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Here are PDFs of the slides of the lectures so far: [[attachment:SearchEnginesWS0910/lecture-1.pdf|Lecture 1]], [[attachment:SearchEnginesWS0910/lecture-2.pdf|Lecture 2]]. | Here are PDFs of the slides of the lectures so far: [[attachment:SearchEnginesWS0910/lecture-1.pdf|Lecture 1]], [[attachment:SearchEnginesWS0910/lecture-2.pdf|Lecture 2]], [[attachment:SearchEnginesWS0910/lecture-3.pdf|Lecture 3]], [[attachment:SearchEnginesWS0910/lecture-4.pdf|Lecture 4]]. |
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Here are PDFs of the exercise sheets so far: [[attachment:SearchEnginesWS0910/exercise-1.pdf|Exercise Sheet 1]], [[attachment:SearchEnginesWS0910/exercise-2.pdf|Exercise Sheet 2]], [[attachment:SearchEnginesWS0910/exercise-3.pdf|Exercise Sheet 3 (tentative version!)]]. | Here are .lpd files of the recordings of the lectures so far (except Lecture 2, where we had problems with the microphone): [[http://vulcano.informatik.uni-freiburg.de/lecturnity/lecture-1.lpd|Lecture 1]] [[http://vulcano.informatik.uni-freiburg.de/lecturnity/lecture-3.lpd|Lecture 3]] [[http://vulcano.informatik.uni-freiburg.de/lecturnity/lecture-4.lpd|Lecture 4]]. |
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Here are your solutions and comments on the previous exercise sheets: [[SearchEnginesWS0910/ExerciseSheet1|Exercise Sheet 1]]. | Here are PDFs of the exercise sheets so far: [[attachment:SearchEnginesWS0910/exercise-1.pdf|Exercise Sheet 1]], [[attachment:SearchEnginesWS0910/exercise-2.pdf|Exercise Sheet 2]], [[attachment:SearchEnginesWS0910/exercise-3.pdf|Exercise Sheet 3]], [[attachment:SearchEnginesWS0910/exercise-4.pdf|Exercise Sheet 4]]. Here are your solutions and comments on the previous exercise sheets: [[SearchEnginesWS0910/ExerciseSheet1|Solutions and Comments 1]], [[SearchEnginesWS0910/ExerciseSheet2|Solutions and Comments 2]], [[SearchEnginesWS0910/ExerciseSheet3|Solutions and Comments 3]] = Exercise Sheet 3 = 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. [[http://www.lecturnity.de/de/download/lecturnity-player|You can download the player for free here]]. |
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= Exercise Sheet 2 = Here are the details about the three servers (UDP, TCP, HTTP) for Exercise 4: All three servers are running on our machine vulcano.informatik.uni-freiburg.de (IP address is 132.230.152.135). The UDP server is running on port 8888 of that machine. You can send it a number and it will then send you back that number of bytes, in packets of 1000 bytes each. (That means you also have to read packets of 1000 bytes each.) The first ten bytes of each packet contain the packet id. That is interesting for checking which packets get lost and in which order packets arrive. The TCP server is running on port 9999 of that machine. You can send it a request of the form GET /<number of bytes> HTTP/1.1, or you can just use a downloading program like wget or curl and time it. The HTTP server is running on port 80, as web servers normally do. Just download the file http://vulcano.informatik.uni-freiburg.de/file_100M and measure the time. You can assume that no data gets lost. For measuring your transfer and error rates, as requested by the exercise, repeat your experiments several times and also at different times, and form the average of these measurements (or report several numbers if you get very different results). You should ask for large amounts of data, like 10 MB or 100 MB. [[SearchEnginesWS0910/ExerciseSheet2|Here you can upload your solutions for Exercise Sheet 2]]. |
[[SearchEnginesWS0910/ExerciseSheet4|Here you can upload your solutions for Exercise Sheet 4]]. |
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Hi Waleed, when you create a conflict, it's your responsibility to remove it and not leave a mess behind. If the instructions given when the conflict occurs do not suffice, try to find more information in the Wiki help pages. '''Hannah 3Nov09 9:00pm''' | To Florian + all: yes, sorry, I forgot to mention this in the lecture. Marjan already explained how to clear the disk cache. Let me add to this an explanation what the disk cache already is. Whenever you read a (part of a) file from disk, the operating system of your computed will use whatever memory is currently unused to store that (part of the) file there. When you read it again and the data hasn't changed and the memory used to store it has not been used otherwise in the meantime, than that data is read right from memory, which is much faster than reading it from disk. Usually that effect is desirable, because it speeds up things, but when you do experiments, it is undesirable, because it leads to unrealistically good running times, especially when carrying out an experiment many times in a row. '''Hannah 15Nov09 8:10pm''' |
