School of Computing and Information Technology. ASSIGNMENT 1 (Individual) CSCI 103 Algorithms and Problem Solving. Session 2, April - June 2017

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1 ASSIGNMENT 1 (Individual) CSCI 103 Algorithms and Problem Solving Session 2, April - June 2017 UOW Moderator: Dr. Luping Zhou (lupingz@uow.edu.au) Lecturer: Mr. Chung Haur KOH (chkoh@uow.edu.au) Total Pages: Seven (7) inclusive of the cover page Total Marks: 10 marks Weightage: 10% of total subject mark Submission Date: 30 April 2017 Copyright SCIT,, 2017 Page 1 of 16

2 Objective In this assignment, students are required to apply the concepts of flowchart, pseudocode, and problem solving, sorting and searching. Flowchart and Pseudocode [2.0 marks] Question 1 (2.0 marks) (a) Alex, the owner of Excellent Tuition Centre, needs a program that his tutors can use to generate a summarized grade report for the subject the tutor teaches. The report is to contain the following information: Subject Title: Functions and graphs Tutor: Diane Miller Grade Number of students A 5 B 9 C 7 F 3 Write a pseudocode algorithm to read the title of the subject, the name of the tutor, and a list of integer values for score, one at a time. The list of integer values is terminated with a negative value. Your algorithm will need to determine the grade and accumulate the count of occurrences for each grade. The grading is done according to the following guideline: Score Grade 80 or more A Less than 80 but 65 or more B Less than 65 but 50 or more C Less than 50 F Suggested solution: (1.0 mark) BEGIN SummarizedGradeReport // Declaration and Initialization DEFINE subjecttitle DEFINE tutorname DEFINE score, counta, countb, countc, countf Copyright SCIT,, 2017 Page 2 of 16

3 // Display message and accept subject title DISPLAY Enter Subject Title: ACCEPT subjecttitle // Display message and accept tutor name DISPLAY Enter Tutor Name: ACCEPT tutorname // Display instruction to enter score DISPLAY Enter the score in integer one at a time. Enter negative score to terminate the process. // Start the loop REPEAT DISPLAY Enter score (negative score to terminate): ACCEPT score // Check score and accumulate the respective counters IF score >= 80 ACCUMULATE counta = counta + 1 ELSE IF score >= 65 ACCUMULATE countb = countb + 1 ELSE IF score >= 50 ACCUMULATE countc = countc + 1 ELSE IF score >= 0 ACCUMULATE countf = countf + 1 ELSE DO NOTHING END IF UNTIL score < 0 // Print the report DISPLAY Subject Title: + subjecttitle // display subject title DISPLAY Tutor: + tutorname // display tutor name DISPLAY // display blank line DISPLAY Grade Number of Student // display sub heading DISPLAY A + counta DISPLAY B + countb DISPLAY C + countc DISPLAY F + countf END SummarizedGradeReport Copyright SCIT,, 2017 Page 3 of 16

4 (b) AAA Packing Company wants a program that the order department can use to calculate and display the price of an order (price * unit). The order clerks will need to enter the number of units ordered, and whether the customer is a retailer or a wholesaler. The price per unit depends on both the customer type, that is wholesaler or retailer, and the number of units ordered, as follows: Wholesaler Retailer Number of units Price per unit ($) Number of units Price per unit ($) and over and over 12 If the customer type is entered incorrectly or the number of units ordered is invalid, the price per unit and the number of units ordered are set to 0. Draw a flowchart for the program requirement specified above. (1.0 mark) Copyright SCIT,, 2017 Page 4 of 16

5 Display priceoforder = priceperunit * numofunitorder Problem Solving and invariant [3.5 marks] Question 2 (0.5 mark) Robert is an intelligent robot and is able to solve problem if given appropriate instructions. Robert is five steps away from a box, and the box is 10 steps away from a chair, as shown below: 5 steps 10 steps Copyright SCIT,, 2017 Page 5 of 16

