The game Wordle has gone pretty viral in the
last month or two, and never one to overlook an opportunity for a math lesson, it occurs to
me that this game makes for a very good central example in a lesson about information theory,
and in particular a topic known as entropy. You see like a lot of people I got kind of sucked
into the puzzle, and like a lot of programmers I also got sucked into trying to write an algorithm
that would play the game as optimally as it could. What I thought I'd do here is just talk through
with you some of my process in that, and explain some of the math that went into it, since the
whole algorithm centers on this idea of entropy. First things first, in case you haven't heard of
it, what is Wordle and to kill two birds with one stone here while we go through the rules of the
game let me also preview where we're going with this, which is to develop a little algorithm
that will basically play the game for us. I haven't done today's Wordle, this is
February 4th, and we'll see how the bot does. The goal of Wordle is to guess a mystery
five-letter word, and you're given six different chances to guess. For example, my
wordlebot suggests that I start with the guess "crane". Each time that you make a guess, you
get some information about how close your guess is to the true answer. Here the gray box is
telling me there's no c in the actual answer, the yellow box is telling me there is
an r but it's not in that position. The green box is telling me that the secret word
does have an a and it's in the third position. And then there's no 'n' and there's no 'e'. Let
me just go in and tell the wordlebot about that information...we started with "crane",
we got gray yellow green grey grey... Don't worry about all the data that it's showing
right now, I'll explain that in due time. Its top suggestion for our second pick is "shtik".
Your guess does have to be an actual five-letter word, but as you'll see it's pretty liberal
with what it will actually let you guess. In this case we try stick and...all right!
Things are looking pretty good. We hit the 's' and the 'h', so we know the first three
letters, and we know that there's an 'r'. So it's going to be like s-h-a something r
or s-h-a-r-something. And it looks like the Wordle-bot knows that it's down to just two
possibilities, either "shard" or "sharp". I's kind of a toss-up between them at this
point, so I guess probably just because it's alphabetical it goes with shard, which...hooray!
It is the actual answer. So we got it in three. If you're wondering if that's any good, the way
I heard one person phrase it is that with Wordle, four is par and three is birdie,
which I think is a pretty apt analogy. You have to be consistently on your game to
be getting four but it's certainly not crazy. But when you get it in three, it just feels great. If you're down for it what I'd like to do here
is just talk through my thought process from the beginning for how I approach the wordlebot.
And like I said really it's an excuse for an information theory lesson, the main goal is to
explain what is information and what is entropy. My first thought in approaching this was to take
a look at the relative frequencies of different letters in the english language. I thought,
okay, is there an opening guess or an opening pair of guesses that hits a lot of these most
frequent letters. One that I was pretty fond of was doing "other" followed by "nails". The
thought is that if you hit a letter, you know you get a green or a yellow, that always feels
good, it feels like you're getting information. But in these cases even if you
don't hit and you always get greys, that's still giving you a lot of information,
since it's pretty rare to find a word that doesn't have any of these letters. But even
still that doesn't feel super systematic, because for example it does nothing to consider
the order of the letters. Why type "nails" when I could type "snail". Is it better to have
that s at the end? I'm not really sure. Now a friend of mine said that he
liked to open with the word "weary", which kind of surprised me because it has
some uncommon letters in there like the 'w' and the 'y'. But who knows, maybe that
is a better opener. Is there some kind of quantitative score that we can give to
judge the quality of a potential guess? To set up for the way that we're going to rank
possible guesses, let's go back and add a little clarity to how exactly the game is set up. There's
a list of words that it will allow you to enter, that are considered valid guesses, that's just
about 13,000 words long. But when you look at it there's a lot of really uncommon things
things like "aahed" or "aalii" and "aargh". The kind of words that bring about
family arguments in a game of Scrabble. But the vibe of the game is that the answer
