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

Recursion

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Recursion

nGeneral Form of Recursive methods

nExamples of Simple Recursion

nHand Simulating Recursive methods

nProving Recursive methods Correct

nModeling Recursion via Onions

nRecursive methods on Naturals/Integers

- 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
- 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.

Problem: Collect $1,000.00 for charity

Assumption: Everyone is willing to donate a penny

- Iterative Solution
- Visit 100,000 people, asking each for a penny

- Recursive Solution
- If you are asked to collect a penny, give a penny to the person who asked you for it
- Otherwise
- Visit 10 people and ask them to each raise 1/10th 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

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: SP1, SP2, ... , SPN

(2) Recursively call Solve (this method) on each subproblem: Solve(SP1), Solve(SP2),..., Solve(SPN)

(3) Combine the solutions to these subproblems into a solution that solves the original Problem}

}

Example using Definition

4! = 4 * 3!

= 4 * 3 * 2!

= 4 * 3 * 2 * 1!

= 4 * 3 * 2 * 1 * 0!

= 4 * 3 * 2 * 1 * 1

1if N = 0

N! =

N*(N-1)!if N > 0

int factorial (int n)

{

if (n == 0)//Non-decomposable

return 1;

else {

int subN = n-1; //Decompose

int subFact = factorial(subN); //Solve Subproblem

return n * subFact;//Combine

}

}

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;

}

Here we combine the 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)

1if N = 0

AN =

A*AN-1if N > 0

double pow(double a, int n)

{

if (n == 0)

return 1.;

else

return a*pow(a,n-1)

}

Example using Definition

A4 = A * A3

= A * A* A2

= A * A* A* A1

= A * A* A* A* A0

= A * A* A* A* 1

Calling pow(a,n)requires exactly n multiplications

The pow in the Java Math library actually computes its answer using logarithms and exponentials

- Assume that a method is computed according to its definition by a person in an apartment complex
- 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 can 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

n =

return =

n =

return =

n =

return =

n =

return =

...

n =

return =

To prove that a recursive method, Solve, is correct:

- Prove that Solve computes (without recursion) the correct answer to any minimal (base case) Problem
- Base cases are simple, so this should be easy

- Prove that the argument to any recursive call of Solve, e.g. SPI, is strictly smaller (closer to the minimal case) than Problem
- 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
- 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)

factorial (recurring on the parameter n)

- factorial correctly returns 1 for the minimal argument 0.
- In factorial(n-1), n-1 is always closer to 0 than n is.
- Assume factorial(n-1) computes the correct answer.
- n times this value is, by definition, the correct value of factorial(n)
pow (recurring on the parameter n)

- n times this value is, by definition, the correct value of factorial(n)
- pow correctly returns 1 for the minimal argument 0.
- In pow(a,n-1), n-1 is always closer to 0 than n is.
- Assume pow(a,n-1) computes the correct answer, an-1.
- a*pow(a,n-1) is a*an-1 is an, the correct value of pow(a,n)

int factorial (int n)

{

if (n == 0)

return 0;//0! is not 0;

else

return n*factorial(n-1);

}

int factorial (int n)

{

if (n == 0)

return 1;

else

return factorial(n+1)/(n+1);//n+1 not closer to 0

}

int factorial (int n)

{

if (n == 0)

return 1;

else

return n + factorial(n-1);//n+(n-1)! is not n!

}

- 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

Note: if N is odd N/2 truncates: so 7/2 = 3

1if N = 0

AN =B2 where B = AN/2 if N > 0 and N is Even

AB2 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;

}

}

Calling fpow(a,n)requires between at least log2n and at most 2log2n 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

- 0 is the smallest Natural
- 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
- If x is a Natural, s(x) is a Natural
- The successor of x is x+1

- If x is a non-0 Natural, p(x) is a Natural
- 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

- If x is a Natural, s(x) is a Natural

s(s(s(0))) = 3

s(s(0)) = 2

s(0) = 1

0

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

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 IllegalArgumentException("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

int sum(int a, int b)

{

if (z(a))

return b;

else

return s(sum(p(a),b));//1 + (a-1 + b)

}

int sum(int a, int b)

{

if (z(a))

return b;

else

return sum(p(a), s(b));//(a-1) + (b+1)

}

In both sum methods, a gets closer to 0 in recursive calls

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

}

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) //Is a < b

{

if (z(b))//Nothing is < 0

return false;

else if (z(a))//If b!=0, a==0, then a<b

return true;

else

return less(p(s),p(b));//Is (a-1) < (b-1)

}

bool even(int a)

{

if (z(a))//True if 0

return true;

else//Opposite of even(a-1)

return !even(p(a))

}

bool odd(int a)

{

if (z(a))//False if 0

return false;

else //Opposite of odd(a-1)

return !odd(p(a));

}

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);

System.out.println(i%10);

}

}

- 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
- Printing i/10 (all digits but the last) and then printing i%10 (the last digit), prints all the digits in i in order

The code above is simplified by bottom factoring and test reversal and noting for i<10, i is i%10

void printReversed (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);

printReversed(i/10);

}

}

- printReversed correctly prints all 1-digit values: 0 - 9.
- In printReversed(i/10), i/10 is closer to 0-9 than i is.
- Assume printReversed(i/10) correctly prints all the digits in i/10 reversed.
- Printing the last digit, i%10, and then printing the i/10 reversed prints all the digits in i reversed.

String toString (int i)

{

if (i < 10)

return ""+(char)(i+’0’);

else

return toString(i/10) + (char)(i%10+’0’);

}

- toString correctly returns all 1-digit values: 0 - 9.
- In toString(i/10), i/10 is one digit closer to 0-9 than i is.
- Assume toString(i/10) correctly returns a String of all the digits in I/10.
- Then, after appending the char equivalent of i%10 at the end of this String correctly stores the String equivalent of i.

- Find the base (non-decomposable) case(s)
- 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
- 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
- Integers : i - something or i/something (digit at a time)
- Linked Lists: l.next

- Write the following methodpublic 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 methodpublic static int toInt (String s) recursively. It is like parseInt, but only for non-negative valuestoInt("138") should return 138

- Write the following methodpublic static boolean equals (String s1, String s2) recursively. Its recursive structure is like that of the equals methods for ints.
- Write the following methodpublic static int compare (String s1, String s2) recursively. Its recursive structure is similar to equals, above.