6 Create a series of instructions, using only the instructions shown below, that will help Robert to sit in the chair. Assume that Robert must jump over the box before he can continue toward the chair. Robert s instruction set: Instruction Stand Walk Sit Turn180 Turn90R Turn90L Jump Fly If reach an obstacle, do this: If reach a chair, do this: Otherwise, do this Repeat x times Repeat until you are directly in front of a chair Repeat until you are directly in front of a box Meaning Robert stands up from a sitting position Robert moves his right foot forward one step, then moves his left foot to meet his right foot Robert sits down Robert turns 180-degree Robert turns 90-degree to the right Robert turns 90-degree to the left Allows Robert to jump over anything in his path Allows Robert to fly over anything in his path) This instruction can be used only in combination with an if-instruction Repeat an action or action(s) x number of times Allows Robert to repeat an action or action(s) until he is in front of a chair Allows Robert to repeat an action or action(s) until he is in front of a box. (0.5 mark) Suggested answer: Copyright SCIT,, 2017 Page 6 of 16

7 Turn90L Repeat 5 times Walk Jump Repeat 10 times Walk Turn180 Sit Or Turn90L Repeat until you are directly in front of a box Walk Jump Repeat until you are directly in from of a chair Walk Turn180 Sit Question 3 (3.0 marks) There is a stack of cards numbered from 1 to 100 uniquely. Randomly choose 20 cards from the stack and put them into a bag. Each time, choose any two cards from the bag. If two odd numbered cards are taken out, put one odd numbered card into the bag. Similarly, if two even numbered cards are taken out, put one odd numbered card into the bag. If one odd and one even-numbered card are taken out, put back an even numbered card into the bag. Eventually, only one card is left inside the bag. (a) Let O, E and N be the number of Odd, Even, and Total cards in the bag. What is the effect on O, E and N after one time of choosing? You need to consider all possible scenario and you may express the effect using the variables O, E and N. (0.5 mark) Suggested solution: Let O be the number of Odd cards E be the number of Even cards N be the total number of cards; that it, N = O + E In a single loop, Case 1: If both cards chosen were Odd E = E // No change in the number of even card O = O 2 // 2 odd cards were drawn O = O + 1 // 1 odd card was put back to the bag N = N 1 // Total number of card is reduced by 1 Case 2: If both cards chosen were Even E = E 2 // 2 even cards were drawn O = O + 1 // 1 odd card was put into the bag Copyright SCIT,, 2017 Page 7 of 16

8 N = N 1 // Total number of card is reduced by 1 Case 3: If one card is Even and the other is Odd E = E // 1 even card was drawn, and 1 was put back O = O 1 // 1 odd card was drawn N = N 1 // Total number of card is reduced by 1 Analysing the above scenario, it is noted that each time cards are chosen, the number of even card (E) remains the same (Case 1 and Case 3) or is reduced by 2 (Case 2). In other words, the number of even card (E) changes by an even number. As for the number of odd card (O), each time cards are chosen, O is reduced by 1 (Case 1 and Case 3) or is increased by 1 (Case 2). In other words, the number of odd card (O) changes by an odd number. Finally, the total number of card in all cases (Case 1, Case 2, and Case 3) is reduced by 1, that is, N = N 1. (b) Based on your answer for part (i), what is the invariant? (1.0 mark) Suggested answer: Noted from Case 1, 2 and 3 described in (i), E = E 2 // E is reduced by 2 if both cards chosen were even (Case 2) E = E // E remains the same if both cards chosen were odd (Case 1) E = E // E remains the same if one of the cards is even and the other is odd // (Case 3) This means E remains the same or reduced by 2 each time cards are chosen. This mean the initial state (number of cards (E) is odd or even) of E remains the same. In other words, if initially, there are 10 even-numbered card, then each time after cards are chosen, the number of even-numbered card left is always even, that is, 8, 6, 4, 2, or 0. With the same reasoning, if the initial number of even-numbered card is odd, for example 9, then each time after cards are chosen, the number of even-numbered card left is always odd, that is, 7, 5, 3, or 1. Hence the invariant in the problem described above is that the initial state (number of cards is odd or even) of the number of evennumbered remains the same. (c) If the initial numbers of odd-numbered and even-numbered cards in the bag is known, can you predict the number of the last card in the bag? (1.0 mark) Suggested answer: Copyright SCIT,, 2017 Page 8 of 16

