A chain cut week

The Sep 11 - Sep 17 week had its contests on the weekend. On Saturday MemSQL Start[c]UP has returned to Codeforces after a three-year absence with Round 1 of its third edition (problems, results, top 5 on the left, my screencast, analysis). moejy0viiiiiv was 25 minutes faster than everybody else to solve all problems, and made sure to protect his lead with a few challenges. Congratulations on the win!

The round had a few nice problems, but let me highlight problem F: there are many contestants taking part in a competition, and top c will be awarded with t-shirts. There are n available t-shirt sizes (n<=200000). Each contestant has indicated one of two things: either they need a t-shirt with some concrete size, or they need a t-shirt with one of the two concrete adjacent sizes. More precisely, for each size i we know the number a_i of contestants (up to 10⁸) that need a t-shirt of size i, and the number b_i of contestants (also up to 10⁸) that need either a t-shirt of size i or a t-shirt of size i+1. What is the minimal number of t-shirts we need to buy in order to be able to give t-shirts to top c contestants, no matter which ones end up in top c?

Apart from being an interesting problem in general, it's a great example for this recent discussion about stories in problem statements. I think that here the t-shirt and contestant story actually makes the problem statement really easy to understand and think about, while a completely formal mathematical statement would be hard to parse.

On Sunday the new season of the Open Cup started with the Grand Prix of Romania (problems, results, top 5 on the left). Team Past Glory did not show any signs of slowing down and won convincingly by a problem — well done! The great performance and second place of the Seoul National U 2 team was a great surprise, and even more so given that this is a different team from the one that earned gold medals at the last ICPC World Finals for Seoul.

Problem L of this contest showed that we can still find new games with non-trivial solutions in a very simple setup: two players have a sequence of 50000 integers, and alternately take numbers from either end of the sequence. When all numbers have been taken, each player computes the bitwise xor of their numbers, and the one with a higher bitwise xor wins. Who will win?

And right after the end of the Grand Prix, Codeforces held its Round 434 (problems, results, top 5 on the left, analysis). The battle for the first place was quite tight. Congratulations to kmjp who came out on top by a mere 25 points!

In my previous summary, I have mentioned a hard TopCoder problem: you are given 50 attractions in a city. You will visit exactly one attraction per day. Each attraction can be visited either in the morning, costing a_i, or in the evening, costing b_i. Every time you visit an attraction in the morning after visiting another attraction the previous evening, you pay an extra cost c. Every time you switch from evening to morning, you pay an extra cost d. Additionally, you are given a directed graph of restrictions — a set pairs of attractions that constrain the order of visits: in each pair, the first attraction must be visited before the second one. What is the minimal cost to perform all visits?

First of all, let's split our visits into groups: a group of morning visits, then a group of evening visits, then a group of morning visits, and so on (we need to consider two cases based on whether the very first visit is in the morning or in the evening). Since we only pay extra costs when we switch groups, the total cost depends on the number of groups and on the assignment of visits to groups, but not on the order of visits within each group.

Now let's try to build a network in which a cost of a cut would correspond to the total cost of the visits, so that we can solve our problem with the minimum cut algorithm. For each attraction, let us build a chain of vertices, with all chains having the source and sink as start and end respectively. A chain for one attraction looks like: s->v₁->v₂->..->v_n->t. Let's add infinite capacity edges going backwards in the chain, to guarantee that any finite cut separates some prefix of this chain from the rest. The size of this prefix will correspond to the number of the group that this attraction belongs to, and thus the capacity of the edges should be alternating a_i and b_i: if the vertex belongs to an even-numbered group, the corresponding edge will contribute a_i to the total capacity of the cut, and otherwise b_i, just as we need.

In order incorporate a restriction "p must be visited before q" into our network, we will add infinite capacity edges from the vertices of the chain for p to the corresponding vertices of the chain for q. This guarantees that with any finite cut, the position we cut the chain for p will be the same or earlier than the position we cut the chain for q.

