Veterans, what level of difficulty would you say Code of the Ring is?

Code of the Ring was the very first game and contest I’ve ever done using programming. I would like to know where I stand, so I would like to know the level of difficulty the puzzle is.

To solve it is really easy even a beginner with basic programming can just use ++++ & --- to print a letter while sitting on one rune.But he’ll probably achieve a score of 10000 or so instructions.
So it really depends on your score.So here is what I think for a given score:

  • 7500 and above would be a newbie.Some one who is totally new to programming
  • 6500 - 7500 would be considered a fairly decent programmer
  • 6000 - 6500 : An average programmer.I saw a hell lot of people at 6000 ish
  • 5000 - 6000 : An really good programmer whos probably been coding for about a year or so.
  • 4000 - 5000 : An awesome advanced programmer who has been coding for a few years
  • 3000 - 4000 : An expert programmer who really has studied tons of pattern matching and optimization.

Lazy programmers can be everywhere in your classification. :whistle:

I think that some of these milestones are weird, given the problem. Actually, I would draw only 4 categories:

  • 100% completion (as you say, that’s very easy)
  • 6001 instructions (that’s what you can get without loops or complicated algorithms, and a lot of people got that; I think that algorithm makes this puzzle on par with the “medium” puzzles on CG)
  • 3900-4500 instructions (there are a few easy loops one can apply to reach that, but implementing them can get tricky)
  • below that is expert, as you say :smiley:

During the contest, I didn’t see any gap between 10k and 6001, and getting from 6001 to 4500 or so was rather fast too (in terms of number of submissions, not time spent on the keyboard!).

Imo any decent programmer with time (and motivation :stuck_out_tongue: ) to dive into the problem should reach the 6001 milestone.

Thanks for the replies. Also, I remember that there was a special rule that states one can replace certain repeated sequence of instructions with something like “[+]”.

I probably wrote it incorrectly. However, you get the idea.

So, in your assessment of the difficulty of the game, did you take this rule into consideration? When you a score of 3000-4000, do you mean someone who has applied this rule in their code?

Also, was it applying this rule hard? I never got the chance to apply the rule.

No it’s not hard, but it requires a little of pattern matching. You know you can still play the game, right?

REALL? I can’t find it. Can you link it to me? thank you

it is on the games page it is called bilbo’s forest

And you can still compete with the others, just not win a prize except glory :slight_smile:

Thank you.

"These actions are also available:
[: If the current rune contains a space, skip all instructions up to the matching closing bracket ], otherwise continue normally.
]: If the current rune contains a space, continue normally, otherwise perform the instructions starting from the matching opening bracket [.
This makes it possible to have Bilbo perform more actions with less characters for your program to output.

For example, AAAAAAAAAAAAAAAAAAAAAAAAAA (A x26) can be achieved with a simple
+… as well as with +>-[<.>-]."

I am confused. Can someone explain the rule to me?

By “loops” do you mean the special rule?

Just google ‘Brainfuck’ :stuck_out_tongue:

The game that I made is based on the brainfuck programming language, if you have trouble you can check tutorial on brainfuck, it’ll help a lot.

Here the example explained:


+ set the current stone to A
>- set the next stone to Z
then [ do nothing since current stone is different from space character.
<. go back to the stone with the ‘A’ character and print it
>- return on the stone with ‘Z’ and make it contain ‘Y’
] since current stone isn’t space, go back to the previous [ which is making the loop

it will then print the A again, then decrements again, and so on until the second stone is equal to space and then it exit the loop.

Since the second stone is initialized at Z (the 26th letter) it will output A 26 times.

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Yes, exactly.