djm03178's blog

By djm03178, history, 4 years ago, In English

I've seen phrases like this in contest announcements many times:

The score distribution is standard: 500 - 1000 - 1500 - 2000 - 2500 - 3000

However, I don't really see that this 'standard' distribution is balanced in practice or anything, so I don't get why this uniform distribution is called standard.

The score decreasing rate (i.e. the number of points lost per minute) is proportional to the initial score of the problem. For example, in a 2-hour contest, if you solve a 1000-point problem in 1:59, you get 524 points, which is still higher than solving a 500-problem in 0:00. But if you solve a 3000-point problem in 1:59, you only get 1572 points. That's much less than solving 2000 or 2500-point problem fast enough.

In most of Div. 2 contests, when I see the standings table, the general tendency of gained score of each participant is like this: A <<< B < C >= D >= E >= F. This includes my recent contest (yes, I underestimated D a lot).

I think the reason we give more points to a harder problem is to reward participants who manage to solve those problems. But in practice, a better strategy is to almost always solve easier problems first, and as a result, most people gain less points by solving harder problems. Another side effect is that when you fail on system test, the most critical one is failing on B or C, and not the harder problems. Let's exclude those who first read harder problems and decide to participate only when they're confident to solve it fast.

This is also related to Div. 1 / 2 parallel rounds. I see most of these rounds tend to use score distribution of 2C~2F = 1A~1D + 1000 when these problems are the same ones. But this means Div. 2 participants are rewarded much less for solving 2E or 2F than Div. 1 participants solving 1C or 1D, compared to how much they get for solving 2C(1A) or 2D(1B).

Of course, I didn't count getting wrong answers to get penalty, but the average number of tries to get AC won't be that much to affect this in general.

Conclusively, I'd like to argue that it will be better not to hesitate setting higher score on harder problems. Let me suggest a 'balanced' score distribution I think: 500 — 750 — 1250 — 1750 — 2500 — 3500.

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4 years ago, # |
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I agree with your idea, but I suggest something extreme. 250-500-1000-1750-2500-4000 It gives way more importance in solving hard problems, rather than solving easy ones fast. I believe solving hard problems is way more important than fast solving.

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4 years ago, # |
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4 years ago, # |
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I don't really see that this 'standard' distribution is balanced in practice or anything, so I don't get why this uniform distribution is called standard.

Standard doesn't mean good or bad, it means it's a default

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    4 years ago, # ^ |
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    But yes, I agree that geometric progression of points would makes more sense than arithmetical one.

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4 years ago, # |
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I think it's actually a benefit that you're rewarded most for solving the easy problems (other than FST, which is always sad).

This makes the optimal strategy clear: it's almost always to go in order. Removing this strategy element of having to decide what order to go in keeps the focus more on problem-solving, and reduces coin-flippiness of whether you picked the right problem to work on first.

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    4 years ago, # ^ |
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    For example, if ABCD gives more points than DCBA, then if D ends up being too difficult, you'll still solve ABC.

    But if the points are structured so that DCBA is better than ABCD, and D ends up being too hard, you'll lose a ton of points from wasting time on D and potentially solving ABC much later (or never). So it results in much higher variance and tests you more on your ability to decide in a minute or two whether to attempt a hard problem or skip it.

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4 years ago, # |
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I would actually prefer that there were no decreasing in points at all and the tiebreaker should be the sum of times just like in ICPC. In this way the hacks could decrease your penalty time. The benefit of this model is that, if someone solves a, b, c and e they would always rank higher than someone who solves a,b,c,d

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    4 years ago, # ^ |
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    While we are at it, why not change the penalty time from "sum of penalties" to "maximum of penalties" like AtCoder? Or at least compromise by using "sum of squared penalties"?

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      4 years ago, # ^ |
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      As far as I heard the logic is to prevent people from doing things like "Couldn't solve D fast enough, not going to make any submissions at all to not lose rating". Sum of penalties (or squared roots of penalties like you suggest) makes it disadvantageous to start with difficult problems with such thoughts in mind

      Not sure how much it is really a problem though

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    4 years ago, # ^ |
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    definitely 100% agree.

    by having no decrease. it doesn't matter which problem you solve first.

    as example from last time grank forces. I could've gain much more point if I solve E first then F and D. I don't think it make sense to reward someone that solve E in 15 minute and then D in 45 minute more than someone that solve D in 45 minute and then E in 15 minute

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4 years ago, # |
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"But if you solve a 3000-point problem in 1:59, you only get 1572 points. That's much less than solving 2000 or 2500-point problem fast enough."

I think you miss out something. Maybe solving the 2000-point problem in 10 minutes requires the similar amount of experience and knowledge to solve the 3000-point problem in 2 hours.

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    4 years ago, # ^ |
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    Except you don't actually solve the 3000-point problem in 2 hours, you solve it maybe 40 minutes after the previous one. Also no one solves really the 2000-point problems in 10 minutes.