How to resolve the algorithm Loops/Break step by step in the Ruby programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Loops/Break step by step in the Ruby programming language

Table of Contents

Problem Statement

Show a loop which prints random numbers (each number newly generated each loop) from 0 to 19 (inclusive). If a number is 10, stop the loop after printing it, and do not generate any further numbers.
Otherwise, generate and print a second random number before restarting the loop.
If the number 10 is never generated as the first number in a loop, loop forever.

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Loops/Break step by step in the Ruby programming language

Code 1:

  • This code is an infinite loop that generates random numbers between 0 and 19.
  • It prints each random number on a new line.
  • If the random number is 10, the loop prints a newline and breaks out of the loop.
  • If the random number is not 10, the loop generates another random number and prints it on the same line.
  • The loop continues until a random number of 10 is generated.

Code 2:

  • This code is also an infinite loop that generates random numbers between 0 and 19.
  • It prints each random number on a new line.
  • If the random number is 10, the loop prints a newline and breaks out of the loop using the or break statement.
  • If the random number is not 10, the loop generates another random number and prints it on the same line.
  • The loop continues until a random number of 10 is generated.

Comparison of the Two Codes:

The two codes have the same functionality, but they use different syntax to implement the loop.

  • Code 1: Uses the loop keyword to define the infinite loop, and the break keyword to exit the loop.
  • Code 2: Uses the loop do shortcut to define the infinite loop, and the or break statement to exit the loop.

The or break statement is a shorthand way to write the following code:

if a == 10
 puts
 break
end

In Ruby, the or operator returns the value of its second operand if the first operand is false. In this case, the first operand is the condition a == 10, and the second operand is the break statement. If the condition is false (i.e., a is not 10), the or operator returns nil. However, if the condition is true (i.e., a is 10), the or operator returns the value of the break statement, which is to break out of the loop.

Performance Considerations:

Both codes have the same time complexity, which is O(1), as they both terminate after generating a random number of 10. However, code 2 may be slightly faster, as it uses a shortcut (or break) to exit the loop, while code 1 uses a separate break statement.

Source code in the ruby programming language

loop do
  a = rand(20)
  print a
  if a == 10
    puts
    break
  end
  b = rand(20)
  puts "\t#{b}"
end


loop do
  print a = rand(20)
  puts or break if a == 10
  puts "\t#{rand(20)}"
end


  

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