CISC-235* Test #2 October 15, 2018

Transcription

1 CISC-235* Test #2 October 15, 2018 Student Number (Required) Name (Optional) This is a closed book test. You may not refer to any resources. This is a 50 minute test. Please write your answers in ink. Pencil answers will be marked, but will not be re-marked under any circumstances. The test will be marked out of 50. Question 1 /10 Question 2 /15 Question 3 /15 Question 4 /10 TOTAL /50 By writing my initials in this box, I give Dr. Dawes permission to destroy this test paper if I have not picked it up by January 15, In the mountains of truth you will never climb in vain. ~ Happy Birthday to Friedrich Nietzche

2 Here are three class definitions: class BT_Vertex: # instance variables value : integer left_child : BT_Vertex right_child : BT_Vertex # constructor def BT_Vertex(x: integer): value = x left_child = nil right_child = nil class Binary_Tree: # instance variables root : BT_Vertex count : integer # constructor def Binary_Tree(): root = nil count = 0 class Binary_Search_Tree: # instance variables root : BT_Vertex count : integer # constructor def Binary_Search_Tree(): root = nil count = 0 # instance methods def BST_Insert(x: integer): count++ root = rec_bst_insert(root,x) def rec_bst_insert(v : BT_Vertex, x : integer): if (v == nil): return new BT_Vertex(x) else if (v.value < x): v.right_child = rec_bst_insert(v.right_child,x) else: v.left_child = rec_bst_insert(v.left_child,x) return v

3 Question 1: (10 Marks) (a) [5 marks] Suppose T is a Binary_Tree with k levels (the root is on level 1, its children are on level 2, etc.) What is the maximum number of vertices T can contain? You can express your answer as a sum. Solution: Level i has vertices, so with k levels T can contain at most vertices This sum = but it is not required to state that. Marking: correct answer 5/5 incorrect answer that shows understanding that the number of vertices on each level can double 3/5

4 (b) [5 marks] The number of levels in a binary tree can be computed using the following pseudo-code method def level(v : BT_Vertex): if (v == nil): return 0 else: return 1 + max(level(v.left_child),level(v.right_child)) called by the statement levels = level(root) What is the big-o complexity of this method on a tree with n vertices? Solution: This algorithm visits every vertex in the tree once and does a constant amount of work for each one the complexity is O(n) Marking: Correct answer 5/5 Incorrect answer that shows understanding of big-o 3/5 Incorrect answer that show poor understanding 1/5 No answer 0/5

5 Question 2 (15 marks): (a) [10 marks] Write a method (or combination of methods) in pseudo-code or in Java, Python, C or C++ that takes a Binary_Tree object as its parameter, and returns a new Binary_Tree containing the same values as the original tree, but as a mirror image. For example if the original tree is the new tree would be (b) [5 marks] What can you say about the complexity of your algorithm? If possible, determine its classification. Please write your solution to both parts on the next page

6 Page for Question 2 solution Solution: def flip(): root = rec_flip(root) def rec_flip(vert): if (vert == nil): return vert else: temp = rec_flip(vert.left_child) vert.left_child = rec_flip(vert.right_child) vert.right_child = temp return vert This algorithm visits every vertex in the tree exactly once and does a constant amount of work at each vertex. It is in, and Marking: Other correct algorithms are certainly possible. Students may build a new tree with new vertices if they wish. Please grade algorithms based on correctness, giving about 8/10 for algorithms that are almost correct, about 6/10 for algorithms with serious flaws such as only flipping part of the tree, about 4/10 for algorithms with critical flaws such as losing some of the vertices, and about 2/10 for incomplete solutions. For the complexity, give full marks if the analysis is correct for the algorithm they use. Give part marks as appropriate for partially correct answers.

7 Question 3 (15 marks) (a) [10 marks] Write a new method (or combination of methods) for the Binary_Search_Tree class that takes an integer parameter and prints all values in the tree that are. The printed values do not have to be in any particular order. Make your method as efficient as possible in terms of its big- O complexity. (b) [5 marks] What can you say about the complexity of your algorithm? Solution: def lte(x): return rec_lte(root,x) def rec_lte(v, x): if (v == nil): return else if (v.value > x): rec_lte(v.left_child, x) else: print v.value rec_lte(v.left_child,x) rec_lte(v.right_child,x) The big-o complexity is O(n) since each vertex is visited at most once and there is a constant amount of work done at each visited vertex. The complexity is since the algorithm can terminate after looking at just one vertex. The complexity is undefined. Marking: as for Question 2

8 Question 4 (10 marks) : The five rules for Red-Black Trees are 1. Every vertex is either Red or Black. 2. The root vertex is Black. 3. All leaves are Black, and contain no data. 4. Every Red vertex has 2 children, both of which are Black. 5. At each vertex, all paths leading down to leaves contain the same number of Black vertices. Prove that every non-leaf vertex in a Red-Black Tree has two children. Solution: Non-leaf vertices can be either Red or Black. By Rule 4, each Red vertex has two children. Let v be a non-leaf Black vertex, and suppose v has only one child. Every path leading down from v through that child must contain at least one Black vertex since each path ends at a leaf. But the path below v on the other side is empty so it it contains 0 Black vertices. This violates Rule 5. BUT: Students could reasonably argue that the empty path is not a path that ends in a leaf, so it doesn t need to be considered. In retrospect this was a poorly worded question. Marking: Any answer that shows an understanding of the balanced structure of RB-trees should get full marks.

9 BONUS QUESTION ( Ø marks) What is the meaning of this figure?