is always going to be a decently common word, and in fact there's another list of around 2,300
words that are the possible answers. This is a human-curated list, I think specifically by the
game creator's girlfriend which is kind of fun. But what I would like to do, our challenge
for this project, is to see if we can write a program solving Wordle that doesn't
incorporate previous knowledge about this list. For one thing there's plenty of pretty common five
letter words that you won't find in that list, so it would be better to write a program that's
a little more resilient and would play Wordle against anyone, not just what happens to be
the official website. And also, the reason that we know what this list of possible answers
is is because it's visible in the source code, but the way that it's visible in the source code
is in the specific order in which answers come up from day to day, so you could always just look up
what tomorrow's answer will be. So clearly there's some sense in which using the list is cheating,
and what makes for a more interesting puzzle and a richer information theory lesson is to instead
use some more universal data, like relative word frequencies in general, to capture this intuition
of having a preference for more common words. So! Of these 13,000 possibilities, how
should we choose the opening guess? For example if my friend proposes "weary", how
should we analyze its quality? Well the reason he said he likes that unlikely 'w' is that he
likes the long shot nature of just how good it feels if you do hit that 'w'. For example if the
first pattern revealed was something like this, then it turns out there are only 58 words in
this giant lexicon that match that pattern, so that's a huge reduction from 13,000. But the
flip side of that, of course, is that it's very uncommon to get a pattern like this. Specifically,
if each word was equally likely to be the answer, the probability of hitting this pattern
would be 58 divided by around 13,000. Of course, they're not equally likely to be
answers, most of these are very obscure and even questionable words, but at least for our
first pass at all of this let's assume that they're all equally likely, then refine that a
bit later. The point is the pattern with a lot of information is by its very nature unlikely to
occur. In fact what it means to be informative is that it's unlikely. A much more
probable pattern to see with this opening would be something like this, where of course
there's not a 'w' in it, maybe there's an 'e' and maybe there's no 'a', there's no 'r', and
there's no 'y'. In this case there are 1,400 possible matches. So if all were equally likely,
it works out to be a probability of about 11% that this is the pattern you would see. So the most
likely outcomes are also the least informative. To get a more global view here, let me show
you the full distribution of probabilities across all of the different
patterns that you might see. Each bar that you're looking at corresponds to a
possible pattern of colors that could be revealed, of which there are 3^5 possibilities.
And they're organized from left to right, most common to least common. So the most common
possibility here is that you get all grays, that happens about 14% of the time. What you're
hoping for when you make a guess is that you end up somewhere out in this long tail, like over
here where there's only 18 possibilities for what matches this pattern, that evidently look
like this. Or if we venture a little farther to the left...you know maybe we go all the way
over here...okay here's a good puzzle for you. What are the three words in the english language
that start with a 'w' end with a 'y' and have an 'r' somewhere in them? It turns out the answers
are... let's see... "wordy" "wormy" and "wrily". To judge how good this word is overall, we want
some kind of measure of the expected amount of information that you're going to get from this
distribution. If we go through each pattern and we multiply its probability of occurring times
something that measures how informative it is, that can maybe give us an objective score. Now
your first instinct for what that something should be might be the number of matches, you
know you want a lower average number of matches, but instead I'd like to use a more universal
measurement that we often ascribe to information, and one that will be more flexible
once we have a different probability assigned to each of these 13,000 words for
whether or not they're actually the answer. The standard unit of information is the bit,
which has a little bit of a funny formula, but it's really intuitive
if we just look at examples. If you have an observation that cuts
your space of possibilities in half, we say that it has one bit of information.
