Monday, July 19, 2021

A reversal week

TopCoder SRM 809 was the first event of the last week (problems, results, top 5 on the left, analysis). This was apparently the first SRM that counts towards the TCO22 qualification, and I have completely overlooked this fact and did not prioritize participating :( Congratulations to those who pay attention, and especially to tourist on solving the hard problem almost twice faster than everybody else and thus earning the clear first place!

The VK Cup 2021 Elimination Round for those who speak Russian and qualified, which was also available as Codeforces Round 733 for everybody else, followed on Saturday (problemsresults, top 5 on the left, parallel round results, top 5 on the right, analysis stream, analysis). Everything went pretty well for me in this round, I got the solution ideas quickly and only had to do heavy debugging once, in problem G. However, Um_nik chose a superior strategy and went for problem H in the end which gave him much more points, and won the round. Well done!

More worryingly for me, despite me feeling pretty good about my performance on the first six problems, some usual suspects in the parallel round were faster to solve those six problems and managed to solve both harder problems to boot! Congratulations to jiangly and ecnerwala. It seems that just having a good day is no longer enough for me to compete for the first place at Codeforces :)

The hardest problem H was a pretty nice example of DP optimization. You have a tape that is infinite in both directions and is split into cells. You're going to write a given random permutation of size n<=15000 onto this tape in the following manner: you write the first number somewhere, then you either stay in the same cell or move one cell to the left or to the right, then write the second number there, then again either stay or move to an adjacent cell, write the third number there, and so on. Whenever you write a number into a cell that already has a number, the old number is overwritten and therefore discarded. After all n numbers have been written, we look at the resulting sequence written on the tape from left to right. What is the maximum possible size of an increasing subsequence of this sequence? Note that the given permutation is guaranteed to have been picked uniformly at random from all permutations of size n (which conveniently makes preparing the testcases for this problem so much easier!)

Thanks for reading, and check back next week!

Sunday, April 25, 2021

A cycle week

Codeforces Round 718 (also known as "Contest 2050") took place on Friday (problems, results, top 5 on the left, analysis). Benq solved the first seven problems 20 minutes faster than Um_nik, who in turn solved them almost half an hour faster than all other contestants. They were the only two top contestants who therefore had time to submit something in the last problem, it did not work out but nevertheless their first and second place were safe with a big margin. Congratulations on the impressive performance!

Google Code Jam 2021 Round 1B finished a few minutes ago (problems, results, top 5 on the left, analysis). neal_wu was the first to score 100 provisional points, but he made one incorrect attempt and therefore had to sit and wait for 4 minutes to see if somebody will overtake him. Nobody else was as fast, though, so Neal kept the first place. Congratulations to him and to all 1500 Round 2 advancers!

In my previous summary, I have mentioned a Codeforces problem: You are given n<=2000 points on the plane such that no three points lie on the same line. Each point has an integer label between 1 and n, and all labels are distinct. Your goal is to rearrange the labels in such a way that the i-th point has label i. You are allowed to take any two points and swap their labels, but when you do that you must draw a straight line segment connecting those two points. The segments you draw must not intersect except at their endpoints (in particular, you are not allowed to draw the same segment twice).

The key trick to make progress in the problem is to find a less general, but still non-trivial case which we can solve. It turns out that such case is the one when the initial labeling of points forms a single cycle, for example: point 1 has label 2, point 2 has label 3, and so on, until point n which has label 1. In this case we can solve the problem using only the swaps that include point 1, and therefore their segments will not intersect: we start with labels 234...n1. We swap the labels of points 1 and 2, and we get 324...n1. We swap the labels of points 1 and 3, and we get 423...n1. We continue with swapping the labels of points 1 and 4, and so on, and eventually we get n234...(n-1)1, and finally we swap the labels of the first and last points to get the identity permutation. Note that our choice of point 1 as the "pivot" of all swaps was arbitrary, and we could do the same with any other point.

What should we do if the labels do not form a single cycle? Let's make some additional swaps before using the above approach to make sure they do! More specifically, let's assume our points are sorted by angle with respect to point 1. The above solution will only draw segments between point 1 and other points. Therefore we are free to swap the labels between adjacent points in this sorted order, as those segments will not intersect each other and the segments to point 1.

