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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T07:13:55+00:00 2026-05-13T07:13:55+00:00

I am writing a simple Python program. My program seems to suffer from linear

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I am writing a simple Python program.

My program seems to suffer from linear access to dictionaries,
its run-time grows exponentially even though the algorithm is quadratic.
I use a dictionary to memoize values. That seems to be a bottleneck.

The values I’m hashing are tuples of points.
Each point is: (x,y), 0 <= x,y <= 50
Each key in the dictionary is: A tuple of 2-5 points: ((x1,y1),(x2,y2),(x3,y3),(x4,y4))

The keys are read many times more often than they are written.

Am I correct that python dicts suffer from linear access times with such inputs?

As far as I know, sets have guaranteed logarithmic access times.
How can I simulate dicts using sets(or something similar) in Python?

edit As per request, here’s a (simplified) version of the memoization function:

def memoize(fun):
    memoized = {}
    def memo(*args):
        key = args
        if not key in memoized:
            memoized[key] = fun(*args)
        return memoized[key]
    return memo
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  1. Editorial Team
    Editorial Team
    2026-05-13T07:13:55+00:00Added an answer on May 13, 2026 at 7:13 am

    See Time Complexity. The python dict is a hashmap, its worst case is therefore O(n) if the hash function is bad and results in a lot of collisions. However that is a very rare case where every item added has the same hash and so is added to the same chain which for a major Python implementation would be extremely unlikely. The average time complexity is of course O(1).

    The best method would be to check and take a look at the hashs of the objects you are using. The CPython Dict uses int PyObject_Hash (PyObject *o) which is the equivalent of hash(o).

    After a quick check, I have not yet managed to find two tuples that hash to the same value, which would indicate that the lookup is O(1)

    l = []
    for x in range(0, 50):
        for y in range(0, 50):
            if hash((x,y)) in l:
                print "Fail: ", (x,y)
            l.append(hash((x,y)))
    print "Test Finished"
    

    CodePad (Available for 24 hours)

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