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I Uploaded my Files and put a new row on table in the excercies sheet 2 page but when i pressed save button it shows me conflict. my version and other version of list. how can i remove conflict? does my assignment is submitted properly or not? '''Waleed''' 3Nov09 | To Florian: Indeed, we were running out of time and there was no room for this in the lecture. I can suggest you few ways how to clear the disk cache: before carrying out your final experiment, read a large amount of data (let's say close to the amount of RAM you have) from disk - this will ensure that your data (the inverted list) is cleared from the disk cache and replaced by something else (thus an actual reading from disk get's timed, and not reading from RAM). Another way is to restart your computer before doing the timing. '''Marjan 15Nov09 7:27pm''' |
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Yes but it is also known that UDP does not guarantee the right order of the packages and that packages can get lost, but we were neverthelesse encouraged to analyze that ourselves. That's why I think that it would have been nice if both servers would have generated the same responses to permit an exact comparison :) '''Florian 3Nov09 11:22am''' | In exercise 4 it says: "Important note: Whenver you measure running times for reading data from disk, you have to clear the disk cache before, as discussed in the lecture". I think that this was not discussed in the lecture? What do we have to do here? '''Florian 15Nov09 7:15pm''' |
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To Florian: Why would you need packet numbers for the TCP protocol? TCP guarantees that they arrive in the original order. '''Marjan 3Nov09 10:44''' | @Bit shifting: The syntax for that is actually the same, irrespectively of whether you use Java, C++, perl, python, or whatever. The >> operator shifts to the right, the << operator shifts to the left, the & operator ands the bits of the two operands and the | operator ors the bits of the two operands. Very simple. You will also find zillions of example programs on the web by typing something like ''java bit shifting'' into Google or whatever your favorite search engine is. '''Hannah 15Nov09 1:16''' |
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I am not being able to delete or replace my zip file uploaded; i have missed one file in the zip file uploaded. '''Paresh 09:20am''' | Hi Marius + all: For Exercise 4, an inverted list of size m with doc ids from the range [1..n] is simply a sorted list of m numbers from the range [1..n]. I leave it to you, whether your lists potentially contain duplicates (as in 3, 5, 5, 8, 12, ...) or whether you generate them in a way that they don't contain duplicates (as in 3, 5, 8, 17, ...). It doesn't really matter for the exercise whether your list has duplicated or not. In any case, consider simple flat lists like in the two examples I gave (and like all the examples I gave in this and past lectures), not lists of lists or anything. '''Hannah 15Nov09 1:12am''' |
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Thx, I missed the fact that the TCP server also answers when I send just the number to it. -EDIT start- The TCP servers were responding again, they just seem not to respond for a while if one requests about 100MB :) -EDIT end-. Furthermore it is a pity that the TCP server does not reply with packets which contain packet numbers like the UDP server does, so it is not possible to compare the order of arriving packets between the two protocols. '''Florian 3Nov09 5:45am''' | @Mirko: Sure, but an inverted list is a list of words where the Doc-IDs are attached to each words in which the words occur. So for Example: If word no. 5 occurs in Doc1, Doc2 and Doc3 and word no. 2 occurs in Doc5, the list would look like: 5 -> Doc1, Doc2, Doc3; 2 -> Doc5. Or am I mistaken? My question then is, how long should these attached lists be in average case? I mean, one could imagine that we got 1mil. documents over 3 words, so these lists could get very large... |
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Hi Florian, yes our TCP server accepts a simple ASCII number, but it also accepts a HTTP GET query. As I said in the lecture, HTTP is a very simple protocol which adds very little on top of TCP, and one point of the exercise was just to realize that again. You realized it, congratulations. By the way, don't you think it's a bit late? Normal people are already sleeping now. '''Hannah 3Nov09 2:04am''' | EDIT: Oh ok. Now, I see your point. It's not an index, it's a list. Okay. So, what is an inverted list with Doc-IDs, then? |