The initial permutation has some number of cycles, and whenever we swap two elements from different cycles they merge into one. While we have more than one cycle, we can find two adjacent points in the sorted order that belong to different cycles, and swap them to merge those cycles. We repeat this until we have just one cycle remaining, and then apply the single-cycle solution.

There are some additional small details to be figured out, which you can find in the official editorial. I could not solve this problem myself, in part because the space of possible approaches is so vast, and yet most of them do not seem to work. I've checked the solutions for this problem from the top finishers, and they all seem to use this approach. In fact, I'm really curious: is some fundamentally different solution possible here? If not, does there exist some intuition why?

Thanks for reading, and check back next week.

Sunday, April 18, 2021

A yuri week

Codeforces Round 715 was the main round of this week (problems, results, top 5 on the left, analysis). The first three problems were quite approachable, and the real fight for the top spots started on the remaining three. ecnerwala went for the 4000-point F instead of the 2250-point D, and this turned out to be exactly the right plan, as there was not enough time to solve all problems. Congratulations on the victory!

I have tried to make the same plan work, but unfortunately I could not solve F in time — my solution was written at the end of the contest, but it had two issues:
  • It did not work on the samples
  • Its complexity was O(n2*log(n)) in the worst case (even though with a 7-second time limit this was not obviously bad)
I could make it work on the samples in just a couple of minutes after the end of the contest, and the solution passed the system tests — only to be hacked as the complexity was in fact too big. I would've probably reached the second place if I had those couple of minutes, and I'm still on the fence whether the fact that I'd have squeezed in an incorrect solution makes me regret this more or less :)

I'd like to highlight another problem though, for which I had completely no working ideas during the contest: problem D. You are given n<=2000 points on the plane such that no three points lie on the same line. Each point has an integer label between 1 and n, and all labels are distinct. Your goal is to rearrange the labels in such a way that the i-th point has label i. You are allowed to take any two points and swap their labels, but when you do that you must draw a straight line segment connecting those two points. The segments you draw must not intersect except at their endpoints (in particular, you are not allowed to draw the same segment twice). Can you see a way to achieve the goal?

In my previous summary, I have mentioned a Code Jam problem: you are given up to 1015 cards, each with a prime number between 2 and 499. You need to split the cards into two piles, such that the sum of the numbers in the first pile is equal to the product of the numbers in the second pile, or tell that it's impossible.

The key idea here is that the product grows very quickly, so the second pile will always be very small. Therefore the sum of the numbers in the second pile will be small. But the sum of the numbers in the first pile is the sum of all given numbers minus the sum of the numbers in the second pile. If the sum of all given numbers is S, we just need to check all numbers in the segment [S-C,S] as the candidates for the sum of the numbers in the first pile (which is equal to the product of the numbers in the second pile) for some relatively small value of C.

How do we check a particular candidate? Here the fact that all numbers are prime comes into play: since we know that the candidate is a product of some subset of the given numbers, and all of them are prime, there is in fact a unique decomposition for it as a product of primes. Factorizing a number this big can still be problematic, but since we're only interested in prime factors up to 499, it is fast enough.

You can check the official analysis for more details, such as what is a reasonable value for C. Thanks for reading, and check back next week!

Sunday, April 11, 2021

An ESP week

Google Code Jam Round 1A was the first round of this week (problems, results, top 5 on the left, analysis). The top 1500 have made it to Round 2, with 3000 further spots up for grabs in the remaining two Round 1s. Nevertheless, the spirit of competition was there, and it is of course a great achievement to get the first place. Congratulations to Um_nik, and his achievement is even more impressive given that the round was probably at 4 in the morning :)

I have prepared the test data for the second problem, Prime Time, and I (humbly, of course :)) think it was a great example of how a not particularly difficult problem can still be nice. The problem went like this: you are given up to 1015 cards, each with a prime number between 2 and 499. You need to split the cards into two piles, such that the sum of the numbers in the first pile is equal to the product of the numbers in the second pile, or tell that it's impossible. Can you see how to handle such a huge amount of cards?

