3 ms·

I made a solver for this and did some comparison with Wordle: https://jacobbrazeal.wordpress.com/2022/02/07/solving-nerdle/ https://jacobbrazeal.wordpress.com/2

by tibbar 5y ago

I made a solver for this and did some comparison with Wordle: https://jacobbrazeal.wordpress.com/2022/02/07/solving-nerdle/ https://jacobbrazeal.wordpress.com/2022/02/07/solving-nerdle.... TL;DR even though there's an order of magnitude more possible words, you can solve it in fewer average guesses than Wordle pretty consistently.

pierrebai 5y ago
That's because there are fewer letters. In wordle, ignoring the fact that the word need to exists, you could make 5 guesses with 5 new letters each time and still not quite exhausting the letter space. With the math one, in two guesses you can figure out the digits. Also, I play wordle in the hardest mode, which forces you to always reuse known letters, which makes it a lot harder. (I also playu in a clone with an extended vocabulary here: https://www.arachnoid.com/wordgame/index.html https://www.arachnoid.com/wordgame/index.html)
- deleted 5y ago
  [deleted]
- dankwizard 5y ago
  weird flex but ok
  - delaaxe 5y ago
    this is not reddit
- gcanyon 5y ago
  Deduce https://apps.apple.com/us/app/deduce/id351558195 https://apps.apple.com/us/app/deduce/id351558195 is not a wordle clone (it's been around over ten years), and is much harder than wordle. It is iOS only and costs $3.
nneonneo 5y ago
You note that the interface accepts a lot more words than are in the dictionary, much like the normal Wordle game. I wonder if you can exploit the full set of acceptable words to narrow down the possibility space even further? I would not be surprised if there's an equation in the full set that performs better than the equation you initially selected.
- tibbar 5y ago
  It’s definitely possible. An exhaustive search would require tapping into the cloud since there are well over 100,000 words that the interface could use, times 17,000 target words, times the grading operation which is moderately expensive. My guess, though, is that the best starter word is number-dense since numbers show up in the middle of the frequency range, and therefore probably wouldn’t change. However, subsequent guesses would very likely benefit from the broader dictionary.
  - nneonneo 5y ago
    Eh, it’s about 1.7 billion operations. I suspect you could churn through those pretty fast in a language like C or Rust.
    - deleted 5y ago
      [deleted]

by tibbar 5y ago

pierrebai 5y ago

That's because there are fewer letters. In wordle, ignoring the fact that the word need to exists, you could make 5 guesses with 5 new letters each time and still not quite exhausting the letter space. With the math one, in two guesses you can figure out the digits. Also, I play wordle in the hardest mode, which forces you to always reuse known letters, which makes it a lot harder. (I also playu in a clone with an extended vocabulary here: https://www.arachnoid.com/wordgame/index.html https://www.arachnoid.com/wordgame/index.html)

deleted 5y ago
[deleted]
dankwizard 5y ago
weird flex but ok
- delaaxe 5y ago
  this is not reddit
gcanyon 5y ago
Deduce https://apps.apple.com/us/app/deduce/id351558195 https://apps.apple.com/us/app/deduce/id351558195 is not a wordle clone (it's been around over ten years), and is much harder than wordle. It is iOS only and costs $3.

nneonneo 5y ago

You note that the interface accepts a lot more words than are in the dictionary, much like the normal Wordle game. I wonder if you can exploit the full set of acceptable words to narrow down the possibility space even further? I would not be surprised if there's an equation in the full set that performs better than the equation you initially selected.

tibbar 5y ago
It’s definitely possible. An exhaustive search would require tapping into the cloud since there are well over 100,000 words that the interface could use, times 17,000 target words, times the grading operation which is moderately expensive. My guess, though, is that the best starter word is number-dense since numbers show up in the middle of the frequency range, and therefore probably wouldn’t change. However, subsequent guesses would very likely benefit from the broader dictionary.
- nneonneo 5y ago
  Eh, it’s about 1.7 billion operations. I suspect you could churn through those pretty fast in a language like C or Rust.
  - deleted 5y ago
    [deleted]