  3. #3
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    Quote Originally Posted by Adak View Post
    Unfortunately "help" isn't a question, and leaves no clue what you are having a problem with. We're not going to extend your program to include another part of it that you haven't finished. That's doing your work for you.

    If you have a specific problem, ask away. "Help" won't do, however.
    I already finished it, sorry...

  4. #4
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    That's the second time you've done this, I believe.

    Please don't post a "problem", if you don't have a problem, or have already solved the problem.

    That's really annoying. I don't know how others feel about it, but I won't be responding to your queries for awhile. I don't understand that kind of thinking.

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