In python I'm already disabled all module imports. I think it much better than source code check.

named 'random1' cause a fail

I think it's ruby.

Do you have any ideas how to disable Random in Java?

Because there was an error in oneTwo function. Now it fixed. Try again

Fixed

The preloaded method oneTwo() actually returns the solution.

Sorry. I need to spend more time to review translations. I've fixed description and oneTwo in Java

I'm unable to approve translation. Please fix description, it must be same as the original.

This is redurant:

    if make_hash(original) == make_hash(other):
        return True
    else:
        return False

You can simply return make_hash(original) == make_hash(other)

Hi, tanks for translation. But threre is a way for cheaters to use built-in functions.
Can you add this code to translation?

def one_two() rand(1..2) end


class Array
  ERROR_STRING_FOR_RANDOM = "Usage of built-in random generators is not allowed"
  RANDOM_METHODS = [:shuffle, :shuffle!, :select, :sample]
  def random_error
    raise ERROR_STRING_FOR_RANDOM
  end
  RANDOM_METHODS.each do |m|
    define_method(m) {|*args| random_error }
  end
end

Try it yourself:

function one_two() { return 1 + Math.random() * 2 | 0}

let counter = {};
for(let i=100000; i>0; --i) {
    value = one_two_three();
    if(!counter[value]) counter[value] = 0;
    counter[value]++;
}

console.log(counter);

I should exdend test cases.

Nice solution 👍

It valid for any random function with uniform distribution that return three elements.

For example P(11) ≈ P(12) ≈ P(13) ≈ P(21) ≈ P(22) ≈ P(23) ≈ P(31) ≈ P(32) ≈ P(33) ≈ (1 / 3)**2 ≈ 0.11111111111
Same for strings with length 3, 4 and so on

Hi! one_two function already defined in a global scope.
You won't need to receive any arguments. You can just call one_two from one_two_three

https://en.wikipedia.org/wiki/Linear_congruential_generator

Do you passed test with consecutive 1's 2's 3's?
Try to generate 60k length string and measure probability to find same length combinations, will it equal expected probability?

P(any combination with length N) ≈ (1 / 3)**N

Thank you for response. I've disabled shuffle, shuffle!, select, sample. Can you describe another ways to cheat?