Cycle optimized solution for Disk 1 Puzzle 7: The Dopplegänger for May 2023 leaderboard (50~69 cycles).
Solution for 60~69 cycles
The idea is to use more lines of code to reduce the processing time. For solutions with 70 cycles or greater, see previous guides.
تلميحات
- We need to store two input values in order to compare them. Solutions in other guides show how to pass a value between registers before storing the next input. That’s not the only way.
- There are more than one register.
- We still need some sort of loop, but a longer one that alternates between two registers.
حل
loop: reg = input; check reg = int; jump if true: print; int = input; check reg = int; jump if false: loop; print: output = int; jump to: loop;
Note that بخار formatting may have inserted extra spaces. This should yield very close to 70 cycles. With a little bit of luck with different input data, or some other tricks (hint:the first input), you can get under 70 cycles more often.
Hint for 50~59 cycles
You would still be trading smaller code size for faster execution speed, just taking that concept to the extreme, at least that’s how I understand it for now.
I used many lines of code.
It can be considered “hacky”.
For puzzle #7 in particular, it’s accurate only probabilistically, but still at a very high rate that passes the game checks. The more lines of code, the less likely it is to make a mistake.
This and other ideas (from sections above) can be adapted to other puzzles as well.
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