solution2.5.2.5.scala

chapter02/worksheets/solution2.5.2.5.scala

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+/** 
+  Exercise 2.5.2.5
+  Same task as in Exercise 2.5.2.4 but using a set of sets.
+
+  Instead of just three sets a, b, c, we are given a value of type
+  Set[Set[Int]].
+
+  The required type signature and a sample test:
+
+  def prodSet(si: Set[Set[Int]]): Set[Set[Int]] = ???
+
+  scala> prodSet(Set(Set(1, 2), Set(3), Set(4, 5), Set(6)))
+  res0: Set[Set[Int]] = Set(Set(1,3,4,6),Set(1,3,5,6),Set(2,3,4,6),Set(2,3,5,6))
+
+  Hint: use foldLeft and flatMap.
+  */
+
+def prodSetIntern(b: Set[Set[Int]] = Set(Set.empty), symbols: Set[Int] = Set.empty): Set[Set[Int]] = {
+  if (b.tail.isEmpty) (b.head).map{ x => Set(x) ++ symbols }
+  else
+    b.head.flatMap{ x => prodSetIntern(b.tail.toSet, Set(x) ++ symbols) }
+}
+
+def prodSet(a: Set[Set[Int]] = Set(Set.empty)): Set[Set[Int]] = {
+  prodSetIntern(a)
+}
+
+val result = prodSet(Set(Set(1, 2), Set(3), Set(4, 5), Set(6)))
+val expected: Set[Set[Int]] = Set(Set(1,3,4,6),Set(1,3,5,6),Set(2,3,4,6),Set(2,3,5,6))
+assert(result == expected)
+
+// scala> :load solution2.5.2.5.scala
+// :load solution2.5.2.5.scala
+// def prodSetIntern(b: Set[Set[Int]], symbols: Set[Int]): Set[Set[Int]]
+// def prodSet(a: Set[Set[Int]]): Set[Set[Int]]
+// val result: Set[Set[Int]] = Set(Set(6, 4, 3, 1), Set(6, 5, 3, 1), Set(6, 4, 3, 2), Set(6, 5, 3, 2))
+// val expected: Set[Set[Int]] = Set(Set(1, 3, 4, 6), Set(1, 3, 5, 6), Set(2, 3, 4, 6), Set(2, 3, 5, 6))