5 ms·

As a point of curiosity, what do you mean by optimal? To me, I can see at least two (potentially different) cost functions that a Wordle strategy can aim to mi

by hgibbs 5y ago

As a point of curiosity, what do you mean by optimal? To me, I can see at least two (potentially different) cost functions that a Wordle strategy can aim to minimise. Firstly, you can try to minimise the expected number of guesses, and secondly you can try to minimise the expected number of guesses conditional on never losing the game. In principle you could optimise for the first but not the second by allowing a small list of hard words to take > 6 guesses on average. Part of my point is the lack of an absolute definition of optimal: strategies are only optimal up to the specified coat function.

hervature 5y ago
Funnily enough, I think neither of those notions are the correct thing to optimize. For me, you want to minimize the maximum number of guesses. This leads to an equilibrium. Otherwise, your opponent can just abuse your strategy.
- pxx 5y ago
  Isn't that just #2, but changing the definition of losing to be the least number of rounds where you can always win (which for Wordle is 5)? These small changes to the definition don't seem to change ev (or the code needed to generate it) that much. If you don't care about optimizing the EV at all, the greedy tree already has the property you're interested in.
- eru 5y ago
  What do you mean by opponent? The entity choosing the word in a game like Wordle or Hang Man is more like the dungeon master in a game of D&D than an opponent in the traditional sense? They are there to help you have fun.
- hgibbs 5y ago
  My point is exactly that though - the choice of cost function is subjective. There is no best cost function, just a mapping from costs functions to optimal strategies.
- noxvilleza 5y ago
  Adversarial strategies are another type of problem entirely (as in, if you're playing Word Mastermind). Wordle is a 1-player game, so I find hgibbs's comment totally reasonable. I would think that 'optimal' is more accurately his second definition - minimizing avg guesses but always winning (as in, losing is +inf guesses).
- hnkain 5y ago
  The (exact) optimal EV-minimizing strategies (assuming each target word is equally likely) always solve in 5 of fewer guesses in normal mode, and always solve in 6 guesses in hard mode. If you wish to guarantee 5 guesses in hard mode, there's a different (optimal) strategy with slightly higher EV. See: http://sonorouschocolate.com/notes/index.php?title=The_best_strategies_for_Wordle http://sonorouschocolate.com/notes/index.php?title=The_best_... I don't think a maximum-number-of-guesses optimizer leads to an equilibrium (if you mean Nash equilibrium), assuming you're playing the game where the score of the game for the setter is the number of guesses. In particular, if the solver's strategy is deterministic, the setter will always pick one of the words that needs 5 guesses. There's no reason to think the nash equilibrium can be easily found.
tshaddox 5y ago
If two players are competing in a single round of wordle, then surely the player who guesses the correct word in the fewer number of guesses wins, right? Then it seems natural to say that the better of two players is the one with the most wins after playing all possible rounds. The only problem is that I think this definition leads to the possibility of non-transitivity, where A beats B, B beats C, and C beats A.
furyofantares 5y ago
They specified that in the post, they were able to constrain to a strategy that guarantees 6 or fewer and then optimize for EV to get 3.55, but could get down to 3.52 by allowing some small number of words to take more than 6. I would be interested to know if 6 is the lowest you can constrain it. Can you guarantee a solution in 5, and if so what is the best EV with that constraint?
- npinsker 5y ago
  Yes, you can -- I believe the EV was around 3.47 for that. The "greedy" strategy (minimizing the worst-case size of the largest set) also ends up using at most 5 words. Perhaps unsurprisingly, you can't guarantee at most 4 words.
  - Sesse__ 5y ago
    You can get EV of 3.46393 with a guaranteed maximum of five tries. (I don't know whether this bound is tight.)
    - pxx 5y ago
      This bound is not tight. http://sonorouschocolate.com/notes/index.php?title=The_best_strategies_for_Wordle http://sonorouschocolate.com/notes/index.php?title=The_best_...
      - Sesse__ 5y ago
        Good to know. Another possible question is: What is the best strategy minimizing (at any point) _first_ the maximum number of tries, and second the EV (average number of tries over all possible words)? That's subtly different from either answer.
svat 5y ago
IMO, never losing the game is the bare minimum for any reasonable cost function: never take > 6 guesses. Beyond that, you can have different optimization functions based on whether you're trying to minimize the worst-case depth of the decision tree (number of guesses) for each position, or the average depth, or some other reasonable function of the distribution over number of guesses. Although you could have arbitrarily many cost functions, a fairly general class is the following: pick some nonnegative constants (w1, w2, w3, w4, w5, w6, w7), and for a certain strategy, define the cost as: w1*n1 + w2*n2 + w3*n3 + w4*n4 + w5*n5 + w6*n6 + w7*n7 where nk (for 1≤k≤6) is the number of hidden (solution) words for which the strategy takes n guesses, and n7 is the number of words for which the strategy loses the game. Then, • Setting w7 infinite / very high is a way of encoding the condition that you not lose the game. (You can set it finite/small if you don't mind occasionally losing the game for some reason!) • Setting w1 = 1, w2 = 2, …, w6 = 6 will simply minimize the average number of guesses, • Having them grow exponentially, e.g. setting wk = c^(k-1), where c is some constant larger than 12972 (the number of acceptable guess words) will make sure that the minimal-cost strategy first minimizes the maximum number of guesses, then break ties by having the fewest words take that many guesses, and so on.
- tshaddox 5y ago
  > IMO, never losing the game is the bare minimum for any reasonable cost function: never take > 6 guesses. I wouldn’t think so. If I get 99.9% of rounds correct with 3 guesses (and fail to find a solution to 0.1% of the rounds), and you get 100% of rounds with 6 guesses, I’d say I’ve soundly defeated you 99.9% of the time.
  - svat 5y ago
    Right, that point of view makes sense if you imagine two players separately playing Wordle and comparing their results, and indeed that's what the OP may have been thinking. But as far as I know, Wordle is primarily a game for one player to play against the computer, where eventually getting to the word (or not) is the most prominent outcome (the only notion of "win" or "defeat" in the game itself), and solving it in fewer guesses is just a bonus. But sure, if you don't care about occasionally losing, you can just set an appropriate weight for w7 within the same family of cost functions.

