Hi, I need help with this code.

This is a discussion on Hi, I need help with this code. within the C Programming forums, part of the General Programming Boards category; This code suppose to check if in every row and column every number appears only once. It only succeeds sometimes.... ...

  1. #1
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    Question Hi, I need help with this code.

    This code suppose to check if in every row and column every number appears only once.
    It only succeeds sometimes....
    Thanks...

    Code:
    #include <stdio.h>
    #define NUM 3
    int main()
    {
    int i,j,flag=1,l ;
    int mat[NUM][NUM] ;
    
    
    printf("please enter %d*%d matrice: \n" ,NUM ,NUM);
    for (i=0 ; i<NUM ; i++)
        for(j=0 ; j<=NUM ; j++)
                scanf("%d" ,&(mat[i][j]));
    
    for (j=0 ; j<NUM && flag!=0; j++)
         for(i=0; i<NUM && flag!=0; i++)
                  for(l=(i+1);l<NUM && flag!=0 ;l++)
                                    if (mat[i][j]!=mat[i+l][j])
                                       flag=1;
                                    else
                                        flag=0;
    
    for (i=0 ; i<NUM && flag!=0 ; i++)
           for(j=0 ; j<NUM && flag!=0; j++)
                   for(l=(j+1) ; l<NUM && flag!=0; l++)
                               if(mat[i][j]!=mat[i][j+l])
                                   flag=1;
                               else
                                   flag=0;
    
    if (flag==1)
       printf("YES\n");
    else 
         printf("NO\n");
    system("pause");
    return 0;
    }

  2. #2
    Deathray Engineer MacGyver's Avatar
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    So besides just dumping your code here and asking us to fix it for free, would you like to pretend to be courteous and try to tell us exactly what you mean when your brain spits out the phrase "It only succeeds sometimes"?

    How about error messages.... output..... something to give us an idea where you messed up?

  3. #3
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    I suggest you to do the following:
    - First, initialize the multidimensional array, and insert its values into a new array:
    Code:
    int newarr[NUM * 2];
    - Sort the array (use bubble-sort or merge sort or w/e).
    - Go over the array and make a loop that has a condition in it which says: (NOTE that you gotta check the condition of your loop so it doesn't reach newarr[NUM * 2]: for(...;i < (num * 2) - 1;...)
    Code:
    if(newarr[i] == newarr[i + 1]) return -1;
    I'm returning -1 because a number is repeated twice (might be more but we don't care).
    If it doesn't return -1 and returns something else it means that the multidimensional array doesn't have "repeating-numbers" (numbers that appear more than one time).
    Last edited by eXeCuTeR; 07-07-2008 at 11:37 AM.

  4. #4
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    hmmmmm well thanks for trying to help, I guess. I thought that was one of the purposes of the forum. And that people just enjoy helping unpaid students with their questions in C Because they like programming and enjoy helping others.

    Maybe I can be more specific...

    For example,
    if I enter the matrix
    1 2 3
    4 5 6
    7 8 9

    it returns "yes"

    If I enter the matrix
    1 1 1
    2 2 2
    3 3 3

    it returns "no"

    but if i enter something like

    1 2 3
    4 4 5
    6 7 8

    it returns "yes" and it shouldn't.

    I appreciate your help.
    Thanks.

  5. #5
    C++ Witch laserlight's Avatar
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    I think the problem is that you are setting the flag on every iteration. You should initialise the flag to 0, then only when you find a match, set it to 1 and break from the loop.
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  6. #6
    ZuK
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    I don't understand your algorithm.
    But your innermost loop accesses elements that don't exist
    Code:
    for(l=(j+1) ; l<NUM && flag!=0; l++)
         if(mat[i][j]!=mat[i][j+l])
    btw: using l as a variable is not very wise ( hard to see a difference to 1 ).
    Kurt

  7. #7
    Just a Human Anuradh_a's Avatar
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    hey Livnat

    your code does not accept 3*3 matrix it accepts 3*4 one. i modified little bit of your code
    here is what i got


