3 ms·

We should avoid certain questions because they're demotivating to some people? Perhaps a compromise would be to start with an easy input that any pile of if st

We should avoid certain questions because they're demotivating to some people? Perhaps a compromise would be to start with an easy input that any pile of if statements can handle, then give a longer input that will break all but the best algorithms as a bonus.

bussierem 5y ago
I think it's more that the puzzles should avoid having a "correct solution" in the form of a known algorithm. If the puzzle can basically only be solved by remembering a specific algorithm, and any other algorithm (or attempt) will grind for hours, then the puzzle isn't about algorithms, or about coding, it's about whether or not you memorized a part of a math textbook and can recognize that it applies to the problem. Granted, I don't know the process of AoC in terms of vetting and building their puzzles, but it seems that the person/people involved all share very similar levels and domain of knowledge. Having more people reviewing or involved in the process that have a wider range of skillsets might avoid these situations. Example from what I was talking about before: Day 13 of AoC2020 was basically "the answer is the Chinese Remainder Theorem". If you remembered that...great you could solve this in 10 seconds (hell some languages have a _function_ for that theorem!). If not....well good luck finding a solution that runs in under 6h. That has nothing to do with _programming_. Its just math, there wasn't even programming involved (literally some solutions were just "parse the input and pass it to mathlib.chinese_remainder()"). That's not satisfying to solve for the majority of people, I feel (opinion, I know).
- Joker_vD 5y ago
  Is this CRT? Not exactly sure, but it does not run for six hours: result = 0 addend = 1 for divisor, desired_offset in [(7, 0), (13, 1), (59, 4), (31, 6), (19, 7)]: # keep trying numbers until we find the one that, with # the desired offset added to it, would divide without # remainder while (result + desired_offset) % divisor != 0: result += addend # adding another addend should not break previous results # so make it divisible by all previous divisors addend *= divisor result %= 7*13*59*31*19 # Is it really needed? Eh. print(result)

bussierem 5y ago

I think it's more that the puzzles should avoid having a "correct solution" in the form of a known algorithm. If the puzzle can basically only be solved by remembering a specific algorithm, and any other algorithm (or attempt) will grind for hours, then the puzzle isn't about algorithms, or about coding, it's about whether or not you memorized a part of a math textbook and can recognize that it applies to the problem. Granted, I don't know the process of AoC in terms of vetting and building their puzzles, but it seems that the person/people involved all share very similar levels and domain of knowledge. Having more people reviewing or involved in the process that have a wider range of skillsets might avoid these situations. Example from what I was talking about before: Day 13 of AoC2020 was basically "the answer is the Chinese Remainder Theorem". If you remembered that...great you could solve this in 10 seconds (hell some languages have a _function_ for that theorem!). If not....well good luck finding a solution that runs in under 6h. That has nothing to do with _programming_. Its just math, there wasn't even programming involved (literally some solutions were just "parse the input and pass it to mathlib.chinese_remainder()"). That's not satisfying to solve for the majority of people, I feel (opinion, I know).

Joker_vD 5y ago
Is this CRT? Not exactly sure, but it does not run for six hours: result = 0 addend = 1 for divisor, desired_offset in [(7, 0), (13, 1), (59, 4), (31, 6), (19, 7)]: # keep trying numbers until we find the one that, with # the desired offset added to it, would divide without # remainder while (result + desired_offset) % divisor != 0: result += addend # adding another addend should not break previous results # so make it divisible by all previous divisors addend *= divisor result %= 7*13*59*31*19 # Is it really needed? Eh. print(result)