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Hi, i am a bit confused about exercise 4: where is the difference between the TCP server and the HTTP server? Looking at the request we have to send to the TCP server and analysing it's response one sees that the TCP server is using the HTTP protocol as well and is therefore also a HTTP server?! So where is the point in handling them separately? '''Florian 3Nov09 1:50am''' | EDIT EDIT: And to your question, Mirko, take a look at [[http://snippets.dzone.com/posts/show/93]]. Especially at Comment no. 2. Maybe this helps... I think, Java supports StreamWriters/Readers that are able to write/read bytes. '''Marius 11/14/2009 08:46pm''' |
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Thanks a lot for the information about the behavior of small and large amounts of data using UDP! Right now i loose less packages and everything seems credible. Even for larger amounts of data, I achieve transfer rates slightly better than tcp/http with "only" 99% loss. This afternoon (when it takes 2minutes to send an email from the place i live), i had error rates of approx. 99.96 which resulted in very strange transfer rates. Http also was slower than now, but at did not suffer as bad as UDP, somehow. I'll try to report it in my PDF, tomorrow. Goodnight everyone. '''Björn 3Nov09 1:20am''' | EDIT EDIT EDIT: Sorry, me again. Well, I bothered Wikipedia which redirects from [[http://en.wikipedia.org/wiki/Inverted_list]] to Inverted Index. So it seems to me, this is being used as a synonym. Actually, I think I'm confused enough, now. I'll better wait for any responses... ;-) '''Marius 11/14/2009 9:08 pm''' |
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To all: what is the ''different'' supposed to mean in some of the columns on the [[SearchEnginesWS0910/ExerciseSheet2|upload page for Exercise Sheet 2]]? Please put the numbers you observed. '''Hannah 3Nov09 1:10am''' | @ Marius: i think we are supposed to generate one inverted __list__ of size m, with doc ids from 1..n (therefore n>=m, because no duplicates?). |
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To all: again, the deadline is (and always will be) 14:00, Tuesday. '''Marjan 3Nov09 00:27am''' | Now a question from my side: ex.4, programming the compression in __java__, is there any __good__ tutorial about how to handle the bit-stuff? (otherwise, i think, it would cost me too much time..) '''Mirko 14Nov09, 19:18''' |
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To Dragos: you should return a selection of the matching documents. For each document, send its id, the title, and maybe the first few lines from its contents. '''Hannah 3Nov09 00:14am''' | Hi, do you have any suggestions what the best numbers for m and n in exercise 4 should look like? Or are we supposed to mess around a bit with ints and longs? And: How long should the list of documents in the inverted index be? '''Marius 14Nov09 6:40pm''' |
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To Björn: as I understood Marjan, I think it's ok to submit your solution until Tuesday around noon. Concerning your UDP error rates: yes, I actually made very similar experiences, and the reason is as follows. If you print out the ids of the packages which actually get true, you will find that it is whole runs of packages which get lost. This, in turn, is because network availability / non-availability happens in bursts. For ten seconds it's there, than for a whole second there is no connectivity and all packets get lost, and so on. That's also why for small messages, they either get lost completely or you get the whole data. For large messages, however, it's quite likely that you hit one or several non-availability periods. At least that is my current understanding of what I have observed. Interesting, isn't it? '''Hannah 3Nov09 00:12am''' | And just to clarify what a single-cycle permutation is. Here is an example for an array of size 5 with a permutation that is a single cycle: 5 4 1 3 2. Why single cycle? Well, A[1] = 5, A[5] = 2, A[2] = 4, A[4] = 3, A[3] = 1. (My indices in this example are 1,...,5 and not 0,...,4.) Here is an example of a permutation with three cycles: 2 1 4 3 5. The first cycle is A[1] = 2, A[2] =1. The second cycle is A[3] = 4, A[4] = 3. The third cycle is A[5] = 5. '''Hannah 12Nov09 8:04pm''' |