by hgibbs 5y ago

hervature 5y ago

Funnily enough, I think neither of those notions are the correct thing to optimize. For me, you want to minimize the maximum number of guesses. This leads to an equilibrium. Otherwise, your opponent can just abuse your strategy.

pxx 5y ago
Isn't that just #2, but changing the definition of losing to be the least number of rounds where you can always win (which for Wordle is 5)? These small changes to the definition don't seem to change ev (or the code needed to generate it) that much. If you don't care about optimizing the EV at all, the greedy tree already has the property you're interested in.
eru 5y ago
What do you mean by opponent? The entity choosing the word in a game like Wordle or Hang Man is more like the dungeon master in a game of D&D than an opponent in the traditional sense? They are there to help you have fun.
hgibbs 5y ago
My point is exactly that though - the choice of cost function is subjective. There is no best cost function, just a mapping from costs functions to optimal strategies.
noxvilleza 5y ago
Adversarial strategies are another type of problem entirely (as in, if you're playing Word Mastermind). Wordle is a 1-player game, so I find hgibbs's comment totally reasonable. I would think that 'optimal' is more accurately his second definition - minimizing avg guesses but always winning (as in, losing is +inf guesses).
hnkain 5y ago
The (exact) optimal EV-minimizing strategies (assuming each target word is equally likely) always solve in 5 of fewer guesses in normal mode, and always solve in 6 guesses in hard mode. If you wish to guarantee 5 guesses in hard mode, there's a different (optimal) strategy with slightly higher EV. See: http://sonorouschocolate.com/notes/index.php?title=The_best_strategies_for_Wordle http://sonorouschocolate.com/notes/index.php?title=The_best_... I don't think a maximum-number-of-guesses optimizer leads to an equilibrium (if you mean Nash equilibrium), assuming you're playing the game where the score of the game for the setter is the number of guesses. In particular, if the solver's strategy is deterministic, the setter will always pick one of the words that needs 5 guesses. There's no reason to think the nash equilibrium can be easily found.

