I’ve been doing this Coursera Discrete Optimization course with my friends. It’s a lot of fun and I’m learning a bunch. The instructor is a total goofball too, which is a plus. I’ve taken a handful of online courses before, but to be honest, the assignments have usually been pretty easy. Not so with this Discrete Optimization (DO (my initials!)) course! Each week, you have to solve 6 problems, and each is graded out of 10 depending on how well you do. I believe the breakdown is: 0/10 if your answer doesn’t even match the output format required, 3/10 if you do basically anything (even if your answer is quite wrong), 7/10 if you have an answer above some threshold but still not perfect, and 10/10 if your answer is optimal (I guess they know the optimal solution for all of them?). Usually, the problems increase in hardness throughout the set; often, the last one is difficult enough that (I believe we saw this said in the course forums by the instructor) it would be a challenge for people who specialize in DO for a living. I think that’s pretty cool! They usually give you a ton of practice problems of various difficulties, and (though I’m not 100% sure) I think the 6 you’re graded on are usually among those.
So what is DO? I certainly didn’t know when I started this course, though I guess I should’ve been able to guess. Optimization is what it sounds like, finding the best solution you can for given problems. The “discrete” part is that the quantities involved are integers or discrete (that’s the name in the title!!!) components. It turns out I had actually heard of many of the problems that DO applies to before, but didn’t know they were DO. I had heard most of them in the context of P vs NP complexity.
Anyway, enough of that. Week 1 didn’t have an assignment, so the first one was in week 2, and it was the famous “knapsack problem”. I’m guessing if you’re reading this (sike, no one reads this!), you’re familiar with it, but in case you aren’t, it’s as follows. You have a knapsack with a certain (weight) capacity, K. There are a bunch of items in front of you, and each one has a certain weight (w) and value (v). You wanna fill your backpack with the most valuable (i.e., largest combined value) set of items you can, and you can only take the items whole (i.e., you can’t break one in half to fill the bag completely).
Because I’ll be referring to it a few times, the format of a given problem (just a text file) they give you looks like this:
n is the number of items there are, and K is the capacity of your bag in this problem. v and w are the value and weight of each item. (I’ll refer to the items by their line number, starting with 0, in the following.)
Here’s where the “discrete-iness” comes into play. If you were allowed to take non-whole pieces of the items, the solution would be trivial: you would calculate the density (d = v/w) of each item, and start taking the most (densely) valuable items in decreasing order. When you get to the point where your bag can’t take the next whole item, you take as much of it as you can fit, and this is the theoretically most valuable set.
Now, for the actual problem (where you can’t take fractional pieces of items) it turns out that this is still often a decent solution, for intuitive reasons; denser items are giving you more bang for your buck. This is called the “greedy solution”, because it just greedily takes the best item it can at any moment. However, the problem gets interesting when that strategy will actually give you sub-optimal results. In fact, helpfully, the simplest problem they give you already has a non-greedy optimal solution (OS):
The items already happen to be ordered by decreasing density. If you did the greedy solution and took item 0 (8, 4) and then item 1 (10, 5), you couldn’t take any more items and your total value would be 18. However, this isn’t OS! You can see that you actually wanna take the least two densest items, 2 and 3, which perfectly fill the bag and also add up to 19, the OS.
So, that’s the problem! Now, how do you actually solve it? (And, what does it mean to solve it?)
The most naive way you can imagine is to try every possible set, and take the largest valid (i.e., all the items fit in the bag) solution. You have a list of n items, so for each one, you can either take it (1) or leave it (0). So there are really only (hah!) 2^n possible sets of items. Naturally, this lends itself to a tree data structure, but it should also be clear that 2^n quickly becomes infeasible for surprisingly low values of n.
Therefore, the real game is in traversing that tree in some clever way such that you don’t have to go through the whole tree. I should make a distinction here. There are two types of solution you can solve for. You can go for the OS, where you’re 100% sure the solution you end up with is optimal. This actually doesn’t necessitate going through the whole tree. Sometimes you can know that you don’t need to go farther down a tree, which allows you to not have to look at the solutions contained within that part of the tree, but still know that a solution you find will be the OS. On the other hand, sometimes this is still infeasible, so you have to go for an approximate solution (AS), where you’re just doing the best you can. However, not requiring the solution you get to be the OS can let you find great solutions way faster, and still sometimes even let you find the OS. The downside is that you won’t know 100% that you got the OS.
So let’s actually start. The way my friends and I originally did it (and I stuck with), we actually don’t explicitly have a tree data structure; however, by dint of doing recursion with two possible recursive calls, this implicitly forms a tree.
The first thing done is reading in the data and storing it in a list of namedtuple Item objects, which I believe was given to us. Then, because we’re going to go through the tree in order of decreasing density, we sort the list by that:
return(sorted(inList, key=lambda item:-item.density))
lines = input_data.split('\n')
firstLine = lines.split()
item_count = int(firstLine)
capacity = int(firstLine)
items = 
for i in range(1, item_count+1):
line = lines[i]
parts = line.split()
items.append(Item(i-1, int(parts), int(parts),float(parts)/float(parts)))
sortedList = sortListDensity(items)