It's Christmas Eve, and Santa is loading up the sleigh for this year's deliveries. However, there's one small problem: he can't get the sleigh to balance. If it isn't balanced, he can't defy physics, and nobody gets presents this year.
No pressure.
Santa has provided you a list of the weights of every package he needs to fit on the sleigh. The packages need to be split into three groups of exactly the same weight, and every package has to fit. The first group goes in the passenger compartment of the sleigh, and the second and third go in containers on either side. Only when all three groups weigh exactly the same amount will the sleigh be able to fly. Defying physics has rules, you know!
Of course, that's not the only problem. The first group - the one going in the passenger compartment - needs as few packages as possible so that Santa has some legroom left over. It doesn't matter how many packages are in either of the other two groups, so long as all of the groups weigh the same.
Furthermore, Santa tells you, if there are multiple ways to arrange the packages such that the fewest possible are in the first group, you need to choose the way where the first group has the smallest quantum entanglement to reduce the chance of any "complications". The quantum entanglement of a group of packages is the product of their weights, that is, the value you get when you multiply their weights together. Only consider quantum entanglement if the first group has the fewest possible number of packages in it and all groups weigh the same amount.
For example, suppose you have ten packages with weights 1 through 5 and 7 through 11. For this situation, some of the unique first groups, their quantum entanglements, and a way to divide the remaining packages are as follows:
Group 1; Group 2; Group 3
11 9 (QE= 99); 10 8 2; 7 5 4 3 1
10 9 1 (QE= 90); 11 7 2; 8 5 4 3
10 8 2 (QE=160); 11 9; 7 5 4 3 1
10 7 3 (QE=210); 11 9; 8 5 4 2 1
10 5 4 1 (QE=200); 11 9; 8 7 3 2
10 5 3 2 (QE=300); 11 9; 8 7 4 1
10 4 3 2 1 (QE=240); 11 9; 8 7 5
9 8 3 (QE=216); 11 7 2; 10 5 4 1
9 7 4 (QE=252); 11 8 1; 10 5 3 2
9 5 4 2 (QE=360); 11 8 1; 10 7 3
8 7 5 (QE=280); 11 9; 10 4 3 2 1
8 5 4 3 (QE=480); 11 9; 10 7 2 1
7 5 4 3 1 (QE=420); 11 9; 10 8 2
Of these, although 10 9 1 has the smallest quantum entanglement (90), the configuration with only two packages, 11 9, in the passenger compartment gives Santa the most legroom and wins. In this situation, the quantum entanglement for the ideal configuration is therefore 99. Had there been two configurations with only two packages in the first group, the one with the smaller quantum entanglement would be chosen.
What is the quantum entanglement of the first group of packages in the ideal configuration?
Your puzzle answer was 11266889531.
That's weird... the sleigh still isn't balancing.
"Ho ho ho", Santa muses to himself. "I forgot the trunk".
Balance the sleigh again, but this time, separate the packages into four groups instead of three. The other constraints still apply.
Given the example packages above, this would be some of the new unique first groups, their quantum entanglements, and one way to divide the remaining packages:
11 4 (QE=44); 10 5; 9 3 2 1; 8 7
10 5 (QE=50); 11 4; 9 3 2 1; 8 7
9 5 1 (QE=45); 11 4; 10 3 2; 8 7
9 4 2 (QE=72); 11 3 1; 10 5; 8 7
9 3 2 1 (QE=54); 11 4; 10 5; 8 7
8 7 (QE=56); 11 4; 10 5; 9 3 2 1
Of these, there are three arrangements that put the minimum (two) number of packages in the first group: 11 4, 10 5, and 8 7. Of these, 11 4 has the lowest quantum entanglement, and so it is selected.
Now, what is the quantum entanglement of the first group of packages in the ideal configuration?
Your puzzle answer was 77387711.
A classical Knapsack problem with some minor twists. Not being aware of any efficient algorithms to solve this type of problem and seeing that the problem size is just about feasible, I tried a naive brute-force solution first. That did work, but it's excruciatingly slow.
So I did what a programmer has to do in such a situation: read up on a suitable algorithm on Wikipedia. With this, a solution is produced almost instantly and the code is even a few bytes shorter. (The end result is a tiny bit longer again, but only because I spent the additional 14 bytes to produce the results for both parts.)
In both implementations, I only tried to generate a single knapsack (the "passenger compartment" in the puzzle's nomenclature); a check whether the remaining items are evenly divisible into the 2 or 3 additional buckets is not performed and not necessary. I can't say if I was just lucky with my input or if this is by construction; in any case, the fact that all weights are prime numbers may have something to do with it ...
- Part 1, Python (brute-force): 232 bytes, ~15 min
- Parts 1+2, Python (proper algorithm): 238 bytes, <100 ms