9 Yes, we can. It is noted that the number of even-numbered cards (E) change by an even number each time cards are chosen and the number of odd-numbered cards (O) change by either +1 or -1. Hence based on the initial number of even-numbered cards (E) and the invariant, we can determine the number of the last card remaining in the bag. Let E be the initial number of even-numbered card and e be the number of evennumbered cards currently in the bag. From the discussion (i) above, each time two cards are chosen from the bag, the total number of card (N) is reduced by1. Hence eventually, there will be only 1 card left in the bag. Starting with an odd number for E, each time after two cards are chosen from the bag, the number e is reduced by 2 or remains the same. Since invariant holds each time two cards are chosen from the bag, e remains an odd number. At the end, the number of cards left in the bag will be 1. Since 1 is odd, hence the number of the card left in the bag must be even-numbered card because the number of even-number card starts with an odd number. Starting with an even number for E, each time after two cards are chosen from the bag, the number e is reduced by 2 or remains the same. Since invariant holds each time two cards are chosen from the bag, e remains and even number. At the end, the number of cards left in the bag will be 1, and since 1 is odd, hence the number of the card left in the bag must be odd-numbered card because the number of even card starts with an even number and 1 is not an even number. In conclusion, the last card remaining in the bag is even-numbered card if the starting number of even-numbered card is an odd number; otherwise the last card remaining in the bag is odd-numbered card. (d) If there are N number of cards are put in the bag, where 2 <= N <= 50, after how many time of choosing will there be one card left? (0.5 mark) Suggested answer: Since each time after two cards are drawn from the bag, the total number of card (N) is reduced by 1. Hence after (N 1) number of times drawing the cards, there will be one card left in the bag. Copyright SCIT,, 2017 Page 9 of 16

10 Sorting and Searching [4.5 marks] Question 4: Sorting (3.0 marks) Eight numbers are stored in an array as shown below. Initial W =? C =? M =? Note: For this exercise, we will assume the following: To swap two numbers the operations involve: Val = A[i]; A[i] = A[j]; A[j] = Val; So, one swap operation is taken as a combined operations that involve 3 assignments operations and 4 indexing. In calculating the number of data movement (M), we will assume it is only one swap operation instead of 3 assignment operations. (a) Using first Insertion Sort, then Selection Sort to sort the eight numbers described above into ascending order. For each of the sorting algorithms, start the state of the numbers from the initial state described above. Trace through each pass of the sorting until the numbers are fully sorted in ascending order (with the smallest number stored at element with index 0.) For each pass, show the start state and specify value of W (wall index), C (the number of comparisons made in that pass), and M (the number of elements that are to be moved in that pass.) At the end of the sorting process, indicate what is the total number of comparisons and total number of data movements for both sorting algorithms. (1.0 mark) Insertion Sort: Initial PASS W = 0 Copyright SCIT,, 2017 Page 10 of 16

11 PASS PASS PASS PASS PASS PASS Final W = 1 W = 2 W = 3 W = 4 W = 5 W = W = 7 The total number of comparisons = sum of number of comparisons for each pass. The total number of comparisons = = 7 (i.e. N-1) Total number of data movement = sum of number of data movement for each pass. Total number of data movement = = 0. Selection Sort: Initial PASS W = 0 Copyright SCIT,, 2017 Page 11 of 16

12 PASS PASS PASS PASS PASS PASS Final C = 7 W = 1 C = 6 W = 2 C = 5 W = 3 C = 4 W = 4 C = 3 W = 5 C = 2 W = W = 7 The total number of comparisons = sum of number of comparisons for each pass. The total number of comparisons = = 28 (i.e. N(N-1) / 2) Total number of data movement = sum of number of data movement for each pass. Total number of data movement = = 0. (b) Repeat the task as described in Part (a) above, but this time, instead of sorting the eight numbers into ascending order, sort the eight numbers into descending order. Trace through each pass of the sorting until the numbers are fully sorted in descending order (with the largest number stored at element with index 0.) For each pass, show the start state and specify value of W (wall index), C (the number of comparisons made in that pass), and M (the number of elements that are to be moved in that pass.) At the end of the sorting process, indicate what is the total number Copyright SCIT,, 2017 Page 12 of 16

