Recursion. Chapter 11. Chapter 11 1

Transcription

1 Recursion Chapter 11 Chapter 11 1

2 Reminders Project 6 is over. Project 7 has begun Start on time! Not much code for milestone, but a great deal of thought Chapter 11 2

3 Exam 2 Problem Problems Problem 2) Be careful about static! Problem 4) Be careful about local assignment (Draw pictures if necessary) Problem 11) All classes are derived from Object Problem 20) Scanner reads text. ObjectInputStream doesn't. Chapter 11 3

4 Recursion Sometimes a problem can be solved by solving a smaller version of itself. e.g. Factorial(n) = (n)(n-1)(n-2)...(3)(2)(1) = (n)factorial(n-1) Chapter 11 4

5 Example: Countdown public void countdown( int number ) { System.out.print( number ); System.out.print( " " ); if ( number > 0 ) countdown( number - 1 ); } What is the result of calling countdown( 10 )? What happens if the print statements and recursive call are transposed? Chapter 11 5

6 Example: Countup public void countdown( int number ) { if ( number > 0 ) countdown( number - 1 ); System.out.print( number ); System.out.print( " " ); } The placement of the recursive call can have subtle effects. Chapter 11 6

7 How Recursion Works countup( 5 ) countup( 4 ) countup( 3 ) countup( 2 ) countup( 1 ) countup( 0 ) Calls Sop( 5 + ) Sop( 4 + ) Sop( 3 + ) Sop( 2 + ) Sop( 1 + ) Sop( 0 + ) Base Case Note: countup( 5 ) doesn't finish until all of its recursive calls do! countup( 0 ) doesn't have any recursive calls! Evaluation & Completion Chapter 11 7

8 Recursion Guidelines A recursive method uses an if-else statement. 1)One branch represents a base case which can be solved directly (without recursion). 3)Another branch includes a recursive call to the method, but with a simpler or smaller set of arguments. The recursive calls must move toward the base case Notice, a loop was not necessary for countdown! Chapter 11 8

9 Infinite Recursion If the recursive call does not solve a simpler or smaller case, a base case may never be reached. Such a method continues to call itself forever (or until a stack overflow). This is called infinite recursion. Chapter 11 9

10 Infinite Recursion public void countdown( int number ) { System.out.print( number ); System.out.print( " " ); // if ( number > 0 ) countdown( number - 1 ); } public void countdown( int number ) { System.out.print( number ); System.out.print( " " ); if ( number > 0 ) countdown( number + 1 ); } Chapter 11 10

11 Recursion vs. Iteration A recursive version of a method is less efficient than the iterative version, but also easier to understand. (Keep track of the recursive calls and suspended computations) It can be much easier to write a recursive method than it is to write a corresponding iterative method. Chapter 11 11

12 Example: Factorial public int factorial( int n ) { if ( n > 1 ) return n * factorial( n - 1 ); else return 1; } Notice that the return value of a recursive call can be used within a recursive method. Chapter 11 12

13 Case Study: Binary Search Find a number in a sorted array If found, return the position of the number If not, return -1 Could search sequentially, but there is a better way! Chapter 11 13

14 Binary Search, cont. Because the array is sorted, we can eliminate whole sections of the array as we search. e.g.: Looking for a 7 and we encounter a location containing a 13, we can disregard from the location containing the 13 through the end Chapter 11 14

15 Binary Search, cont. Other situations to handle: Looking for a 7 and we encounter a location containing a 3, we can disregard from the location containing the 3 through the beginning Looking for a 7 and we encounter a location containing a 7, we are finished Chapter 11 15

16 Binary Search, cont. first eventually becomes larger than last and we can terminate the search if not found. Why do we use the middle element? mid = (first + last)/2 if (first > last) return -1; else if (target == a[mid]) return mid; else if (target < a[mid] search a[first] through a[mid-1] else search a[mid + 1] through a[last] Chapter 11 16

17 Binary Search, cont. Chapter 11 17

18 Binary Search, cont. Each recursive call eliminates about half of the remaining array. The number of calls to either find an element or to determine that the item is not present is log n for an array of n elements. Thus, for an array of 1024 elements, only 10 recursive calls are needed. (instead of 1024) Chapter 11 18

19 Merge Sort Merge sort takes a divide and conquer approach. 1)The array is divided in halves and the halves are sorted recursively. 2)Sorted subarrays are merged to form a larger sorted array. Chapter 11 19

20 Pseudocode: Merge Sort, cont. If the array has only one element, stop. Otherwise Copy the first half of the elements into an array named front. Copy the second half of the elements into an array named tail. Sort array front recursively. Sort array tail recursively. Merge arrays front and tail. Chapter 11 20

21 Merging Sorted Arrays Front Tail Take the smallest unread element from the top of both arrays. 1, 3, 4, 5, 7, 8, 14, 15, 16, 17 Chapter 11 21

22 Merging Sorted Arrays, cont. Remove the smaller of the elements from the beginning of (the remainders) of the two arrays and placing it in the next available position in a larger collector array. When one of the two arrays becomes empty, the remainder of the other array is copied into the collector array. You consume the smallest elements Chapter 11 22

23 Merging Sorted Arrays, cont. int frontindex = 0, tailindex = 0, aindex = 0; while ((frontindex < front.length) && (tailindex < tail.length)) { if(front[frontindex] < tail[tailindex]} { a[aindex] = front[frontindex]; aindex++; frontindex++; } Chapter 11 23

24 Merging Sorted Arrays, cont. } else { a[aindex] = tail[tailindex]; aindex++; tailindex++ } Chapter 11 24

25 Merging Sorted Arrays, cont. If either array front or array tail becomes empty, any remaining elements from the other array need to be copied into array a. These elements are sorted and are larger than any elements already in array a. Chapter 11 25

26 Merge Sort, cont. Chapter 11 26

27 Merge Sort, cont. Chapter 11 27

28 Merge Sort, cont. Chapter 11 28