CP222 Computer Science II. Recurrence Relations and Basic Sorting

Transcription

1 CP222 Computer Science II Recurrence Relations and Basic Sorting

2 Amazon raising Prime subscription price to $119 Tech News!

3 Tech News! Amazon raising Prime subscription price to $119 Golden State Killer identified partially through the use of onilne genealogy databases

4 Running times of: - bag methods - linked list methods - stack methods - queue methods

5 Fingerprinters?

6 class Stopwatch { private long start_time; public Stopwatch() { start_time = System.currentTimeMillis(); } } public double elapsedtime() { long now = System.currentTimeMillis(); return (now start) / ; }

7 Recursion Some problems can be solved elegantly by first solving a smaller, related problem We can model this technique using recursive methods: methods that call themselves

8 The Two Rules of Recursion Rule 1: The recursion has to stop somewhere Infinite recursion is not helpful

9 The Two Rules of Recursion Rule 1: The recursion has to stop somewhere Infinite recursion is not helpful Rule 2: Each recursive call should get the program closer to the stopping point Recursive calls solve a smaller version of the original problem

10 The Two Rules of Recursion Rule 1: Base case Rule 2: Recursive case

11 Recursive Example: Factorial What is 7 factorial?

12 Recursive Example: Factorial What is 7 factorial? 7 * 6!

13 Recursive Example: Factorial What is 6 factorial? 6 * 5!

14 Factorial Base case (n == 1) Recursive case (n > 1)

15 Factorial Base case (n == 1) : result = 1 Recursive case (n > 1) : result = n * factorial(n -1)

16 Recursively print out a Linked List function printlisthelper(node): if node is not null: print node printlisthelper(node.getneighbor()) function printlist(): printlisthelper(first)

17 Recursively print out a Linked List Time complexity? function printlisthelper(node): if node is not null: print node printlisthelper(node.getneighbor()) function printlist(): printlisthelper(first)

18 Time Complexity of Recursion Based on two things: Number of operations in the recursive method Number of recursive calls

19 Time Complexity of Recursion Based on two things: Number of operations in the recursive method Number of recursive calls Define a recurrence relation

20 Recurrence Relation T(n) = T(n-1) + 3 T(0) = 0

21 Recurrence Relation T(n) = T(n-1) + 3 T(0) = 0 Number of operations for input size, n

22 Recurrence Relation Number of operations for input size, n - 1 T(n) = T(n-1) + 3 T(0) = 0

23 Recurrence Relation There are 3 extra steps (check if, print node, get neighbor) T(n) = T(n-1) + 3 T(0) = 0

24 Recurrence Relation Base case: no nodes, no steps. T(n) = T(n-1) + 3 T(0) = 0

25 Find closed form solution for the relation T(n) = T(n-1) + 3

26 Find closed form solution for the relation T(n) = T(n-1) + 3 = [T(n-2) + 3] + 3 Why can we rewrite this way?

27 Find closed form solution for the relation T(n) = T(n-1) + 3 = [T(n-2) + 3] + 3 = T(n-2) + 6

28 Find closed form solution for the relation T(n) = T(n-2) + 6 = [T(n-3) + 3] + 6 = T(n-3) + 9

29 Find closed form solution for the relation T(n) = T(n-i) + 3i

30 Find closed form solution for the relation T(n) = T(n-i) + 3i Let i = n = T(n-n) + 3n

31 Find closed form solution for the relation T(n) = T(n-i) + 3i = T(n-n) + 3n = T(0) + 3n O(n) T(n) = 3n

32 Recurrence Relation? public void guessnumber(int bottom, int top) { int guess = (top + bottom) / 2; System.out.println( Is it + guess +? ); String response = myscanner.next(); if (response.equals( y )) { System.out.println( I win. Good game. ); } else if (response.equals( h )) { guessnumber(guess, top); } else { guessnumber(bottom, guess); } }