13 of comparisons and total number of data movements for both sorting algorithms. (1.0 mark) Insertion Sort: Initial PASS PASS PASS PASS PASS PASS PASS Final W = 0 M = 1 W = 1 C = 2 M = 2 W = 2 C = 3 M = 3 W = 3 C = 4 M = 4 W = 4 C = 5 M = 5 W = 5 C = 6 M = 6 W = 6 C = 7 M = W = 7 The total number of comparisons = sum of number of comparisons for each pass. The total number of comparisons = = 28 (i.e. N(N-1) / 2) Total number of data movement = sum of number of data movement for each pass. Total number of data movement = = 28 (i.e. N(N-1) / 2) Copyright SCIT,, 2017 Page 13 of 16

14 Selection Sort: Initial PASS PASS PASS PASS PASS PASS PASS Final W = 0 C = 7 M = 1 W = 1 C = 6 M = 1 W = 2 C = 5 M = 1 W = 3 C = 4 M = 1 W = 4 C = 3 W = 5 C = 2 W = W = 7 The total number of comparisons = sum of number of comparisons for each pass. The total number of comparisons = = 28 (i.e. N(N-1) / 2) Total number of data movement = sum of number of data movement for each pass. Total number of data movement = = 4 (i.e. N / 2) (c) From the results of C and M you obtained from Part (a) and (b), discuss on the running time complexity of the two sorting algorithms in term of the number of comparison (C) and the number of data movement (M) with respect to the order of the data (8 numbers) in the initial state. (1.0 mark) Copyright SCIT,, 2017 Page 14 of 16

15 Suggested answer: The 8 numbers used in this question are initially already in ascending order. This leads to two scenarios for discussions best-case scenario and worst-case scenario. Best-case scenario: Insertion sort Number of comparisons (C) = 7 Number of data movement (M) = 0 Selection sort Number of comparisons (C) = 28 Number of data movement (M) = 0 For insertion sort, the total number of comparisons (C) equals (N-1). The algorithm compares (n-1) data and since the data are already in correct order, there is no swapping (data movement) needed. For selection sort, the total number of comparisons (C) equals N*(N-1). The algorithm compares N*(N-1) times of the data to determine the largest or the smallest data. Even though the data are already in correct order, the algorithm will still do the comparisons. Since the data are already in correct order, there is no swapping (data movement) needed. Worst-case scenario: Insertion sort Number of comparisons (C) = 28 Number of data movement (M) = 28 Selection sort Number of comparisons (C) = 28 Number of data movement (M) = 4 For insertion sort, the total number of comparisons (C) equals N*(N-1). In each pass, the algorithm compares (N-1) data that are already in sorted order to search for a correct location to insert the next data. There are altogether N data, hence N*(N-1) time of comparisons. Since the data are in reverse order (worst-case), every data comparison, required one data swapping (data movement) to provide a correct position for the next data to be inserted. Hence the total number of data movement is N*(N-1). For selection sort, the total number of comparisons (C) equals N*(N-1) because the algorithm compares N*(N-1) times of the data to determine the largest or the smallest data. Once the largest or the smallest data is obtained, the data is swapped with the current wall position data. Since the data are in total reverse order (worstcase), mid through the sorting, the data will be resulted in correct order, hence half of the data do not need to be swapped. The number of data movement is then N/2. Copyright SCIT,, 2017 Page 15 of 16

16 Question 5: Searching (1.5 marks) Given a list of sorted data: 1, 7, 11, 18, 23, 27, and 51 stored in an array. Perform a binary search on the data to find the values 23 and 3. clearly the values of first, mid, last, and the comparison made for each step. Take note that first, mid and last are numbers that represent array indices. (1.5 mark) Suggested answer: First Mid Last Comparison with target [0] = 1 [3] = 18 [6] = > 18 [3+1] = 23 [5] = 27 [6] = < 27 [4] = 23 [4] = 23 [5-1] = = 23 (found) first 1, mid 18, last > 18 first 23, mid 27, last < 27 first 23, mid 23, last == 23 target found First Mid Last Comparison with target [0] = 1 [3] = 18 [6] = 51 3 < 18 [0] = 1 [1] = 7 [3-1] = 11 3 < 7 [0] = 1 [0] = 1 [1-1] = 1 3 > 1 [First] = [Mid] = [Last] (cannot be found) [0+1] = 7 [0] = 1 [First] > [Last] (not found) first 1 mid 18, last 51 3 < 18 first - 1, mid- 7, last 11 3 < 7 first 1, mid 1, last 1 3 > 1 3 cannot be found. Copyright SCIT,, 2017 Page 16 of 16