I've been thinking about evaluating the ideal seed word for a long time. @OmCheetoshared his method and got me thinking.
Indeed, the correct measure is remaining entropy. The aim of the game is to decrease entropy to zero.
jack action said:
This idea of evaluating the patterns came to me. I'm still trying to figure out a proper way to weigh them.
This is a question I asked myself as well. How to weigh a seed word that gives only a single valid pattern with 1 possible solution with another giving 150 valid patterns with 400 possible solutions each?
Expected remaining entropy after the seed seems like a reasonable idea.
jack action said:
But, for the first option, isn't it ##10\cdot1\log_2(1) = 0##? Which gives us the same problem.
It is ##10\log_2(1) + 1000 \log_2(1000)##. It is never an issue because I am not dividing by those zeros. A zero in the sum indicates a pattern where the puzzle is solved - no entropy left in that pattern.
jack action said:
So I gave your method a try and the results are (comparing with the previous post):
I like this one better because it has more popular letters in the first words and all the last words have very few letters, which seems logical. The list also looks more like the one from @OmCheeto . The one in the previous post had a lot of b's in the first words which seems odd.
If you want to normalize the comparator, it should also be divided by the total number of remaining words. This will give the expected entropy after using the seed. This would be more relevant to compare seeds between days as the normalization would do nothing for the order on a particular day (all comparators receive the same normalising factor). As the base entropy keeps decreasing as words are used up, the expected entropy after one guess should also generally decrease.
No guess based on likelihood? Could have had it in 3?
If I had done it, it would have probably been the same word, although it would have been a very tough choice. But the true reason I selected my third guess is that this word guaranteed the answer in 4. (Now that I look at it again, I realize that OPTIC and WITCH would also have done the job.)
yeah, I had jerk, DITCH, HITCH, PITCH, WITCH. So I eliminated jerk (following @jack action 's human input theory) and went with PITCH. But my second guess had been STRAP, so I shouldn't have even thought PITCH to be viable. A dumb mistake. Instead of PITCH I should have used a filter maybe DWEEB. Hindsight, you know, lol.
That's the downside of being a recluse. I don't pay for movies anymore, so I may catch that one only in a few years, on free TV ... if it is successful enough.
So I couldn't find anything positive about that word unless we were near Halloween.
That's the downside of being a recluse. I don't pay for movies anymore, so I may catch that one only in a few years, on free TV ... if it is successful enough.
So I couldn't find anything positive about that word unless we were near Halloween.
I mean, it is an a posteriori construction for me as well. I also went PITCH for the third guess. I probably should have thought more about it because it was not an actual filter for me. I was risking a 5.
This is the first time I had to exclude half the alphabet before getting any letter in the correct place only to get all 5 simultaneously in the end. The incorrectly placed letter in guess 1 position 2 is the same as in guess 3 position 5. This illustrates the usefulness of filters with "orthogonal" letters.
I used up most of the alphabet but I finally got there.
I think these are the types of wordle games that turn people to the dark side and start using spreadsheets and computer programs to figure out ways of battling such beasts.