33 Guess Number Recurrence Relation T(n) = T(n/2) + 5 T(1) = 1

34 Guess Number Recurrence Relation T(n) = T(n/2) + c T(1) = c

35 Solve the recurrence relation? T(n) = T(n/2) + c T(1) = c

36 Solve it! T(n) = T(n/2) + c = [T(n/4) + c] + c = T(n/4) + 2c

37 Solve it! T(n) = T(n/4) + 2c = [T(n/8) + c] + 2c = T(n/8) + 3c

38 Step Expression 1 T(n/2) + c 2 T(n/4) + 2c 3 T(n/8) + 3c k???

39 T(n) = T(n/2 k ) + kc Solve it!

40 Solve it! T(n) = T(n/2 k ) + kc Set k = log 2 (n) Why?

41 Solve it! T(n) = T(n/2 k ) + kc Set k = log 2 (n) So 2 k = n

42 Solve it! T(n) = T(n/2 k ) + kc Set k = log 2 (n) So 2 k = n T(n) = T(1) + c log 2 (n) T(n) = c + c log 2 (n)

43 Solve it! T(n) = T(n/2 k ) + kc Set k = log 2 (n) So 2 k = n T(n) = T(1) + c log 2 (n) O(log 2 (n)) T(n) = c + c log 2 (n)

44 Searching and Sorting Given a data structure, how do we find a particular element? How long does it take? How can we put data into sorted order? How can we do it quickly?

45 Suppose we want to write a program to check if a bingo is legal in Scrabble. We have a dictionary of legal Scrabble words.

46 Is Bingo Legal? // Assume we have an ArrayList of legal words // called words public boolean islegal(string possiblebingo) { for (int i = 0; i < words.size(); i++) { if (possiblebingo.equals(words.get(i))) { return true; } } return false; }

47 What is the time complexity of islegal()? Best case? Average case? Worst case?

48 This process is called linear search.

49 Do humans use linear search when checking if words are legal when playing Scrabble?

50 If we assume the wordlist is sorted, can we search with fewer comparisons?

51 Binary Search // Assume words is in sorted order islegal(possiblebingo, start, end): if start == end: return false else: nextindex = (start + end) / 2 current = words[nextindex] if (current == possiblebingo) { return true; } else if (current < possiblebingo) { return islegal(possiblebingo, nextindex + 1, end) } else { return islegalhelper(possiblebingo, start, nextindex - 1) }

52 OK, binary search is faster. How do we get the wordlist in sorted order in the first place?

53 Sorting Problem [ 8, 3, 1, 7, 9, 2, 4 ] [ 1, 2, 3, 4, 7, 8, 9 ]

54 How do humans sort?

55 Selection Sort Algorithm Find smallest element left to be sorted This is done by linear search Put it at the end of the sorted part of the list For the first smallest, put it at the very beginning of the list The next smallest goes at position 2, etc.

56 Selection Sort [ 8, 3, 1, 7, 9, 2, 4 ]

57 Selection Sort [ 8, 3, 1, 7, 9, 2, 4 ] Scan list (linearly) for smallest element

58 Selection Sort [ 8, 3, 1, 7, 9, 2, 4 ] Here's the smallest.

59 Selection Sort [ 8, 3, 1, 7, 9, 2, 4 ] Flip the position of the smallest with the next unsorted value.

60 Selection Sort [ 1, 3, 8, 7, 9, 2, 4 ] Flip the position of the smallest with the next unsorted value.

61 Selection Sort [ 1, 3, 8, 7, 9, 2, 4 ] Part of the list is now sorted.

62 Selection Sort [ 1, 3, 8, 7, 9, 2, 4 ] Find the smallest element not in the sorted part of the list.

63 Selection Sort [ 1, 3, 8, 7, 9, 2, 4 ] Here's the next smallest.

64 Selection Sort [ 1, 3, 8, 7, 9, 2, 4 ] Flip it with the first unsorted element.

65 Selection Sort [ 1, 2, 8, 7, 9, 3, 4 ] Flip it with the first unsorted element.

66 Selection Sort [ 1, 2, 3, 4, 7, 8, 9 ] How long does the algorithm take with a length n array?

67 Insertion Sort Algorithm Maintain two parts of the list: unsorted and sorted For each unsorted element, insert it into the sorted part of the list in the correct location This insertion may require shifting values over to make room

68 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ]

69 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ] The sorted part of the array starts out as just the first value.

