For a word lock with 8 faces and 5 letters on each face, is it

>>8977714
>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?

>>8977717
Yeah, that would make the lock pointlessly hard to open for no reason.

>>8977714
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.

>>8977736
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?

>>8977792
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.

>>8977828
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.

>>8977714
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

>>8977851
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.

>>8977936
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.

>>8977936
nice bro. cheers

>>8977988
this interpretation of the problem is trivial

>>8978007
Kek bro, list out the 8 words then.

If it's so "trivial"

>>8977988
>>8978007
>>8978166
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

>>8977936
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

>>8979341
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.

>>8978166
Your interpretation is literally "list 8 five letter words"

>>8979361
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

>>8979361
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 >>8979321

>>8979361
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.

>>8979321
>>8979358
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.

>>8979321
>DREAR

ehhh, close enough

>>8979441
https://www.merriam-webster.com/dictionary/drear

>>8979439
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

>>8979441
you could replace it with clear

>>8979489
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

>>8979558
for the first one there's already an example, the dial turning one is what seems impossible

>>8979564
whoops, should have turned the name field off

>>8979564
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.

>>8979358
Basically we're trying to get an alphabetical "Gray Code" with some really fancy properties...

>>8979769
Am I a winrar?
ABORT
COEDS
GIVEN
LUMPY
MARIO
SHAME
TRILL
WENCH

>>8979797
of the one that doesn't involve dial turning, yeah

>>8979358
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.