dcdz

ftp://ftp.ets.org/pub/gre/994994.pdf concurs with scores and percentage in the topic above this one.

700 40% and I got accepted to NYU, which doesn't require GRE CS for master degree. I was going to apply to Columbia, with score like that, my chance is slim I guess.

and what does arp accomplish?

answer for the practice book form GR9629 (online copy http://www.cs.cornell.edu/~pradu/CompSci.pdf), the answer is at http://groups.yahoo.com/group/agre-cs-2003/files/ click link csgretestsolved.doc I found this doc somewhere, can't remember. I wish I was the author ;) someone please make this post to be sticky.

depends on the school. it's a mandatory for applying to Phd, but it's not required for MS in many schools, and they don't clarify it on their website, so you have to call the admission office to find out.

well, before I had a hard time of accepting failures. I expect and demand success in most of things I do, and if things didn't turn out as I expected, I get really depressed and feel worthless. Now I carry the philosophy of just doing your best, and don't care too much of the outcome. I'm no long dictated by the test score, the rating of annual performance review, or anything alike. I only have to answer to myself. As long as I've tried my best, I'm content. Agonizing over the past is waste of time and is not going to help you for the future. About applying for school, I don't think school measure each credential with pre assigned weighting. I thinks it depends on each school, admission committee. Some of them may prefer scores, others may prefer recommendations, who knows. I believe as long as your whole package is within their comfort zone and not totally out of wack, they should admit you. Generally, school looking for certain characteristics from candidates, whether its ability of doing independent research, or ability of creative problem solving, or whatever. You have to exhibit such ability through one or more of your credential, through score, letter, statement, one way or another. If you can make your point and that's what they looking for, then you hit the jackpot.

I think the test was easier than the sample test I received upon registration. Not like the some of the questions on sample test I have no idea what the question is asking for, at least I can understand all today's questions by just reading them once or twice. Some are easy, and some I just guessed, even when I'm not certain. I only left one question blank. For first 30 questions I was able to do 1 question a minute. After that, my brain became thick and took me longer to answer a question, but I did have good 15-20 minutes left after I finished all questions, and then I went back to unanswered questions. I found this way works out better, you get a clean start to approach the same old question, so you might not get stuck in the same old dead end and waste time. so when I stuck on a question for more than 3-4 minutes, I moved on the next question. anyway, I feel so much relieved now, no matter how well or how bad I did.

ya, finally it's over. yeeeeeaaaaahhhh!!! I feel so much better now.

partial 2001 test questions http://www.geocities.com/chengliucc/writings/Dump_my.htm

I thank everyone who made this forum a great place for preparing the test. I specially thank wood, AlbaLed, and nonevent99 for your massive quantity of posts, I couldn't learn so much in last few weeks without your around the clock posts. I'll be in Rutgers testing center Saturday. Good luck to all.

http://www.cs.oberlin.edu/classes/dragn/labs/analysis/analysis15.html gave 1st case answer, but no proof.

Can someone give mathmatical proof for the runtime of following? 1. T(n) = 2 T(n/2) + n*log(n) runtime is Theta(n * (log n)^2), Master theorem does not apply in this case. 2. T(n) = T(n/2) + T(n/4) 3. T(n) = T(n/2) + 2T(n/4)

proof of: a ^ (logb N) = N ^ (logb a) a ^ (logb N) = X loga X = logb N logb X / logb a = logb N / logb b logb X / logb a = logb N logb X / logb N = logb a logN X = logb a N ^ (logb a) = X N ^ (logb a) = a ^ (logb N)

disregard my last post. a loga N = N, is there any simplification for: a logb N?

is there a formula for loga (logb N)? example: log3 81 = 4 log4 (log 3 81) = log4 4 = 1, where a=4, b=3, N=81