Finally, in order to incorporate the morning/evening switching costs, let us add an extra attraction that must be visited after all others, and set up the capacities on its chain not as alternating a_i and b_i, but rather as 0, c, c+d, 2c+d, 2c+2d, ...

The finite cuts in the resulting network correspond to valid ways of visiting the attractions, and the capacity of the cut is equal to the total visit cost, so now we just need to find the minimum cut. Also note that we have built our network using two relatively standard building blocks: infinite edges can give us constraints on the cut, and a chain of vertices with infinite backwards edges implements a choice from several alternatives with specific costs.

Thanks for reading, and check back soon!

An almost there week

The Sep 4 - Sep 10 week started with two Codeforces rounds. On Monday, Codeforces Round 432 took place (problems, results, top 5 on the left, analysis). AngryBacon has started with the hardest problem F and solved it very fast, which gave them a clear path to victory. Congratulations!

On Wednesday, Codeforces hosted the next Round 433 (problems, results, top 5 on the left, analysis). In a round with 4 problems solvable within an hour and the fifth one too hard to crack, Um_nik has combined excellent speed with 3 successful challenges for a clear first place. Well done!

On the weekend TopCoder hosted not one but two rounds that finalized the list of finalists for its flagship tournament TopCoder Open 2017. Round 3B took place on Saturday (problems — unavailable now, but hopefully the TopCoder website will be fixed soon, results, top 5 on the left). In order to advance one simply needed to solve either the first two (like qwerty787788, Um_nik and ikatanic did), or the third problem (like rng_58 and RRx did). Congratulations to all advancers!

With 1.5 minutes to go in the challenge phase, I had 344.14 points which would be enough to advance with just the easy problem solved, only to make an incorrect challenge that brought me down to 319.14 which was not enough. Of course I didn't know that 344.14 would be enough during the round itself, but seeing the final scoreboard made me quite annoyed at myself for that -25 :)

Here is the easy problem that brought the challenge phase excitement about: you are given two strings of the same length, which is at most 50. You are allowed to swap corresponding characters in the strings (i-th character of the first string with the i-th character of the second string) as many times as you want. Your goal is to obtain two strings with as large as possible longest common substring. Can you see the right approach?

In between the TopCoder rounds, Russian Code Cup 2017 Final Round saw the top contestants compete for glory and prizes (problems, results, top 5 on the left, my screencast, analysis). xudyh was about half an hour faster than everybody else on the first 5 problems. I have tried to leapfrog him by trying to squeeze in a suboptimal solution for the last problem that is fast to code, but it did not pass and xudyh has maintained his first place — congratulations on the win!

I have found the easiest problem of this round to be the cutest. You are given a set 100 distinct integers a_i each up to a million, and need to find any set of 100 integers b_j also each up to a million, such that all 100² pairwise sums a_i+b_j are distinct. What is the nicest way to achieve that?

And finally, the Wildcard Round of TCO 2017 determined the last two finalists on Sunday (problems — unavailable now, but hopefully the TopCoder website will be fixed soon, results, top 5 on the left, my screencast). Only kuniavski and xudyh have solved the hard problem, so it is only fitting that they qualified for the TCO onsite — great job!

I have once again tried to compensate the poor coding phase performance with lots of successful challenges, but was unable to find enough and got eliminated from the TCO. The hard problem that I couldn't crack stated: you are given 50 attractions in a city. You will visit exactly one attraction per day. Each attraction can be visited either in the morning, costing a_i, or in the evening, costing b_i. Every time you visit an attraction in the morning after visiting another attraction the previous evening, you pay an extra cost c. Every time you switch from evening to morning, you pay an extra cost d. Additionally, you are given a directed graph of restrictions — a set pairs of attractions that constrain the order of visits: in each pair, the first attraction must be visited before the second one. Can you see the way to use max flow to find the minimal cost to perform all visits?

