Finding all possible variations in a Bin-Packing Problem - c++

Question

For my task I had to write a bin-packing algorithm where there are N objects with different volumes. They all had to be packed into boxes of volume V. Using Decreasing Sorting I successfully wrote the algorithm. But another task includes writing out all possible variations of bin-packing in an amount of boxes that I previously found most effective. So for example:
There are 4 objects with volumes: 4, 6, 3, 2. Volume of boxes is 10. Using the bin-packing algorithm I find that I will need 2 boxes.
All possible variations would be:
4,6 and 3,2
4,3 and 6,2
4,2 and 6,3
6 and 4,3,2
I'm having trouble coming up with an appropriate algorithm for this problem, where should I start ? Any help would be greatly appreciated.

Answer 1

The general algorithm for solving this problem goes like this:
Try to fit all objects in n bins by creating all possible split configurations into n groups and test if any such configuration fits in the bins.
If not, increase n and try again.
Now, how do you find all possible split configurations?
Consider putting a tag on each object to decide into which bin it belongs. If you have 3 objects and 2 bins, then each object can get the tag 0 or 1 (for any of the two bins). This makes 2^3 = 8 combinations:
000
001
010
...
Now it also becomes clear how to create all combinations. You can use a counter and convert it into the base of the number of bins (2 in this case) and use the digits as tags. There are other options, e. g. you could use a recursive solution. I prefer that.
When you have a solution you just need to check that for each bin the volume sum of the objects of this tag is not greater than the bin size.
Here would be some pseudo code for creating a list of all the combinations recursively:
combinations(object_counter, bin_counter) {
if (object_counter == 0) {
return [[]] // a list of one empty list
}
result = [] // empty list
for i in 0 .. bin_counter-1 {
sub_results = combinations(object_counter-1, bin_counter)
for sub_result in sub_results {
result.append([i] + sub_result)
}
}
return result
}