[joakleaf]

No...

If you use a "bonehead" data structure each memory reference will force a cache-miss, i.e. taking q CPU cycles instead of 1 CPU cycles.

If the algorithms has a running time of O(N^3), it means it is using some number of instructions <= c * N^3 to complete. Let's for simplicity sake say 1 operation is identical to 1 CPU-cycle.

Now, if each operation takes q CPU cycles instead of 1 cycles due to cache-miss on every operation, the running time of your algorithm is still bounded by q * (c * N^3) operations.

For the example with N=100 -> The fast version takes c * 1,000,000 operations to complete, for some value c. For the slow version, as each operation is now q times slower, it will take q * c * 1,000,000 operations to complete for values q and c.

I.e. it takes q times as long, not q * q * q.

What you were saying is that, if you run a O(N) algorithm on a 100 MHZ CPU, it will take 10 times as long as on a 1000 MHZ CPU, but an O(N^3) algorithm would take 1000 times as long. And that is obviously, not true. Both will take 10 times longer.

[pacaro]

No

I'm saying that a poor choice of data structure will have an amplified impact on an O(N^3) algorithm.

Giving it as an example of where you could have the different implementations of the same algorithm have massively different runtimes on the same processor.

[joakleaf]

If the algorithm has a running time of O(N^3), that implies that it takes O(N^3) < c * N^3 operations to complete. Regardless of datastructures. That's the definition of asymptotic running time.

If we disregard c, and set N = 100, that is 1,000,000 operations.

We have two situations. 1) Either we use a good layout where each operation is fast because each memory access is from cache. 2) Or we use a bad layout where each operation is slow because we have a cache-miss and have to access the RAM. These access times are independent of the specific algorithm, and only depends on CPU architecture.

Let say that the fast operations each take 1 second. And the slow operations each take f seconds.

In the fast case, with good memory layout, the algorithm completes in 1,000,000 seconds. In the slow case, with bad memory layout, the algorithm completes in f * 1,000,000 seconds.

The time of each operation we perform does not depend on the size of N. However, in practice, you'll typically have a lot of cache hits even with a bad memory layout, so the difference may be less than a factor of f for smaller values of N (where you'll have more cache hits in a cubic algorithm). However, it will be upper bounded by f. I.e. there is an upper bound on how much slower you can make an algorithm by changing the memory layout.

[pacaro]

Obviously N and f are independent (although they might not be, your N may dictate how often a cache miss occurs).

Right, so now let's assume that our O(N^3) algorithm is simply a triple nested loop. And that we have an expensive operation f, but that through choice of data structures we can choose in which layer that operation occurs. All other costs are constant.

If f is in the outer loop, it happens N times, in the middle loop N^2, in the inner loop N^3.

So our choice of data structure impacts our running times by any factor from linear through cube of N.

I was simply answering the question of is it really possible to have two implementations of the same algorithm have such disparate run times, and it's trivially possible.

Another way of saying it, is that the mistake need not look expensive on cursory inspection, a simple way to have very different runtimes is through the accumulation of a large number of small costs.

Maybe the important point is to use a profiler.

I've recently gained a 60x improvement in real-world run time of what is considered in the literature to be the best known algorithm for a particular problem. The speed up didn't come from improving on the algorithm, it came from improving data locality, removing memory allocations, and general modern code hygiene (to be fair the canonical implementation is ~20 years old)