Recursion n n n General Form of Recursive

Recursion n n n General Form of Recursive methods Examples of Simple Recursion Hand Simulating Recursive methods Proving Recursive methods Correct Modeling Recursion via Onions Recursive methods on Naturals/Integers

Recursion n n Recursion is a programming technique in which a call to a method appears in that method’s body (i. e. , a method calls itself: this is called direct recursion) Once you understand recursive methods, they are often simpler to write than their iterative equivalents l n In modern programming languages, recursive methods may run a bit slower (maybe 5%) than equivalent iterative methods; in a typical application, this time is insignificant (most of the time will be taken up elsewhere anyway) We will begin by studying the form of general recursive methods; then apply this form to methods operating on int values; and finally apply that form to methods operating on linked lists (where they are most useful). In both cases we will discuss how values of these types are recursively defined. 15 -111 2

Fund Raising: Iteration vs. Recursion 15 -111 3 Problem: Collect $1, 000. 00 for charity Assumption: Everyone is willing to donate a penny n Iterative Solution l n Visit 100, 000 people, asking each for a penny Recursive Solution l l If you are asked to collect a penny, give a penny to the person who asked you for it Otherwise s s s Visit 10 people and ask them to each raise 1/10 th of the amount of money that you have been asked to raise Collect the money that they give you and combine it into one bag Give it to the person who asked you to collect the money

General Form of Recursive methods Solve(Problem) { if (Problem is minimal/not decomposable: a base case) solve Problem directly; i. e. , without recursion else { (1) Decompose Problem into one or more similar, strictly smaller subproblems: SP 1, SP 2, . . . , SPN (2) Recursively call Solve (this method) on each subproblem: Solve(SP 1), Solve(SP 2), . . . , Solve(SPN) (3) Combine the solutions to these subproblems into a solution that solves the original Problem } } 15 -111 4

factorial: In Mathematics and Java N! = 1 N*(N-1)! if N = 0 if N > 0 Example using Definition 4! = 4 * 3 * 2! = 4 * 3 * 2 * 1 * 0! =4*3*2*1*1 int factorial (int n) { if (n == 0) //Non-decomposable return 1; else { int sub. N = n-1; //Decompose int sub. Fact = factorial(sub. N); //Solve Subproblem return n * sub. Fact; //Combine } } 15 -111 5

Simplified and Iterative factorial int factorial (int n) { if (n == 0) return 1; else return n * factorial(n-1); } int factorial (int n) { int answer = 1; for (int i=1; i<=n; i++) answer *= i; return answer; } 15 -111 6 Here we combine three steps in the previous else block into a single return statement (one expression including the decompose, solve, and combine steps). Compare this method to the one below it, which operates by iteration. The iterative version requires the declaration of two variables and the use of state change operators (which are always difficult to reason about)

pow: Raising a Number to a Power AN = 1 if N = 0 A*AN-1 if N > 0 double pow(double a, int n) { if (n == 0) return 1. ; else return a*pow(a, n-1) } The pow in the Java Math library actually computes its answer using logarithms and exponentials Example using Definition A 4 = A * A 3 = A * A 2 = A * A * A 1 = A * A * A 0 = A * A* A* A* 1 Calling pow(a, n)requires exactly n multiplications 15 -111 7

Simulation of Recursive Methods n Assume that a method is computed according to its definition by a person in an apartment complex l n n 15 -111 8 That person can be called to compute an answer; once he/she computes the answer (possibly helped by calling another person in the apartment complex), he/she places a return call, and reports back the answer to the person who called him/her Assume that each person knows the phone number of the person living in the apartment underneath them, who also has the same set of instructions: so any person call the one underneath to solve a simpler problem The first method call initiates a sequence of calls to people living in lower level apartments (each person solving a simpler problem), whose answers percolate back to the top, finally solving the original problem

