### .o.'s blog

By .o., 7 years ago,

Hi, I'm the author of Good Bye 2014. Good Bye 2014 will be held for 2.5 hours with 7 problems, and it will be division conbined, like last year. However, I have a doubt about the normal score distribution 500-1000-1500-2000-2500-3000-3500.

This post already brought up the problems of the normal distribution. Solving Div1E at the last moment of the contest may be worse than solving Div1A very quickly. So I asked Zlobober, whether the score leak (currently 0.4% per minute) could be changed or not, and unfortunately, there is no option to decrease the score leak currently.

So, I would like to know what would be the best score distribution for Good Bye 2014. I will consider all your opinions and set the score distribution. Thanks in advance! :D

• I don't know how to write or read Russian, so please write in English.

• +106

 » 7 years ago, # |   +41 What about 1k-2k-4k-8k-16k-32k-64k ?
•  » » 7 years ago, # ^ |   +79 It is good for topcoder, but not for codeforces.
•  » » » 7 years ago, # ^ |   +9 I agree with izban.
•  » » 7 years ago, # ^ |   +9 person A solve first 4 problem and person B only solve 5. problem , person B on the person A,is it make a sense?
 » 7 years ago, # |   +36 Well,you should tell us problems,and then we will know what is the best options :-)
•  » » 7 years ago, # ^ |   +80 All the problems will start with "New Year".
•  » » » 7 years ago, # ^ |   +64 with this information I can tell you that all problem scores should end with two zeros (from right)
 » 7 years ago, # |   0 Score distribution highly depend on problem difficulties.I think the best way will be something like:1k,2k,3k,4k,4.5k,5k,7k.
 » 7 years ago, # | ← Rev. 2 →   +58 What about 250-500-750-1250-2000-2750-3500?It will be 1:2:3:5 for Div2 participants who can't solve Div2 E, and participants who can solve Div2E is good enough to be in Div1.It will be 1.11:1.85:2.96:4.07:5.18 for Div1 participants. It is near 1:2:3:4:5.
•  » » 7 years ago, # ^ |   +16 Sounds quite reasonable. Thanks for your opinion.
•  » » » 7 years ago, # ^ |   0 How about 250-500-750-1500-2000-3000-3500? I agree with gs12117's opinion and I think div1 BC problems deserves better score. And considering the score leak, Div1 D needs higher score.
 » 7 years ago, # |   -16 I think 2.5 hour is not enough for 7problems.maybe increase it to 3 ?
•  » » 7 years ago, # ^ |   +3 Then the score leak will be even bigger :D Also, for Div1 participants, Div2A and Div2B will not be a great deal.
 » 7 years ago, # |   +23 Of course, nobody except you can give right score distribution, because nobody knows difficulty of problems.As I remember, Codeforces now can do scores for the problems as multiplier of 250. If I take an ideal round, where all tasks have strictly increasing difficulty, I would make something 250-500-750-1000-1500-2000-2500, or even 250-500-750-1250-2000-2500-3000. Why? Almost all rounds with 7 problems, where I participated, were rather disbalanced: there were 3-4 blitz problems with equal difficulty from red's participant view, and few hard problems. Even when you solved hard problem and you expect good place, if your 1500 task (div1A normally) fail, you can fall down in standings.Another minuse of standard (500-1000-1500-2000-2500-3000-3500) is that challenges are almost useless (compare standard round, when you solve 3 problems with score 2500-3000, and round with 7 tasks, when you have around 6000 points with 5 tasks).
•  » » 7 years ago, # ^ |   0 OK, thanks for your opinion.
 » 7 years ago, # |   -8 My opinion for score distribution 300-500-800-1100-1500-2000-2500. Why?On codeforces one important thing is hack.If the points are higher then all people must have more good hacks for annulled one unsolved problem.Also I think that the hardest problem not worth more then 25 good hacks.Third problem is is easier a lot of then 2 times of 7th.
•  » » 7 years ago, # ^ |   +23 I strongly disagree. 1) Hardest problem is not equal to 35 hacks, it will be much cheaper because no one can solve it instantly, 2) Hacks on Div1+Div2 contest are way too noisy (if you have lots of "greenies" in your room, you get lots of hacks, if you have an "overlord" in your room (tourist, Petr, Egor, ...) — you don't get any hacks).
•  » » » 7 years ago, # ^ | ← Rev. 4 →   0 2) Sorry, but your list of "overlords" should begin: anta, mosiomohsen, kmjp, ... (standings). Every room should have random collections of accounts, which increases probability that there will be similar amount of strong hackers in every room.
 » 7 years ago, # |   +8 What if you asked testers to measure/estimate for each problem how much time they spent/would spend solving it and then set scores proportional to these times?
•  » » 7 years ago, # ^ |   +11 There are not that many testers to set scores according to their performance..