In our example the space of possibilities is all possible words, and it turns out about
half of the five letter words have an 's', a little less than that but about half. So that
observation would give you one bit of information. If instead a new fact chops down that space of
possibilities by a factor of four, we say that it has two bits of information. For example it turns
out about a quarter of these words have a 't'. If the observation cuts that space by a factor of
eight, we say it has three bits of information, and so on and so forth. Four bits cuts it into
a sixteenth, five bits cuts it into a 32nd. So now is when you might want to take a moment to
pause and ask for yourself, what is the formula for information, for the number of bits in
terms of the probability of an occurrence? Well, what we're saying here is basically that
when you take one half to the number of bits, that's the same thing as the probability, which
is the same thing as saying 2 to the power of the number of bits is 1 over the probability,
which rearranges further to saying the information is the log base 2 of 1 divided by the
probability. And sometimes you see this with one more rearrangement still, where the information
is the negative log base 2 of the probability. Expressed like this it can look a little bit
weird to the uninitiated, but it really is just the very intuitive idea of asking how many
times you've cut down your possibilities in half. Now if you're wondering, you know, I thought
we were just playing a fun word game why are logarithms entering the picture? One reason this
is a nicer unit is it just a lot easier to talk about very unlikely events. Much easier to say
that an observation has 20 bits of information than it is to say that the probability
of such and such occurring is 0.00000095. But a more substantive reason that this
logarithmic expression turned out to be a very useful addition to the theory of probability
is the way that information adds together. For example if one observation gives you two bits
of information, cutting your space down by four, and then a second observation, like your second
guess in Wordle, gives you another three bits of information, chopping you down further by
another factor of eight, the two together give you five bits of information. In the same
way that probabilities like to multiply, Information likes to add. So as soon as we're in
the realm of something like an expected value, where we're adding a bunch of numbers up,
the logs make it a lot nicer to deal with. Let's go back to our distribution for weary
and add another little tracker on here showing us how much information
there is for each pattern. The main thing I want you to notice
is that the higher the probability, as we get to those more likely patterns, the
lower the information, the fewer bits you gain. The way we measure the quality of this guess will
be to take the expected value of this information, where we go through each pattern, we say how
probable is it, and then we multiply that by how many bits of information do we get. And in the
example of weary, that turns out to be 4.9 bits. So on average, the information you get
from this opening guess is as good as chopping your space of possibilities
in half about five times. By contrast, an example of a guess with a higher expected
information value would be something like "slate". In this case you'll notice the distribution
looks a lot flatter, in particular the most probable occurrence of all grays only has about
a 6% chance of occurring. So at minimum you're getting, evidently, 3.9 bits of information.
But that's a minimum, more typically you'd get something better than that. And it turns out
when you crunch the numbers on this one and you add up all of the relevant terms, the average
information is about 5.8. So in contrast with weary your space of possibilities will be about
half as big after this first guess, on average. There's actually a fun story about the name for
this expected value of information quantity. You see information theory was developed by
Claude Shannon, who was working at Bell labs in the 1940s. He was talking about some of his
yet-to-be-published ideas with John von Neumann, who was this intellectual giant of the time,
a very prominent in math and physics and the beginnings of what was becoming computer
science. And when he mentioned that he didn't really have a good name for this
expected value of information quantity, von Neumann supposedly said, so the story
goes, "well you should call it Entropy, and for two reasons. In the first place your
uncertainty function has been used in statistical mechanics under that name, so it already has a
name. And in the second place, and more important, nobody knows what entropy really is, so in
a debate you'll always have the advantage." So if the name seems a little bit mysterious,
and if this story is to be believed, that's kind of by design. Also, if you're
wondering about its relation to all of that second law of thermodynamics stuff from physics, there
definitely is a connection, but in its origins Shannon was just dealing with pure probability
theory. And for our purposes here, when I use the word entropy, I just want you to think the
expected information value of a particular guess. You can think of entropy as
measuring two things simultaneously. The first one is how flat is the distribution.
