I suppose this could be broken by injecting in a unique visitor id that would hash to something with an absurd amount of zeroes? That's assuming that the user has control over their user id and that I'm understanding the algorithm correctly.
I suppose this could be broken by injecting in a unique visitor id that would hash to something with an absurd amount of zeroes? That's assuming that the user has control over their user id and that I'm understanding the algorithm correctly.
Just skimmed through it and seems pretty interesting. I'll read it more in depth later.
I noticed that you used 5000 buckets to store the frequency of 7000 non-unique words in the section on 'Counting Bloom Filters'. How is that better than using 7000 buckets and a uniformly distributed hash function, which would maintain frequencies perfectly? We would be using fewer buckets by an order of magnitude in a real-world implementation to save memory.
The usual trick for preventing folks _maliciously_ sending in outliers is to use a hash with a secret key such as SipHash so that folks on the outside can't trivially figure out what inputs will lead to hashes with a lot of leading zeroes.
HyperLogLog is different because you have the entire population (not just a sample), and it's a multiset (the same element can appear more than once). Getting the size of a (non-multi) set is easy, you just keep a counter and increment it for each element; it only takes enough memory to maintain the counter. Counting the distinct members of a multiset takes a lot more memory because you have to remember whether you've already seen a particular element or not.
The tl;dr is that the German Tank Problem is about making an estimate of size when you have imperfect information, and HyperLogLog gives you an estimate when you have perfect information, but it's too expensive to make an exact calculation.