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I'm trying to build a data tree like the one shown below, and I need an efficient matching algorithm that can do the following.

You can think of this tree as a list of prerequisites for taking courses. For example, course1 has prerequisite 3 and 4, and course 3 has prerequisites 7 and 8. If one wants to take course1, he/she must take both 3 and 4, or all prerequisites for course3 or course4 (so if he/she took 7,8,4, it's equivalent to having taken 3 and 4).

Now a kid comes and says he wants to take course2, providing that he took course 8,9 and 6 in advance. How can I quickly check if he is eligible?

The only way I can think of right now is to construct a look-up table that contains all combinations of prerequisites, and checks through it to find the match. However, as the tree grows bigger(I'm trying to build one with potentially >10,000 nodes), this method is going to blow up my computer.

Does anyone have any advice for this? Or better yet, is there a well-defined searching algorithm that can handle this type of task already? Thanks in advance. Jim

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Fig: Some arbitrary data tree

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  • Sounds like a relational database concept to me. I know a lot of databases use B-Trees but I don't know how much help that'll be without a lot of augmentation. Commented May 30, 2012 at 23:03
  • Actually, now that I think about it. It seems more like a Graph than a Tree, with the edges between node being the prereqs. Commented May 30, 2012 at 23:04
  • "providing that he took course 8,9 and 6 in advance" - did you mean 9,10, and 6? Commented May 30, 2012 at 23:38

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Performance will probably be fine just doing it in a simple way. If the student wants 2, first check whether they satisfy the requirements imposed by the 5 prerequisite, then check whether they satisfy the requirements imposed by the 6 prerequisite. You requirements-checking function is going to involve a recursive call or similar.

If this doesn't work for some reason (cycles in the graph?!), you could go in the other direction to what you've described: start from the courses the student has taken, and flood-fill to get a list of all the courses the student qualifies for. Check whether the target is in that list (or rather: stop immediately if you encounter the target while building the list).

The flood fill is somewhat complicated by the fact that you're not just following arrows: I can qualify for 1 in four different ways: (3 or (7 and 8)) and (4 or 9). But the basic idea is the same: keep testing the parents of my reachable nodes to see whether I can add anything to my set, until I can't any more.

Applicable to both methods: I'm not entirely clear what the rules are, whether this "satisfying the prerequisites of 3 instead of taking 3" is transitive or not. Given:

      1
     / \
    2   3
   /
  4
 / \
5   6

I get that having (5,6) qualifies me for 2, but does (5,6,3) qualify me for 1?

If so then I think the process is a bit easier, because as a simplifying assumption I can just pretend that if I've taken (5,6) then I've also taken 4, and the question "am I eligible for 2?" becomes, "have I taken 2?".

It seems to me more sensible if it doesn't, since I don't suppose I can really take a whole bunch of elementary classes and then jump straight into post-graduate work 10 levels higher up. Then there's a functional difference between courses that I've taken, vs courses whose prerequisites I've taken. That difference needs to be tracked. In the case of the flood-fill, I think that means 2 flags per node to consider rather than 1. In the case of the requirements-checking function, I think it means you need 2 checks (for direct and indirect satisfaction), and no recursion is needed.

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"(so if he/she took 7,8,4, it's equivalent to having taken 3 and 4)." --original poster

This isn't very much like prerequisites; normal classes do not function like this. What this is, is a list of possible substitutions. You would like to have a 2 for example, but you are also allowed to have all the substitutions for 2, which are 5,6. You can further substitute 9,10 for 5, or 11 for 6.

This is a recursive definition of substitutability.

In python:

def canMake(desired, withIngredients):
    if desired in withIngredients:
        return True
    else:
        return all(canMake(s,withIngredients) for s in desired.substitutions)

sidenote:

You will need to be able to query what the parent of a node is in O(1) time. This works just fine if you have your classes defines as:

2 requires 5,6
5 requires 9,10
...

It will require that you add backreferences if you store it like:

5 allows 2
6 allows 2
9 allows 4,5

We assume the first situation where course.prereqs is a list of required classes (in the second situation, you can build this list easily by defining: course.parents = [] for each course, then appending child.parents.push(child) for each course's child).

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An intuitive would be traverse the tree by either BFS or DFS.

Whenever you reach a node denoting a course you already took, cut all the nodes under that node.

So there are only two cases where the eligibility check fails:

  1. after you reach all the nodes denoting courses you already took, there are still other nodes to traverse.
  2. you reach a leaf node which is not a node denoting a course you already took
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