Steal 10 Treasures (ParserComp 2023): determining precise ranking cutoffs

This isn’t so much a hint to find all 10 treasures, as to point out a way to see at what ranks you get more points. Also, it shows you how to get points outside of the treasures. You may be able to guess this.

It is a pretty trivial exercise in modular arithmetic, but I wanted to share it, because I think even non-parser fans will enjoy S1T, and maybe someone is curious where the precise rankings are. And who knows, this could make an odd puzzle of their own for someone, down the line.

(Note: you can look in the source code! But that’s cheating, sort of.)

You get 5 points for each treasure, and you get 3 points for auxiliary actions: getting various helper items, or making treasures available, or passing an obstacle.

Seven treasures can be picked up without solving any puzzles.

  • 1 entering the castle
  • 2 taking the pepper
  • 3 freeing the owl
  • 4 getting the screwdriver
  • 5 making the rose take-able

Three treasures require one of these actions, and one requires two.

Now we can get 13 points by (1) and finding two treasures.
We can get 12 points by (1)-(4).

We can get 25 points, if we just do (1)-(5) and get two treasures. And we can get 26 with (1-2) and 4 treasures.

We can get 38 points by getting the 7 treasures not tied to points. Some are actually tougher to get than these, but the supporting actions score you no points. The 3 treasures tied to previous point-scoring actions are the griffin, the spike and the rose.

We can’t get 51 points, since that’d mean we would leave 3 3-pointers and a 1-pointer. But we can get50 by doing all the supporting actions and getting 7 treasures.

52 points is 8 treasures and 4 actions. We could just leave (4) for last, as well as another. But we can’t get 51. That would require leaving 3 3-pointers and a 5-pointer undone, but the best we can do is 2 and 1.

The lower-end scores I found unreachable were 1, 2, 4, 7, and 10. The higher-end scores were 43, 48, 53, 58, 63, 51, 55, 59, 61, 62 and 64.

Now it would be interesting to see how to brute-force through all the ways to achieve each score. I think it could be done with Python and recursion, where we use a list of lists to take less time (e.g. we only need to look through one treasure-gaining path with 5 points, not all 7.) But I thought this was an interesting numerical experiment & hope you-all do too. S1T seems to lend itself to that sort of whimsy.