You have 90 minutes to complete the exam of InformatikIIb. The following rules apply:

Transcription

1 Department of Informatics Prof. Dr. Michael Böhlen Binzmühlestrasse Zurich Phone: AlgoDat Nachholtest Spring Name: Advice You have 90 minutes to complete the exam of InformatikIIb. The following rules apply: Answer the questions on the exam sheets or the backside. Mark clearly which answer belongs to which question. Additional sheets are provided upon request. If you use additional sheets, put your name and matriculation number on each of them. Check the completeness of your exam (19 numbered pages). Use a pen in blue or black colour for your solutions. Pencils and pens in other colours are not allowed. Solutions written in pencil will not be corrected. Stick to the terminology and notations used in the lectures. For the exam Informatik IIb, only the following materials are allowed for the exam: One A4 sheet (2-sided) with your personal handwritten notes, written by yourself. Sheets that do not conform to this specification will be collected. A foreign language dictionary is allowed. The dictionary will be checked by a supervisor. No additional items are allowed. Notably calculators, computers, pdas, smart-phones, audio-devices or similar devices may not be used. Any cheating attempt will result in a failed test (meaning 0 points). Put your student legitimation card ( Legi ) on the desk. Signature: Correction slot Please do not fill out the part below Exercise Total Points Achieved Maximum Points

2 Exercise 1 20 Points Sorting & Lists 1.1 [3 points] Given array A = [8, 4, 2, 9, 3, 1, 6]. Apply selection sort on A. Complete the following matrix where you need to show the content of A after each execution of the outer for loop. The first line of the matrix shows the initial array A. i A[1] A[2] A[3] A[4] A[5] A[6] A[7]

3 Name: 1.2 [4 points] Consider the following doubly linked list and pointers a and b to its nodes. Assume also a function swap(a, b) that receives two pointers to nodes of a list and swaps them by modifying the pointers next and previous of list nodes instead of modifying the key values. head a b tail i) Draw the pointers of the list after applying the functions swap for the given pointers a and b. head a b tail

4 ii) Specify all special cases for a and b that must be considered and provide examples of the input data for each of them. Use the given list for your examples. head tail

5 Name: # Case a b Results (node key) (node key) 1 a=head (mirror case: b=head) a=tail (mirror case: b=tail) a=b a.next=b (mirror case: b.next=a) a=null (mirror case: b=null) (swap is not done or error is returned 5

6 1.3 [5 points] Consider a doubly linked list defined as follows: struct node { int key ; struct node next ; // p o i n t e r t o n e x t node struct node prev ; // p o i n t e r t o p r e v i o u s node ; struct node head ; struct node t a i l ; Use C or pseudocode to write a function that performs selection sort on the linked list defined above. You can use the function swap of Task 1.2. void s e l e c t i o n S o r t ( ) { struct l i s t N o d e i = head ; struct l i s t N o d e j, k, temp ; while ( i!= NULL) { k = i ; j = i >next ; while ( j!= NULL) { i f ( j >key < k >key ) k=j ; j = j >next ; swap ( k, i ) ; i = k >next ; 6

7 Name: 1.4 [5 points] Given an array A of n integers, use C or pseudocode to write a recursive version of selection sort to sort A in ascending order. void s e l e c t i o n S o r t R e c ( int A[ ], int i, int n ) { int k, j ; int temp ; i f ( i < n 1) { k = i ; for ( j=i +1; j<n ; j++) { i f (A[ j ] < A[ k ] ) k = j ; temp = A[ i ] ; A[ i ] = A[ k ] ; A[ k ] = temp ; s e l e c t i o n S o r t R e c (A, i +1, n ) ; 7

8 1.5 [3 points] Write the recurrence for the complexity of your algorithm in Task 1.4 and solve the recurrence. { T (n) = 1, if n = 0 T (n 1) + n, if n > 0 T (n) = T (n 1) + n = T (n 2) + 2n = T (n 3) + 3n = T (n i) + in i = n : T (n) = T (0) + nn = 1 + n 2 = O(n 2 ) 8

9 Name: Exercise 2 20 Points Hash Tables & Trees 2.1 [5 points] Consider the following definition of a hash table using the hash function h(k) = k mod 11 and linear probing to solve conflicts. struct element { int key ; int s t a t u s ; / 0 : OCCupied, 1: EMPty, 2: DELeted / ; struct element HT[ 1 1 ] ; The following operations are performed in the given order. Insert 20 Insert 4 Insert -2 Delete 20 Draw (a) Insert -1 Insert 0 Delete 4 Insert 31 Draw (b) At the positions marked by Draw (table), draw the resulting hash table HT using the tables below. Hint: In order to compute h(k) when k < 0 you can use the following property of the modulo function: k mod n = (k + n) mod n 9

