Ruby - creating a grid - ruby

Question

This is very basic, but can someone explain in plain in english what exactly is happening in this code?
3.times do |row_index|
board[row_index] = []
3.times do |column_index|
board[column_index] = []
board[row_index][column_index] = nil
end
end
end

Answer 1

I will first correct your code and then will show you how to improve it with increasing simplifications.
Presumably the array board is initialized before your code and, because of the extra end is probably in a method, we need:
def initialize_board(n, val)
board = Array.new(n)
n.times do |row_index|
board[row_index] = Array.new(n)
n.times do |column_index|
board[row_index][column_index] = val
end
end
board
end
initialize_board(3, nil)
#=> [[nil, nil, nil], [nil, nil, nil], [nil, nil, nil]]
You see that, to make the method more robust, I've made the board's size (n) and initialization value (val) variables. The method must return board, so we need board as the next-to-last line. (Because it is the last line of the method that is executed, return board is not needed.)
Firstly, since you have board[row_index] =..., board must be created as an array with n elements. That's what Array.new(n) does. Similarly, since you have board[row_index][column_index] =..., board[row_index] (for each value of row_index) must be created as an array with n elements:
board[row_index] = Array.new(n)
This works, but it's not very Ruby-like. Better would be to write:
def initialize_board(n, val)
board = []
n.times do |row_index|
row = []
n.times { |column_index| row << val } # or row.push(val)
board << row # or board.push(val)
end
board
end
initialize_board(3, nil)
#=> [[nil, nil, nil], [nil, nil, nil], [nil, nil, nil]]
Notice that board is initialized to an empty array, is then filled with rows, then board is returned. Similarly, row is initialized to an empty array, filled with copies of val and then appended to board. We can tighten that up by using Enumerable#each_with_object:
def initialize_board(n, val)
n.times.with_object([]) do |row_index, board|
board << n.times.with_object([]) { |column_index, row| row << val }
end
end
initialize_board(3, nil)
#=> [[nil, nil, nil], [nil, nil, nil], [nil, nil, nil]]
We can now use Array.new with a block to create each row with the default value:
def initialize_board(n, val)
n.times.with_object([]) do |row_index, board|
board << Array.new(n) { val }
end
end
initialize_board(3, nil)
#=> [[nil, nil, nil], [nil, nil, nil], [nil, nil, nil]]
and then do it again:
def initialize_board(n, val)
Array.new(n) { Array.new(n) { val } }
end
arr = initialize_board(3, nil)
#=> [[nil, nil, nil], [nil, nil, nil], [nil, nil, nil]]
There's one last thing I'd like to mention. Suppose we set:
arr[1][1] = 'cat'
Then
arr #=> [[nil, nil, nil], [nil, "cat", nil], [nil, nil, nil]]
as expected.
If, however, we had written:
def initialize_board(n, val)
Array.new(n, Array.new(n, val))
end
Then:
arr = initialize_board(3, nil)
#=> [[nil, nil, nil], [nil, nil, nil], [nil, nil, nil]]
arr[1][1] = 'cat'
arr #=> [[nil, "cat", nil], [nil, "cat", nil], [nil, "cat", nil]]
which clearly is not what you want.

Answer 2

Try this way in Ruby
Input n defined as size of n x n matrix
Example
if n = 3
Matrix size is 3 x 3
Code
n = gets.chomp.to_i
# Array('A'..'Z').sample is random value from A to Z
matrix = Array.new(n) { Array.new(n) { Array('A'..'Z').sample } }
Output
[["D", "A", "M", "V"], ["X", "Q", "A", "E"], ["P", "D", "L", "S"],
["V", "M", "P", "Z"]]