Hand Simulation of Factorial n = return =. . . n = return = 15 -111 9

Proof Rules for Recursive Methods To prove that a recursive method, Solve, is correct: n Prove that Solve computes (without recursion) the correct answer to any minimal (base case) Problem l n Prove that the argument to any recursive call of Solve, e. g. SPI, is strictly smaller (closer to the minimal case) than Problem l n Base cases are simple, so this should be easy The notion of strictly smaller should be easy to understand for the recursive argument, so this should be easy Prove that these answers are combined to compute the correct answer for Problem l l In this proof, you may assume that each recursive call, Solve(SPI), correctly computes its answer The assumption makes this proof easy Recursion is computable induction (a math concept) 15 -111 10

Applying the Proof Rules 15 -111 11 factorial (recurring on the parameter n) n factorial correctly returns 1 for the minimal argument 0. n In factorial(n-1), n-1 is always closer to 0 than n is. n Assume factorial(n-1) computes the correct answer. l n times this value is, by definition, the correct value of factorial(n) pow (recurring on the parameter n) n pow correctly returns 1 for the minimal argument 0. n In pow(a, n-1), n-1 is always closer to 0 than n is. n Assume pow(a, n-1) computes the correct answer, an-1. l a*pow(a, n-1) is a*an-1 is an, the correct value of pow(a, n)

Proof Rules and Bad Factorials int factorial (int n) { if (n == 0) return 0; else return n*factorial(n-1); } int factorial (int n) { if (n == 0) return 1; else return factorial(n+1)/(n+1); } int factorial (int n) { if (n == 0) return 1; else return n + factorial(n-1); } //0! is not 0; //n+1 not closer to 0 //n+(n-1)! is not n! 15 -111 12

Proof Rules and Bad Factorials (cont) n n 15 -111 13 In the first method, the wrong answer (0) is returned for the base case; since everything depends on the base case, ultimately this method always returns 0 In the second method, n+1 is farther away from the base case: this method will continue calling factorial with ever larger arguments, until the maximum int value is exceeded: a runaway (or “infinite”) recursion (actually, each recursive call can take up some space, so eventually memory is exhausted). In the third method, the wrong answer is returned by incorrectly combining n and the solved subproblem; this method returns one more than the sum of all the integers from 1 to n (an interesting method in its own right) not the product of these values In the first method, it always returns the wrong answer; the second method never returns an answer, the third method returns the correct answer only in the base case

fpow: a faster version of pow 1 if N = 0 AN = B 2 where B = AN/2 if N > 0 and N is Even AB 2 where B = AN/2 if N > 0 and N is Odd int fpow(double a, int n) { if (n == 0) return 1. ; else { double b = fpow(a, n/2); if (n%2 == 0) return b*b; else return a*b*b; } } 15 -111 14 Note: if N is odd N/2 truncates: so 7/2 = 3 Calling fpow(a, n) requires between at least log 2 n and at most 2 log 2 n multiplications Compare this complexity to calling pow(a, n)requiring n multiplications Contemporary cryptography raises large numbers to huge (hundred digit) powers; it needs a fast method

Recursive Definition of Naturals n 0 is the smallest Natural l l n It is minimal, and cannot be decomposed z(x) is true if x is zero and false if x is non-zero The successor of a Natural is a Natural l If x is a Natural, s(x) is a Natural s l The successor of x is x+1 If x is a non-0 Natural, p(x) is a Natural s s The predecessor of x is x-1 p(0) throws an exception Note: z(s(x)) is always false p(s(x)) = x s(p(x)) = x only if X 0 15 -111 15

Onion Model of Recursive Naturals 15 -111 16 Every Onion is either Base Case : The core Recursive Case: A layer of skin around a smaller onion (which may be the core, or some deeper layer) The value of an onion is the number of layers of skin beyond the core 0 s(0) = 1 s(s(0)) = 2 s(s(s(0))) = 3

Computation with Naturals 15 -111 17 Let’s Define 3 methods that operate on Naturals in Java int s(int x) {return x+1; } int p(int x) { if (x==0) throw new Illegal. Argument. Exception("p with x=0"); return x-1; } bool z(int x) {return x == 0; } We can define all common operators (both arithmetic and relational) on Naturals by writing simple recursive methods that use these three methods