The closer a distribution is to uniform, the higher that entropy will be. In our
case, where there are 3^5 total patterns, for a uniform distribution, observing any one
of them would have information log_2(3^5), which happens to be 7.92. So that is the
absolute maximum that you could possibly have for this entropy. But entropy is also kind of a
measure of how many possibilities there are in the first place. For example if you happen to have
some word where there's only 16 possible patterns, and each one is equally likely, this entropy, this
expected information, would be four bits. But if you have another word where there are 64 possible
patterns that could come up, and they're all equally likely, then the entropy would work out to
be six bits. So if you see some distribution out in the wild that has an entropy of six bits, it's
sort of like it's saying there's as much variation and uncertainty in what's about to happen
as if there were 64 equally likely outcomes. For my first pass at the worldbot,
I basically had it just do this. It goes through all of the different
possible guesses that you could have, all 13,000 words. It computes the entropy for
each one, or more specifically the entropy of the distribution across all patterns that you might
see for each one, and then it picks the highest, since that's the one that's likely to chop down
your space of possibilities as much as possible. And even though I've only been talking about the
first guess here it does the same thing for the next few guesses. For example, after you see
some pattern on that first guess, which would restrict you to a smaller number of possible words
based on what matches with that, you just play the same game with respect to that smaller set of
words. For a proposed second guess, you look at the distribution of all patterns that could
occur from that more restricted set of words. You search through all 13,000 possibilities,
and you find the one that maximizes that entropy To show you how this works in action let me
just pull up a little variant of Wordle that I wrote that shows the highlights of this
analysis in the margins. So after doing all its entropy calculations, on the right here it's
showing us which ones have the highest expected information. It turns out the top answer, at
least at the moment we'll refine this later, is "tares", which means...um...of
course, a vetch the most common vetch. Each time we make a guess here, where maybe I kind
of ignore its recommendations and go with slate, because I like slate, we can see how much
expected information it had. But then on the right of the word here it's showing
us how much actual information we got given this particular pattern. So here
it looks like we were a little unlucky. We were expected to get 5.8, but we happened
to get something with less than that. And then on the left side here it's showing us all of the
different possible words given where we are now. The blue bars are telling us how likely it
thinks each word is, so at the moment it's assuming each word is equally likely to
occur, but we'll refine that in a moment. And then this uncertainty measurement is
telling us the entropy of this distribution across the possible words, which right
now, because it's a uniform distribution, is just a needlessly complicated way
to count the number of possibilities. For example, if we were to take 2 to the power
of 13.66, that should be around the 13,000 possibilities. It' a little bit off here, but only
because I'm not showing all the decimal places. At the moment that might feel redundant, and like
it's overly complicating things, but you'll see why it's useful to have both numbers in a minute.
Here it looks like it's suggesting the highest entropy for our second guess is "ramin", which
again...just really doesn't feel like a word. So to take the moral high ground here I'm going
to go ahead and type in "rains". Again it looks like we were a little unlucky, we were expecting
4.3 bits and we only got 3.39 bits of information. So that takes us down to 55 possibilities.
And here maybe I'll just actually go with what it's suggesting, which is "kombu",
whatever that means. Okay! This is actually a good chance for a puzzle. It's telling us
this pattern gives us 4.78 bits of information, but over on the left before we see
that pattern there were 5.78 bits of uncertainty. So as a quiz for you, what does that
mean about the number of remaining possibilities? Well it means that we're reduced
down to 1 bit of uncertainty, which is the same thing as saying that there's
two possible answers, it's a 50/50 choice. And from here, because you and I know which
words are more common, we know that the answer should be "abyss". But as it's written right now
the program doesn't know that, so it just keeps going trying to gain as much information as it
can until there's only one possibility left, and then it guesses it. So obviously we need a
better endgame strategy, but let's say we call this version one of our Wordle solver and then we
go and run some simulations to see how it does. The way this is working is it's playing every
possible Wordle game, it's going through all of those 2,315 words that are the actual Wordle
answers, it's basically using that as a testing set, and with this naive method of not considering
how common a word is and just trying to maximize the information at each step along the way until
it gets down to one and only one choice, by the end of the simulation the average score works
out to be about 4.124. Which...you know it's not bad. To be honest I kind of expected to do worse.
But the people who play Wordle will tell you that they can usually get it in four. The real
challenge is to get as many in three as you can. It's a pretty big jump between the score four and
the score of three. The obvious low-hanging fruit here is to somehow incorporate whether or not a
word is common, and how exactly do we do that? The way I approached it is to get a list
of the relative frequencies for all of the words in the english language. I just used
Mathematica's word frequency data function, which itself pulls from the google books english
n-gram public dataset. And it's kind of fun to look at, for example if we sort it from the
most common words to the least common words, evidently these are the most common five letter
words in the english language. Or rather, "these" is the eighth most common. First is "which" after
which there's "there" and "their". "First" itself is not first but ninth, and it makes sense that
these other words could come about more often, where those after "first" are "after," "where",
and "those", being just a little bit less common. Now, in using this data to model how likely
each of these words is to be the final answer, it shouldn't just be proportional to the
frequency. Because for example "which" is given a score of 0.002 in this data set,
whereas the word "braid" is in some sense about a thousand times less likely. But both
of these are common enough words that they're almost certainly worth considering,
so we want more of a binary cutoff. The way I went about it is to imagine
taking this whole sorted list of words, and then arranging it on an x-axis, and then
applying the sigmoid function, which is the standard way to have a function whose output is
basically binary, it's either zero or it's one, but there's a smoothing in between
for that region of uncertainty. So essentially the probability that I'm assigning
to each word for being in the final list will be the value of the sigmoid function
above wherever it sits on the x-axis. Now obviously this depends on a few parameters,
for example how wide a space on the x-axis those words fill determines how gradually or
steeply we drop off from one to zero, and where we situate them left to right
determines the cut off. And to be honest the way I did this was kind of just licking my
finger and sticking it into the wind. I looked through the sorted list and tried to find a window
where when I looked at it, I figured about half of these words are more likely than not to be
the final answer. And I use that as the cutoff. Now once we have a distribution like this across
the words, it gives us another situation where entropy becomes this really useful measurement.
