1. Chapter 20 RecursionLiang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 1
2. MotivationsSuppose you want to find all the files under adirectory that contains a particular word. How doyou solve this problem? There are several ways tosolve this problem. An intuitive solution is to userecursion by searching the files in thesubdirectories recursively. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 2
3. MotivationsH-trees, depicted in Figure 20.1, are used in a very large-scale integration (VLSI) design as a clock distributionnetwork for routing timing signals to all parts of a chipwith equal propagation delays. How do you write aprogram to display H-trees? A good approach is to userecursion. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 3
4. Objectives To describe what a recursive method is and the benefits of using recursion (§20.1). To develop recursive methods for recursive mathematical functions (§§20.2– 20.3). To explain how recursive method calls are handled in a call stack (§§20.2–20.3). To solve problems using recursion (§20.4). To use an overloaded helper method to derive a recursive method (§20.5). To implement a selection sort using recursion (§20.5.1). To implement a binary search using recursion (§20.5.2). To get the directory size using recursion (§20.6). To solve the Towers of Hanoi problem using recursion (§20.7). To draw fractals using recursion (§20.8). To discover the relationship and difference between recursion and iteration (§20.9). To know tail-recursive methods and why they are desirable (§20.10). Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 4
5. Computing Factorialfactorial(0) = 1;factorial(n) = n*factorial(n-1);n! = n * (n-1)! ComputeFactorial Run Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 5
6. animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 6
7. animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 7
8. animation Computing Factorial factorial(0) = 1; factorial(n) = n*factorial(n-1); factorial(3) = 3 * factorial(2) = 3 * (2 * factorial(1)) Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 8
25. factorial(4) Stack Trace 5 Space Required for factorial(0) 4 Space Required for factorial(1) Space Required for factorial(1) 3 Space Required for factorial(2) Space Required for factorial(2) Space Required for factorial(2) 2 Space Required for factorial(3) Space Required for factorial(3) Space Required for factorial(3) Space Required for factorial(3)1 Space Required for factorial(4) Space Required for factorial(4) Space Required for factorial(4) Space Required for factorial(4) Space Required for factorial(4)6 Space Required for factorial(1) Space Required 7 Space Required for factorial(2) for factorial(2) Space Required Space Required 8 Space Required for factorial(3) for factorial(3) for factorial(3) Space Required Space Required Space Required 9 Space Required for factorial(4) for factorial(4) for factorial(4) for factorial(4) Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 25
26. Other Examplesf(0) = 0;f(n) = n + f(n-1); Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 26
29. Characteristics of RecursionAll recursive methods have the following characteristics: – One or more base cases (the simplest case) are used to stop recursion. – Every recursive call reduces the original problem, bringing it increasingly closer to a base case until it becomes that case.In general, to solve a problem using recursion, you break itinto subproblems. If a subproblem resembles the originalproblem, you can apply the same approach to solve thesubproblem recursively. This subproblem is almost thesame as the original problem in nature with a smaller size. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 29
30. Problem Solving Using RecursionLet us consider a simple problem of printing a message forn times. You can break the problem into two subproblems:one is to print the message one time and the other is to printthe message for n-1 times. The second problem is the sameas the original problem with a smaller size. The base casefor the problem is n==0. You can solve this problem usingrecursion as follows: public static void nPrintln(String message, int times) { if (times >= 1) { System.out.println(message); nPrintln(message, times - 1); } // The base case is times == 0 } Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 30
31. Think RecursivelyMany of the problems presented in the early chapters canbe solved using recursion if you think recursively. Forexample, the palindrome problem in Listing 7.1 can besolved recursively as follows: public static boolean isPalindrome(String s) { if (s.length() <= 1) // Base case return true; else if (s.charAt(0) != s.charAt(s.length() - 1)) // Base case return false; else return isPalindrome(s.substring(1, s.length() - 1)); } Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 31
32. Recursive Helper MethodsThe preceding recursive isPalindrome method is notefficient, because it creates a new string for every recursivecall. To avoid creating new strings, use a helper method: public static boolean isPalindrome(String s) { return isPalindrome(s, 0, s.length() - 1); } public static boolean isPalindrome(String s, int low, int high) { if (high <= low) // Base case return true; else if (s.charAt(low) != s.charAt(high)) // Base case return false; else return isPalindrome(s, low + 1, high - 1); } Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 32
33. Recursive Selection Sort1. Find the smallest number in the list and swaps it with the first number.2. Ignore the first number and sort the remaining smaller list recursively. RecursiveSelectionSort Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 33
34. Recursive Binary Search1. Case 1: If the key is less than the middle element, recursively search the key in the first half of the array.2. Case 2: If the key is equal to the middle element, the search ends with a match.3. Case 3: If the key is greater than the middle element, recursively search the key in the second half of the array. RecursiveBinarySearch Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 34
