Wordle Lovers - Play the NYT Daily Game

[jack action]

Wordle 1,255 3/6

[PARSE] 47 left
[BITOU] filter - 1 left
[BROWN]

If the solution to a daily puzzle is ever CLONS, BIGHA, or WAKFS

I don't expect them to be solutions, there are only filter words to eliminate the most common letters.

If not then it is unclear how you have defined $x_i$.

Example:

For the pattern I have one possible solution - the actual answer to the puzzle - so $x = 1$ thus $\frac{1}{x} = 1$ for that pattern.

Say for the pattern I have 47 possible solutions for that same seed word - as it did today for my seed word - then $x = 47$ thus $\frac{1}{x} = 0.0213$ for that pattern.

If a pattern gives you 0 possible solutions then all known solutions - say 1000 of them - are still possible, thus $\frac{1}{x} = 0.001$ for that pattern.

Hence, for that word, with those 3 patterns only, the sum is $1.0223$. With 238 patterns, the "value" (i.e. $\sum \frac{1}{x_i}$) for a possible seed word, with a list of 1000 possible solutions, should lie somewhere between 1.237 and 238.

Here are the head and tail of my list of possible solutions (1063 words) as calculated today:

56.87877176330782981142 parse
56.71509589392361046686 bleat
55.43888935213718560553 shale
55.01052218765906827879 cleat
54.88989328229117523048 baron
53.81442942634878178095 bloat
53.79423003532885470371 leach
53.22460980294216326536 pulse
52.84561657818863628216 pried
52.75236703440116855539 purse

[...]

16.26406338557719068169 banjo
16.16046932891694134750 buxom
16.15133115383457283515 quack
15.96338486087747459938 juicy
15.92016037764995554693 issue
15.62426833882845554579 kitty
15.29790757004372842349 squib
15.22805152679214042192 jewel
14.10969827716669384071 puffy
12.53200693341326199116 fuzzy

There may be better seed words that are not possible solutions, but it is an additional 10,000+ words to check, and a list for 1000 words already takes ~40 minutes, so ...

 

[kuruman]

Wordle 1,255 4/6

 

[gmax137]

ugh, I haven't lost a game in many months
Wordle 1,255 X/6





A mistake, this has a letter I already knew to be unused

 

[Orodruin]

Example:

For the pattern I have one possible solution - the actual answer to the puzzle - so $x = 1$ thus $\frac{1}{x} = 1$ for that pattern.

Ok.

Say for the pattern I have 47 possible solutions for that same seed word - as it did today for my seed word - then $x = 47$ thus $\frac{1}{x} = 0.0213$ for that pattern.

Ok

If a pattern gives you 0 possible solutions then all known solutions - say 1000 of them - are still possible, thus $\frac{1}{x} = 0.001$ for that pattern.

But now the logic broke. You cannot have 1000 solutions satisfying the pattern at the same time as you have 0 solutions satisfying the pattern.

As per your previous two examples, if there are $x_i$ words giving that pattern, then you add $1/x_i$. If there are 0 words giving a specific pattern, then you should get a 1/0. Unless you choose to disregard that pattern as it will never show up? (It cannot possibly show if there are zero words giving the pattern)

 

[dwarde]

Wordle 1,255 4/6

 

[jack action]

But now the logic broke. You cannot have 1000 solutions satisfying the pattern at the same time as you have 0 solutions satisfying the pattern.

If no solutions satisfy the pattern, then all solutions are still possible. No? The "good" patterns only restrict it even more. It is not about how many solutions satisfy the pattern but more about how many solutions are possible.

You have to remember also that my list is not complete, as discussed earlier, here and here. So a pattern not finding a solution might be misleading.

In any case, I just want to have a number a lot smaller than the other ones. Zero might have been more realistic (because these patterns will never show up), but I don't think it would change the result significantly. For example, the word ABBOT value is 34.57 with 169 patterns having no hit. Setting these values to 0 instead of 1/1063 just removes 0.16 from the final value.

 

[Orodruin]

If no solutions satisfy the pattern, then all solutions are still possible. No?

No. If you would get a pattern that no solutions satisfy, then none of the solutions are possible anymore. This could technically happen if the solution is a word that satisfies the pattern, but is not in your list of possible solutions.

You have to remember also that my list is not complete, as discussed earlier, here and here. So a pattern not finding a solution might be misleading.

Exactly, all I am saying that - as you presented your model - it would mean division by zero. If your actual model is adding 1/1000 (or zero) for those patterns that give x = 0 but 1/x for all others, then that is a better model in the sense that it won’t contain 1/0s.

Still, the model seems improvable. Ideally you should make some information argument and weight by the probability of falling into each pattern.

The information left to deduce would typically be logarithmic in the number of words left so I would instead use $s_i =\log_2(x_i)$ as the remaining entropy for a given pattern. The probability of falling into that pattern is also proportional to $x_i$ so I would weight the sum by that to make a score based on expected entropy remaining
$$
S = \sum_i x_i s_i.
$$
Larger entropy is bad so you would want to minimise this score.

If you want to get really fancy with your own hypothesis you should also give each word a separate weight based on its probability of being chosen such that $x_i$ is no longer the number of words in each pattern but the sum of their weights.