For example let's say we were playing a game and we start with my old openers which were "other"
and "nails", and we end up with a situation where there's four possible words that match it. And
let's say we consider them all equally likely. Let me ask you, what is the entropy of this
distribution? Well the information associated with each one of these possibilities is going
to be the log_2(4), since each one is 1/4, and that's 2. It's two bits of information,
4 possibilities, all very well and good. But! What if I told you that actually there
are more than four matches. In reality, when we look through the full word list, there
are 16 words that match it. But suppose our model puts a really low probability on those other
12 words of actually being the final answer, something like one in a thousand,
because they're really obscure. Now let me ask you, what is the
entropy of this distribution? If entropy was purely measuring the number of
matches here, then you might expect it to be something like the log_2(16), which would be 4.
Two more bits of uncertainty than we had before. But of course, the actual uncertainty is
not really that different from what we had before. Just because there's these 12 really
obscure words doesn't mean that it would be all that more surprising to learn that the
final answer is "charm", for example. So when you actually do the calculation here, and
you add up the probability of each occurrence times the corresponding information, what you get
is 2.11 bits. It's saying it's basically two bits, it's basically those four possibilities, but
there's a little more uncertainty because of all of those highly unlikely events. though if you did
learn them you'd get a ton of information from it. So zooming out, this is part of what makes
Wordle such a nice example for an information theory lesson. We have these two distinct feeling
applications for entropy, the first one telling us what's the expected information we'll get
from a given guess, and the second one saying can we measure the remaining uncertainty
among all of the words that are possible. And I should emphasize, in that first case where
we're looking at the expected information of a guess, once we have an unequal weighting to the
words, that affects the entropy calculation. For example let me pull up that same case we were
looking at earlier of the distribution associated with "weary", but this time using a non-uniform
distribution across all possible words. So let me see if I can find a part here
that illustrates it pretty well...uh okay, here, this is pretty good. Here we have two
adjacent patterns that are about equally likely but one of them, we're told, has
32 possible words that match it. And if we check what they are, these are those
32, which are all just very unlikely words. As you scan your eyes over them it's hard to find any
that feel like plausible answers. Maybe "yells"? But if we look at the neighboring pattern in
the distribution, which is considered just about as likely, we're told that it only has eight
possible matches. So a quarter as many matches, but it's about as likely. And when we pull
up those matches, we can see why. Some of these are actual plausible answers
like "wring" or "wrath" or "wraps". To illustrate how we incorporate all that, let
me pull up version 2 of the wordlebot here. There are two or three main differences from
the first one that we saw. First off, like I just said, the way that we're computing these
entropies, these expected values of information, is now using the more refined distributions across
the patterns that incorporates the probability that a given word would actually be the answer.
As it happens, "tares" is still number one, though the ones following are a bit different.
