Recently, I was attempting the problem 1692H - Азартные игры. The details of the problem are not important for this post, it is just sufficient to know that I was making use of a dictionary to store some data.
I ran into a TLE verdict (2000ms), which I found strange, since my solution was linear as far as I could tell. After a ton of debugging, I was able to deduce that the following lines
inside a for loop were being incredibly slow (again, what these lines do isn't important, only note that there are a few dictionary accesses). Around 20k iterations were taking about a second to execute.
I suspected that there might be too many hash collisions, so I tried multiplying i by 10^20, and later dividing it by 10^20, and this solution passed.
163665232 — TLE Submission
163664864 — AC Submission
Now in Python, for upto fairly large numbers (~10^18), hash(x) = x, which is why I was multiplying by 10^20. However, even for larger numbers, as far as I can tell, hash(x+1) = hash(x)+1. In this case, I thought the distribution of the hashes should be pretty similar with and without multiplying by 10^20.
However, this is speculation, and I could be wrong. Does anyone know a little more about this topic, and how I can avoid facing these issues in the future?
I'd make a wrapper class of integer and implement your own hash and eq.
I don't know why double underbars are replaced.
https://mirror.codeforces.com/blog/entry/101817
In short, Python hashing method is deterministic, and hackers specifically put in tests that force hash collisions for that problem. When you multiply by $$$10^{20}$$$, you change the malicious input values to different ones that don't cause a hash collision. Of course, the constant you use should be random, otherwise you don't address the underlying issue as a strong enough adversary could still figure out what values when multiplied by $$$10^{20}$$$ collide.
Ahh, I see. Thanks!