Xellos's blog

Invitation to IPSC 2014
By Xellos, 11 years ago, In English

Badum badum...

The Internet Problem Solving Contest takes place again in 2014! The registration's started (those of you who competed last year should've gotten an e-mail about it) and will be open until the end of the contest.

The contest will take place from 15.6.2014 10:00 UTC 15.6.2014 11:00 UTC to 15.6.2014 15:00 UTC 15.6.2014 16:00 UTC and will be preceded by a short practice session.

You can find the registration form, archive and everything at the contest site.

The rules can be found at the contest site or in my blog post about last year's IPSC. That's why this post is short — you can just read that one and replace the dates.

Also, if you think one blog post isn't a good enough emphasis, here are two more :D

*CONTEST HAS ENDED!* You can find the solutions in the archive.

Here is a summary of some registered teams by CF handles. It's not very long, but better than nothing.

Team name Member 1 Member 2 Member 3 Place
Zenith I_love_Hoang_Yen flashmt ngfam_kongu 39
MSU Trinity Zlobober sankear malcolm 31
cutie, dropout & useless woman vadimmm Rubanenko baba_beda 113
XZ Team eatmore AlexFetisov winger 26
KPZ eduardische popoffka Alex_2oo8 28
Break a leg Olja Oleg805 andgein 270
Charles University Legion fhlasek k21 simsa.st 12
SPb SU 4 Dmitry_Egorov PavelKunyavskiy yeputons 9

And separately for single-person teams.

Team name Member 1 Place
Alexander Udalov udalov 28
duolCauqA Aquacloud 170
Xellos Xellos 15
Snowbear marat.snowbear 44
Futures Leaguer gs12117 12
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10 years ago, # |
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This post should stay in recent actions.

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10 years ago, # |
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Check up the link to the contest site

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    10 years ago, # ^ |
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    Hm, the link's acting strange: it was written correctly in HTML, but http://codeforces.me/ was added before it when displayed on the page. Adding http:// in it solves the problem.

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10 years ago, # |
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I have Final exam At that time :(

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    10 years ago, # ^ |
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    Shit happens. I had an onsite competition during CF round 248.

    Maybe you could write an alternative one later? It's already bad enough when there aren't multiple dates for exams.

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10 years ago, # |
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It would be nice if we start posting our teams associated with CF handles.

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10 years ago, # |
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If authors read this topic, please, publish pdf version of statements on some mirror (maybe with link here on CF) when contest starts. Every year there are problems with site accessibility.

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    10 years ago, # ^ |
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    It's best to write right to the official contact mailbox — if it doesn't get read there, then it's hopeless anyway :D

    I'll try to dropbox or something the statements and archive when they're available and I download them, at least.

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    10 years ago, # ^ |
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    They allow to download full package (problem set, in files, etc.) before contest starts and after that just wait for a short text password. It's easy and we don't have any problems with that (this and previous year)

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10 years ago, # |
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XZ team: eatmore AlexFetisov winger

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KPZ: eduardische popoffka Alex_2oo8

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10 years ago, # |
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Count me in.

single-person team: duolCauqA

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10 years ago, # |
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As usual, very nice and fun contest. Huge thanks to the organizers!

The only technical downside in my opinion was problem E — we've realized too late in the contest that it gives full information about cases and that is exploitable to get the full solution, but by that time the queue was too large and we had to wait 2-5 minutes for each test to evaluate. Sure, this problem is unavoidable, but the system also didn't let us submit the next code without waiting for the result for the previous one (which would help massively, since all 6-9 submits to get the information needed are totally independent of each other), so we couldn't make it in time. So, I would really prefer if this restriction wasn't present in the future.

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10 years ago, # |
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How to solve problem I hard?

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    10 years ago, # ^ |
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    The solutions have been published already. Short idea: fix a vertex; we need to assign a binary string to each son of this vertex, so that the sum of (substring length * son's subtree size) is minimal; the optimal solution is known as Huffmann encoding.

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      10 years ago, # ^ |
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      Thanks for the response! Did you know before the contest about Huffmann encoding?

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        10 years ago, # ^ |
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        Yeah, I did. But during the contest, it was more of a guesswork: I tried various greedies and then realized that what we're trying to do (build a binary tree whose leaves are substrings with cost = depth * number of occurences) is really similar to what Huffmann encoding does — I guessed that it could be Huffmann and then realized why it works.

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      10 years ago, # ^ |
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      The solution for I is not published yet.