Thanks for reading, and check back tomorrow for more backfills!

A Chihuahua week

Codeforces hosted its Round 431 during the Aug 28 - Sep 3 week (problems, results, top 5 on the left, analysis). dotorya has managed to solve all four solvable problems in just over one hour, more than 20 minutes faster than the second-placed Reyna and more than 50 minutes faster than everybody else. Well done!

Also during this week, the biannual Petrozavodsk Training Camp for teams taking part in ICPC has completed (results, top 5 on the left). With 9 5-hour contests packed into just 11 days, and many world class teams participating, this is the first detailed comparison of top student teams this season. Congratulations to the Moscow SU team Chihuahua for topping the overall standings with quite a margin!

Thanks for reading, and check back for more.

A yosupo week

Miss me?

The Aug 21 - Aug 27 week started of with TopCoder SRM 720 on Thursday (problems — unavailable now, but hopefully the TopCoder website will be fixed soon, results, top 5 on the left). There was a tense challenge phase battle at the top, but yosupo has made sure it was only for the second place by solving the 1000-pointer — congratulations on the victory!

The said 1000-pointer was concerned with somewhat theoretical isotonic regression in L₁ metric. On the outside, it looked like this: you are given an undirected weighted graph with at most 50 vertices and at most 1000 edges, and two spanning trees in it. You have to modify the weights of the edges in such a way that the first given spanning tree is minimal, and the second given spanning tree is maximal, while minimizing the total change of weights.

A few solutions for this problem are described and linked in this comment thread.

Later on the same day we had AIM Tech Round 4 on Codeforces (problems, results, top 5 on the left, analysis). yosupo has continued his great day, this time winning without an extra problem but thanks to solving everything fast enough. Well done!

Finally, Saturday presented us with AtCoder Grand Contest 019 (problems, results, top 5 on the left, my screencast, analysis). This time yosupo was only 6th, and the victory instead went to Um_nik who has dominated the round by finishing all problems with half an hour to spare. Congratulations!

In my previous summary, I have mentioned a Codeforces problem: you are given an array of 300000 numbers, and 300000 queries of the following form: a segment [l;r] and a number k, for which you need to find the smallest number that appears on at least 1/k of all positions in the given segment of the array. Note that k is at most 5, so the number must be quite frequent within the segment.

The first idea is to use Mo's algorithm to handle the segments; we'll keep count of how many of each number are there in the current segment in a simple array, and thus can handle the counting for all segments in O(n*sqrt(n)). In order to find the answer for each segment, we'll need to find the counts that are above 1/5 of their sum. One way to do that would be to maintain a priority queue, but that adds an extra log factor both for queries and for the Mo's part, and some coding complexity.

A nicer way to find the high counts is to use randomization: let's just pick 120 random numbers within our segment, and hope that all numbers that occupy at least 1/5 of the segment will be found. Indeed, the probability of that not occurring for a number is less than (1-1/5)¹²⁰ , or about 2.3e-12. We have at most 5 interesting numbers, 300000 queries, and around 100 testcases, so the overall probability of an incorrect answer does not exceed 5*100*300000*2.3e-12=3.45e-4, which is good enough.

Thanks for reading, and check back soon as I crawl towards the present :)

Commentary stream for TopCoder Open 2017 Finals

TopCoder Open 2017 final round is today at 19:30 CEST. Tune in at https://youtu.be/qrlDoP3JQkI to hear me, pashka and Endagorion discuss! This time we should have problem statements at the beginning of the round :)

Commentary stream for TopCoder Open 2017 Semifinal 2 with Egor

We'll try to do live commentary of TopCoder Open 2017 Semifinal 2 with Egor on https://youtu.be/AqpjM86iZlc  - tune in at 19:30 CEST!

A randomized week

The Aug 14 - Aug 20 week contained the usual suspects: a TopCoder SRM and a Codeforces round. First off, TopCoder SRM 719 took place in the early Tuesday hours (problems, results, top 5 on the left, analysis). yanQval was a lot faster than everybody else and won the round - well done!

