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Recursion. Chapter 10. Chapter Contents. What Is Recursion? Tracing a Recursive Method Recursive Methods That Return a Value Recursively Processing an Array Recursively Processing a Linked Chain The Time Efficiency of Recursive Methods Time Efficiency of countDown

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## Recursion

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**Recursion**Chapter 10**Chapter Contents**• What Is Recursion? • Tracing a Recursive Method • Recursive Methods That Return a Value • Recursively Processing an Array • Recursively Processing a Linked Chain • The Time Efficiency of Recursive Methods • Time Efficiency of countDown • A Simple Solution to a Difficult Problem • A Poor Solution to a Simple Problem**What Is Recursion?**• It is a problem-solving process involves repetition • Breaks a problem into identical but smaller problems • Eventually you reach a smallest problem • Answer is obvious or trivial • Using the solution to smallest problem enables you to solve the previous smaller problems • Eventually the original big problem is solved • An alternative to iteration • An iterative solution involves loops**What Is Recursion?**New Year Eve: counting down from 10.**What Is Recursion?**• A method that calls itself is a recursive method • Base case: a known case in a recursive definition. The smallest problem • Eventually, one of the smaller problems must be the base case /** Task: Counts down from a given positive integer.* @param integer an integer > 0 */ public static void countDown(intinteger) { System.out.println(integer);if (integer > 1)countDown(integer - 1);} // end countDown**When Designing Recursive Solution**• Four questions to ask before construct recursive solutions. If you follow these guidelines, you can be assured that it will work. • How can you define the problem in terms of a smaller problem of the same type? • How does each recursive call diminish the size of the problem? • What instance of the problem can serve as the base case? • As the problem size diminishes, will you reach this base case?**When Designing Recursive Solution**• For the method countDown, we have the following answers. • countDown displays the given integer as the part of the solution that it contribute directly. Then call countDown with smaller size. • The smaller problem is counting down from integer -1. • The if statement asks if the process has reached the base case. Here, the base case occurs when integer is 1.**Recursive Solution Guidelines**• Method definition must provide parameter • Leads to different cases • Typically includes an if or a switch statement • One or more of these cases should provide a non recursive solution( infinite recursion if don’t) • The base or stopping case • One or more cases includes recursive invocation • Takes a step towards the base case**Tracing a Recursive Method**public static void countDown(int integer) { System.out.println(integer);if (integer > 1) countDown(integer - 1);} // end countDown • Given: The effect of method call countDown(3)**Tracing a Recursive Method**Tracing the recursive call countDown(3)**Compare Iterative and Recursive Programs**//Iterative version public static void countDown( int integer) { while ( integer >= 1) { System.out.println(integer); integer--; } } //Recursive version public static void countDown( int integer) { if ( integer >= 1) { System.out.println(integer); countDown(integer -1); } }**Question?**• Could you write an recursive method that skips n lines of output, where n is a positive integer. Use System.out.println() to skip one line. • Describe a recursive algorithm that draws a given number of concentric circles. The innermost or outmost circle should have a given diameter. The diameter of each of the other circles should be three-fourths the diameter of the circle just outside it.**Answer to skipLines**pubic static void skipLines ( int givenNumber) { if (givenNumber >=1) { System.out.println(); skipLines(givenNumber – 1); } } void drawConcentricCircle( givenNumber, givenDiameter) { // what should be put inside??……. }**Tracing a Recursive Method**• Each call to a method generate an activation record that captures the state of the method’s execution and that is placed into a ADT stack. • The activation-record stack remembers the history of incompleted method calls. A snapshot of a method’s state. • The topmost activation record holds the data values for the currently executing method. • When topmost method finishes, its activation record is popped • In this way, Java can suspend the execution of a recursive method and re-invoke it when it appears on the top.**Tracing a Recursive Method**The stack of activation records during the execution of a call to countDown(3)… continued→**Tracing a Recursive Method**Note: the recursive method will use more memory than an iterative method due to the stack of activation records ctd. The stack of activation records during the execution of a call to countDown(3)**Tracing a Recursive Method**• Too many recursive calls can cause the error message “stack overflow”. Stack of activation records has become full. Method has used too much memory. • Infinite recursion or large-size problems are the likely cause of this error.