Exercise Sheet 5

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Along with your files please also fill in the rate in elements per second of the two algorithms when intersecting two lists of size 10^6 = 1 million. Please give these two numbers with the same precision as in the rows already there, for example, 8.5 million, but not 8.48834 million. For comparison, I also implemented the two algorithms and put my two numbers. Can you beat them? I did not do anything particularly fancy, expect paying attention that the code in the inner loop is as simple as possible, as discussed in the lecture. Please also add information about the processor of your machine. On a linux system you get this information with cat /proc/cpuinfo. The machine I used is quite new. I also ran the same code on a machine that is a few years old (AMD Opteron 2220 SE) and there the rates are about half of the ones on the new machine.

These were the questions and comments on Exercise Sheet 5

Hi Marius, in principle youo are right, but as a rule of thumb, I would say that speed is roughly proportional to the processor speed, so processor speed is a good single number. Hannah 24Nov09 3:06pm

Hi, I would appreciate the Bus-speed of the systems in the list of the exercises, too, because the higher the access to the RAM, the better the values would be. Marius 11/24/2009 10:40am

Hi Dragos, yes, just iterate over the doc ids and pick each with probability m/n. This will give you a sorted list of doc ids of average length m. It does not have to be of size exactly m. Hannah 24Nov09 0:37am

About generating the docID lists: I am guessing I'm not wrong when I say that we don't have to make sure we have exactly m values in each list, we should just generate each element with a m/n probability, right? Thank you ! Dragos 24Nov 12.30 AM

Hi Claudius, yes, the inequality actually holds for "<", too. Hannah 23Nov09 4:05pm

In Exercise 5, could it be, that the inequality holds actually for "<", too? Because k may not be 0 and k <= m. Or am I wrong? Claudius 23Nov09 3:58pm

Hi Marius + all: I mean only "real" comparisons, that is, comparisons of list elements. If you use sentinels, then sentinels count as list elements. Hannah 22Nov 10:38

Hi, just to tie in with the last question: Do you mean by comparisons only the comparisons of list elements (such as A[i] <= B[j]) or are guarding conditional comparisons also to be counted (such as "if (i < list1.size())"? Marius 11/22/2009 10:32pm

Hi Björn + all: good question. One simple way to deal with this would be to use a comma expression like (numComparisons++, A[i] <= B[j]). A list of expressions, separated by commas, gets evaluated from left to right, and the value of the whole expression is simply the value of the last expression. At least that is the case in C++, but most programming languages have the same or a very similar construct. An alternative that essentially does the same thing, would be to do the comparison in a separate function, which, besides doing the comparison, also increase the comparisons counter. If you do that, you should absolutely make that that function is inline though since otherwise you pay the price of a function call for every comparison, which is likely to spoil your performance. Hannah 22Nov09 6:30pm

Hi, is there a good way to count comparisons if locial connectives are used? As far as I know, && will only check until the first comparison is false and || if it until some is true. Unfortunately I cannot think of a way to count those comparisons without rewriting the code and nesting the if statements. I'm not too familiar with compilers but I hope that this changes to the code won't effect the performance.Björn 22Nov09 6:20pm

Hi Claudius + all: you are right, it's not very precise, the intended meaning is something like "compare the tables for the two algorithms" or "for each of the four measurements, compare the tables for the two algorithms". Hannah 22Nov09 3:39pm

I am confused: In Exercise 4, you write: "Compare the two tables...", but when I take both algorithms and all 4 measurements (running time, ratio, ...), I get 8 4x4 tables (for all the combinations of 10³, 10⁴, 10⁵, 10⁶). What I'm doing wrong? Claudius 22Nov09 3:30pm

Hi Björn + all: very good question and thanks for pointing that out. You should indeed always search the elements of the smaller list in the larger list, and the first thing your (advanced) list intersection algorithm should do is figure out which of the two lists is the smaller one. That is, your 4 x 4 table will be symmetric, and actually only contains 10 different values (the 6 below the diagonal, which are the same as the ones above the diagonal, and the 4 on the diagonal). Hannah 22Nov09 2:50pm

For the exp/bin-search intersection algorithm it clearly matters that it searches for the elements of the smaller list in the larger one. A good implementation will certainly take care of that. Should our implementation also do that or ignore it in order to get 16 measurements that are really different? Björn 22Nov09 1:00pm

Ok, no problem, I'm happy when it's clear now. Hannah 22Nov09 0:24am

You're right, I misread your comment, sorry. I was thinking of 10MB per lists processed in 1 second, resulting in 20MB/s and was wondering where the 100MB/s are coming from. Thomas 22Nov09 00:20am

Hi Thomas, I am at a loss of words here. I am saying a car is driving 20 kilometers and it needs 10 minutes for that, so its average speed was 120 km / hours. And you are saying how can the speed of a car be 120 km / hours, when it only drives 20 kilometers. Well, what should I say. Besides, in my example I clearly said that the two lists together occupy 10 MB, not 10 MB per list. Please read again what I wrote. Hannah 22Nov09 0:16am

Why should two lists of 10MB size result in 100MB processed, if each list is only iterated over once to do the intersection (O(m+n) complexity)? The data processed after all is just 20MB, no matter how the algorithm is implemented (even if it iterates a thousand times over every list, it still just processed 20MB of data). Thomas 21Nov09 12:00am

By the way, whenever I talk about "lists" here or on the exercise sheets or in the lecture, I am not referring to a particular data structure (in particular I am NOT talking about a linked list), but "list of elements" is just "series of elements". And well, "inverted list" is just common terminology. To implement a "list of doc ids" or anything like that you should of course always use an array or a vector or a data structure like that. Hannah 21Nov09 8:30pm

Hi Marius + all, let me explain it by an example. Your two input lists occupy a certain amount of memory. Every programming language has built-in functions for this. For example, if your list entries are ints, then for C++ you can use sizeof(int) to get the number of bytes occupied by one entry. Multiply by the number of list elements to get the number of bytes occupied by one list. One Megabyte (MB) is 1024 * 1024 bytes. Now assume your two lists together occupy 10 MB. Assume your code takes 0.1 seconds to intersect these two lists. Then the "MB processed per second" is 100 MB / second. Hannah 21Nov09 8:26pm

Hi, in exercise 3, what do you mean by "MB processed per second"? Is a MB the equivalent to 4096 processed integers? And when is a MB to be considered as processed? When it's written to the intersected list or in the comparisons, already? Marius 21Nov09 7:33pm

The slides + all my hand-writing on it are now online, see the link Recording Lecture 5 (no audio) above. Hannah 20Nov09 3:24am

The recording of todays lecture again did not work. I am very sorry for that (and very angry that there are so many problems with this software). Anyway, the end result of the lecture, that is the slides with all the writing on it are available and I will put them online as soon as possible. Hannah 19Nov09 11:23pm

There is a typo in Exercise 5 of the new sheet. The two occurrences of n should be m. Hannah 19Nov09 11:22pm