I dont think the greedy approach gives the optimal solution here. Take the
below case -

Items are - 5, 4X2, 3, 2X2 and bins are of size 10. Greedy approach will
make choice in order -
bin1 - 5 + 4
bin2 - 4 + 3 + 2
bin3 - 2
Total bins required - 3

While in optimal solution -
bin1 - 5 + 3 +2
bi2 - 4 + 4 + 2
Total bins required - 2

Need to used DP approach similar to 0-1 knapsack.

Feedback welcome :-),
- Ravindra

On Sat, Oct 29, 2011 at 7:52 PM, icy` <[email protected]> wrote:

> yea, i'd go with greedy also.   Fill bin with biggest size s1 as much
> as possible (and same for other bins),  then try to squeeze in next
> biggest size s2, etc.
>
> On Oct 29, 7:17 am, teja bala <[email protected]> wrote:
> > Greedy knapsack algorithm will work fine in this case as in each bin
> >
> > n1s1+n2s2+..nrsr<=C gives the optimal solution...........
> >
> >
> >
> >
> >
> >
> >
> > On Sat, Oct 29, 2011 at 4:34 AM, SAMMM <[email protected]> wrote:
> > > Suppose u have n1 items of size s1,
> > > n2 items of size s2 and n3 items of size s3. You'd like to pack
> > > all of these items into bins each of capacity C, such that the
> > > total number of bins used is minimized.
> >
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