🔒 Find strings in a hashmap

[molten sage]

I'm having trouble understanding what I'm supposed to do in this question

"""
You are given the size N of a hash table. 
Your task is to find two strings that map to the same location in the hash table based on Python's hash function. 
In other words, you need to find strings x and y such that hash(x) % N == hash(y) % N. 
You can assume that N is at most 100. 
Your solution should work efficiently in these cases. 
Implement in the file samehash.py a function find that returns the desired strings as a pair.

Note that the behavior of the hash function changes each time the Python interpreter is started. Therefore, the function find should give a different solution each time it is run
"""
def find(N):    
    return 0
if __name__ == "__main__":
    print(find(42)) # e.g. ('abc', 'aybabtu')

I though maybe it was the ascii values or something but that wasn't the case. How are abc and aybabtu related in any way??
According to the question they have the same location in this arbitrary hash table, but why?? To me it looks like these are randomly generated strings and there's no way to verify if that output is correct or not, because the condition isn't even satisfied

[strange steppe]

@molten sage

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[molten sage]

all I have as an input is a hashtable size, no actual hash table

there is literally a 1/42 chance of getting equal output when running this

print(hash('abc')%42,hash('aybabtu')%42)

!e
print(hash('abc')%42,hash('aybabtu')%42)

[strange steppe]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

28 28

[molten sage]

!e
print(hash('abc')%42,hash('aybabtu')%42)

[strange steppe]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

12 38

[molten sage]

I am probably looking at this exercise the wrong way, but I'm just not seeing it

[sly tusk]

maybe find out how python's hash work and try to generate 2 strings that can collide

or you can just generate like 101 words (given N is atmost 100) by pigeon hole principle atleast 2 words will collide you can find them by bruteforce

!e

from collections import defaultdict
def find(N):
  hashes = defaultdict(set)
  for i in map(str, range(101)):
    hashes[hash(i)%N].add(i)
    if len(hashes[hash(i)%N]) > 1:
      print(f"Collision Found: {hashes[hash(i)%N]} with hash {hash(i)%N}")
      break
find(50)

[strange steppe]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

Collision Found: {'6', '7'} with hash 1

[molten sage]

!solved

[strange steppe]

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