So that (post #1090) shows one thing... it won't pass if two letters from the same group are in the word (because the only other two letters are vowels and they basically don't matter). So perhaps the rule is something like, you can't have more than one letter from each group... something like that?

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zuiq

I still think it has something to do with the vowels.

nope... i don't think it's that... because 'clashed' (and probably many others) has more than one from the same group

(April 6, 2017 at 10:24 am)emjay Wrote: nope... i don't think it's that... because 'clashed' (and probably many others) has more than one from the same group

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Ninja'd you... "I don't think it's that" was in reference to my post, not your idea

(April 6, 2017 at 10:24 am)Grandizer Wrote: I still think it has something to do with the vowels.

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The problem with that is the vowels are totally interchangeable as far as we've seen. So whatever the function is seems to strip them out entirely and/or give them all the same value if they're still relevant.

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hatch

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patch

that said, it does seem that a specific combination of consonants predicts a certain number of vowels... for instance why does watch pass and waatdh pass? In that case it's plus one to the c to make it d, and plus one vowel to the word. But as promising as that looked, I can't work out why.

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What I found interesting is that doubling a passing word doesn't yield one, e.g watchwatch doesn't pass.