Step by Step solution about How to resolve the algorithm Count in octal step by step in the Ruby programming language

The code you provided is a simple program that prints octal numbers (base 8) from 0 to infinity. It uses three different methods to do this: a loop with a format string, a for loop with the to_s method, and the upto method.

The first method uses a loop with a format string to print the octal representation of the number n. The format string %o tells the printf function to print the number in octal format. The loop increments the value of n by 1 each time through the loop, so it will print the octal numbers from 0 to infinity.

n = 0
loop do
 puts "%o" % n
 n += 1
end

The second method uses a for loop with the to_s method to print the octal representation of the number n. The to_s method converts the number to a string, and the 8 argument tells the method to convert the number to octal format. The loop increments the value of n by 1 each time through the loop, so it will print the octal numbers from 0 to infinity.

for n in (0..)
 puts n.to_s(8)
end

The third method uses the upto method to print the octal representation of the number n. The upto method takes two arguments: the starting number and the ending number. It will print the octal representation of each number between the starting and ending numbers, inclusive. In this case, the starting number is 0 and the ending number is 1/0.0, which is infinity. So, this method will print the octal numbers from 0 to infinity.

0.upto(1/0.0) do |n|
 printf "%o\n", n
end

The fourth method uses the step method to print the octal representation of the number n. The step method takes two arguments: the starting number and the step size. It will print the octal representation of each number between the starting number and infinity, incrementing the number by the step size each time. In this case, the starting number is 0 and the step size is 1. So, this method will print the octal numbers from 0 to infinity.

0.step do |n|
 puts format("%o", n)
end

n = 0
loop do
  puts "%o" % n
  n += 1
end

# or
for n in (0..)
  puts n.to_s(8)
end

# or
0.upto(1/0.0) do |n|
  printf "%o\n", n
end

# version 2.1 later
0.step do |n|
  puts format("%o", n)
end