Codeforces Round 429 followed on Friday (problems, results, top 5 on the left, analysis). anta and LHiC have both solved their final fifth problem with only a few minutes to spare and thus obtained quite a margin at the top of the scoreboard - way to go!

Problem D in this round allowed for a cute randomized solution that in my view is a bit easier than the one from the official analysis. The problem was: you are given an array of 300000 numbers, and 300000 queries of the following form: a segment [l;r] and a number k, for which you need to find the smallest number that appears on at least 1/k of all positions in the given segment of the array. Note that k is at most 5, so the number must be quite frequent within the segment. Can you see how to use that fact to implement a simple randomized solution?

And finally, I've checked out CodeChef August 2017 Cook-Off on Sunday (problems, results, top 5 on the left, my screencast). EgorK was at the top of the standings with 4 problems for quite some time, but gennady.korotkevich has then submitted his fourth and fifth problems in rapid succession and won the competition. Congratulations!

In my previous summary, I have mentioned a problem that I have set for Google Code Jam 2017 finals: you are given a number k between 3 and 10000. Output any simple undirected graph with at most 22 vertices that has exactly k spanning trees.

The approach that works the most frequently in such problems is to learn several transitions that help cover all required numbers. For example, if we could come up with a way to transition from a graph with k spanning trees to graphs with 2k and 2k+1 spanning trees by adding one vertex, the problem would be solved. Alas, this does not seem possible, as it's not clear how to achieve that extra "+1" spanning tree.

As explicit construction turns out to be hard to impossible, we have to turn to more experimental approaches. One thing that stands out is that there's an enormous amount of graphs on 22 vertices, and while they have at most 22²⁰ spanning trees, many have under 10000 spanning trees, and thus we can hope that for each k in the given range there are many possible answers. So the next idea is to try generating random graphs and checking how many spanning trees those graphs have using the matrix tree theorem, hoping to find all numbers between 3 and 10000 at least once.

It turns out that some amounts of trees are more difficult to achieve than others, so this approach by itself is not sufficient to generate all answers within reasonable time. However, there are numerous different ideas that help it find all required numbers faster, and thus help get the problem accepted. I won't be surprised if everybody who got this problem accepted at the Code Jam has used a different approach :) My experimentation led me down the following path:

• Since some numbers are hard to get, we try to steer the random graphs towards those numbers. Suppose we have a "goal" number we're trying to achieve. The easiest way to do that is to avoid recreating the graph from scratch, but instead take the graph from the previous attempt, and check if it has less trees than the goal, or more. In the former case, we add a random edge to it, and otherwise remove one. As a result, we're more likely to hit the goal, and as soon as we do it, we pick another goal from the yet unseen numbers and repeat the process.
• This process has an unfortunate bad state: when we remove an edge, we can make the graph disconnected, thus having 0 spanning trees. At this point we start adding random edges, but if the graph stays disconnected, it keeps having 0 spanning trees. Finally, we add an edge that connects the graph back, but since we have added so many other edges in the meantime, suddenly the graph has a lot of spanning trees, way more than we need. So we need to remove a lot of edges to get back to reasonable numbers, and may well jump back to 0 in the process. Having noticed this happening, I've modified the process to never remove an edge that would make the graph disconnected.
• Finally, we get almost all numbers we need! After a minute or so, my solution has generated all numbers between 3 and 10000 except 13 and 22 (if I remember correctly). The final idea is that small numbers can still be generated by an explicit construction: a cycle with k vertices has k spanning trees. Since we're allowed at most 22 vertices, 13 and 22 can be done using a cycle.

That's all for my solution, but I have one more question for you: why is 22 so hard to get? In particular, does there exist a simple undirected graph with at most 21 vertices that has 22 spanning trees? I strongly suspect that the answer is no, but don't know for sure.

Thanks for reading, and check back later!