--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/TAs Tue Nov 15 23:04:52 2016 +0000
@@ -0,0 +1,14 @@
+ daniil.baryshnikov@kcl.ac.uk,
+ andrew.coles@kcl.ac.uk,
+ oliver.hohn@kcl.ac.uk,
+ fahad.ausaf@icloud.com,
+ fares.alaboud@kcl.ac.uk,
+ sara.boutamina@kcl.ac.uk,
+ mark.ormesher@kcl.ac.uk,
+ clarence.ji@kcl.ac.uk,
+ andrei.nae_-_stroie@kcl.ac.uk,
+ alexander.hanbury-Botherway@kcl.ac.uk,
+ rosen.dangov@kcl.ac.uk,
+ diana.ghitun@kcl.ac.uk,
+ andrei.juganaru@kcl.ac.uk,
+ ainur.makhmet@kcl.ac.uk
\ No newline at end of file
--- /dev/null Thu Jan 01 00:00:00 1970 +0000
+++ b/progs/collatz_sol2.scala Tue Nov 15 23:04:52 2016 +0000
@@ -0,0 +1,55 @@
+// Part 1 about the 3n+1 conceture
+//=================================
+
+//(1) Complete the collatz function below. It should
+// recursively calculate the number of steps needed
+// until the collatz series reaches the number 1.
+// If needed you can use an auxilary function that
+// performs the recursion. The function should expect
+// arguments in the range of 1 to 1 Million.
+
+def collatz(n: Long): List[Long] =
+ if (n == 1) List(1) else
+ if (n % 2 == 0) (n::collatz(n / 2)) else
+ (n::collatz(3 * n + 1))
+
+
+// an alternative that calculates the steps directly
+def collatz1(n: Long): Int =
+ if (n == 1) 1 else
+ if (n % 2 == 0) (1 + collatz1(n / 2)) else
+ (1 + collatz1(3 * n + 1))
+
+import scala.annotation.tailrec
+
+@tailrec
+def collatz2(n: Long, acc: Long): Long =
+ if (n == 1) acc else
+ if (n % 2 == 0) collatz2(n / 2, acc + 1) else
+ collatz2(3 * n + 1, acc + 1)
+
+
+//(2) Complete the collatz bound function below. It should
+// calculuate how many steps are needed for each number
+// from 1 upto a bound and return the maximum number of
+// steps and the corresponding number that needs that many
+// steps. You should expect bounds in the range of 1
+// upto 1 million.
+
+def collatz_max(bnd: Long): (Long, Long) = {
+ (1L to bnd).view.map((i) => (collatz2(i, 1), i)).maxBy(_._1)
+}
+
+
+// some testing harness
+//val bnds = List(10, 100, 1000, 10000, 100000, 1000000)
+val bnds = List(10, 100, 1000, 10000, 100000, 1000000, 10000000, 100000000, 1000000000)
+
+for (bnd <- bnds) {
+ val (steps, max) = collatz_max(bnd)
+ println(s"In the range of 1 - ${bnd} the number ${max} needs the maximum steps of ${steps}")
+}
+
+
+//val all = for (i <- (1 to 100000).toList) yield collatz1(i)
+//println(all.sorted.reverse.take(10))