35. Recursive Implementation/** Use binary search to find the key in the list */public static int recursiveBinarySearch(int[] list, int key) { int low = 0; int high = list.length - 1; return recursiveBinarySearch(list, key, low, high);}/** Use binary search to find the key in the list between list[low] list[high] */public static int recursiveBinarySearch(int[] list, int key, int low, int high) { if (low > high) // The list has been exhausted without a match return -low - 1; int mid = (low + high) / 2; if (key < list[mid]) return recursiveBinarySearch(list, key, low, mid - 1); else if (key == list[mid]) return mid; else return recursiveBinarySearch(list, key, mid + 1, high);} Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 35
36. Directory SizeThe preceding examples can easily be solved without using recursion. This section presents a problem that is difficult to solve without using recursion. The problem is to find the size of a directory. The size of a directory is the sum of the sizes of all files in the directory. A directory may contain subdirectories. Suppose a directory contains files , , ..., , and subdirectories , , ..., , as shown below. directory f1 f2 ... fm d1 d2 ... dn 1 1 1 1 1 1 Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 36
37. Directory SizeThe size of the directory can be defined recursively as follows:size( d ) size( f1 ) size( f 2 ) ... size( f m ) size(d1 ) size(d 2 ) ... size( d n ) DirectorySize Run Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 37
38. Towers of Hanoi There are n disks labeled 1, 2, 3, . . ., n, and three towers labeled A, B, and C. No disk can be on top of a smaller disk at any time. All the disks are initially placed on tower A. Only one disk can be moved at a time, and it must be the top disk on the tower. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 38
39. Towers of Hanoi, cont. A B C A B C Original position Step 4: Move disk 3 from A to B A B C A B C Step 1: Move disk 1 from A to B Step 5: Move disk 1 from C to A A B C A B C Step 2: Move disk 2 from A to C Step 6: Move disk 2 from C to B A B C A B C Step 3: Move disk 1 from B to C Step 7: Mve disk 1 from A to B Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 39
40. Solution to Towers of HanoiThe Towers of Hanoi problem can be decomposed into threesubproblems. n-1 disks n-1 disks . . . . . . A B C A B C Original position Step2: Move disk n from A to C n-1 disks n-1 disks . . . . . . A B C A B C Step 1: Move the first n-1 disks from A to C recursively Step3: Move n-1 disks from C to B recursively Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 40
41. Solution to Towers of Hanoi Move the first n - 1 disks from A to C with the assistance of tower B. Move disk n from A to B. Move n - 1 disks from C to B with the assistance of tower A. TowersOfHanoi Run Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 41
42. Exercise 20.3 GCDgcd(2, 3) = 1gcd(2, 10) = 2gcd(25, 35) = 5gcd(205, 301) = 5gcd(m, n)Approach 1: Brute-force, start from min(n, m) down to 1, to check if a number is common divisor for both m and n, if so, it is the greatest common divisor.Approach 2: Euclid’s algorithmApproach 3: Recursive method Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 42
43. Approach 2: Euclid’s algorithm// Get absolute value of m and n;t1 = Math.abs(m); t2 = Math.abs(n);// r is the remainder of t1 divided by t2;r = t1 % t2;while (r != 0) { t1 = t2; t2 = r; r = t1 % t2;}// When r is 0, t2 is the greatest common// divisor between t1 and t2return t2; Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 43
44. Approach 3: Recursive Methodgcd(m, n) = n if m % n = 0;gcd(m, n) = gcd(n, m % n); otherwise; Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 44
45. Fractals?A fractal is a geometrical figure just liketriangles, circles, and rectangles, but fractalscan be divided into parts, each of which is areduced-size copy of the whole. There aremany interesting examples of fractals. Thissection introduces a simple fractal, calledSierpinski triangle, named after a famousPolish mathematician. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 45
46. Sierpinski Triangle1. It begins with an equilateral triangle, which is considered to be the Sierpinski fractal of order (or level) 0, as shown in Figure (a).2. Connect the midpoints of the sides of the triangle of order 0 to create a Sierpinski triangle of order 1, as shown in Figure (b).3. Leave the center triangle intact. Connect the midpoints of the sides of the three other triangles to create a Sierpinski of order 2, as shown in Figure (c).4. You can repeat the same process recursively to create a Sierpinski triangle of order 3, 4, ..., and so on, as shown in Figure (d). Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 46
47. Sierpinski Triangle Solution p1midBetweenP1P2 midBetweenP3P1 p2 p3 midBetweenP2P3 SierpinskiTriangle Run Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 47
48. Sierpinski Triangle Solution p1 Draw the Sierpinski triangle displayTriangles(g, order, p1, p2, p3) p2 p3 SierpinskiTriangle Run Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 48
49. Sierpinski Triangle Solution Recursively draw the small Sierpinski triangle p1 displayTriangles(g, order - 1, p1, p12, p31) )Recursively draw the small p12 p31Sierpinski triangle Recursively draw thedisplayTriangles(g, small Sierpinski triangle order - 1, p12, p2, p23) displayTriangles(g, order - 1, p31, p23, p3) p2 p3 ) p23 SierpinskiTriangle Run Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 49
50. Recursion vs. IterationRecursion is an alternative form of programcontrol. It is essentially repetition without a loop.Recursion bears substantial overhead. Each time theprogram calls a method, the system must assignspace for all of the method’s local variables andparameters. This can consume considerablememory and requires extra time to manage theadditional space. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 50
51. Advantages of Using RecursionRecursion is good for solving the problems that areinherently recursive. Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 51
52. Tail RecursionA recursive method is said to be tail recursive ifthere are no pending operations to be performed onreturn from a recursive call. Non-tail recursive ComputeFactorial Tail recursive ComputeFactorialTailRecursion Liang, Introduction to Java Programming, Ninth Edition, (c) 2013 Pearson Education, Inc. All rights reserved. 52
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