Lab 11: practice recursion

Introduction
- This lab will reinforce recursion which is traditionally a pretty complicated technique for students to master in Computer Science.
Preparation
- Copy-and-paste these commands to your terminal. They copy the files and cd to the lab directory:
  We will learn to write 2 recursive algorithms in this lab.
  But first, some theory on how to think about recursive algorithms.

How to write recursive methods

As discussed in class, a recursive method will invoke itself, but the invocation always uses a parameter value that is smaller that the original value.

Example: factorial

public static int factorial( int n ) { int sol1, mySol; if ( n == 0 ) { return 1; // A solution is available for a base case } else { sol1 = factorial ( n - 1 ); mySol = n * sol1; return mySol; } }

The original parameter is n, and factorial(n) invokes itself with a smaller parameter value (n−1)

The key in understanding recusive methods is recognizing that you are:
So:

Example: compute 20!

We know that: 20! = 20 × 19!
Suppose we know that: 19! = 121645100408832000, then we can use the solution of this smaller problem (19!) to solve the original problem 20!:

The Base case(s)
- In every recursive method, there is at least one or more base cases that must have been solved (i.e., you know the solution)
- General rule:

Structure of a recursive method

The easiest recursive method uses one smaller problem to solve the original problem.

The structure of a recursive method to solve a problem of size n using the solution of a problem of size n−1 is:

public static ReturnType solveProblem( int n ) { ReturnType sol1, mySol; if ( n is a base case ) { return solution; // A solution is available for a base case } else { sol1 = solveProblem ( n - 1 ); mySol = solve problem of size n using solution sol1; return mySol; } }

Example: factorial

public static int factorial( int n ) { int sol1, mySol; if ( n == 0 ) { return 1; // A solution is available for a base case } else { sol1 = factorial ( n - 1 ); mySol = n * sol1; return mySol; } }

Your first recursion exercise

Problem description:

Write a recursive method power(int a, int n) that computes aⁿ

Steps:

Edit the file Power.java and write the method power(int a, int n ). (See hints on how to do it below)

Compile with the command:

javac Power.java (or click compile in BlueJ)

Run with the command:

java Power (or select the main() method to run in BlueJ)

How to approach the problem:

What is the problem solved by the method ?

Answer:

power( a, n ) computes aⁿ (example: power( 2, 10 ) computes 2¹⁰)

What is a smaller problem of the same nature ?

Answer:

power( a, n-1 ) computes a^n-1 (example: power( 2, 9 ) computes 2⁹)

How can you use the solution of power( a, n−1 ) to solve the original problem ?

Answer:

If we are solving 2¹⁰ and we know the solution 2⁹ = 512, we can solve 2¹⁰ as follows: 2×512
In general: If we are solving aⁿ and we know the solution a^n-1 = sol1, we can solve aⁿ as follows: a×sol1

What is a base case that we have a ready solution ?
Answer:
Use the above information to write the recursive method power(int a, int n)

Other kinds of smaller problems
- The most common smaller problem that a recursive method solves within itself uses the parameter n−1
  But, that is not always the case.
- Problems involving strings that are solved by recursive method usually uses a shorter length string.
  The most common smaller problem that a "string typed" recursive method solves within itself uses a string of shorter length n−1.

A recursion exercise using "string" typed input

Problem description:

Given a non-negative number n.
Write a recursive method sumDigits(int n) that computes sum of the digits in the number n.

Steps:

Edit the file SumDigits.java and write the method sumDigits(int n ). (See hints on how to do it below)

Compile with the command:

javac SumDigits.java (or click compile in BlueJ)

Run with the command:

java SumDigits (or select the main() method to run in BlueJ)

How to approach the problem:

It is easier to describe the method by listing out the digits of a number.
So I will use the notation:
to represent the fact that the number n consists of a number of digits xyz (it does not mean 3 digits, what I mean is: "some digits")

What is the problem solved by the method ?

Answer:

sumDigits( n = xyz ) computes x+y+z (example: sumDigits( 3412 ) computes 3+4+1+2)

What is a smaller problem of the same nature ?

Answer:

sumDigits( n/10 = xy ) computes x+y (example: sumDigits( 341 ) computes 3+4+1) (Note that: 341 = 3412/10 So the operation to shorten a number by 1 digit is "dividing by 10" !)

How can you use the solution of sumDigits(n/10) to solve the original problem ?

Answer:

If we are solving sumDigit( 3412 ) and we know the solution sumDigit( 341 ) = 8, we can solve sumDigit( 3412 ) as follows: 8 + 2 (2 = the last digit of 3412) or: 8 + (3412 % 10)
In general: If we are solving sumDigit( n = xyz ) and we know the solution sumDigit( n/10 = xy ) = sol1, we can solve sumDigit( n = xyz ) as follows: sol1 + z (z = the last digit of n = xyz) or: sol1 + (n % 10)

What is a base case that we have a ready solution ?
Answer:
Use the above information to write the recursive method sumDigits(int n)

Turnin
- Use the following command to turn in your lab 11:
For the Lead TA: