"The Genuine Sieve of Eratosthenes" in Python - why is heapq slower than dict?
heap, performance, python, sieve-of-eratosthenes
Solution
Actually, it should be expected that a dictionary-based approach would be faster than a heap-queue-based approach. Heap insertion and replacement operations are O(log n), while dictionary insertion and replacement operations are O(1).
Indeed, I was surprised to hear that the author of the paper was claiming otherwise. But in fact that's not the case. You've assumed that a `Data.Map` is implemented as a hash map, but actually, it's a size balanced binary tree. So its performance characteristics are quite similar to the performance characteristics of a heap queue. The difference is that retrieving the minimum key from a heap queue is O(1), which speeds up parts of the sieve code -- but a hash map is faster still.
Problem
Following M. O'Neill's great paper, I tried implementing some lazy, infinite versions of the Sieve of Eratosthenes in Python. I was surprised to find that the heap-based version, that the paper claims should run faster, was actually over two times slower for me. The paper contains two examples, one based on a dict, which I have translated (from Haskell) thus: ``` from itertools import count def dict_sieve(): yield 2 yield 3 candidates = count(5, 2) composites = {9:{3}} # map composites to their prime factors for candidate in candidates: try: factors = composites.pop(candidate) except KeyError: # if it's not in the dict, it's prime yield candidate composites[candidate**2] = {candidate} # Euler's optimization: start from prime**2 else: for prime in factors: # go through the prime factors and increment their keys try: composites[candidate+prime*2].add(prime) # use prime*2 because we are ignoring evens except KeyError: composites[candidate+prime*2] = {prime} ``` The second example in the paper demonstrates use of a priority queue as data structure. It also uses lazy lists, rather than a simple increment, which I have not done in the interests of a fair test. (Besides, I was using instances of `itertools.count` for my lazy lists and I found it to run slightly slower). ``` from itertools import count from heapq import heappush, heapreplace def heap_sieve(): yield 2 yield 3 candidates = count(5,2) composites = [(9, 3)] # a priority queue of composite/factor pairs, keyed by composite for candidate in candidates: prime_flag = True while composites[0][0] == candidate: # loop because there may be duplicates prime_flag = False # signal to the if statement below composite, prime = composites[0] heapreplace(composites, (composite + prime*2, prime)) if prime_flag: yield candidate heappush(composites, (candidate**2, candidate)) ``` I timed the two, along with an 'eager' version, not reproduced here, which produces a list of all primes below a limit: ``` In [44]: from itertools import islice In [45]: %timeit list(islice(dict_sieve(), 100000)) ...: %timeit list(islice(heap_sieve(), 100000)) ...: %timeit eager_sieve(1299710) # 1299709 is the 100,000th prime ...: 1 loops, best of 3: 2.12 s per loop 1 loops, best of 3: 4.94 s per loop 1 loops, best of 3: 677 ms per loop ``` It's not surprising that the 'eager' version is much quicker - it's basically a trade-off between memory use, the inconvenience of having to specify an upper limit, and CPU time. However, I did find it surprising that the `heapq` version is much slower, when the paper claims it to be more efficient. Is it a problem with my implementation? Or is it simply that, as we all know, dicts are super-fast (and `heapq` is comparatively slow)?