samskivert: Euler 018

18 January 2008

Problem 018:

object Euler18 extends Application {
  val triangle = List(
     4, 62, 98, 27, 23,  9, 70, 98, 73, 93, 38, 53, 60,  4, 23,
      63, 66,  4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31,
        91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48,
          70, 11, 33, 28, 77, 73, 17, 78, 39, 68, 17, 57,
            53, 71, 44, 65, 25, 43, 91, 52, 97, 51, 14,
              41, 48, 72, 33, 47, 32, 37, 16, 94, 29,
                41, 41, 26, 56, 83, 40, 80, 70, 33,
                  99, 65,  4, 28,  6, 16, 70, 92,
                    88,  2, 77, 73,  7, 63, 67,
                      19,  1, 23, 75,  3, 34,
                        20,  4, 82, 47, 65,
                          18, 35, 87, 10,
                            17, 47, 82,
                              95, 64,
                                75
  );

  def fold (triangle :List[Int], max :List[Int]) :Int = {
    if (max.length == 1) {
      return max(0);
    } else {
      return fold(triangle.drop(max.length-1), List.range(0, max.length-1).map(
        (i) => (triangle(i) + Math.max(max(i), max(i+1)))));
    }
  }
  var base = (Math.sqrt(1+8*triangle.length) - 1) / 2; // 15
  println(fold(triangle.drop(base), triangle.slice(0, base)));
}

The interesting thing about this one (other than my excellent triangle array formatting) is that we can simply fold the triangle up on itself computing the larger of each pair of elements in the bottom row and adding that larger value to the row above it.

A reduction step takes something like this:

91, 71, 52, 38, 17, 14, 91, 43, 58, 50, 27, 29, 48,
      63, 66,  4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31,
     4, 62, 98, 27, 23,  9, 70, 98, 73, 93, 38, 53, 60,  4, 23,

and turns it into:

91,  71,  52,  38,  17,  14,  91,  43,  58,  50,  27,  29,  48,
125, 164, 102,  95, 112, 123, 165, 128, 166, 109, 122, 145, 100, 54,

Then we fold that up into the next line and so on until we've got our solution in the top row.

©1999–2015 Michael Bayne