3 ms·

Which is why you have to evaluate in a model with an explicit cache width. In such a model, b is not a constant. The cache-oblivious model [1] is a fairly well-

Which is why you have to evaluate in a model with an explicit cache width. In such a model, b is not a constant. The cache-oblivious model [1] is a fairly well-known example, but that won't work in this case, since we need to know b to set the tree width. Any of the other external memory models will do. [1] https://en.wikipedia.org/wiki/Cache-oblivious_algorithm https://en.wikipedia.org/wiki/Cache-oblivious_algorithm

ChuckMcM 10y ago
I think the notation is imprecise. The quantity the author describes is commonly called the working set. His explicit thesis is that the time to process the working set is related to the size of the working set regardless of computational complexity. He spends the entire first part of his discussion belaboring this point with a linked list example. He goes on to do a systems analysis of the results and concludes that time of execution is system dependent not complexity dependent. Again using the same algorithm engaging with a larger and larger memory system. Generally engineers would experience this important principal in a more binary way, once their program started swapping to disk, its performance would fall through the floor. If you hold the size of the datum constant (and that is important), then processing fewer of them will take less time than processing more of them. The argument you make, as I understand it is as follows; Consider two algorithms that process N and P datums respectively to achieve the same computational result, and where N > P. There exists an algorithm such that P datums takes less per unit time to process than N. Resulting in a violation of my Lemma because Oi(P) > Oi(N) but Ot(P) < Ot(N). And you cite cache obliviousness as the principle that enables that violation. Would you agree that I have accurately summed up your argument?
- DanWaterworth 10y ago
  It's roughly right, but it's not cache obliviousness that enables the violation. For the argument to work out, you have to evaluate it within a model that has an explicit cache and hence, treats big O results as being parametrized over the cache width.

ChuckMcM 10y ago

I think the notation is imprecise. The quantity the author describes is commonly called the working set. His explicit thesis is that the time to process the working set is related to the size of the working set regardless of computational complexity. He spends the entire first part of his discussion belaboring this point with a linked list example. He goes on to do a systems analysis of the results and concludes that time of execution is system dependent not complexity dependent. Again using the same algorithm engaging with a larger and larger memory system. Generally engineers would experience this important principal in a more binary way, once their program started swapping to disk, its performance would fall through the floor. If you hold the size of the datum constant (and that is important), then processing fewer of them will take less time than processing more of them. The argument you make, as I understand it is as follows; Consider two algorithms that process N and P datums respectively to achieve the same computational result, and where N > P. There exists an algorithm such that P datums takes less per unit time to process than N. Resulting in a violation of my Lemma because Oi(P) > Oi(N) but Ot(P) < Ot(N). And you cite cache obliviousness as the principle that enables that violation. Would you agree that I have accurately summed up your argument?

DanWaterworth 10y ago
It's roughly right, but it's not cache obliviousness that enables the violation. For the argument to work out, you have to evaluate it within a model that has an explicit cache and hence, treats big O results as being parametrized over the cache width.