AtCoder has returned with its Grand Contest 053 a few hours later (problems, results, top 5 on the left, analysis). I have solved the first three problems in less than 50 minutes, which was great, but then I could not score any additional points in the remaining 2 hours and 10 minutes, which was a bit frustrating :) I've tried to approach all three remaining problems in the beginning, but eventually focused on the problem E, and it turns out I was really close: I did identify the two cases from the official analysis, I have correctly (I think) handled the first case and tried various formulas that looked very similar to the correct one for the second case, but I could neither figure out all details and come up with a formula on paper, nor get the correct answers for the sample cases by trying small changes to the formulas. It turns out that my problem was that I was inserting the segments in the increasing order of the starting time, instead of the decreasing order of the ending time. These two ideas look symmetric on the surface, but they actually are not equivalent because the problem asks about local maxima, and the symmetric case would be local minima. However, my idea can still handle the first case just as well (because the first case is in fact symmetric, as we have n-1 local minima there as well), which made it hard to move away from it for the second case.

Congratulations to the seven contestants who have managed to handle all details correctly and solve one of the harder problems, and especially to jiangly on the win!

Thanks for reading, and check back next week.

Monday, April 5, 2021

A no-regret week

TopCoder SRM 803 last week wrapped up the third stage of the race for TCO21 qualification (problems, results, top 5 on the left, TCO21 race results, analysis). Even though tourist had already guaranteed himself the first place and the direct ticket (or a direct login, in case it takes place online again :)) to TCO21, he has won the round with a big margin once again. This was his fourth SRM win in a row, which has happened before, but nobody else could win five in a row in the past. tourist has also claimed the all-time high rating during this streak. Congratulations Gennady, and no pressure at all for SRM 804 ;)

Codeforces Round 712 followed on Saturday (problems, results, top 5 on the left, analysis). The last problem kind of screamed that some sort of a greedy approach for embedding each cycle should work, but figuring out all details of the approach, as well as implementing it, was still extremely hard. Very well done to ksun48 on doing that and getting a clear first place!

Finally, the Northern Eurasia region of the ICPC held its onsite finals for the 2020-21 season on Sunday (problems, results, top 12 on the left, online mirror resultsanalysis). The top 50 teams from the online finals were invited to this round (how does one actually call the round between a semifinal and a final? A 2/3-final? A 1/sqrt(2)-final? :)). In the end, solving the 6 relatively easier problems was the bar for getting a medal, while getting a gold medal required also solving either the traditionally cactus-themed problem C, or the also traditionally interactive problem I. Congratulations to all medalists and especially to the team "Insert your name" from ITMO on the victory!

I set that interactive problem I for this round. Given that the contestants were onsite and did not have internet access during the round, it was a good opportunity to give a problem involving a beautiful but googleable algorithm: the randomized weighted majority algorithm. Preparing the test cases for this problem was extremely tricky, as the space of possible approaches is virtually unlimited, and there seems to be no single way to fail all of them. It turned out that the testcases were good enough for the onsite round, with the only two passing solutions using the provably correct approach, and with the top two teams probably accumulating quite some love for me with their -35 and -29 on this problem.

In the online mirror, some more questionable, but at the same time really creative, approaches passed all tests. In fact, the ML approach pwns the test set, making several times less mistakes than b (the smallest the number of mistakes of experts) in most test cases. It only has difficulties on test cases with a really small b (say, 7 out of 10000), where the allowed leeway of 100 extra mistakes is barely enough for the neural network training to converge.

It looks like I need to add some simple gradient descent and boosted decision forest algorithms to my prewritten code library for the future :)

In my previous summary, I have mentioned a Codeforces problem: there is a hidden array of n b-bit integers (in other words, each element is between 0 and 2b-1). You do not know the elements of the array, but you can ask questions about them: in one question, you can ask "is it true that the i-th element of the array is greater than j?" for some value of i between 1 and n and j between 0 and 2b-1. Your goal is to find the value of the maximum element in the array, and you need to do it in O(n+b) queries.

The first question is kind of expected: let's compare the first element of the array with 2b-1-1, in other words let's learn the highest bit of the first element. If that bit is 1, we're all good: we then know that the highest bit of the answer is also 1, therefore we have effectively reduced b by 1 using just one query, so we're on track for O(n+b) queries overall.

