you can search for box stacking problem in google. There is a DP method. On Sep 22, 12:11 am, Dave <[email protected]> wrote: > Certainly having a smaller volume is necessary for a box to fit in > another box, but it is not sufficient. E.g., a box of size 1 x 1 x 1 > will not fit in a box of size 2 x 2 x 1/2. > > Dave > > On Sep 21, 1:16 pm, rajess <[email protected]> wrote: > > > > > find the volume of boxes as v=l*b*h > > sort boxes in volumes in descending order and this is the way to > > insert boxes one into another > > > On Sep 21, 7:55 pm, Rashmi Shrivastava <[email protected]> wrote: > > > > If there are n number of boxes and each with different dimensions and your > > > job is to insert one box having lesser dimension than that to another. > > > Consider size of boxes as, > > > b1->s1(h1,l1,w1) > > > b2->s2(h2,l2,w2) > > > . > > > . > > > . > > > bn->sn(hn,ln,wn)- Hide quoted text - > > > - Show quoted text -- Hide quoted text - > > - Show quoted text -
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