70 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ] The rest of the list is unsorted.

71 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ] Start with the next unsorted element.

72 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ] Insert it in the correct location of the sorted part of the list. Here it needs to go before the 8.

73 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ] There isn't space before the 8, so we have to shift the 8 up.

74 Insertion Sort [ 8, 3, 1, 7, 9, 2, 4 ] Before we shift the 8, we store the 3 somewhere else so it is not lost. key = 3

75 Insertion Sort [ 8, 8, 1, 7, 9, 2, 4 ] 8 is now shifted. key = 3

76 Insertion Sort [ 3, 8, 1, 7, 9, 2, 4 ] Now put the key value in its correct location. key = 3

77 Insertion Sort [ 3, 8, 1, 7, 9, 2, 4 ] Now this part of the list is sorted.

78 Insertion Sort [ 3, 8, 1, 7, 9, 2, 4 ] Repeat the process for the next unsorted element.

79 Insertion Sort [ 3, 8, 1, 7, 9, 2, 4 ] key = 1 Set the key.

80 Insertion Sort [ 3, 8, 1, 7, 9, 2, 4 ] key = 1 Find its insertion point.

81 Insertion Sort [ 3, 8, 1, 7, 9, 2, 4 ] key = 1 Shift both 3 and 8 up. Start by shifting 8. Why?

82 Insertion Sort [ 3, 8, 8, 7, 9, 2, 4 ] key = 1 8 was shifted.

83 Insertion Sort [ 3, 3, 8, 7, 9, 2, 4 ] key = 1 3 was shifted.

84 Insertion Sort [ 1, 3, 8, 7, 9, 2, 4 ] key = 1 Insert the key.

85 InsertionSort.java

86 InsertionSort.java vs. SelectionSort.java

87 Bubble Sort Algorithm Bubble up larger values to the end of the list Scan through the unbubbled elements two at a time If the second of the two elements is smaller, flip their positions Now make the second element the new first and get the next element from the list as the new second

88 Bubble Sort [ 8, 3, 1, 7, 9, 2, 4 ] Compare the first two elements.

89 Bubble Sort [ 8, 3, 1, 7, 9, 2, 4 ] The second one is smaller, so flip them.

90 Bubble Sort [ 3, 8, 1, 7, 9, 2, 4 ] The second one is smaller, so flip them.

91 Bubble Sort [ 3, 8, 1, 7, 9, 2, 4 ] Compare the next group of two.

92 Bubble Sort [ 3, 8, 1, 7, 9, 2, 4 ] The second one is smaller, so flip them.

93 Bubble Sort [ 3, 1, 8, 7, 9, 2, 4 ] The second one is smaller, so flip them.

94 Bubble Sort [ 3, 1, 8, 7, 9, 2, 4 ] See how 8 is bubbling up?

95 Bubble Sort [ 3, 1, 8, 7, 9, 2, 4 ] Compare next two. Second is smaller.

96 Bubble Sort [ 3, 1, 7, 8, 9, 2, 4 ] Flip them.

97 Bubble Sort [ 3, 1, 7, 8, 2, 4, 9 ] At the end of the process the largest element has bubbled up.

98 Bubble Sort [ 3, 1, 7, 8, 2, 4, 9 ] Now bubble up the next largest element.

99 Bubble Sort [ 1, 3, 7, 2, 4, 8, 9 ] Stop bubbling up when you reach the values that have already bubbled.

100 Obama Bubble

101 BREAK!