Move 2015 out of the way.

2015/day-17/README.md

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+# Day 17 efficient solution
+
+Day 17 is an instance of the subset sum problem. This problem asks whether for
+a (multi)set of integers *V*, there is a non-empty subset of integers summing up to
+exactly *s*. This problem is NP-complete.
+
+The brute force approach of this is trying every possible set in the powerset
+of *V* to see if they match. This is inpractical however, because a powerset of
+a set of size *n* contains 2<sup>2</sup> sets.
+
+In the exercise, we have 20 buckets, and 2<sup>20</sup> is still
+brute-forcable. There is a smarter approach.
+
+We split the list of buckets in two lists of (approximately) the same size. We
+then take the powersets of those two lists and compute the sum for each entry.
+This leaves us with a total of 2<sup>n / 2 + 1</sup> entries. We then sort both
+sublists on the total value of each entry.
+
+Finally, we iterate of the first list, and use binary search to see whether
+there is an appropriately sized entry in the second list. This gives us a final
+complexity of *n* times 2<sup>*n*/2</sup>, allowing the solution to be computed
+instantly.
+
+The algorithm above can be modified to find all combinations, not just one, in
+time proportional to the number of solutions. This is implemented in the final
+program.