Is it the case that a logarithmic gap is shown for a specific problem ? Why is this a big deal ? Doesn't Grover's algorithm already demonstrate a sqrt(n) gap ?
For Groover algorithm, which is applied to the problem of search you can prove that there is no classical algorithm faster than O(N) - looking at each item in the worst case scenario.
>
For Groover algorithm, which is applied to the problem of search you can prove that there is no classical algorithm faster than O(N) - looking at each item in the worst case scenario.
Assuming that you cannot "look into the black box of the oracle" (i.e. we are only allowed to use a given oracle to evaluate an item at an index).