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by Thorrez
1610 days ago
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What you described is not what joshbuckler described. joshbuckler wants an algorithm that has the fewest words that take at least 6 guesses, and also among all algorithms that tie for that number of words that take at least 6 guesses, has the fewest number of words that take at least 5 guesses, and also among all the algorithms that tie for that number of words that take at least 6 and at least 5 guesses, has the fewest number of words that take at least 4 guesses, and also... The page you linked to doesn't seem to say it does that. |
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If you removed just 1 of those 2 words (leaving 2312 possible Wordles) all Wordles can be solved in max 4 guesses.
Ignoring the 2 scenarios where your first two "optimal" guesses are the Wordle
X% of the time you have 1 word left (guesses= 3)
Y% of the time you have 2 words left (guesses =3.5)
(100-Y-X)% of the time there is a pool of words left, requiring one more "optimal" guess
And either your optimal guess #3 is the Wordle Z% of the time (guess=3) or there is now one word remaining (guess ≈ 4)
Where Z in this case is just a function of the number of words remaining (ie Y is just a special case of Z)
So you want to maximize X and the weighted average of Z across pools.
By sheer brute force you can do so
1) for each 2 guess combo 2) discard any "suboptimal" combo where for any remaining response state word pool there is no optimal guess #3 (ie not possible to definitively guess in 4 guesses) 3) calculate avg remaining guesses 4) identify optimal word 1 (minimum of sum of step 2 per first guess) 5) within combos with word 1, identify optimal guess 2 for each response state
And the weighted average of step 3 for these combos is the global minimum for Wordle.
I believe this is the algorithm you're after? In this case, we're making a first guess that maximizes the chance we will get 3 guesses instead of 4.
Per this algorithm, optimal guess 1 is RAVED.