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I have a question regarding Exercise 2 (I just want to make sure I get it right this time:) ). The server should respond with all the pairs of two-word-queries with one hit, or it should receive two-word-queries and then answer with the documents in which they can be found. Sorry if it sounds stupid, but I misunderstood the exercise from the previous sheet, and unfortunately I didn't manage to attend the course last week. Thank you in advance. '''Dragos 3Nov 12.03am''' | Hi Daniel + all, I don't quite understand your question and your example (if your array is 1 5 3 4 2, why is A[1] = 3?). In case you refer to the requirement of the exercise that the permutation consists only of a single cycle. That is because your code should go over each element exactly once (it should, of course, stop after n iterations, where n is the size of the array). If your permutation has more than one cycle, it is hard to achieve that. Also note that for both (1) and (2), the sum of the array values should be sum_i=1,...,n i = n * (n+1) / 2. '''Hannah 12Nov09 7:54pm''' |
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When was the deadline for exercise submission again? I'm only missing the TCP tests but I got other important stuff to do, so i would love to do it tomorrow before the tutorial. Unfortunately I'm not sure when on tuesday we had to submit our solutions. Oh, and one more thing: My Test series (REALLY bad connection here with my shared media cable connection) for UDP see to have very low error rates for few requested packages (less than 100) and incredibly high error rates if i request something like a million packets. Am I doing something wrong and should work to fix my code smell or is this possible somehow? '''Björn 2Nov09 11:59pm''' Wow, interesting piece of information, Zhongjie. There is now an additional TCP server, running on port 8889. Maybe that helps. '''Hannah 2Nov09 11:53pm''' To Jonas: In old times, many viruses and trojans are using TCP port 9999 to communicate. So some network systems, especially LAN based networks, will block this port forever. I'm not sure whether this is a two way block or not. But at least you don't want to open your own 9999 port for TCP use. I hope this information helps somehow. '''Zhongjie 2Nov09 23:19''' To Jonas again: I checked from one of my machines in Saarbrücken, that port 8888 and 9999 are reachable from there. I checked that already several days ago. So yeah, it looks like a problem of your network, though very strange that port 8888 works but port 9999 doesn't. '''Hannah 2Nov09 11:06pm''' To Jonas: This is not a problem, just say a port that will work for you and we will start another TCP server on this port. '''Marjan 02Nov09 10:04''' Good evening. After hours of trying to debug my TCP client I am now sure that I can't use port 9999. UDP works fine if I establish a vpn channel. If not, neither TCP nor UDP may connect to the server. Inside the university network both versions of the client and the curl test went well. Probably the 9999 port is blocked inside this annoying Wohnheim-network. '''Jonas 2Nov09 9:50pm''' Hi Christian + all: yes, you are right, it would have been nicer to provide a multi-threaded server and we should have announced that our servers are not multi-threaded, sorry for that. About the null-termination: I wasn't aware of that myself. But I actually believe that it is an important part of the exercise to get your feet dirty. Such stupid details are part of the game. I am still doing a lot of implementation work myself and even after 30 years of developing software I still spend a lot of time on such stupid details. But hey, that is how you actually learn stuff. The knowledge of all these details is a big part of what eventually makes the difference between a master and an apprentice. This is not a bad excuse, I really mean it. And, hey, we have the Wiki for talking about this stuff, asking questions, sharing something we have found out with the others, this is also an important part of the learning experience. '''Hannah 2Nov09 8:02pm''' Hi Björn + all: yes, transfer rate is amount of data received divided by time. '''Hannah 2Nov09 7:58pm''' To lecture team: No offense intended, but knowing that the servers are not multi-threaded would have been nice to know in the first place, just like the fact that null-termination is required for the UDP server. I would imagine that a lot of people would have had less frustration with testing and debugging their code. I realize, you had your hands full and writing a server is not a trivial task, and maybe a non-asychronous server may have been intentional, but the next time you might want to check out the asnychat and asyncore modules of Python: http://docs.python.org/library/asynchat.html '''Christian 2Nov09 7:49p.m.''' Hey, how is the "transfer rate" defined in the case of UDP? Should we only count packets that have actually been recieved?'''Björn 2Nov09 7:42pm''' To Markus: Note that the server is not able to handle multiple connections in the same