Recursive methods on Naturals int sum(int a, int b) { if (z(a)) return b; else return s(sum(p(a), b)); } int sum(int a, int b) { if (z(a)) return b; else return sum(p(a), s(b)); } //1 + (a-1 + b) //(a-1) + (b+1) In both sum methods, a gets closer to 0 in recursive calls 15 -111 18

Recursive methods on Naturals II int product(int a, int b) { if (z(a)) return 0; else return sum(b, product(p(a), b)) } //b + (a-1)*b int power(int a, int b) { if (z(b)) return 1; else return product(a, power(a, p(b))); //a * ab-1 } 15 -111 19

Recursive methods on Naturals III 15 -111 20 bool equal(int a, int b) //Is a == b { if (z(a) || z(b)) //If either is 0. . . return z(a) && z(b); //return whether both are 0 else return equal(p(a), p(b)); //Is (a-1) == (b-1) } bool less(int a, int b) { if (z(b)) return false; else if (z(a)) return true; else return less(p(a), p(b)); } //Is a < b //Nothing is < 0 //If b!=0, a==0, then a<b //Is (a-1) < (b-1)

Recursive methods on Naturals IV bool even(int a) { if (z(a)) return true; else return !even(p(a)) } bool odd(int a) { if (z(a)) return false; else return !odd(p(a)); } //True if 0 //Opposite of even(a-1) //False if 0 //Opposite of odd(a-1) 15 -111 21

Printing an Integer’s Digits 15 -111 22 void print (int i) { if (i < 10) //if (i >= 10) System. out. print(i); // print(i/10); else { //System. out. print(i%10); print(i/10); The code above is simplified by System. out. println(i%10); bottom factoring and test reversal } and noting for i<10, i is i%10 } n n n print correctly prints all 1 -digit values: 0 - 9. In print(i/10), i/10 is one digit closer to 0 - 9 than i is. Assume print (i/10) correctly prints all the digits in i/10 in the correct order l Printing i/10 (all digits but the last) and then printing i%10 (the last digit), prints all the digits in i in order

Printing an Integer’s Digits in Reverse 15 -111 23 void print. Reversed (int i) { if (i < 10) //System. out. print(i%10); System. out. print(i); //if (i >= 10) else { // System. out. print(i/10); System. out. print(i%10); print. Reversed(i/10); } } n n n print. Reversed correctly prints all 1 -digit values: 0 - 9. In print. Reversed(i/10), i/10 is closer to 0 -9 than i is. Assume print. Reversed(i/10) correctly prints all the digits in i/10 reversed. l Printing the last digit, i%10, and then printing the i/10 reversed prints all the digits in i reversed.

Converting an int to a string 15 -111 24 String to. String (int i) { if (i < 10) return ""+(char)(i+’ 0’); else return to. String(i/10) + (char)(i%10+’ 0’); } n n n to. String correctly returns all 1 -digit values: 0 - 9. In to. String(i/10), i/10 is one digit closer to 0 -9 than i is. Assume to. String(i/10) correctly returns a String of all the digits in I/10. l Then, after appending the char equivalent of i%10 at the end of this String correctly stores the String equivalent of i.

Synthesizing Recursive Methods n Find the base (non-decomposable) case(s) l n Write the code that detects the base case(s) and returns the correct answer for them without using recursion Assume that you can decompose all non base-case(s) and then solve these smaller subproblems via recursion l n 15 -111 25 Choose the decomposition There should be some natural decomposition Write code that combines these solved subproblems (often there is just one) to solve the original problem Typical decompositions l l Integers : i - something or i/something (digit at a time) Linked Lists: l. next

Problems n n 15 -111 26 Write the following method public static String reverse (String s) recursively. Choose an appropriate base case, recursive call (hint: substring method using one parameter), and a way to combine (hint: +) solved smaller problems to solve the original problem. reverse("abcd") should return "dcba" Write the following method public static int to. Int (String s) recursively. It is like parse. Int, but only for non-negative values to. Int("138") should return 138

Problems (continued) n n Write the following method public static boolean equals (String s 1, String s 2) recursively. Its recursive structure is like that of the equals methods for ints. Write the following method public static int compare (String s 1, String s 2) recursively. Its recursive structure is similar to equals, above. 15 -111 27