 » 7 years ago, # |   +19 How it is possible that "solving Div1E at the last moment of the contest may be worse than solving Div1A very quickly"? If you solve A instantly, you get only 500 points; while solving E at the last minute results in 2500 * (1 — 0.004 * 150) = 1000 points (well, you can get less points if you have multiple incorrect submissions, but you cannot get less than 2500 * 0.3 = 750 points).IMO score distribution is good enough as is. If you increase score(G) / score(A) ratio, this will essentially enforce strategy "solve problems in reverse order", because score drop with time for hard problems will be way too large.
•  » » 7 years ago, # ^ |   +8 I think that he is talking about problems of standard 7-tasks scoring. Div1A costs 1500, Div1E costs 3500, and 1500 > 3500 * 0.3 = 1050.
•  » » 7 years ago, # ^ |   +24 In here, Div1A is worth 1500, and Div1E is worth 3500. I get 1500 points if I solve Div1A instantly, but if I solve Div1E at the last moment, I get only 3500 * (1 — 0.004 * 150) = 1400 points.
•  » » » 7 years ago, # ^ |   +8 I see. So this is essentially the comparison between tasks C and G. I still think this is not a big deal — no one will be able to solve C instantly (most of the people will solve the problems in "natural" order and spend some time on A and B), so it will be cheaper than G submitted at the last minute.Fast D is still better than slow G, but I think this makes some sense. Anyway, this is better than the alternative of enforcing G->A solving order.
•  » » » » 7 years ago, # ^ |   +16 However, fast D is not better than slow G in Div1. Also, in the standard distribution, the difference between the two scores is quite big, even if I consider the time needed to solve A,B,C. This is why I don't like this score distribution. This makes people who solved hard problems but failed easy problrmd by a silly mistake under people who only solved easy problems.
•  » » » » » 7 years ago, # ^ |   +8 Ok, you have a point. If difficulty(D) == difficulty(Div1B) in usual contest, probably this scoring is not the best option. You can try to reproduce normal Div1 scoring by doing smth like "166-333-500-1000-1500-2000-2500" (then the prefix ABCDE is similar to normal Div2 contest, and CDEFG — to normal Div1 contest), but this probably needs some scaling because of hacks.
•  » » » » 7 years ago, # ^ |   +8 > no one will be able to solve C instantlyA,B,C: 4,10,20 minutes seems reasonable for a red coder. That gives total point loss of 168 over all 3 problems. Still considerably less than the difference 1500-1050.
•  » » 7 years ago, # ^ | ← Rev. 3 →   0 In this contest, the normal score distribution will be 500-1000-1500-2000-2500-3000-3500 because we have 7 problems. Div1 A is worth 1500 points, Div1 E is worth 3500 points. If you solve Div1 A instantly, you get 1500 points, and if you solve Div1 E at last moment, you get 3500*(1-0.004*149)=1414 points.Edit: sorry, I didn't saw other people's reply before writing this...
•  » » 7 years ago, # ^ |   +16 "IMO score distribution is good enough as is. If you increase score(G) / score(A) ratio, this will essentially enforce strategy "solve problems in reverse order", because score drop with time for hard problems will be way too large."And what's bad about enforcing strategy "solve problems in reverse order"? I believe that solving harder problems helps improving your skills much more than solving easy ones. IMO if the score distribution encouraged solving harder problems first, contests would become much more interesting.
•  » » » 7 years ago, # ^ |   +8 Well, but when you don't know what problem you are able to solve it becomes just more random, doesn't it?(You should choose the right problem to start with and go in reversed order)
•  » » » » 7 years ago, # ^ |   0 I agree that choosing the optimal order of solving would become much less trivial. If you don't like the 'strategical' aspect of contests, then it is just more random for you. If you do, then it's much more interesting.
•  » » » » 7 years ago, # ^ |   +8 I suppose that'd be the "interesting" part. But since not everyone is competing for lulz, that wouldn't be such a good idea.
•  » » » 7 years ago, # ^ |   +16 First, why enforcing the strategy is bad. In normal codeforces contests, scores are well-balanced and participants are able to use whatever strategy they like. Some people solve problems in reverse order, some — in the "A-E" order (many people actually like to have some warm-up before solving hard problems), some — in mixed order, solving nicer problems first. Reducing strategy diversity and forcing all top participants to solve problems in reverse order makes contests less interesting IMO. I believe that solving harder problems helps improving your skills much more than solving easy ones That's of course true, but from the point of view of contest results, it is much more noisy. And noise kills the purpose of rating system. So, yes, it is more interesting to solve hard problems, but for the purpose of public contests, one needs both easy and hard problems to make a good contest.
