@Arun
If you read the post in which i have explained the process properly,
the following is also present:
while(1)
{
If x <= min (b,d ),
/* here b is nothing but the element placed next to 'x' on the same
row..
d is the element placed right below 'x' in the same column...
then we are done...*/
break;
else
swap ('x', min (b,d))
}
If you see in the comments i have mentioned that b and d are not
exactly the same b and d as shown in the matrix.. but they are current
right and current bottom elements of 'x'..
Hence, the swaps go on till the condition " x <= min (b,d ) " is not
satisfied..
On Jan 23, 3:44 am, Arun Vishwanathan <aaron.nar...@gmail.com> wrote:
> @lucifer:nice explanation !... just to make a small clarification, in your
> stabilisation part u jus compare x with min (b,d) , make a swap if
> necessary and then next time u compare it shud be <=min(b,d) and so u
> break.
>
> x b c
>
> d e f
>
> g h i
>
> so now after breaking x is less than both b and d but present b could be
> greater than e right? for example initally it cud be
> 8 5....
> 6 7.....
> .
> .
> .
> and we swap 8 and 5....now 8 is above 7 after swap ...but is this taken
> care of next iteration when we do swaps of a[row][col] with a[row+1][0]??
> so is heapify sep in all just comparison of x with b and d only and swap if
> needed??
>
>
>
>
>
>
>
>
>
> On Sat, Jan 14, 2012 at 1:48 AM, Gaurav Kalra <gvka...@gmail.com> wrote:
> > Bases on algorithm suggested by Lucifer, I have coded the problem in C
> > (please see attachment).
>
> > The code has been tested against the following test cases:
>
> > 1 3 4 8 9
> > 2 5 18 25 50
> > 6 7 22 45 55
>
> > and
>
> > 1 2 7
> > 3 5 8
> > 4 6 9
>
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