Commit 2bd712b

@@ -0,0 +1,102 @@
+<<-DOC
+Consider a special family of Engineers and Doctors. This family has the following rules:
+
+Everybody has two children.
+The first child of an Engineer is an Engineer and the second child is a Doctor.
+The first child of a Doctor is a Doctor and the second child is an Engineer.
+All generations of Doctors and Engineers start with an Engineer.
+We can represent the situation using this diagram:
+
+                E
+           /         \
+          E           D
+        /   \        /  \
+       E     D      D    E
+      / \   / \    / \   / \
+     E   D D   E  D   E E   D
+Given the level and position of a person in the ancestor tree above, find the profession of the person.
+Note: in this tree first child is considered as left child, second - as right.
+
+Example
+
+For level = 3 and pos = 3, the output should be
+findProfession(level, pos) = "Doctor".
+
+Input/Output
+
+[time limit] 4000ms (rb)
+[input] integer level
+
+The level of a person in the ancestor tree, 1-based.
+
+Guaranteed constraints:
+1 ≤ level ≤ 30.
+
+[input] integer pos
+
+The position of a person in the given level of ancestor tree, 1-based, counting from left to right.
+
+Guaranteed constraints:
+1 ≤ pos ≤ 2(level - 1).
+
+[output] string
+
+Return Engineer or Doctor.
+http://www.geeksforgeeks.org/find-profession-in-a-hypothetical-special-situation/
+DOC
+
+describe "#find_profession" do
+  # [e]
+  # [e,d]
+  # [e,d,d,e]
+  # [e,d,d,e,d,e,e,d]
+  # [e,d,d,e,d,e,e,d,d,e,e,d,e,d,d,e]
+  # [e,d,d,e,d,e,e,d,d,e,e,d,e,d,d,e,d,e,e,d,e,d,d,e,e,d,d,e,d,e,e,d]
+  def find_profession(level, position)
+    return :Engineer if level == 1
+
+    parent_position = position.odd? ? (position + 1) / 2 : position / 2
+    parent = find_profession(level - 1, parent_position)
+    position.odd? ? parent : parent == :Doctor ? :Engineer : :Doctor
+  end
+
+  def bits_for(n)
+    #count = 0
+    #while n > 0
+      #n &= n - 1
+      #count += 1
+    #end
+    #count
+    n.to_s(2).count('1')
+  end
+
+  # [1]
+  # [1,0]
+  # [1,0,0,1]
+  # [1,0,0,1,0,1,1,0]
+  # [1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1]
+  # [1,0,0,1,0,1,1,0,0,1,1,0,1,0,0,1,0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0]
+  def find_profession(level, position)
+    bits_for(position - 1).odd? ? :Doctor : :Engineer
+  end
+
+  def find_profession(level, position)
+    (position - 1).to_s(2).count('1').odd? ? :Doctor : :Engineer
+  end
+
+  [
+    { level: 3, pos: 3, x: "Doctor" },
+    { level: 4, pos: 2, x: "Doctor" },
+    { level: 1, pos: 1, x: "Engineer" },
+    { level: 8, pos: 100, x: "Engineer" },
+    { level: 10, pos: 470, x: "Engineer" },
+    { level: 17, pos: 5921, x: "Doctor" },
+    { level: 20, pos: 171971, x: "Engineer" },
+    { level: 25, pos: 16777216, x: "Engineer" },
+    { level: 30, pos: 163126329, x: "Doctor" },
+  ].each do |x|
+    it do
+      expect(find_profession(x[:level], x[:pos])).to eql(x[:x].to_sym)
+    end
+  end
+end