10 Index Key Status Index Key Status (a) (b) 10

11 Name: 2.2 [4 points] Consider a hash table HT and the functions insert(ht, key) and search(ht, key) to insert and search an integer in HT. i) Specify the exact return types and values of insert and search. Return type of insert: integer. Return values: -1, if the hash table is full and the insertion fails. the number of the slot where key is inserted, if the insertion succeeds. Return type of search: integer. Return values: -1, if key is not in the hash table. the number of the slot where key is stored, if key is in the hash table. ii) Consider two arrays of integers, A[n A ] and B[n B ], that have no duplicates. Use C or pseudocode to describe an algorithm that uses the hash table HT (initially empty) and prints the intersection of A and B, which is a subset of elements existing both in A and B. You must not use any data structure other than the arrays A and B and the provided hash table HT. You can use functions search and insert of HT. Calculate the asymptotical upper bound for the runtime of your algorithm. Algorithm: intersection(s, T) 1 Hash Table HT ; 2 for i = 1 to n A do 3 insert(ht,a[i]); 4 for i to n B do 5 if search(ht,b[i]) 0 then 6 print(b[i]); Time Complexity: O(n A + n B ) 11

12 2.3 [3 points] The values are inserted in the given order into the following red-black tree Draw the red-black tree after both numbers are inserted. Write down any of the following operations that have been done during the insertions: Assignement of color to a node. Left or right rotation of a node. For each operation determine the key of the node to which it is applied and the arguments of the operation (if any). Operation Node key Argument Insert 55: assign color 55 red assign color 99 black assign color 10 black assign color 38 red assign color 38 black Insert 85: assign color 85 red left rotate 55 assign color 85 black assign color 99 red right rotate 99 assign color 38 black 12

13 Name: 2.4 [5 points] Write an algorithm that does left rotation on a node x of a red-black tree. You are allowed only to change nodes pointers, all other attributes in a node must remain the same. You can use C or pseudocode. void r b l e f t R o t a t e ( struct r b t r e e tree, struct rb node t ) { struct rb node s ; s = t >r i g h t ; t >r i g h t = s >l e f t ; s >parent = t >parent ; i f ( s >l e f t!= tree >n i l ) ( s >l e f t ) >parent = t ; i f ( t >parent == tree >n i l ) tree >root = s ; else i f ( t == ( t >parent) > l e f t ) ( t >parent) > l e f t = s ; else ( t >parent) >r i g h t = s ; s >l e f t = t ; t >parent = s ; 13

14 y x γ α β Figure 1: A tree 2.5 [3 points] Assume that a is a node in subtree α, b is a node in subtree β, and c is a node in subtree γ of the tree in Figure 1. The depths of these nodes are given in the following table. Determine the depths of the same nodes after performing a right rotation on node y. Subtree Depth before rotation Depth after rotation α 7 6 β 6 6 γ

15 Name: Exercise 3 20 Points Recursion and Dynamic Programming Given an array A consisting of n integers, the maximum contiguous subarray (MCS) problem is defined as finding a contiguous subarray within A that has the largest sum. For example, given A = [4, 2, 7, 9, 1], then [ 2, 7, 9] and [9, 1] are two possible contiguous subarray of A. However, neither of them is a MCS. 3.1 [2 points] Given an array A = [ 5, 4, 2, 1, 2, 6, 3, 9, 5, 2, 10], determine the MCS and its sum. MCS: [5, 2, 10] sum: [4 points] Given an array A of size n containing integers, define the recursive relation f(a, n, i) for the maximum sum of a contiguous subarray ending at the i th element of A. 0, if i < 0 f(a, n, i) = max(f(a, n, i 1) + A[i], A[i]), if i 0 15

16 3.3 [5 points] Given an array A of size n that contains integers, write a recursive algorithm that calculates the recursive relation of Task 3.2. Determine how this algorithm must be called to calculate the sum of a MCS of A. Use C or pseudocode for your solution. int mcsrecursive ( int a [ ], int n, int i ) { i f ( i == 0) return a [ 0 ] ; return max( mcsrecursive ( a, n, i 1) + a [ i ], a [ i ] ) ; int callmcsrecursive ( int a [ ], int n ) { int i, temp ; int maxsum = a [ 0 ] ; for ( i = 1 ; i < n ; i ++) { temp = mcsrecursive ( a, n, i ) ; i f ( temp > maxsum) maxsum = temp ; return maxsum ; 16

17 Name: 3.4 [6 points] When computing the sum of the MCS of an array A using dynamic programming, we need to create another array int M[n]. Element M[i] contains the maximum sum of a contiguous subarray ending at the i th element of A. i) Given the array A = [ 5, 4, 1, 2, 6, 4], fill in the array M below: M[]: i ii) Given an array A of size n that contains integers, write a dynamic programming algorithm that uses an array M[n] and calculates the sum of A s MCS. You can use C or pseudocode for your solution. int mcsdynamic ( int a [ ], int n ) { int i ; int maxsum = a [ 0 ] ; int currentmax = a [ 0 ] ; for ( i = 1 ; i < n ; i++) { currentmax = max( currentmax + a [ i ], a [ i ] ) ; i f (maxsum < currentmax ) maxsum = currentmax ; return maxsum ; 17

18 3.5 [3 points] In order to specify also the elements of a MCS of A, based on M, we need a matrix boolean sol[n][n], where element sol[j][i] corresponds to whether or not the j th element of array A is part of the corresponding contiguous subarray that has sum M[i]. Given the array A = [ 5, 4, 1, 2, 6, 4], and the array M that you filled in in Task 3.4, fill in the array sol below: sol[][]: j \ i

19 Name: kfjlksjfl 19