    Code:
    #include <stdio.h>
    #include <stdlib.h>
    #define NUM 3
    int main()
    {
    int i,j,flag=1,l ;
    int mat[NUM][NUM] ;
    
    
    printf("please enter %d*%d matrice: \n" ,NUM ,NUM);
    for (i=0 ; i<NUM ; i++)
        for(j=0 ; j<NUM ; j++)
                scanf("%d" ,&(mat[i][j]));
    
    	i=j=l=0;
    for ( j=0 ; j<NUM && flag!=0; j++)
         for( i=0; i<NUM && flag!=0; i++)
            for( l=1;l<=NUM && flag!=0 ;l++)
     if ((mat[i][j]!=mat[i+l][j])&&(mat[i][j]!=mat[i][j+l]))
                                       flag=1;
    								
                                    else 
                                       flag=0;
    
    
    if (flag==1)
       printf("YES\n");
    else 
         printf("NO\n");
    system("pause");
    return 0;
    }

  8. #8
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    Thank you all...
    It was all very helpful
    Last edited by Livnat; 07-07-2008 at 12:53 PM.

  9. #9
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    Quote Originally Posted by Livnat View Post
    Thank you all...
    It was all very helpful
    This is what I did eventually...


    Code:
    #include <stdio.h>
    #define NUM 3
    int main()
    {
    int i,j,flag=0,l ;
    int mat[NUM][NUM] ;
    
    
    printf("please enter &#37;d*%d matrix: \n" ,NUM ,NUM);
    for (i=0 ; i<NUM ; i++)
        for(j=0 ; j<NUM ; j++)
                scanf("%d" ,&(mat[i][j]));
    
    for (j=0 ; j<NUM && flag==0; j++)
         for(i=0; i<NUM && flag==0; i++)
                  for(l=(i+1);l<NUM && flag==0 ;l++)
                                    if (mat[i][j]==mat[l][j])
                                       flag=1;
                                  
    
    for (i=0 ; i<NUM && flag==0 ; i++)
           for(j=0 ; j<NUM && flag==0; j++)
                   for(l=(j+1) ; l<NUM && flag==0; l++)
                               if(mat[i][j]==mat[i][l])
                                   flag=1;
                             
    if (flag==0)
       printf("YES\n");
    else
    if (flag==1)
         printf("NO\n");
    system("pause");
    return 0;
    }
    That's really not efficient, that works on O(n^3), you could have done it way better (look at my post)
    BTW, why did you post these 6 loops, only 3 is needed (not including the initialization)
    Code:
    for(int i = 0; i < NUM && flag == 0; i++)
     for(int j = 0; j < NUM && flag == 0; j++)
       for(int k = 0; k < NUM && flag == 0; k++)
              if(mat[i][j] == mat[i][k])
                     flag = 1;
    Last edited by eXeCuTeR; 07-07-2008 at 12:58 PM.

  10. #10
    Deathray Engineer MacGyver's Avatar
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    Are you trying to check for duplicates across all rows and columns, or just individually? For example:

    Code:
    please enter 3*3 matrix:
    0
    1
    2
    0
    3
    4
    5
    6
    7
    NO
    Press any key to continue . . .
    Code:
    please enter 3*3 matrix:
    0
    1
    2
    3
    0
    4
    5
    6
    7
    YES
    Press any key to continue . . .
    Is this intended behavior? Or should both examples show duplicates?

  11. #11
    C++ Witch laserlight's Avatar
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    Quote Originally Posted by eXeCuTeR
    That's really not efficient, that works on O(n^3), you could have done it way better (look at my post)
    Indeed, though bubble sort was the wrong thing to suggest for improving the complexity.

    Quote Originally Posted by MacGyver
    Are you trying to check for duplicates across all rows and columns, or just individually?
    It seems quite clear that the check is for duplicates across each row and each column, as that is how I interpret "if in every row and column every number appears only once".
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  12. #12
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    Quote Originally Posted by laserlight View Post
    Indeed, though bubble sort was the wrong thing to suggest for improving the complexity.


    It seems quite clear that the check is for duplicates across each row and each column, as that is how I interpret "if in every row and column every number appears only once".
    Yep, bubble sort is n^2 which is still faster than his code but anyways he could have used merge sort or any other good sort (heap sort and stuff..)

  13. #13
    C++ Witch laserlight's Avatar
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    Yep, bubble sort is n^2 which is still faster than his code
    It is not faster than his code in terms of complexity since the code needs to check for duplicates across each row and column, so a sort merely speeds up the check across a particular row or column. Of course, if the duplicates are to be checked across the entire grid then it would indeed be faster, but a naive check for uniqueness that checks each element against every other element would also take quadratic time (and this is currently the case).
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