Second, when it ranks its top picks, it's now going to keep a model of the probability that each
word is the actual answer, and it'll incorporate that into its decision, which is easier to
see once we have a few guesses on the table. Again ignoring its recommendation, because
we can't let machines rule our lives... And I suppose I should mention another
thing different here is over on the left, that uncertainty value, that number of bits, is no
longer just redundant with the number of possible matches. Now if we pull it up, and you know,
we calculated, say, 2 to the 8.02, which would be a little above 256...I guess 259. What it's
saying is even though there are 526 total words that actually match this pattern, the amount of
uncertainty it has is more akin to what it would be if there were 259 equally likely outcomes. You
could think of it like this, it knows "borks" is not the answer same with "yortz" and "zoril" and
"zorus". So it's a little less uncertain than it was in the previous case, this number of bits will
be smaller. And if I keep playing the game, I'll refining this down with a couple guesses that are
apropos of what I would like to explain here... By the fourth guess, if you look over at its top
picks, you can see it's no longer just maximizing the entropy. At this point, there's technically
seven possibilities, but the only ones with a meaningful chance are "dorms" and "words", and
you can see it ranks choosing both of those above all of these other values that, strictly
speaking, would give more information. The very first time I did this I just added
up these two numbers to measure the quality of each guess, which actually worked better than
you might suspect. But it really didn't feel systematic. I'm sure there are other approaches
people could take, but here's the one I landed on. If we're considering the prospect of a next guess,
like in this case "words", what we really care about is the expected score of our game if we
do that. And to calculate that expected score, we say "what's the probability that 'words'
is the actual answer?", which at the moment it ascribes 58% to. So we say with a 58%
chance, our score in this game would be four, and then with the probability of one minus that
58 percent, our score will be more than that four. How much more? We don't know, but we can estimate
it based on how much uncertainty there's likely to be once we get to that point. Specifically, at
the moment there are 1.44 bits of uncertainty, if we guess "words" it's telling us
the expected information we'll get is 1.27 bits, so if we guess "words" this difference represents how much uncertainty
we're likely to be left with after that happens. What we need is some kind of function, which I'm
calling f here, that associates this uncertainty with an expected score. The way I went about
this was to just plot a bunch of the data from previous games based on version one of the bot, to
say "hey what was the actual score after various points with certain very measurable amounts of
uncertainty?" For example, these data points here that are sitting above a value that's
around 8.7 or so are saying "for some games, after a point at which there were 8.7 bits
of uncertainty, it took two guesses to get the final answer. For other games it took three
guesses, for other games it took four guesses." If we shift over to the left here, all the
points over zero are saying "whenever there are zero bits of uncertainty, which is
to say there's only one possibility, then the number of guesses required is
always just one", which is reassuring. Whenever there was one bit of uncertainty, meaning
it was essentially just down to two possibilities, then sometimes it required one more guess
sometimes it required two more guesses, and so on and so forth. Here, maybe a slightly
easier way to visualize this data is to bucket it together and take averages. For example,
this bar here is saying "among all the points where we had one bit of uncertainty, on average
the number of new guesses required was about 1.5." And the bar over here saying "among all of the
different games were at some point the uncertainty was a little above 4 bits, which is like
narrowing it down to 16 different possibilities, then on average it requires a little more
than two guesses from that point forward". And from here I just did a regression to fit
a function that seemed reasonable to this. And remember the whole point of doing any of
that is so that we can quantify this intuition that the more information we gain from a
word, the lower the expected score will be. So with this as version 2.0, if we go back and we
run the same set of simulations, having it play against all 2,315 possible Wordle answers, how
does it do? Well in contrast to our first version, it's definitely better, which is reassuring.
All said and done, the average is around 3.6. Although unlike the first version, there are
a couple times that it loses and requires more than six in this circumstance, presumably
because there are times when it's making that trade-off to actually go for the
goal, rather than maximizing information. So can we do better than 3.6? We definitely can.
I said at the start that it's most fun to try not incorporating the true list of Wordle
answers into the way that it builds its model. But if we do incorporate it, the best performance
I could get was around 3.43. So if we try to get more sophisticated than just using word frequency
data to choose this prior distribution, this 3.43 probably gives a max at how good we could get with
that. Or at least how good I could get with that. That best performance essentially just uses
the ideas that I've been talking about here, but it goes a little farther. Like it does a
search for the expected information two steps forward, rather than just one. Originally
I was planning on talking more about that, but I realize we've actually gone quite long as
it is. The one thing I'll say is after doing this two-step search, and then running a couple
sample simulations in the top candidates, so far for me at least it's looking like "crane"
is the best opener. Who would have guessed? Also if you use the true word list to determine
your space of possibilities then the uncertainty you start with is a little over 11 bits. And
it turns out, just from a brute force search, the maximum possible expected information
after the first two guesses is around 10 bits, which suggests that, best case
scenario after your first two guesses, with perfectly optimal play, you'll be left with
around one bit of uncertainty, which is the same as being down to two possible guesses. So I think
it's fair, and probably pretty conservative, to say that you could never possibly write an
algorithm that gets this average as low as 3, because with the words available to you there's
simply not room to get enough information after only two steps to be able to guarantee the answer
in the third slot every single time without fail.
