ICPCNews's blog

The 2023 ICPC World Finals Luxor
By ICPCNews, 2 years ago, In English

text Hello Codeforces!

The 2023 ICPC World Finals Luxor will begin on April 18, 2024 at 10:00 UTC. This time, we have a double World Finals. You can distinguish them by colors: blue (fish emoji) and green (crocodile). Join us on the live broadcast for this awesome double event of the year in competitive programming; commentators will include ecnerwala, SecondThread, Egor, and more!

ICPC World Finals Luxor is hosted by The Arab Academy for Science, Technology, & Maritime Transport (AASTMT), the ICPC World Finalist teams for two seasons will compete for World Championship awards, prizes, and bragging rights. More than 130 teams of each season represent the best of great universities of eight ICPC regions.

Teams qualified to ICPC WF Luxor:

Some useful links:

Scoreboards:

46 47

All available broadcasts:

MAIN AR RU ES PT CN ID

Also, you can observe teams' monitors and web cameras on the separate broadcast. Sent team's hashtag to the chat to see your favorite! All hashtags for both seasons were collected in this list

Split screen 46 Split screen 47

We wish good luck to all competing teams to have a great time spending, and to do the best to get amazing results!

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2 years ago, hide # |
 
Vote: I like it +10 Vote: I do not like it

Why is the Mirror linking to tourist's twitch?

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2 years ago, hide # |
 
Vote: I like it +38 Vote: I do not like it

Will be there any live contest mirror? Can you please share the link?

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2 years ago, hide # |
 
Vote: I like it +34 Vote: I do not like it

On their main website icpc.global, they have a link to an online mirror: https://onlinejudge.icpc.global/. The icpc.global website top-left menu has pretty useful links :).

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2 years ago, hide # |
 
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what is the meaning of double world finals?

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2 years ago, hide # |
 
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Is there a prize for winning an ICPC world final? or just normal contest?

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2 years ago, hide # |
 
Vote: I like it +3 Vote: I do not like it

For all official and mirror participants, good luck and don't forget to have fun!

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2 years ago, hide # |
 
Vote: I like it +5 Vote: I do not like it

fix the live stream plz!

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2 years ago, hide # |
 
Vote: I like it +85 Vote: I do not like it

here we go again:
https://zibada.guru/finals/

separate pages:
https://zibada.guru/finals/2022/
https://zibada.guru/finals/2023/

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2 years ago, hide # |
 
Vote: I like it +21 Vote: I do not like it

Looks like the intersection of the problemsets is 5 problems. Shame. Now for training purposes you have to choose one or the other.

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2 years ago, hide # |
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Who won? Can anyone tell from stream who won?

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2 years ago, hide # |
 
Vote: I like it +16 Vote: I do not like it

I want to congratulate the [HSE] FFTilted team on their well-deserved victory at the 47th ICPC World Final!

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2 years ago, hide # |
 
Vote: I like it +40 Vote: I do not like it

I wanna go home :((

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2 years ago, hide # |
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Vote: I like it +54 Vote: I do not like it

Please feed cowboys to the sharks and let someone else announce the results.

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2 years ago, hide # |
 
Vote: I like it +67 Vote: I do not like it

jiangly won 1st place.

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2 years ago, hide # |
 
Vote: I like it +17 Vote: I do not like it

kudos to Um_nik for being such a wonderful host

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2 years ago, hide # |
 
Vote: I like it +68 Vote: I do not like it

If antontrygubO_o has a million fans, then I am one of them. If antontrygubO_o has ten fans, then I am one of them. If antontrygubO_o has only one fan, then that is me. If antontrygubO_o has no fans, then that means I am no longer on earth. If the world is against antontrygubO_o, then I am against the world.

Idol.

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2 years ago, hide # |
 
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Congratulations to the Peking Universe and National Research University Higher School of Economics！And thanks to all the competitors for performing such an excellent game！

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2 years ago, hide # |
 
Vote: I like it +17 Vote: I do not like it

Congratulations to jiangly!!!

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2 years ago, hide # |
 
Vote: I like it +5 Vote: I do not like it

Are solution sketches available anywhere? In particular, I'm wondering what's the intended solution to problem E.

My Approach
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    2 years ago, hide # ^ |
     
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    We had the same approach. But we couldn't code it clearly and we didn't have enough time. The author's solution is also interesting. If you find an editorial, please share.

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    2 years ago, hide # ^ |
    ← Rev. 3  
    Vote: I like it +38 Vote: I do not like it

    That looks largely correct. The main idea to speed up the solution and reduce its memory consumption is: for each continuation, instead of just counting the number of recurrences that begin with that continuation, compute all possible next values to this continuation, along with their frequencies. This lets you prune out a bunch of useless continuations since the values get very sparse after the first few values in the sequence.

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      2 years ago, hide # ^ |
       
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      Would you mind providing solution code? Don't quite understand how the pruning works.

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        2 years ago, hide # ^ |
         
        Vote: I like it +36 Vote: I do not like it

        SnapDragon has uploaded his code to GitHub.

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          2 years ago, hide # ^ |
           
          Vote: I like it +36 Vote: I do not like it

          Great, I'll take a look.

          Btw, anyone know the proofs for Zoo Management / Schedule? I came up with the necessary condition / some reasonable construction, though I wasn't able to show that the condition is sufficient / the construction is the best possible when the answer is odd.