tshaddox 5y ago

If two players are competing in a single round of wordle, then surely the player who guesses the correct word in the fewer number of guesses wins, right? Then it seems natural to say that the better of two players is the one with the most wins after playing all possible rounds. The only problem is that I think this definition leads to the possibility of non-transitivity, where A beats B, B beats C, and C beats A.

furyofantares 5y ago

They specified that in the post, they were able to constrain to a strategy that guarantees 6 or fewer and then optimize for EV to get 3.55, but could get down to 3.52 by allowing some small number of words to take more than 6. I would be interested to know if 6 is the lowest you can constrain it. Can you guarantee a solution in 5, and if so what is the best EV with that constraint?

npinsker 5y ago
Yes, you can -- I believe the EV was around 3.47 for that. The "greedy" strategy (minimizing the worst-case size of the largest set) also ends up using at most 5 words. Perhaps unsurprisingly, you can't guarantee at most 4 words.
- Sesse__ 5y ago
  You can get EV of 3.46393 with a guaranteed maximum of five tries. (I don't know whether this bound is tight.)
  - pxx 5y ago
    This bound is not tight. http://sonorouschocolate.com/notes/index.php?title=The_best_strategies_for_Wordle http://sonorouschocolate.com/notes/index.php?title=The_best_...
    - Sesse__ 5y ago
      Good to know. Another possible question is: What is the best strategy minimizing (at any point) _first_ the maximum number of tries, and second the EV (average number of tries over all possible words)? That's subtly different from either answer.

svat 5y ago

IMO, never losing the game is the bare minimum for any reasonable cost function: never take > 6 guesses. Beyond that, you can have different optimization functions based on whether you're trying to minimize the worst-case depth of the decision tree (number of guesses) for each position, or the average depth, or some other reasonable function of the distribution over number of guesses. Although you could have arbitrarily many cost functions, a fairly general class is the following: pick some nonnegative constants (w1, w2, w3, w4, w5, w6, w7), and for a certain strategy, define the cost as: w1*n1 + w2*n2 + w3*n3 + w4*n4 + w5*n5 + w6*n6 + w7*n7 where nk (for 1≤k≤6) is the number of hidden (solution) words for which the strategy takes n guesses, and n7 is the number of words for which the strategy loses the game. Then, • Setting w7 infinite / very high is a way of encoding the condition that you not lose the game. (You can set it finite/small if you don't mind occasionally losing the game for some reason!) • Setting w1 = 1, w2 = 2, …, w6 = 6 will simply minimize the average number of guesses, • Having them grow exponentially, e.g. setting wk = c^(k-1), where c is some constant larger than 12972 (the number of acceptable guess words) will make sure that the minimal-cost strategy first minimizes the maximum number of guesses, then break ties by having the fewest words take that many guesses, and so on.

tshaddox 5y ago
> IMO, never losing the game is the bare minimum for any reasonable cost function: never take > 6 guesses. I wouldn’t think so. If I get 99.9% of rounds correct with 3 guesses (and fail to find a solution to 0.1% of the rounds), and you get 100% of rounds with 6 guesses, I’d say I’ve soundly defeated you 99.9% of the time.
- svat 5y ago
  Right, that point of view makes sense if you imagine two players separately playing Wordle and comparing their results, and indeed that's what the OP may have been thinking. But as far as I know, Wordle is primarily a game for one player to play against the computer, where eventually getting to the word (or not) is the most prominent outcome (the only notion of "win" or "defeat" in the game itself), and solving it in fewer guesses is just a bonus. But sure, if you don't care about occasionally losing, you can just set an appropriate weight for w7 within the same family of cost functions.