      Also, fun fact, we somehow managed to solve it without Huffman encoding by sorting the sons' subtree sizes in descending order and then calculating the minimal sum with following DP: a[i][j][k] — max sum if we processed i sons, and are left with k different strings of length j. So, if (N-i) <= k, we just fill the remainder with j, otherwise we can go from a[i][j][k] to a[i+1][j][k-1] (k > 1) and a[i][j+1][min(N-i, 2*k)].

      This is O(N^3), but in reality we don't use most of the cases — the only thing we couldn't solve by this was the root of the final test case in I2, but the sons' subtree sizes are 12479 and 1 (4999 times), so this bit is solvable by hand.

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        10 years ago, # ^ |
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        or you can wait for the run for 2-3 minutes.

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          10 years ago, # ^ |
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          We used map for storing DP values for faster execution, and we ran out of memory for it, so we received exception rather than the code was executed for too long.

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        10 years ago, # ^ |
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        I tried to solve I with a similar O(N^3) DP, except i wasn't able to solve the last case, after some wrong answer, i manage to remove [j] from my DP state by precompute the suffix sum of the subtree size, so a[i][k] go to a[i+1][k-1] (k > 1) and s[i]+a[i][min(N-i, 2*k)] which run in O(N^2).

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How to solve G2 and M2?

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10 years ago, # |
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I'm surprised at the low number of solutions of F hard, considering it was just about googling the first 107 digits of π (slightly more doesn't matter, a lot more isn't a good idea anyway), finding all first occurences of permutations in it and writing a backtrack. I could've solved it if I had just a bit more time...

The condition that we want the smallest occurence actually helps optimize the naive backtrack — since the order of the 3 triples of lines and that of lines in each triple doesn't matter, then we can limit ourselves to searching sudokus with o[0] > o[3] > o[6] and o[3k] > o[3k + 1] > o[3k + 2] (o[i] is the offset of the i-th row). This way, if we find a solution, we know that it's optimal.

Plus, there are just 3.6 thousand different permutations among the first 107 digits of π, so we can precompute the conditions "can lines x and y be rows of 1 sudoku?" and "can lines x and y be rows of 1 triple?" (triple: lines which form 3 neighboring squares) to make the code faster.

My code has these as only observations (the rest is raw parsing) and runs in 5 minutes on my computer.

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    10 years ago, # ^ |
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    Nowadays, most contests only set problems with fancy efficient algorithmic solutions. Maybe the art of writing good brute force solutions is getting lost :)

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      10 years ago, # ^ |
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      That's why I love IPSC :)

      It's really a PROBLEM SOLVING contest, rather than (academic) algorithmic/programming contest.

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    10 years ago, # ^ |
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    Well, we had first 100 millions digits of pi. So I got 46K rows, which was enough to solve F1, but then I figured out that we could group rows into triples, which could solve F2, by iterating over 3 triples. But, of course, with 46K rows my triple computation was taking too long and I didn't thought of reducing 46K to 1K (like in solution).

    So, indeed, sometimes less is better.

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Can anyone explain why the solution for D1 works? I meant how could "cat easy.in | bunzip2 | bunzip2 | tr -d ’\000’" not take 1TB of hard disk? What is the mechanism behind this? Thanks :)

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    10 years ago, # ^ |
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    The piped programs run in parallel, not sequentially. The pipe can be seen as a temporary buffer. On my Linux, the size of this buffer is 64 kB. Roughly, it works as follows: whenever a program wants to read from an empty pipe or write into a full one, it is suspended until it can do so. In this way, you get a simple synchronization between the piped programs:

    • whenever cat easy.in writes a bunch of data, the first bunzip2 reads it
    • whenever the first bunzip2 reads an entire compressed block of data, it decompresses it and outputs the result
    • detto for the second bunzip2
    • tr operates character by character, so whatever it reads, it outputs immediately

    So there never is the entire unpacked file in memory anywhere. At any moment, each of the bunzips only sees a small part of the file it decompresses.

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    10 years ago, # ^ |
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    It's pipe technology. Every input/output happening in three inner pipes can be made using temporary buffer of little size (64kb on linux by default). For example, cat easy.in writes 64kb of output then freezes until bunzip2 starts reading this buffer producing his output tp the next pipe and so on.

    The main idea is that cat, bunzip2 and tr are stream-processing algorithms that need only O(1) of input data to produce O(1) of corresponding output data.

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