How good can you be at Codenames without knowing any words?
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Re: How good can you be at Codenames without knowing any words?
#2Re: How good can you be at Codenames without knowing any words?
#3On the topic of word game recommendations, I’ve been playing So Clover recently and highly recommend it. It’s a cooperative game and has almost no downtime (all players write clues at the same time, then all players guess together, so the possible downtime is the difference between the fastest clue writer and the slowest clue writer). https://www.rprod.com/en/games/so-clover
What I mean by there not being a game to it is that there's no winning or losing, or even evaluating success. Everyone gives their best clues (by some metric; sometimes people optimize for fun rather than clarity), you solve the boards and get a score, but the score is totally meaningless. It can't even be compared across sessions.
It turns out to be quite hard to add a game into it. I thought about the problem for a year before deciding that there was no viable way to do it within the constraints of the physical game, and it'd need to be online. For the last few weeks I've been working on an async online word-game inspired by So Clover, where one of the tihngs I wanted to do was to add an actual game (as a competitive mode), and so far the feedback from my playtesters has been that they're not interested in any of those plans, and just want to use it as a chill word game sandbox and share puzzle links over WhatsApp. (The site isn't ready for public consumption, but if anyone here wants to try the game out with some friends, hit me up with an email and I'll send you a link. Contact details in HN profile.)
Re: How good can you be at Codenames without knowing any words?
#4But it's very similar to chess, where positional sense is crucial
Re: How good can you be at Codenames without knowing any words?
#5But it does mean that defeating the bot would only mean creating a custom grid, which sounds very practical virtually, and possible (but harder) in person.
Re: How good can you be at Codenames without knowing any words?
#6On the topic of word game recommendations, I’ve been playing So Clover recently and highly recommend it. It’s a cooperative game and has almost no downtime (all players write clues at the same time, then all players guess together, so the possible downtime is the difference between the fastest clue writer and the slowest clue writer). https://www.rprod.com/en/games/so-clover
+1, but one thing about So Clover is that not really any game in it. It's more of an activity. A very engaging activity, and one that has worked with every group I've tried it on, gamers and non-gamers alike, and my current #1 choice if I was asked to bring some games to some non-gaming social event. What I mean by there not being a game to it is that there's no winning or losing, or even evaluating success. Everyone…
Re: How good can you be at Codenames without knowing any words?
#7After some failed experiments - it performed worse than I thought it will - I've googled the subject, and... it turns out there's a whole paper about ML and codenames :)
https://arxiv.org/abs/2105.05885 (Playing Codenames with Language Graphs and Word Embeddings) - fun to read
Re: How good can you be at Codenames without knowing any words?
#8For completeness, I wondered how many cards would be required to give the complete set of patterns.
If we number the positions in the 5x5 grid such that the top row has positions 1-5, the second row has 6-10 and so on, the grid positions can be converted to a sequence and we can use the permutation formula to find the number of arrangements. To account for rotations, we can divide the final value by 4 since every arrangement can be rotated and is therefore valid.
Of the 25 cards, there are 7 white, 8 red, 8 blue, 1 black and 1 double agent that can be red or blue, also deciding which team goes first. We can treat this final card as one of a kind, then double the formula output to account for cases where it is swapped to the other team.
Permutations of a multiset has a standard formula [0] that calculates a result from these values (rolling in the double agent factor of 2 and rotation division factor of 4):
25! / (7!8!8!1!1! * 2) = 946,551,177,000
(edit: as pointed out, this is 9 times too large as the double agent can indistinguishably replace each of the other 8 cards - a corrected value is 105,172,353,000)
This is (edit: still) more layout cards than have ever been printed across all production runs of Codenames, and would probably not fit into the current box size.
[0] https://en.wikipedia.org/wiki/Multinomial_theorem#Number_of_...
Re: How good can you be at Codenames without knowing any words?
#9This is an interesting view of how random number based security is compromised for economic practicalities (though the meanings of "security" and "compromised" might be overstretched here). For completeness, I wondered how many cards would be required to give the complete set of patterns. If we number the positions in the 5x5 grid such that the top row has positions 1-5, the second row has 6-10 and so on, the grid po…
Note that this is only approximately correct, since some layouts will have nontrivial symmetries. (Edit: Actually no they won't due to parity reasons! Oops. So that step is exact after all.)
Edit: Actually, this isn't correct either, and in a more serious way:
> Of the 25 cards, there are 7 white, 8 red, 8 blue, 1 black and 1 double agent that can be red or blue, also deciding which team goes first. We can treat this final card as one of a kind, then double the formula output to account for cases where it is swapped to the other team.
On the layout cards, nothing distinguishes the double agent -- the double agent is purely a matter of representation, it's not part of the actual layout. So doing things this way will give you a number that's too large by a factor of about 8 or 9.
What you want to do here (ignoring rotations) is take 25!/(9!8!7!) to get the count with 9 red and 8 blue, then double it to include the count with 8 red and 9 blue, then divide by 4 to account for rotation (contrary to what I said earlier no need to account for symmetries because the numbers mean that none of the layouts can be rotationally symmetric), so you get a total of
25!/(9!8!7!2) = 105,172,353,000
So yeah about 9 times smaller than what you wrote. For what it's worth, anyway. :P
Re: How good can you be at Codenames without knowing any words?
#10The methods in the post are interesting, but technically against the rules of the game.
EDIT: Though I didn't figure out whether he means "clues about the position" or "clues about the words, until you can open enough cards to narrow down to one memorized position".