time (it's not multi-threaded), so if somebody else is using the server that means in that very moment the server won't be available for you. '''Marjan 2Nov09 5:44pm''' Hi. I have some problems with the TCP server, wget got for some hours 26 Bytes but nothing else, but with the URL from below wget does nothing, it waits. In my own program I get a connection but no data, it waits too. But good news, the UDP works for me. '''Markus 2Nov09 5:42pm''' No, the UDP server is not down. The way it is now written, it simply restarts itself after it crashes, but it hasn't even crashed once in the last two days. Connection refused does not mean that the server is down, telnet is just an incompatible protocol. '''Hannah 2Nov09 5:16pm''' Is the UDP-server actually down? When I try to telnet it, I get a response saying: $ telnet vulcano.informatik.uni-freiburg.de 8888 Trying 132.230.152.135... telnet: connect to address 132.230.152.135: Connection refused. TCP seems to work although I don't get any packet via telnet. '''Marius Nov 2nd, 5:00 p.m.''' Hi Björn, just numbers in ASCII. For example, you can ask the TCP server via ''curl http://vulcano.informatik.uni-freiburg.de:9999/150'' (or just type that URL into your browser), which will effectively send the string ''GET /150 HTTP/1.1 ...'' and you will get 150 bytes in return. To the UDP server you just send the number in ASCII, for example ''150'', but make sure that you null-terminate your string, that is, it should have a zero-byte at the end (as C strings naturally have). '''Hannah 2Nov09 4:17pm''' What kind of "numbers" do the servers from exercise 4 expect? UTF-8 encoded Strings? Byte values? Anything else? Exercises 1-3 were fun to do but I'm completely stuck at ex4 right now. I constantly fail to get any proper response. '''Björn 2Nov09 4:13pm''' One more comment: if possible, for Exercise 4, run your clients from a machine ''outside'' of the university. That way you get more interesting results, in particular, you should then see a very marked difference between UDP and TCP, whereas within the Uni or even Informatik network that difference might be very small. '''Hannah 2Nov09 2:37pm''' Dear Daniel + all: if you have one binary that can do it all, that is fine. If you have three separate binaries that is also fine. Just make sure to avoid code duplication, that is, if you have three separate binaries (which you, Daniel, have not), make sure that the common code is on commonly used classes and not just copied and pasted. Copying and pasting code is the ultimate evil, believe me. If you have one binary, make sure that the code is well modularized with the various functionalities in appropriately chosen and named classed and methods, and that not all the code is in one big main function or in a single function named solve_exercise_2 or things like that. If this does not fully answer your question, don't hesitate to ask again. '''Hannah 2Nov09 2:33pm''' Hi, sorry to ask again about code writing, but I'm not sure I got it right. Do we have to submit 3 binaries for the exercises? I have written a web server with 3 available URLs (sentence/search/index), which provides an interface for repeating a sentence, searching for keywords with GET parameters and showing a search form to enter keywords respectively. For the last exercise I have standalone code for the clients then. Is this also ok? '''Daniel 2Nov09 2:19pm''' Ok, sorry, I just see that I indeed gave contradicting information. Above I wrote, for the TCP server, "... or you can just use a downloading program like wget or curl". So, yes, feel free to just use wget or curl to ask the TCP server, and in that case you can assume that no data gets lost. It would be great though (and not much additional) work, if your client can talk with both the UDP and the TCP server. That way you really make the experience that TCP never drops a packet, while for UDP this is a frequent event. '''Hannah 1Nov09 6:49pm''' Hi Mirko + all. All I said is that you can *test* the TCP server via wget or curl. For the exercise you should implement your own client, but since that is very similar to the UDP client, that is not much additional work. For the HTTP part of the exercise, you can, if you want, indeed just use wget or curl and assume that the number of lost packets is zero. '''Hannah 1Nov09 6:35pm''' Hi, I am confused, the exercise-sheet says write a client which can communicate over TCP and UDP and for comparison query the HTTP server via wget/curl. Here you are saying we can download the TCP-part via wget/curl. Therefore i wanted to download the files over TCP via wget, but i couldn't find a way to measure the amount of lost packets, does anyone know? '''Mirko 1Nov09 17:23pm''' To all: Now both