— 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)
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— 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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