 » 7 years ago, # |   +80 What about making a large score gap between D and E with reasonably gradual changes in the A-D and E-G groups, for example 250-500-1000-1500-3000-3500-4000?Take Codeforces Round #270 as an example. If E-G do have very few solutions and D has over 9000 1000, then solving E is already better than solving C (assuming no pretest fails and worst case) and often better than solving D. Solving G is always better than solving D. It doesn't matter so much which hard problem you solve, but if you are able to solve even one quickly, then you have a practically guaranteed high place.Of course, different scorings for different difficulties. If E isn't that hard or D is harder, then the gap may be between E or F instead... but don't let yourselves be limited by arithmetic (or close to arithmetic) sequences.
•  » » 7 years ago, # ^ |   +9 Oh, it's a creative idea! Thanks for your comment.
 » 7 years ago, # |   0 What about 250 500 750 1500 2250 3250 4250 ?Or multiply these by two, but 500 1000 1500 3000 4500 6500 8500 will make hacking worthless.
 » 7 years ago, # |   +10 Is this good? 500-750-1250-2000-3000-4250-5750The harder the problems is, the more difference it has with the easier ones!
•  » » 7 years ago, # ^ |   +8 The differences form an arithmetic sequence :)
 » 7 years ago, # |   +1 Is it possible to have both dynamic and normal scoring ? i.e. dynamic scoring for hard problems and normal scoring for the easy problems.
•  » » 7 years ago, # ^ |   0 And what about dynamic scoring for all problems?
•  » » » 7 years ago, # ^ |   0 I think it is the best idea for rounds with many problems.
•  » » » » 7 years ago, # ^ |   +50 One comment doesn't mean only one sentence:-)
 » 7 years ago, # |   -26 In my opinion, since 2015 is coming year and 2014 is ending so I think, Score for a problem could be 2014 or 2015.
•  » » 7 years ago, # ^ |   0 It can't be. Scores must be multiples of 250.
 » 7 years ago, # | ← Rev. 2 →   0 250-500-1000-1500-2000-3000-4000 ? Price of easiest A is twice smaller than usually, price of hardest G — 1.6 bigger than usually. Maximum score — sum of prices — 12250 is 1.633 more than usually, and there are 1.4 more problems, participants can hack at suitable price.
 » 7 years ago, # | ← Rev. 3 →   +53 how about Fibonacci sequence 250 500 750 1250 2000 3250 5250(just kidding)
•  » » 7 years ago, # ^ |   +19 +1. This actually isn't a bad idea to be honest..
 » 7 years ago, # |   +4 Normally score of Div1E is equal to 5 times of Div1A. And for Low Candidate Masters and High Experts it's needed a great effort to solve Div2D than Div2C. So,Div2D should get a lot more scores than Div2C as it actually gets in Div1,in fact double. 250 — 500 — 750 — 1500 — 2250 — 3000 — 3750 would be good if we consider this i think.
 » 7 years ago, # |   -8 dynamic distribution with start point 4 000
 » 7 years ago, # |   0 Hah, nice that my post is mentioned :). I even PMed Zlobober yesterday asking whether score drain can be smaller and suggesting that score distribution should be appropriately adjusted :P.In that blog entry I wrote I suggested distribution 250-500-750-1000-1500-2000-2500. In my opinion it's pretty cool, because for Div2 guys it's like scaled 500-1000-1500-2000-3000 and for Div1 guys it's something which everyone gets accepted + 750-1000-1500-2000-2500, so it also pretty resembles standard distribution. But I think that what Xellos suggested is even better, because it also artificially handles with big drain problem. Even if drain will do the bad thing than 250-500-1000-1500-3000-3500-4000 distribution assures that solving harder problems will guarantee a good place.
 » 7 years ago, # |   +3 It can be whatever but not dynamic scoring...
 » 7 years ago, # |   0 You can multiply all scores you wrote by 2 (1000,2000,3000,4000,5000,6000,7000). It should be good enough.
 » 7 years ago, # |   +6 Solving Div1E at the last moment of the contest may be worse than solving Div1A very quickly Is it a big problem? There is the same situation in TC: solving DIV1-500 in 45 mins may be worse, than solving DIV1-250 in 10 mins. But solving BOTH of them is better than solving DIV1-250 in less than minute. Roughly speaking, according this distribution, the winner is the one who can solve as many problems as possible, and the winner is not the one who can solve the hardest problem in 74 mins and nothing more. Who are you tipping as the winner? Thereby the score distribution can be defined. It's up to you.
 » 7 years ago, # |   -8 500 — 1000 — 1500 — 2500 — 3750 — 5000 ????
•  » » 7 years ago, # ^ |   0 There are 7 problems.