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."
For a word lock with 8 faces and 5 letters on each face...
Lucas Diaz
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Anthony Butler
>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?
Carson Torres
Yeah, that would make the lock pointlessly hard to open for no reason.
Cameron Foster
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.
Matthew Reyes
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?
Gavin White
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.
Anthony Turner
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.
Jacob Walker
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
Nathaniel Reed
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.
Samuel Diaz
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.