I really enjoyed this analysis. Wordle reminds me of learning Mastermind years ago. In attempts to make Mastermind more challenging I would allow duplicates, blanks, and eventually programmed it with the 16 MS-DOS colors. But a key challenge was making every guess conform to all previous guesses. For this reason, I play Wordle the same way.
This is interesting. I’ve been going with a Wheel of Fortune based start word (r, s, t are most popular) and using “stare”.
Wonderful video, but I am sceptical whether this is actually the optimal strategy, or whether it is a good heuristic.
I tackled this problem last week by considering it as an adversarial game where player A picks a guess word, player B then picks the largest class of possible final words (where a class is just the words that would produce the same green-yellow-gray clue), and so forth. Then the problem boils down to a minimax problem, which as far as I could tell would give you a true optimal strategy (if you pretend to play as player A).
I would be interested whether these strategies would boil down to the same strategy. Or, perhaps, we have some slightly different definitions of what "optimal" Wordle play entails.
This is a good video, but I disagree a bit with how it defines optimal strategy as the word which results in the lowest average guesses for a solution as for me the optimal strategy is one which always guarantees a victory is possible in six guesses or fewer, and also minimises the maximum number of guesses.
On this front, there's a great post on Stack Overflow which details this (https://puzzling.stackexchange.com/questions/114316/whats-the-optimal-strategy-for-wordle) and contains a 'solution' to Wordle (although the second aspect as to which starting word minimises the average maximum number of guesses required still seems to be open).
I have two questions. (He may have already addressed them to certain degree in the video but I didn't catch since I didn't pay 100% attention in the whole 30 minutes. Please let me know if it's the case.)
In general, would "only use the word that fulfill the patterns revealed in previous guesses" a better strategy than picking the one to maximize (average) new information?
Seems to be obviously no at least for the first few tries, but I'm still kinda curious (like, if the answer set is much smaller, surely it would help to try to directly guess the answer earlier right?)
In his second strategy, he uses frequency of words (not literally, but fitting them into a sigmoid function) as a weight.
The idea here is that the answer is more likely to be a common words, but from a game design practice perspective (even with no knowledge about the word list in source code) , it makes more sense that there would be a cutoff somewhere than a smooth transition, and after that all the words are equally likely. i.e. a "more common" words is not going to be more likely to be the answer than a, "slightly less common" words as soon as both made into the list of possible answers. We may still want to assign a non-zero low probability with other words since we don't know where the cutoff is at.
In other words, the probability should be a step function. I guess he uses sigmoid function simply for ease of computation? Anyway, in one of his demo there are only two choices left (
wordsanddorms), and one has significant higher P, which sounds wrong. But then again, in these "pick 1 from 2" scenarios it doesn't really matter. I'm just curious if it may affect the strategy in other cases?I always open with penis.
My first two words contain the 10 most popular letters in the English language. I recognize this isn’t perfect, and considered doing some additional work on it, but thought better of it since I’m already solving at a 100% rate.
Great video. The way Entropy is used here doesn't guarantee an optimal strategy. The problem is that not every group of words is the same. You could have a relatively large group of words with lots of distinct letters between them that can be completely disambiguated with a single word, and on the other hand you might have a set of relatively fewer words that all share similar letters like [bills, dills, fills, hills, kills, ...] which can't be disambiguated in a single guess and therefore at least some of those words require three or more guesses. Therefore the high entropy/low probability event (the group of words is small) doesn't exactly correspond the target "will be solved in the fewest possible guesses"
One way to find the best solution is to use a search tree, where you evaluate the top N candidates based on the above heuristic at each branch. Here's a nice write up
Here's a fun idea:
Suppose Wordle is a 2 player game (player A and player B). A picks a word for B to guess and B picks a word for A to guess. The winner is the player who guesses the word in the least number of turns.