Recursion learning hour

These are exercises for a learning hour about recursion. They are available in both JavaScript and Ruby versions.

The goal of each exercise is to make tests pass by implementing the appropriate recursive function in javascript/lib or ruby/lib respectively.

In each case ask yourself:

In what way can I make the problem somehow “smaller”?
What is the solution to the “smallest” possible version of the problem?
If I am given the solution to a “smaller” version of the problem, how can I use it to solve the original problem?

If you’re having trouble understanding the problem itself then you may find it helpful to implement an iterative solution first, although this isn’t always easier.

Getting started

For JavaScript:

$ cd javascript
$ yarn install
$ yarn jest test/sumUpTo.test.js # or whatever

For Ruby:

$ cd ruby
$ bundle install
$ bundle exec rspec spec/sum_up_to_spec.rb # or whatever

Doing the exercises

The exercises are organised into three difficulty levels:

Introductory
Intermediate
- Fibonacci numbers
- anagrams
Advanced
- Tower of Hanoi

Introductory

sum numbers from 1 to n

JavaScript: yarn jest test/sumUpTo.test.js
Ruby: bundle exec rspec spec/sum_up_to_spec.rb

This exercise is about summing all the numbers from 1 to n for some number n. For example, the sum of all the numbers from 1 to 5 is 1 + 2 + 3 + 4 + 5 = 15.

An iterative Ruby solution would have roughly this shape:

def sum_up_to(number)
  result = … # some initial value

  while … # some condition involving the argument
    result = … # update result
  end

  result
end

Whereas a recursive solution would look more like:

def sum_up_to(number)
  if … # some condition involving the argument
    … # some initial value
  else
    … sum_up_to(…) … # something involving a recursive call
  end
end

factorial

JavaScript: yarn jest test/factorial.test.js
Ruby: bundle exec rspec spec/factorial_spec.rb

This exercise is about finding the factorial of a number n — that is, all the numbers from 1 to n multiplied together. For example, the factorial of 5 is 1 × 2 × 3 × 4 × 5 = 120.

An iterative JavaScript solution would have roughly this shape:

const factorial = number => {
  let result = …; // some initial value

  while (…) { // some condition involving the argument
    result = …; // update result
  }

  return result;
};

Whereas a recursive solution would look more like:

const factorial = number => {
  if (…) { // some condition involving the argument
    return …; // some initial value
  } else {
    return … factorial(…) …; // something involving a recursive call
  }
};

sum array

JavaScript: yarn jest test/sumArray.test.js
Ruby: bundle exec rspec spec/sum_array_spec.rb

This exercise is about summing the numbers in an array. For example, the sum of the numbers in the array [1984, 451, 1138, 22] is 1984 + 451 + 1138 + 22 = 3595.

Remember:

In what way can I make the problem somehow “smaller”?
What is the solution to the “smallest” possible version of the problem?
If I am given the solution to a “smaller” version of the problem, how can I use it to solve the original problem?

reverse string

JavaScript: yarn jest test/reverseString.test.js
Ruby: bundle exec rspec spec/reverse_string_spec.rb

This exercise is about reversing a string. For example, 'hello' reversed is 'olleh'.

Remember:

In what way can I make the problem somehow “smaller”?
What is the solution to the “smallest” possible version of the problem?
If I am given the solution to a “smaller” version of the problem, how can I use it to solve the original problem?

Intermediate

Fibonacci numbers

JavaScript: yarn jest test/fibonacci.test.js
Ruby: bundle exec rspec spec/fibonacci_spec.rb

This exercise is about calculating the Fibonacci numbers. The first two numbers in the Fibonacci sequence are 0 and 1, and each subsequent number is calculated by adding the previous two. For example, the next numbers in the sequence are 1 (= 0 + 1), 2 (= 1 + 1), 3 (= 1 + 2) and 5 (= 2 + 3).

Bonus points: the simplest recursive implementation of this is quite slow for large inputs. Why? Can you speed it up?

anagrams

JavaScript: yarn jest test/anagramsOf.test.js
Ruby: bundle exec rspec spec/anagrams_of_spec.rb

This exercise is about finding all the anagrams of a string. For example, all the anagrams of 'cat' are 'act', 'atc', 'cat', 'cta', 'tac' and 'tca'.

Remember:

In what way(s) can I make the problem somehow “smaller”?
What is the solution to the “smallest” possible version of the problem?
If I am given the solution(s) to “smaller” version(s) of the problem, how can I use that to solve the original problem?

Advanced

Tower of Hanoi

JavaScript: yarn jest test/moveDisks.test.js
Ruby: bundle exec rspec spec/move_disks_spec.rb

This exercise is about solving a puzzle called the Tower of Hanoi.

The puzzle starts with a stack of differently-sized disks in ascending order sitting on a base. There are two other empty bases and you are allowed to move the top disk from one base to another as long you don’t put a large disk on top of a smaller one. The challenge is to move the entire stack of disks from the first to the last base in a series of valid moves.

An implementation of the puzzle is provided. Each base has an identifier (a, b or c) and the tower object provides a move method which moves the top disk from one base to another and returns the new state of the puzzle.

You can get a feel for this puzzle by trying to solve it manually first. Here’s how you can “play” it on the JavaScript console, visualised with some slightly naff ASCII art:

$ yarn babel-node
> var Hanoi = require('./lib/hanoi')
undefined

> var tower = Hanoi.Tower.create(5) // or try 3 for an easier challenge
undefined

> tower
    **
   ****
  ******
 ********
**********
---------- ---------- ----------
a          b          c

> tower.move('a', 'c')

   ****
  ******
 ********
**********                **
---------- ---------- ----------
a          b          c

> tower.move('a', 'c').move('a', 'b').move('c', 'b')


  ******
 ********      **
**********    ****
---------- ---------- ----------
a          b          c

(If you want command history, brew install rlwrap and then use NODE_NO_READLINE=1 rlwrap --always-readline yarn babel-node to start the console.)

And in Ruby:

$ irb
>> require_relative 'lib/hanoi'
=> true

>> tower = Hanoi::Tower.create(5) # or try 3 for an easier challenge
=>
    **
   ****
  ******
 ********
**********
---------- ---------- ----------
:a         :b         :c

>> tower.move(:a, :c)
=>

   ****
  ******
 ********
**********                **
---------- ---------- ----------
:a         :b         :c

>> tower.move(:a, :c).move(:a, :b).move(:c, :b)
=>


  ******
 ********      **
**********    ****
---------- ---------- ----------
:a         :b         :c

Your goal is to write a function which automatically solve the puzzle in the fewest possible moves. In practice this means implementing a function that can move any number of disks from any base to any other base, assuming the third base is available. The number of moves required increases exponentially with the number of disks; for example, to move a stack of 10 disks requires over 1000 moves.

Remember:

In what way can I make the problem somehow “smaller”?
What is the solution to the “smallest” possible version of the problem?
If I am given the solution to a “smaller” version of the problem, how can I use it to solve the original problem?

tuzz / recursion-workshop

Recursion learning hour

Getting started

Doing the exercises

Introductory

sum numbers from 1 to n

factorial

sum array

reverse string

Intermediate

Fibonacci numbers

anagrams

Advanced

Tower of Hanoi

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