servers should run fine. '''Marjan 1Nov09 17:03pm''' To Matthias: Unlike the TCP server, the UDP server is still working fine, I just checked. '''Marjan 1Nov09 16:17pm''' Hi, would you please check the UDP as well? It looks like it isn't returning any data now altough the same implementation worked 2 hours ago. ''' Matthias 1Nov09 3:58pm''' Preliminary fix for the TCP server problem: the TCP server is now automatically restarted as soon as it crashed. So you should be able to work with it properly now. '''Hannah 1Nov09 2:54pm''' Dear all, the TCP is currently crashing whenever the client aborts, and then it's down before we restart it. Marjan is working on solving this problem, and we will tell you as soon as it's done. The UDP server does not have this problem. '''Hannah 1Nov09 2:33pm''' Hi Zhongjie + all, as it says above "The first ten bytes of each packet contain the packet id ...". (But it only does that if a packet is larger than 10 bytes.) For example, if you ask for 10000 bytes, the server will send you 10 packets with 1000 bytes each, with ids from 0 to 9. This is interesting information, because your client can use it to print the packet id of each package it receives and see how many packets arrive out of order. You don't have to do this for the exercise, but it's interesting and easy to do. And as you will see then, out of order arrival indeed happens. '''Hannah 1Nov09 2:07pm''' Well, problem solved... You need to send a package end with a '\0' char to server, otherwise server will not respond... But here is another problem: when I send a UDP package like "5\0" to server, I will receive reply package like "xxxxx". If I send "10\0", the reply is "xxxxxxxxxx". And if it is "20\0" I send, it is "0000000000\0xxxxxxxxx" I receive. Confused... '''Zhongjie 1Nov09 11:20am''' I had the same problem. But try to send your query with two linefeeds at the end, like this: send_data = '50\n\n' This makes the UDP server a lot more responsive... '''Christian 1Nov09 10:26am''' hello Marjan, Now its work, thanks. what do you mean about skipping one ex.sheet without loosing any points? I wonder whether my exercise uploaded on 26 of Oct is still counted? '''Triatmoko 1Nov09 10:19''' Hello! I still could not get any response from both the UDP and the TCP server port by now, but HTTP server works fine. If anyone could get some result, please tell me that you can get response from servers, so that I will know it's my own problem... Thank you! '''Zhongjie 1Nov09 08:43''' Hi, I need some clarification on this term "HTTP result header" in Excercise 2 in sheet 2. Will HTTP header contains generic http information or something related to Results? Offcourse our Result will in HTML form. '''Waleed 1Nov09 5:52AM''' I am sorry for the downtime, these maintenance works were announced already several weeks ago, but then I forgot about them because they were scheduled on a Saturday which I thought would not affect me. The downtime also killed our servers, but now they are running again. About the corrections: of course you should get comments on what you did wrong and why you got less points for what. Sorry, if that didn't happen for the first exercise sheet. I will talk with Marjan. '''Hannah 1Nov09 00:38am''' Yes, Eric. Today afternoon and evening was nearly the whole Uni-Net Offline because of intended maintenance. I have a question: Will there be any correction or comments or a sample solution or something like that for every exercise because I now only know how many points I have for every Exercise in Exercise Sheet 1 but I don't know why I have a lack of 1/2 point in one exercise and in another. I think it would be good if anybody will know what he has done wrong or/and what he could have done better. '''Waldemar 31Oct09 21:08''' I cannot connect to vulcano.informatik.uni-freiburg.de:9999 and :8888, neither from outside, nor from inside (logged into pool account). Also the whole informatik.uni... was down a few minutes ago. It's hard to solve exercise 4 without the servers running. '''Eric 31Oct09 19:00''' Hi, I wonder whether those three servers for exercise 4 of sheet 2 online or not? '''Zhongjie 31Oct09 11:49''' To all: Please note that (almost) everybody will get +2 points for Exercise Sheet 1 (previously I did not assign any points to the first and the last problem). '''Marjan 31Oct09 10:03''' To Triatmoko and Ahmed + all: You obviously haven't created your wiki page. Please go to your link and click "Create a new page" and then "Save changes". You should then upload your solutions there and put the link on the wiki like everybody else did. If it is still not clear please ask some of your fellow students. I don't remember if it is mentioned, but you can skip one Exercise Sheet