For a word lock with 8 faces and 5 letters on each face...

For a word lock with 8 faces and 5 letters on each face, is it mathematically possible for all alignments to spell eight words in the English language? Aside from the trivial "HORSE-in-all-fields."

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>and 5 letters on each face
>Aside from the trivial "HORSE-in-all-fields."
so the catch is that no peice can have the same letter on it twice?

Yeah, that would make the lock pointlessly hard to open for no reason.

Seems very unlikely. There would be 8^5 = 32768 distinct words that you could make with the lock.

The total number of 5 letter words is (according to my half assed google search) only 158390

So if this was possible, then roughly 20% of all 5-letter words would be possible on the lock.

How did you arrive at 20%?
It's tricky counting problem I think.
We have to consider all valid combinations of 5 letter words such than no two words have matching letters in a place and all words are valid english. How would you begin counting that without a computer?

well the total number of all words with 5 letters is (according to google again) 158390. Sure, some of those probably shouldn't be counted. But it's an upper bound on the number of words we are interested in.

I doubt those words are evenly distributed by letters in each place. The question wasn't theoretical, so we need to know for sure it's a possibility given the limited choices we have.

Ok, so the same ring can't have the same letter twice.

But if it's on a different ring, I'm going to assume that's fine, so you can have words like "ROBOT"

Ok let me run a program to see

It's a messy, annoying problem because words are not nicely distributed.

Most likely a proof of the following form can work: it involves counting the number of 5-letter words that start with a specific two letters.

Basically you just show (by brute force) that there are no sets X, Y of 8 letters each, such that there are enough ("enough" being 8^5) words in the English language whose first letter is in X and second letter is in Y.

If that doesn't work, then consider three sets X, Y, Z of 8 letters each and words starting XYZ...

It won't work if you only focus on the first letter though. You can definitely choose a set X of 8 letters such that there are 8^5 English words that start with X.

If no dial can have the same letter twice, then every dial has a consonent. Every combination has to make an English word, including the combinations of 5 consonents. Impossible.

no, the point is there is one arrangement where each of the 8 faces forms a word. It doesn't need to be every combination.

nice bro. cheers

this interpretation of the problem is trivial

Kek bro, list out the 8 words then.

If it's so "trivial"

What the fuck? This was the problem the whole time? That's definitely trivial:

EXTRA
DREAR
ROBOT
BADLY
FUCKS
ANIME
STRUM
HIKED

I even got six of them to almost make a sentence

Nice

New question: What is the largest number of faces on each dial such that a 5-dial lock will have a 5-letter English word at every possible alignment?
5 is probably an upper bound for this, but I"d like to see what examples people can come up with

Even 2 faces is extremely difficult, as there are 32 words to consider.

I've been trying to come up with one for a while now and can't even come close. I can't even do it with 4 dials.

SONS
TONS

SINS
TINS

SOTS
TOTS

SITS
TITS

But then when can I replace the last S with? Too hard.

Your interpretation is literally "list 8 five letter words"

Well, it's a wee bit less trivial than that. You have to make sure none of the first letters are the same, none of the second letters are the same, etc.
But still fairly trivial

No, because every first letter needs to be different from every other first letter, every second letter needs to be different, from every other second etc, like in

DO IT THEN.

With no ring with more than one letter.

We have already proved that the "every single combination" interpretation of the problem is impossible, so there's no point talking about that anymore.

Let's solve THIS interpretation.

Not OP, I'm still trying to interpret the problem. I thought this was a counting problem. cool that it's been proven one exists, now how many possible combinations are there? Assume one combination is any set of 8 distinct words.

>DREAR

ehhh, close enough

merriam-webster.com/dictionary/drear

the two posts you quoted are solutions (or attempts at solutions) to different versions of the problem

one simply states that every nth letter of every word must be different from the nth letter of every other word

the other problem requires you to be able to turn the dials and still have a word every time, with the previous requirement stacked on top

you could replace it with clear

OK, I guess I've been interpreting the problem differently from everyone the entire time. If it's a problem about finding a single example, I'm gonna crack out my pen and paper

for the first one there's already an example, the dial turning one is what seems impossible

whoops, should have turned the name field off

Counting, there's 5^5 = 3125 possible alignments. since each dial needs distinct letter, I think the order of the letters on each dial is important and is going to complicate this a lot. I give up.

Basically we're trying to get an alphabetical "Gray Code" with some really fancy properties...

Am I a winrar?
ABORT
COEDS
GIVEN
LUMPY
MARIO
SHAME
TRILL
WENCH

of the one that doesn't involve dial turning, yeah

OK, I spent a few hours of concentrated autism algorithmically going through word lists to try and find two five-letter English words
(a1, a2, a3, a4, a5) and (b1, b2, b3, b4, b5)
s.t. any permutation of ({a1,b1}, {a2,b2}, {a3,b3}, {a4,b4}, {a5,b5}) is a valid five-letter English word.

There aren't any such pairs. I'll try out four-letter words next, that should be easier.