NONE (in particular, no calculator)

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1 Duration: Aids Allowed: 110 minutes NONE (in particular, no calculator) Student Number: Last (Family) Name: First (Given) Name(s): Do not turn this page until you have received the signal to start. (In the meantime, please fill out the identification section above, and read the instructions below carefully.) This term test consists of 5 questions on 9 pages (including this one), printed on both sides of the paper. When you receive the signal to start, please make sure that your copy of the test is complete. Answer each question directly on the test paper, in the space provided. If you need more space for one of your solutions, use the reverse side of the last page and indicate clearly the part of your work that should be marked. In your answers, you may use without proof any fact covered in lecture, tutorial, or on assignments. You must justify all other facts required for your solution. General Hint: We were careful to leave ample space on the test paper to answer each question. Guide # 1: / 9 # 2: /10 # 3: / 3 # 4: / 3 # 5: / 5 TOTAL: /30 Good Luck! Total Pages = 9 Page 1 over...

2 Question 1. [9 marks] Consider the following Python code which creates a list of the n integers from 0 to n-1, shuffles the list and then passes it to the function dosomething. Assume that the shuffle function is equally likely to return any ordering of the list elements. We are interested in the number of time the getitem method is called on the list. Recall that getitem is called whenever you read the value of the array at a particular index. Notice that we are not concerned with what happens inside the function dosomethingelse since it is passed the list element not the entire list. from random import shuffle def dosomething(a,head): for i in range(len(a)): if a[i] == head: return "found" else: dosomethingelse(a[i]) #this access counts #this access counts a = range(n) shuffle(a) first_head = a[0] shuffle(a) dosomething(a,first_head) # make list [0,1,... n-1] # shuffe it # this access doesn t count. It is outside dosomething # shuffle again Part (a) [1 mark] Perform a best-case analysis of the performance of the method dosomething when called in the context shown. The best case occurs when a[0] == head i.e. when the second shuffle leaves the first element in the first position. In this case exactly 1 getitem call is made and then the function returns. This is O(1). You can t do better than that. 1/2 mark for describing case and 1/2 for saying O(1) or constant time. 0 for saying that best case is when n = 1. Part (b) [3 marks] Perform a worst-case analysis of the performance of the method dosomething when called in the context shown. Worst-case occurs when head==a[n-1] which is when the item of interest is in the last list element. For each of n passes we check a[i]==head which is 1 getitem call each. For each of n 1 passes we dosomethingelse(a[i]) which each have 1 getitem call. On the last pass we return and do not execute the else clause. Total is 2n 1 calls to getitem. Because shuffle only rearranges items in the list, the item we are searching for must be one of the elements. If it comes at any point before the last item, we can only return sooner. So this must be the worst case. 1 mark each for these 3 ideas Student #: Page 2 of 9 over...

3 Part (c) [5 marks] Perform an average-case analysis of the performance of the method dosomething when called in the context shown. Consider that head is found at position i in list where 0 i n 1 This takes i + 1 calls from the line a[i]== head plus i calls from the line dosomethingelse... total = 2i + 1 calls to getitem The average number of evaluations T is calculated as follows. T = = 1 n n 1 i=0 n 1 p(head found at i) calls(head found at i) i=0 (2i + 1) = 1 n 1 n (2 i + n) i=0 = 1 1)(n) (2(n + n) n 2 = n = n 5 marks total: needs to be perfect for full marks including at least some explanation that 2i + 1 is the number of calls (or they might say accesses or evaluations) when head is found at position i. If there is just the math and no explanation at all but it is perfect give 4. For part marks use your own judgement and keep track of what you did so that you can be consistent for all tests. Student #: Page 3 of 9 over...

4 Question 2. [10 marks] A binary search tree stores records containing keys and other information. The tree is ordered by the keys and the normal search, insert and delete methods must have complexity in O(h) where h is the height of the tree. Additionally we would like a method replace(x,i) which will replace the other information of the i th record with x. Use the convention that i begins at 1. In this question you will explain how to augment a binary search tree to provide this new operation in O(h). Part (a) [2 marks] What new information must be maintained in the tree? Explain with a formal written explanation and then show the new information on the binary tree below. New Information: At every node store the size of the subtree rooted at that node including the node itself. B size = 7 size = 1 A C size = 5 F size = 4 size = 1 size = 2 D E G size = 1 Part (b) [2 marks] Provie an algorithm for the delete operation. You may make reference to the standard bst deletion algorithm and just indicate the differences. Use the standard BST deletion algorithm. Let d be the node to be deleted. When you are finding d decrement the sizes along the path from the root to d. If d is a leaf you are done. If d is an internal node with only one child you are also done. Just make the parent pointer that previously pointed to d, now point to d s only child. The sizes of the child (and its descendants) do not get changed. When d is an internal node with two children, you swap d node with a leaf node x (either d s predecessor or successor) and then delete x. As you go down the tree to find x continue to decrement the sizes of the nodes along the path from d to x. So in the end all the nodes along the path from the root to x will be decremented by 1. Student #: Page 4 of 9 over...

