Craig Warnick

Greenhorn

Posts: 8

posted 5 years ago

Why not use Comparable? You can use compareTo for the same causes. For

You can also use <= and >= of course.

`a.compareTo(b)``< 0`means that

`a`is less than

`b`

`> 0`means that

`a`is greater than

`b`

`== 0`means that

`a`is equal to

`b`

You can also use <= and >= of course.

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posted 5 years ago

Well, here's a link to the Wikipedia article about it: Partially ordered set. I must say, I don't really see how to implement a partial ordering over a domain in Java; perhaps you throw a runtime exception if you try to compare a pair of incomparable elements?

But anyway if you're just going to compare the volume of boxes, then that's really a Total order. In that case it's much easier to implement a greaterThan method.

But anyway if you're just going to compare the volume of boxes, then that's really a Total order. In that case it's much easier to implement a greaterThan method.

Craig Warnick

Greenhorn

Posts: 8

posted 5 years ago

class BoxVolSort implements Comparator<Box> {

public int compare(Box b1, Box b2, Box b3) {

if(b1.getVolume() == b2.getVolume()==b3.get volume)

return 0;

else if (b1.getVolume() < b2.getVolume() > b3.getVolume)

return -1;

else

return 1;

This is what I came up with off the top of my head, but the statement that caused my helmet fire is:

box b > box c returns 1 whenever box c fits inside box box b. Note that c fits inside b if there is a way to arrange the dimensions of each box so

the corresponding dimensions of b are each larger than c.

No clue how to do that. Thanks for your earlier reply Saloon Keeper.

public int compare(Box b1, Box b2, Box b3) {

if(b1.getVolume() == b2.getVolume()==b3.get volume)

return 0;

else if (b1.getVolume() < b2.getVolume() > b3.getVolume)

return -1;

else

return 1;

This is what I came up with off the top of my head, but the statement that caused my helmet fire is:

box b > box c returns 1 whenever box c fits inside box box b. Note that c fits inside b if there is a way to arrange the dimensions of each box so

the corresponding dimensions of b are each larger than c.

No clue how to do that. Thanks for your earlier reply Saloon Keeper.

Craig Warnick

Greenhorn

Posts: 8

posted 5 years ago

Well, first off, java.util.Comparator works on two objects, not three, so I'd start off by amending your class accordingly.

Second, as you said yourself:

should return a negative number, and

should return a positive one; otherwise, your comparator should return 0 because the two objects have no 'order' with respect to each other in terms of "fitting in".

Then it's just a matter of calling the method for combinations of pairs for larger numbers of boxes.

[Edit] Note that while the above would work for providing a rudimentary "fits in" ordering, I suspect there's quite a bit more logic required to actually create a POS for these boxes. In terms of structure, I think I might look at an array of linked lists, but I suspect you're also going to need to come up with a heuristic to ensure that as many boxes are "fitted" as possible.

HIH

Winston

Craig Warnick wrote:box b > box c returns 1 whenever box c fits inside box box b. Note that c fits inside b if there is a way to arrange the dimensions of each box so

the corresponding dimensions of b are each larger than c.

No clue how to do that. Thanks for your earlier reply Saloon Keeper.

Well, first off, java.util.Comparator works on two objects, not three, so I'd start off by amending your class accordingly.

Second, as you said yourself:

So, in terms of a "fits inside" order, I would say that if Box a fits inside Box b, thenNote that c fits inside b if there is a way to arrange the dimensions of each box so the corresponding dimensions of b are each larger than c.

`compare(a, b)`

should return a negative number, and

`compare(b, a)`

should return a positive one; otherwise, your comparator should return 0 because the two objects have no 'order' with respect to each other in terms of "fitting in".

Then it's just a matter of calling the method for combinations of pairs for larger numbers of boxes.

[Edit] Note that while the above would work for providing a rudimentary "fits in" ordering, I suspect there's quite a bit more logic required to actually create a POS for these boxes. In terms of structure, I think I might look at an array of linked lists, but I suspect you're also going to need to come up with a heuristic to ensure that as many boxes are "fitted" as possible.

HIH

Winston

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posted 5 years ago

It sounds like the exercise is to define an interface with one method, greaterThan(). You can do that in just 3 lines of code or so.

See What Is an Interface? in Oracle's Java Tutorials.

Craig Warnick wrote:I need some help understanding what a partial order interface with one method greaterThan(). I am working on an exercise to compare the volume of three boxes and this statement is the next step. Thank you,

It sounds like the exercise is to define an interface with one method, greaterThan(). You can do that in just 3 lines of code or so.

See What Is an Interface? in Oracle's Java Tutorials.

Craig Warnick

Greenhorn

Posts: 8