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            2 years ago, hide # ^ |
             
            Vote: I like it +10 Vote: I do not like it

            For zoo management:

            There are a lot of results you can get from this: https://en.wikipedia.org/wiki/Jordan%27s_theorem_%28symmetric_group%29 (assume we are talking about a BCC that is not just a cycle--I think the permutation group generated by the cycles is always primitive and transitive)

            • There is a three-cycle we can get by taking the commutator of two cycles that share a single vertex; so this is enough for the "even permutations only" case and it is enough whenever two cycles meet at just a vertex.

            • I think we should be able to get a three-cycle whenever two cycles meet at 2+ vertices, but the most interesting thing I can construct (also by the commutator of these two) is a pair of transpositions that are "displaced". My guess is that you can produce either a single transposition or a three-cycle by some conjugation and three of these moves.

            Please let me know if you can finish the second bullet, since I'm also struggling to finish this.

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              2 years ago, hide # ^ |
               
              Vote: I like it +48 Vote: I do not like it

              I think we should be able to get a three-cycle whenever two cycles meet at 2+ vertices

              The nice idea here is that in this case you actually have three simple cycles in a graph.

              So if you draw a picture of 2 cycles like OO with intersection in the middle, you can rotate these 2 cycles clockwise, and then rotate the outer cycle counterclockwise (the outer being formed by a set of edges from symmetric difference of 2 "O" cycles). And that gives you a single trasposition.

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2 years ago, hide # |
 
Vote: I like it +19 Vote: I do not like it

Did tourist, petr and ksun manage to solve all the problems during their stream?

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    2 years ago, hide # ^ |
     
    Vote: I like it +3 Vote: I do not like it

    yes they aked with 2min left

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      2 years ago, hide # ^ |
       
      Vote: I like it +8 Vote: I do not like it

      Pretty amazing stuff. It seemed that they started a bit slow (relative to some of the competitors), but they have caught up during the contest (I missed the latter part of their stream). Glad that they finished in time!

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2 years ago, hide # |
 
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Link for results table?

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2 years ago, hide # |
 
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Where is solution?I don't know why my solution to problem U got WA...

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2 years ago, hide # |
 
Vote: I like it +14 Vote: I do not like it

What would be pretty cool is if someone can create a scatter plot of the average Codeforces rating of each team vs the number of problems they have solved.

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2 years ago, hide # |
 
Vote: I like it +22 Vote: I do not like it

As a member of Jiangly Fan Club, I'm so excited that jiangly won the 46th ICPC world finals!

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2 years ago, hide # |
 
Vote: I like it +18 Vote: I do not like it

Will there be an editorial? I'm really curious Alea Iacta Est and three kinds of dice.

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    2 years ago, hide # ^ |
     
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    They've uploaded solution videos here

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      2 years ago, hide # ^ |
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      Yeah it was helpful for most of the problems that I was curious about, but Alea Iacta Est is still confusing to me. I am having trouble visualizing the state graph, and how the cycles happens, and how using the min heap helps with cycles. I wonder if someone can provide a larger example of state graph than the video editorial. I recognize the problem is beyond my skill level, but it I feel I'm just too close to understanding it now. If I could just understand the state graph I think I would get the transitions and how min heap applies.

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        2 years ago, hide # ^ |
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        I think I have a general picture of it a little now, but this part I'm having trouble rationalizing. best_expectation = (pw + current_expectation[mask]) / counts[mask] This is how I roughly interpret it.

        • pw is a variable that represents if I select subset of dice to roll {1, 3}, this calculates the number of possible configurations from that subset, so for this it is 6^2. And it represents the number of expected rolls to achieve the current target configuration.

        • current_expectation[mask] is the accumulated expected number of rolls to take a given configuration with some dice fixed to a face value, and picking the subset of dice to roll {1, 3} to achieve a configuration in the dictionary.

        • counts[mask] is the total number of visited complete configurations (all face values fixed) that are rolled from this current mask with some fixed face values.

        Some observations

        • counts[mask] = 6^number_of_dice_to_roll, after it finishes exploring everything, so like if mask represents A?B?, then it will have 6^2 = 36 counts at the end.

        • It first explores complete configurations that are in dictionary, and in those cases the current_expectation[mask] = 0, and the best_expectation = pw / counts[mask] really, so if there are 3 complete configurations that can be reached from some state of dice to roll selection, then it will be pw / 3, which seems reasonable.

        • Then it explores the lowest expectation values for incomplete states, and finds all the unvisited complete configurations that are intermediate, that is they could lead to config in dictionary. Or in other words, for all the possible selection states of rolling -> PACB -> ?ACB -> KACB (in dictionary). I think this is where my understanding really starts to drop off.

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2 years ago, hide # |
 
Vote: I like it -13 Vote: I do not like it

jiangly orz !!!!!

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2 years ago, hide # |
 
Vote: I like it +21 Vote: I do not like it

Congratulations to the Peking Universe and National Research University Higher School of Economics！And thanks to all the competitors for performing such an excellent game！

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2 years ago, hide # |
 
Vote: I like it +8 Vote: I do not like it

When will editorials be made available?

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2 years ago, hide # |
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Vote: I like it -36 Vote: I do not like it

_*_*_

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2 years ago, hide # |
 
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Is there any way I can resubmit for the problems? I was busy during the mirror contest (which is ended) and now I really want to try.

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