 

[Orodruin]

To put the above into context with an example.

Consider the comparison between two possible seeds. Both seeds are possible solutions but the first gives 10 additional patterns with a single remaining word and one pattern with the rest. The second seed gives 20 patterns with 50 remaining words each.

Your comparator would give roughly 11 in score for the first, but 20/50 + 1 = 1.4 for the second. However, the second is generally going to be a better seed in most cases. Sure, if you are lucky you will get a 1% chance to get the solution on the next guess with the first seed, but in 99% of the cases you are basically none the wiser than before the seed. For the second seed, you guarantee at most 50 remaining words.

In the comparator of post 5817 the first option would score around $1000\log_2(1000) = 10000$ whereas the second would score $20\cdot 50 \log_2(50) = 1000 \log_2(50) = 5600$.

 

[ChaseLess]

Wordle 1,256 5/6






#GuessTheLogo #71
®
#IconIdentifier
https://guessthelogo.wtf/p/71

 

[jack action]

I've been thinking about evaluating the ideal seed word for a long time. @OmCheeto shared his method and got me thinking.

Intuitively, what I don't like in his method is that a correctly positioned letter weighs as much as one present in the word.

This idea of evaluating the patterns came to me. I'm still trying to figure out a proper way to weigh them.

Your comparator would give roughly 11 in score for the first, but 20/50 + 1 = 1.4 for the second. However, the second is generally going to be a better seed in most cases. Sure, if you are lucky you will get a 1% chance to get the solution on the next guess with the first seed, but in 99% of the cases you are basically none the wiser than before the seed. For the second seed, you guarantee at most 50 remaining words.

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?

In the comparator of post 5817 the first option would score around $1000\log_2(1000) = 10000$ whereas the second would score $20\cdot 50 \log_2(50) = 1000 \log_2(50) = 5600$.

But, for the first option, isn't it $10\cdot1\log_2(1) = 0$? Which gives us the same problem.

So I gave your method a try and the results are (comparing with the previous post):

4588.20531194210523224391 arise
4675.54314301487405749215 saner
4676.11281919344526423557 arose
4721.76625519139482748979 least
4728.89845461676390824165 stare
4758.30369784855998400051 later
4776.94838870606320728379 parse
4787.55291969866344838355 alert
4830.69125524560430809960 roast
4845.61333562902289041874 siren

[...]

8803.49726239582453041463 kappa
8814.80630391203441924483 femme
8939.14471626754972523066 slyly
9069.06715927188421580731 ninny
9237.85859957449918736648 eerie
10004.90304370215078150383 mamma
10198.61998557413705737079 geese
11435.74223379053281061499 emcee
11507.15016864005646684804 melee
11625.47238092071591300699 tepee

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.

 

[Orodruin]

I've been thinking about evaluating the ideal seed word for a long time. @OmCheeto shared his method and got me thinking.

Indeed, the correct measure is remaining entropy. The aim of the game is to decrease entropy to zero.

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.

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.

So I gave your method a try and the results are (comparing with the previous post):

4588.20531194210523224391 arise
4675.54314301487405749215 saner
4676.11281919344526423557 arose
4721.76625519139482748979 least
4728.89845461676390824165 stare
4758.30369784855998400051 later
4776.94838870606320728379 parse
4787.55291969866344838355 alert
4830.69125524560430809960 roast
4845.61333562902289041874 siren

[...]

8803.49726239582453041463 kappa
8814.80630391203441924483 femme
8939.14471626754972523066 slyly
9069.06715927188421580731 ninny
9237.85859957449918736648 eerie
10004.90304370215078150383 mamma
10198.61998557413705737079 geese
11435.74223379053281061499 emcee
11507.15016864005646684804 melee
11625.47238092071591300699 tepee

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.

 

[OmCheeto]

Wordle 1,255 3/6

 

[Orodruin]

I should also say:

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.

 

[kuruman]

Wordle 1,256 4/6

 

[jack action]

Wordle 1,256 4/6

[ARISE] 62 left
[LYTED] filter - 4 left: MOTIF OPTIC PITCH WITCH
[PITCH] filter - 1 left
[WITCH]

 

[Orodruin]

[WITCH]

No guess based on likelihood? Could have had it in 3?

 

[gmax137]

Wordle 1,256 6/6




another dumb mistake
filter


I need to slow down and think more before leaping

 

[Orodruin]

another dumb mistake
filter


I need to slow down and think more before leaping

This is more or less what happened last time I lost my streak except I managed to misspell the solution… 581 days ago …

 

[dwarde]

Wordle 1,256 6/6




another dumb mistake
filter


I need to slow down and think more before leaping

Me too.

 

[dwarde]

Wordle 1,256 6/6

 

[jack action]

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.)

 

[gmax137]

it would have been a very tough choice

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.

 

[OmCheeto]

Wordle 1,256 3/6

 

[Orodruin]

If I had done it, it would have probably been the same word, although it would have been a very tough choice.

Spoiler: In the news cycle

 

[jack action]

Spoiler: In the news cycle

View attachment 353903

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.

 

[Orodruin]

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.