How to design an algorithm to calculate countdown style maths number puzzle

Question

I have always wanted to do this but every time I start thinking about the problem it blows my mind because of its exponential nature.

The problem solver I want to be

Answer 1

By far the easiest approach is to intelligently brute force it. There is only a finite amount of expressions you can build out of 6 numbers and 4 operators, simply go through all of them.

How many? Since you don't have to use all numbers and may use the same operator multiple times, This problem is equivalent to "how many labeled strictly binary trees (aka full binary trees) can you make with at most 6 leaves, and four possible labels for each non-leaf node?".

The amount of full binary trees with n leaves is equal to catalan(n-1). You can see this as follows:

Every full binary tree with n leaves has n-1 internal nodes and corresponds to a non-full binary tree with n-1 nodes in a unique way (just delete all the leaves from the full one to get it). There happen to be catalan(n) possible binary trees with n nodes, so we can say that a strictly binary tree with n leaves has catalan(n-1) possible different structures.

There are 4 possible operators for each non-leaf node: 4^(n-1) possibilities The leaves can be numbered in n! * (6 choose (n-1)) different ways. (Divide this by k! for each number that occurs k times, or just make sure all numbers are different)

So for 6 different numbers and 4 possible operators you get Sum(n=1...6) [ Catalan(n-1) * 6!/(6-n)! * 4^(n-1) ] possible expressions for a total of 33,665,406. Not a lot.

How do you enumerate these trees?

Given a collection of all trees with n-1 or less nodes, you can create all trees with n nodes by systematically pairing all of the n-1 trees with the empty tree, all n-2 trees with the 1 node tree, all n-3 trees with all 2 node tree etc. and using them as the left and right sub trees of a newly formed tree.

So starting with an empty set you first generate the tree that has just a root node, then from a new root you can use that either as a left or right sub tree which yields the two trees that look like this: / and . And so on.

You can turn them into a set of expressions on the fly (just loop over the operators and numbers) and evaluate them as you go until one yields the target number.