5 Download Part (c) [5 marks] Provide pseudo code for the operation replace(x,i). Optional Hint Begin by defining replace(node,x,i) so that you can call this method on subtrees rooted at node. replace(node,x,i): // first check to see if i is in node s left subtree if node.left!= null and node.left.size >= i: replace (node.left,x,i) else // get size of left child - 0 if no left child exists if node.left == null: leftsize = 0 else leftsize = node.left.size // check to see if i is in node itself if leftsize + 1 == i: node.other = x // found the node replace the info return else: // look in the right subtree // decrease the rank by the size of nodes skipped over replace(node.right,x,i-leftsize-1) replace(x,i): replace(root,x,i) One mark for each of these basic ideas: size of left i then go to left subtree size of left + 1 == i then found so change the data szie of left +1 < i then go right and when you go right change i correctly remember that if no leftsubtree exists that size is 0. They might check for null explicitly in algorithm or just say that the convention is thatany trees which are null pointers have size 0. Remember that if they defined the size differently in part a, the constants in this algorithm will be different. The five basic points should still apply. Part (d) [1 mark] Briefly justify why your algorithm is O(h). The method itself contains no loops only at most one recursive call. Each call is on a child so at most there can be h calls. Student #: Page 5 of 9 over...

6 Question 3. [3 marks] Consider the 2,3,4 tree T shown below. Assume that item x from the list [1,2,...,10] is selected at random with equal probability. Calculate the average height of tree T after removing item x. Assume that the deletion algorithm swaps an internal node with its predecessor rather than its successor. Show all your work For elements in the set {5, 6, 7, 8, 9, 10} the height of the tree does not decrease with the deletion. Notice that 7 swaps with 6 so it is included in this set. For elements in the set {1,2,3,4}, the height of the tree is 2 after the deletion. average height of T = ( 4 10 )(2) + ( 6 10 )(3) = = marks if they get the correct final answer and you can follow what they have done. 2 marks if the answer is correct but you can t follow the explanation or it is not present. 1 mark if they have the incorrect final answer but you can see that much of it is correct and they have made a mathematical error or have made a single wrong mistake. For example, if they mixed up precessor and successor and assumed that removing 4 does not decrease the tree size the final answer they would calculate is 2.7. This is worth 1/3. Student #: Page 6 of 9 over...

7 Question 4. [3 marks] Perform DFS on the graph below choosing the start vertices in alphabetical order. Indicate on the graph the DFS forest produced by running the algorithm. Also provide a list of edges in the resulting DFS forest. For each vertex, indicate the discovery time and finish time. A B E D C Vertex v discovery: d(v) finish: f(v) A 1 4 B 5 10 C 2 3 D 6 9 E 7 8 List the edges in the DFS forest here. (A,C) (B,D) (D,E) 1 mark for indicating the correct DFS forest edges on the graph 1 mark for the values in the table (all or nothing) 1 mark for listing the edges in the forest. These can be listed in any order since technically this is a set of edges. I don t care if they accidentally use lowercase vertex names. Edges must be indicated with regular parenthesis not curly braces. So {(B,D),(A,C),(d,e)} is fine but {B,D},{A,C},{D,E} is not. Student #: Page 7 of 9 over...

8 Question 5. [5 marks] Consider a database for a hockey league with k player records each containing the following information: Name: unique for every player Team number: not unique Jersey number: unique within a given team League Registration Number: unique for every player Other data: potentially very large You need to provide the following ADT. All methods must run in O(log k) time. search(n): returns record x which has name n search(t,j): returns record x which wears jersey j on team t search(r): returns record x which has league registration number r insert(x): insert record x into the database delete(x): given a pointer to record x, delete it from the database Part (a) [3 marks] Describe an efficient data-structure to implement this ADT. Consider space efficiency because the other data for each player is potentially very large. Use 3 balanced search trees and a separate file of the complete records. The key in the first tree is name and for each record in the tree, the value is a pointer to the full player record in the separate file. The key in the second tree is a combination of jersey number and team and again the values are pointers to the full records file. The third tree has registration numbers as keys and pointers to the full records file. When you insert a into the data-structure, the full record gets put in the file and a pointer to this record is inserted with the appropriate key into each of the tree trees. When you delete from this data-structure, the corresponding entry is removed from each of the 3trees and the spot in the records file is marked as available. This operation can not affect the values of the pointers to the other records in the file. You can search for a record using any one of the 3 trees. Part (b) [2 marks] On the next page draw a picture of your data-structure using the following four records. name team jersey reg num other Michelle L left wing Tom K goalie Craig B defence Kejie J centre Student #: Page 8 of 9 over...

9 Draw your picture here. Mich 2,2 Craig Kejie Tom 1,2 1,3 2,5 Michelle l leftwing Tom k goalie Craig B defence Kejie K centre 434K 117B 145L 516J Student #: Page 9 of 9 End of Scheme