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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 a
by 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 agoIt’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 agoEh, 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]