Message from JavaScript discussions

May 2017

— This lets us know if words in the set use any of the unique letters without knowing which words they are

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Adding "Helion" will make it get to that check, and it will notice the L's in "Hello" are more than are in the set, and will return false (definitely not in set)

— Wat

— Couldn't it still have "Helion"?

— If L is 3, it could have both

— Oh right, just needed to re-read

— Add "Helion",
Run has for "Hello"

— It will see 1 L in the set, but 2 in Hello

— And thus we have a contradictory proof that Hello does not exist in the set

— Without ever looking at the actual set :D

— When the dataset gets diverse enough the first few checks will always pass, so this will gradually decrease in speed

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— So is it better than a linear lookup? We'll find out when it's working in a bit if you want to throw some tests at it

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