The case where that bit is 0 is more interesting. We can't really afford to keep asking about the first bit of all numbers, since in case they are all 0, we would've spent n queries to reduce b by 1, which does not lead to a linear number of queries. This issue kind of points us to a better approach: in case the answers are always "less than", we want to be left with a really small range after asking the n queries. Therefore let's compare the second number with 2b-2-1, in other words let's ask "is it true that the two highest bits of the second number are 0"?

In case we keep getting "less than" answers, after n queries we will know that the first number starts with 0, the second with 00, the third with 000, and so on. Now let's go from right to left, and ask "does the n-1-th number start with n zeros?". If not, then we know that the n-th number is smaller than the (n-1)-th number, and can be discarded for the rest of the solution, and we continue by asking "does the (n-2)-th number start with n-1 zeros?" If the (n-1)-th number does start with n zeros, we continue by asking "does the (n-2)-th number start with n zeros"? After we complete this process going from right to left, we will have discarded some amount k of all numbers, and will know that the remaining numbers all start with n-k zeros. Therefore we have reduced n by k, and b by n-k, so n+b was reduced by n using 2n queries, which is good enough for a linear solution!

Finally, we need to figure out what to do in case we get some "greater" answer after a few "less than" answers, for example when we learn that the first number starts with 0, the second with 00, the third with 000, but the fourth does not start with 0000. We will then ask: does the fourth number start with 0001? If the answer is also no, then we know that the fourth number starts with at least 001, therefore it's greater than the third number which can be discarded, and we continue by asking if the fourth number really starts with 001, potentially discarding the second number if not, and so on. If the fourth number does start with 0001, then we continue with the fifth number, but instead of asking if it starts with 00000, we ask if its prefix is at least 00010 (since we're not really interested in numbers smaller than 0001, given that we already have evidence of a number that starts with 0001).

At any moment, our algorithm therefore maintains a stack of numbers, where for each number we know a prefix that is equal to the prefix we know for the previous number in the stack plus one more bit. When we run out of bits, we do the backwards pass as described above, and obtain a problem of reduced size. Just like in the all-zeros case, we spend O(n) queries to reduce n+b by n, therefore achieving a linear solution.

This is the main idea of the solution. There are still some small details to be figured out, which you can find in the official editorial.

Thanks for reading, and check back next week!

Monday, January 4, 2021

A Samara week

"Good Bye 2020" was the last round of the year on Codeforces (problems, results, top 5 on the left, analysis). The round has left me with mixed feelings, as I've spent almost the whole 3 hours solving relatively standard problems, and did not have enough time to solve the very exciting problem I. The main reason for that was that I forgot how to count DAGs, and tried to reinvent the wheel in problem H for a very long time. tourist, on the other hand, did not struggle so much with H, and won the round. Well done! Also well done to scott_wu, fivedemands, mnbvmar and qwerty787788 who were able to solve the last problem.

Here's that exciting problem: there is a hidden array of n b-bit integers (in other words, each element is between 0 and 2b-1). You do not know the elements of the array, but you can ask questions about them: in one question, you can ask "is it true that the i-th element of the array is greater than j?" for some value of i between 1 and n and j between 0 and 2b-1. Your goal is to find the value of the maximum element in the array, and you need to do it in O(n+b) queries. The problem actually asked to do it in at most 3*(n+b) queries, but I think just doing it in a linear number of queries is the most interesting part.

Note that simply finding each element with binary search would lead to n*b queries, and it's not clear initially how we can do any better as each query only asks about one element, and the elements are somewhat independent. Can you see the way? n and b are up to 200, and the interactor is adaptive (so your solution most likely needs to be deterministic).

The new year pause was not long, and Open Cup 2020-21 Grand Prix of Samara took place on Sunday (results, top 5 on the left, analysis, overall Open Cup standings). Unlike last year, which saw a fierce competition at the top of the overall Open Cup scoreboard between three teams, this year team USA1 really dominates the proceedings, and they won their 7th Grand Prix out of 8 this time, finishing all problems in 3.5 hours. Congratulations!

The Prime New Year Contest is another staple of the holidays. It is running until the end of this week, and features the problems from 2020 which were not solved by anybody during respective contests. Good luck getting something accepted there, and huge respect to those who already did!

