Thread: program to find the saddle point of a matrix ??

So write it.

Nobody here is going to do your homework for you. If you put in some effort and show your code, people will guide you along and provide hints to fix it and make it work.

Please read the Homework policy for this site.

Even if today is the last day to submit it if you start working on it now, it will probably only take you about 1-2 hours to get it done, and that includes the time it takes to post your intermediate code here.

i will post the code here

Code:

saddlepointa(int a1[10][10],int rowa,int cola)
{
 int big[10],small[10],i,j,max,min;
 clrscr();
 big[0]=a1[0][0];
 for(i=0;i<rowa;i++)
 {
  for(j=0;j<cola;j++)
  {
   if(a1[i][j] > big[i])
   {
     big[i] = a1[i][j];
   }
  }
 }
 for(j=0;j<cola;j++)
 {
  small[j]=a1[0][j];
  for(i=0;i<rowa;i++)
  {
   if(a1[i][j] < small[j])
   {
    small[j]=a1[i][j];
   }
  }
 }
 min=big[0];
 for(i=0;i<rowa;i++)
 {
  if(big[i] < min)
  {
   min = big[i];
  }
 }
 max=small[0];
 for(j=0;j<cola;j++)
 {
  if(small[j] > max)
  {
   max = small[j];
  }
 }
 if(max==min)
 {
  printf("\n\n\n\t\t\tTHE SADDLE POINT IS --> %d ",min);
 }
 else
 {
  printf("\n\n\n\nNO SADDLE POINT");
 }
getch();
return 0;
}
void main()
{
 int a[10][10],i,j;
 printf("Enter the matrix elements\n");
 for(i=0;i<3;i++)
 {
  for(j=0;j<3;j++)
  {
    scanf("%d",&a[i][j]);
  }
 }
saddlepointa(a,3,3);
getch();
}

Post the code using CODE TAGS! Nobody will read that.

Post is OK, (much better!).

Let's see...

I don't understand your logic, but here's one way:

sort the array, in descending order.

The value in the last row, zero column, will always be the saddlepoint.

Originally Posted by intermediatech

is the post ok ?

if we enter

1 2 3
4 5 6
7 8 9

The program will return 3 as saddle point.
but according to question the answer is 7

pls help to correct this

The correct answer is 7, because 7 is the smallest on its row (row 3) and largest on its col(col 1).

I'm not sure why you have so many loops. Loop once through the provided column. Store the biggest, or smallest or whatever it is you're looking for, the index of that value. Do the same thing for the row. Return the value that is at the coordinates those two indexes make.

Code:

for each element in this column
    if here < or > (whatever) than previous low or high
        update stored column index

for each element in this row
    if here < or > (whatever) than previous low or high
        update stored row index

return array[ stored row index ][ stored column index ];

You have twice as many loops as you need, unless I'm misunderstanding what you're really trying to do. Also, your code doesn't actually spit out 3, because your code doesn't actually ever call your saddle function.


Quzah.