Welcome to the Wiki page of the course Search Engines, WS 2009 / 2010. Lecturer: Hannah Bast. Tutorials: Marjan Celikik. Course web page: click here.

Here are PDFs of the slides of the lectures so far: Lecture 1, Lecture 2, Lecture 3, Lecture 4, Lecture 5.

Here are .lpd files of the recordings of the lectures so far (except Lecture 2, where we had problems with the microphone): Recording Lecture 1, Recording Lecture 3, Recording Lecture 4, Recording Lecture 5 (no audio).

Here are PDFs of the exercise sheets so far: Exercise Sheet 1, Exercise Sheet 2, Exercise Sheet 3, Exercise Sheet 4, Exercise Sheet 5.

Here are your solutions and comments on the previous exercise sheets: Solutions and Comments 1, Solutions and Comments 2, Solutions and Comments 3, Solutions and Comments 4.

Exercise Sheet 5

The recordings of all lectures are now available, see above. Lecture 2 is missing because we had technical problems there. To play the recordings (it's .lpd files) you need the Lecturnity Player. You can download the player for free here.

Here are the rules for the exercises as explained in Lecture 2.

Here you can upload your solutions for Exercise Sheet 5.

Questions or comments below this line, most recent on top please

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