without losing any points. '''Marjan 31Oct09 09:38''' Hi, I have some question about my exercise Sheet 1, I saw in my exercise page that my name, my upload solution and code in gray color and other persons in blue color. I try to click my attachment file in my exercise page, and I have some message that there are no attachment. any body know about this issue?because my exercise uploaded on Monday 26oct '''Triatmoko 30Oct09 22:18''' Hi Björn + all: For Exercise 4 from Exercise Sheet 1 you had to write code that is at least able to process 2-word queries. If your code can indeed only handle 2-word queries and not an arbitrary number of query words, that is also fine for this exercise, you won't get less points because of that. Your second question is also very valid. You should put the various functionalities into modules / classes of their own, so that you can easily combine them for the three different binaries required for Exercises 1 - 3. Each of your three programs will then be quite short, just putting together the right things. I should have added it to my list of evil coding NoNos: never ever duplicate code, but instead put it in a class / module of its own. I hope this answers your questions, if not please ask again. Sorry for the late answer, but I was super busy until now, hardly had time to breathe. '''Hannah 30Oct09 19:08''' I have a question concerning exercise 2. There was no concrete task to produce "query processing code" on ex sheet 1. Are there any requirements that have to be fulfilled? Should it be able to handle two word queries? n-word queries? Additionally there is something else I want to ask: I think it surely isn't bad practice to write a more generic webserver and use it for exercises 1-3. Apart from that it says "change your code" some times in the exercises. How should your submission behave w.r.t. the exercises? Different src files / executables for each exercise? One program that solve each exercises depending on startup parameters? Anything else? '''Björn 30ct09 2:25pm''' I now reorganized the page. Old stuff went to separate pages (links above). The idea is that the front page is always for the current lecture / exercises. The problem with your exercise page should be solved now, Ivo. '''Hannah 30ct09 00:05am''' Having problems to access my exercise page after loging in - IvoChichkovExercises, '''Ivo 29Oct 22:56pm''' Sorry to bother you. I added the Link to exercise sheet 2 with the linked pdf. Needed this to find the sheet as fast as possible. '''Marius 29Oct 10:04 p.m.''' Is there a webpage for exercise sheet 2 somewhere? '''Johannes 29Oct 07:45 pm''' |
Hi, I just looked at the new exercise sheet 4, in exercise 1 we should generate a permutation and sum the resulting array up, am I wrong or doesn't iterating method two iterate throw the whole array in every situation. for ex.: n= 5 permutation: 1 5 3 4 2, then A[1] = 3, A[A[1]]= A[3] = 1, A[1] = 3 ... '''Daniel 12Nov09 19:44pm''' |
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.
Here are .lpd files of the recordings of the lectures so far (except Lecture 2, where we had problems with the microphone): Lecture 1 Lecture 3 Lecture 4.
Here are PDFs of the exercise sheets so far: Exercise Sheet 1, Exercise Sheet 2, Exercise Sheet 3, Exercise Sheet 4.
Here are your solutions and comments on the previous exercise sheets: Solutions and Comments 1, Solutions and Comments 2, Solutions and Comments 3
Exercise Sheet 3
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 4.
Questions or comments below this line, most recent on top please
To Florian + all: yes, sorry, I forgot to mention this in the lecture. Marjan already explained how to clear the disk cache. Let me add to this an explanation what the disk cache already is. Whenever you read a (part of a) file from disk, the operating system of your computed will use whatever memory is currently unused to store that (part of the) file there. When you read it again and the data hasn't changed and the memory used to store it has not been used otherwise in the meantime, than that data is read right from memory, which is much faster than reading it from disk. Usually that effect is desirable, because it speeds up things, but when you do experiments, it is undesirable, because it leads to unrealistically good running times, especially when carrying out an experiment many times in a row. Hannah 15Nov09 8:10pm
To Florian: Indeed, we were running out of time and there was no room for this in the lecture. I can suggest you few ways how to clear the disk cache: before carrying out your final experiment, read a large amount of data (let's say close to the amount of RAM you have) from disk - this will ensure that your data (the inverted list) is cleared from the disk cache and replaced by something else (thus an actual reading from disk get's timed, and not reading from RAM). Another way is to restart your computer before doing the timing. Marjan 15Nov09 7:27pm
