Project Euler in Clojure – Problem 15

Here is my solution to Problem 15 of Project Euler. As always, my progress you can tracked on GitHub at

Problem 15 of Project Euler is to find the starting number with the longest Collatz sequence, summarized from the problem page as:

Starting in the top left corner of a 22 grid,
there are 6 routes (without backtracking) to the bottom right corner.

How many routes are there through a 2020 grid?

I started this problem, by trying to tracing and counting the routes through grids of 2×2, 3×3, and 4×4, and even setled in and did a 5×5 square. Having these numbers, and knowing I had two choices for ever position I was in, except for when the path got to the far edge and bottom, I had a hint at the growth rate of the problem. I tried some powers of 2 with the relationship of the numbers, and some factorials with the numbers. After seeing some possible relationships with the factorials that might be leading me in the right direction, I tried a number of permutation calculations, and the combination calculations. Having seen the numbers show up in different combination results, I then spent time back-calculating from those numbers into my ns, and found that the pattern seemed to match 2n Choose n.

The source code to this was the factorial function:

(defn factorial [n]
  (multiply (range 1M (inc (bigdec n)))))

And, I could have done it recursively, but I figured I would just operate against the sequence of numbers, especially now that the reducers are available in the Clojure 1.5-Alpha 3 release (at the time of this writing) of Clojure. After I get through a few more problems (of which I am working ahead of these posts), I am thinking it would be interesting to run the same Project Euler Problems against 1.4 and 1.5 using the reducers library, just substituting map/reduce for the reduce/combine functionality, and seeing how much effort it takes to move them over, as well as the differences in the timings of the different problems.

The other base function I needed was a combination function:

(defn combination [n k]
  (cond (zero? n) 0
        (zero? k) 1
        :else (/ (factorial n) (* (factorial (- n k)) (factorial k)))))

This function just does the basic calculation for combinations, from the formula:

\frac{n!}{\big((n-k)! * k!\big)}

With that, and my stumbling upon the matching of the fact that {2n}\choose{n} is the solution to the number of paths through the square the function problem15 is defined as:

(defn problem15
  ([] (problem15 20))
  ([n] (combination (+ n n) n)))

As always, any feedback you have for me is greatly appreciated.


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