In my previous summary, I have mentioned an AtCoder problem: you are given an undirected graph with four vertices and four edges that looks like a square (the picture from the statement on the right). You know that a walk started from vertex S, finished at vertex S, has passed the ST edge (in any direction) a times, the TU edge b times, the VU edge c times, and the SV edge d times. How many different walks fit that description? a, b, c and d are up to 500000, and you need to compute the answer modulo 998244353.

If we replace the ST edge with a parallel edges, the TU edge with b parallel edges, and so on, then we're looking for the number of Euler tours in the resulting graph modulo dividing by some factorials to account for the fact that all parallel edges are equivalent. However, counting Euler tours in undirected graphs is hard.

But given the simple structure of our graph, we can actually reduce our problem to counting Euler tours in directed graphs, which can be done using the BEST theorem! We can iterate over the number of times we pass the ST edge in the direction from S to T, in other words over how many ST arcs would our directed graph have. This determines the number of TS arcs by subtracting from a, then the number of SV and VS arcs from the constraint that the in-degree and out-degree of S must be the same, and so on until we know the number of times we pass each edge in each direction, and we can then count the Euler tours in the resulting graph in constant time (because the graph has 4 vertices and 8 "arc types", and the actual number of parallel arcs does not affect the running time of the BEST theorem). Since a is up to 500000, we have to do this constant time computation 500000 times, which is fast enough.

Thanks for reading, and check back next week!

Tuesday, December 29, 2020

A rng_58 week

Last week wrapped up the year for two major contest platforms — TopCoder and AtCoder — and with it the races to qualify to the corresponding onsites.

TopCoder SRM 796 was the first round of the week (problems, results, top 5 on the left, analysis). maroon_kuri was leading after the coding phase but failed the system tests; none of myself, tourist and Egor could find a challenge opportunity even though there were lots of them available, including in the 500-pointer which screamed "look for challenges here!" as it had just one sample case.

This round concluded the three-month race for the second TCO21 spot (results, top 5 on the left). Even though six SRMs is quite a lot, the race came down to the wire and to a lot of very close calls: in SRM 792, neal_wu produced an amazing 200-point challenge phase to get a 3-point advantage and deny Um_nik the 5 race points; in SRM 793, tourist's easy was challenged but he would still win the round had ksun48 not taken his turn to earn 200 challenge points; you can see the story of SRM 796 above.

AtCoder held its two last rounds of the year on the weekend, starting with AtCoder Grand Contest 050 on Saturday (problems, results, top 5 on the left, analysis). Um_nik was the fastest on the four relatively easier problems, and protected his lead by finding the main insight and figuring out all details in the very tricky problem F. Congratulations on the win! duality and newbiedmy were also able to solve the hardest problem, and rounded up the top three. newbiedmy in particular must've had one hell of a contest: he started with this problem, and did not give up after 11 incorrect attempts. This unusual strategy was richly rewarded!

AtCoder Grand Contest 051 on Sunday was the last contest of rng_58 as AtCoder admin (problems, results, top 5 on the left, analysis). Even though the point values were different, the scoreboard looks very similar: the first four problems were solved by many, while the last two were very hard and only got accepted solutions in the last 45 minutes of the 4.5-hour contest. semiexp and yutaka1999 went for problem F which gave them more points and the first two places. Congratulations!

I'd like to highlight problem D: you are given an undirected graph with four vertices and four edges that looks like a square (the picture from the statement on the right). You know that a walk started from vertex S, finished at vertex S, has passed the ST edge (in any direction) a times, the TU edge b times, the VU edge c times, and the SV edge d times. How many different walks fit that description? a, b, c and d are up to 500000, and you need to compute the answer modulo 998244353.

Given that Makoto (rng_58) is retiring as AtCoder admin, I'd like to thank him for bringing AtCoder to the English-speaking world, and for making it the place for solving awesome problems. Extremely well done!

The two rounds have concluded the race for the 8 AtCoder World Tour Finals spots (results, top 8 on the left). Congratulations to all finalists! This is the third season of the AtCoder points system. The top 8 has quite a big intersection with those from the two previous years (2018, 2019), with tourist, Um_nik and myself qualifying all three times, and ecnerwala and mnbvmar qualifying last year. It will be the first WTF for Benq, Stonefeang and endagorion, assuming it does actually take place of course :)

Thanks for reading, and Happy New Year!