In exercise 4 it says: "Important note: Whenver you measure running times for reading data from disk, you have to clear the disk cache before, as discussed in the lecture". I think that this was not discussed in the lecture? What do we have to do here? Florian 15Nov09 7:15pm
@Bit shifting: The syntax for that is actually the same, irrespectively of whether you use Java, C++, perl, python, or whatever. The >> operator shifts to the right, the << operator shifts to the left, the & operator ands the bits of the two operands and the | operator ors the bits of the two operands. Very simple. You will also find zillions of example programs on the web by typing something like java bit shifting into Google or whatever your favorite search engine is. Hannah 15Nov09 1:16
Hi Marius + all: For Exercise 4, an inverted list of size m with doc ids from the range [1..n] is simply a sorted list of m numbers from the range [1..n]. I leave it to you, whether your lists potentially contain duplicates (as in 3, 5, 5, 8, 12, ...) or whether you generate them in a way that they don't contain duplicates (as in 3, 5, 8, 17, ...). It doesn't really matter for the exercise whether your list has duplicated or not. In any case, consider simple flat lists like in the two examples I gave (and like all the examples I gave in this and past lectures), not lists of lists or anything. Hannah 15Nov09 1:12am
@Mirko: Sure, but an inverted list is a list of words where the Doc-IDs are attached to each words in which the words occur. So for Example: If word no. 5 occurs in Doc1, Doc2 and Doc3 and word no. 2 occurs in Doc5, the list would look like: 5 -> Doc1, Doc2, Doc3; 2 -> Doc5. Or am I mistaken? My question then is, how long should these attached lists be in average case? I mean, one could imagine that we got 1mil. documents over 3 words, so these lists could get very large...
EDIT: Oh ok. Now, I see your point. It's not an index, it's a list. Okay. So, what is an inverted list with Doc-IDs, then?
EDIT EDIT: And to your question, Mirko, take a look at http://snippets.dzone.com/posts/show/93. Especially at Comment no. 2. Maybe this helps... I think, Java supports StreamWriters/Readers that are able to write/read bytes. Marius 11/14/2009 08:46pm
EDIT EDIT EDIT: Sorry, me again. Well, I bothered Wikipedia which redirects from http://en.wikipedia.org/wiki/Inverted_list to Inverted Index. So it seems to me, this is being used as a synonym. Actually, I think I'm confused enough, now. I'll better wait for any responses... Marius 11/14/2009 9:08 pm
@ Marius: i think we are supposed to generate one inverted list of size m, with doc ids from 1..n (therefore n>=m, because no duplicates?).
Now a question from my side: ex.4, programming the compression in java, is there any good tutorial about how to handle the bit-stuff? (otherwise, i think, it would cost me too much time..) Mirko 14Nov09, 19:18
Hi, do you have any suggestions what the best numbers for m and n in exercise 4 should look like? Or are we supposed to mess around a bit with ints and longs? And: How long should the list of documents in the inverted index be? Marius 14Nov09 6:40pm
And just to clarify what a single-cycle permutation is. Here is an example for an array of size 5 with a permutation that is a single cycle: 5 4 1 3 2. Why single cycle? Well, A[1] = 5, A[5] = 2, A[2] = 4, A[4] = 3, A[3] = 1. (My indices in this example are 1,...,5 and not 0,...,4.) Here is an example of a permutation with three cycles: 2 1 4 3 5. The first cycle is A[1] = 2, A[2] =1. The second cycle is A[3] = 4, A[4] = 3. The third cycle is A[5] = 5. Hannah 12Nov09 8:04pm
Hi Daniel + all, I don't quite understand your question and your example (if your array is 1 5 3 4 2, why is A[1] = 3?). In case you refer to the requirement of the exercise that the permutation consists only of a single cycle. That is because your code should go over each element exactly once (it should, of course, stop after n iterations, where n is the size of the array). If your permutation has more than one cycle, it is hard to achieve that. Also note that for both (1) and (2), the sum of the array values should be sum_i=1,...,n i = n * (n+1) / 2. Hannah 12Nov09 7:54pm
Hi, I just looked at the new exercise sheet 4, in exercise 1 we should generate a permutation and sum the resulting array up, am I wrong or doesn't iterating method two iterate throw the whole array in every situation. for ex.: n= 5 permutation: 1 5 3 4 2, then A[1] = 3, A[A[1]]= A[3] = 1, A[1] = 3 ... Daniel 12Nov09 19:44pm