**Recursive Methods That Return a Value**• Task: Compute the sum1 + 2 + 3 + … + n for an integer n > 0 public static int sumOf(int n){ int sum;if (n = = 1) sum = 1; // base caseelsesum = sumOf(n - 1) + n; // recursive callreturn sum;} // end sumOf**Recursive Methods That Return a Value**The stack of activation records during the execution of a call to sumOf(3)**Recursively Processing an Array**• When processing array recursively, divide it into two pieces • Last element one piece, rest of array another • First element one piece, rest of array another • Divide array into two halves • A recursive method part of an implementation of an ADT is often private • Its necessary parameters make it unsuitable as an ADT operation**Recursively Processing a Linked Chain**• To write a method that processes a chain of linked nodes recursively • Use a reference to the chain's first node as the method's parameter • Then process the first node • Followed by the rest of the chain public void display(){ displayChain(firstNode); System.out.println();} // end display private void displayChain(Node nodeOne) { if (nodeOne != null) { System.out.print(nodeOne.data + " "); displayChain(nodeOne.next); }} // end displayChain**Recursively Divide the Array in Half**publicstatic voiddisplayArray( int array[], int first, int last) { if (first == last) System.out.print(array[first]); else { int mid = (first + last) /2; displayArray(array, first, mid); displayArray(array, mid+1, last); } }**A Simple Solution to a Difficult Problem**The initial configuration of the Towers of Hanoi for three disks**A Simple Solution to a Difficult Problem**Rules for the Towers of Hanoi game • Move one disk at a time. Each disk you move must be a topmost disk. • No disk may rest on top of a disk smaller than itself. • You can store disks on the second pole temporarily, as long as you observe the previous two rules.**A Simple Solution to a Difficult Problem**The sequence of moves for solving the Towers of Hanoi problem with three disks. Continued →**A Simple Solution to a Difficult Problem**(ctd) The sequence of moves for solving the Towers of Hanoi problem with three disks**A Simple Solution to a Difficult Problem**The smaller problems in a recursive solution for four disks**A Simple Solution to a Difficult Problem**• Algorithm for solution with 1 disk as the base case Algorithm solveTowers(numberOfDisks, startPole, tempPole, endPole) if (numberOfDisks == 1) Move disk from startPole to endPole else { solveTowers(numberOfDisks-1, startPole, endPole, tempPole) Move disk from startPole to endPole solveTowers(numberOfDisks-1, tempPole, startPole, endPole) }**Recursion Efficiency**• How many moves occur for n disks? • m(1) = 1 for n>1, two recursive calls to solve problems that have n-1 disks. • m(n) = m(n-1) + m(n-1) +1 = 2*m(n-1) +1 • Let’s evaluate the recurrence for m(n) for a few values of n: • m(1) =1; m(2) = 3; m(3) = 7;m(4) = 15; m(5) = 31; m(6) = 63…. • m(n) = 2^n -1**Mathematical Induction**• Prove this conjecture m(n) = 2^n -1 by using mathematical induction: • We know that m(1) =1, which equals to 2^1-1=1, so the conjecture is true for n =1. • Now assume that it is true for n=1,2,…,k, and consider m(k+1). • m(k+1) = 2*m(k) +1 (use the recurrence relation) • =2*(2^k-1) +1 = 2^(k+1) -1 ( we assume that m(k) = 2^k-1) • Since the conjecture is true for n=k+1, it is truefor all n>=1**Mathematical Induction**• Assume you want to prove some statement P, P(n) is true for all n starting with n = 1. The Principle of Math Induction states that, to this end, one should accomplish just two steps: • 1). Prove that P(1) is true. • 2). Assume that P(k) is true for some k. Derive from here that P(k+1) is also true. • look P(1) is true and implies P(2). Therefore P(2) is true. But P(2) implies P(3). Therefore P(3) is true which implies P(4) and so on.**Multiplying Rabbits (The Fibonacci Sequence)**• “Facts” about rabbits • Rabbits never die • A rabbit reaches sexual maturity exactly two months after birth, that is, at the beginning of its third month of life • Rabbits are always born in male-female pairs • At the beginning of every month, each sexually mature male-female pair gives birth to exactly one male-female pair**Multiplying Rabbits (The Fibonacci Sequence)**• Problem • How many pairs of rabbits are alive in month n? • Month Month No calculation rabbit couples • January 1 1 + 0 = 1 • February 2 1 + 0 = 1 • March 3 1 + 1 = 2 • April 4 2 + 1 = 3 • May 5 3 + 2 = 5 • June 6 5 + 3 = 8 • July 7 8 + 5 = 13 • August 8 13 + 8 = 21 • Recurrence relation rabbit(n) = rabbit(n-1) + rabbit(n-2)**A Poor Solution to a Simple Problem**• Fibonacci numbers • First two numbers of sequence are 1 and 1 • Successive numbers are the sum of the previous two • 1, 1, 2, 3, 5, 8, 13, … • This has a natural looking recursive solution • Turns out to be a poor (inefficient) solution**A Poor Solution to a Simple Problem**• The recursive algorithm Algorithm Fibonacci(n) if (n <= 1)return 1else return Fibonacci(n-1) + Fibonacci(n-2)**A Poor Solution to a Simple Problem**Time efficiency grows exponentially with n, which is k^n int fib(int n) { int f[n+1]; f[1] = f[2] = 1; for (int i = 3; i <= n; i++) f[i] = f[i-1] + f[i-2]; return f[n]; } Iterative solution is O(n) The computation of the Fibonacci number F6(a) recursively; (b) iteratively

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