@ lucifer: thank you !

On Sun, Jan 22, 2012 at 4:12 PM, Lucifer <[email protected]> wrote:

> @Arun,
>
> Nope.. the loop exits only when there are no more swaps possible...
>
> Let me explain with an example..
> x  b  c
> d  e   f
> g   h  i
>
> say x > min(b,d) , where min(b,d) = b,
>
> Hence, swap happens..
>
> b  x  c
> d  e   f
> g   h  i
>
> say, x > min(c, e), where min(c,e) = e..
> Hence, swap takes place..
>
> b   e  c
> d   x  f
> g   h  i
>
> Now say,
> x <= min(f,h)..
>
> Hence, we hit the break statement and exit from the loop..
>
>
> On Jan 23, 5:03 am, Arun Vishwanathan <[email protected]> wrote:
> > @lucifer: yes I get that...I was just saying that after a swap has
> happened
> > within the while loop ( since x > min(b,d) might have been the case ) ,
> > then in the next looping within while,  break wud happen right? meaning
> it
> > doesnt stay in the while after a swap happens...
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > On Sun, Jan 22, 2012 at 3:25 PM, Lucifer <[email protected]> wrote:
> > > @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 <[email protected]> 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 <[email protected]>
> 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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