## Wednesday, February 27, 2019

### A WTF week

AtCoder World Tour Finals 2019 in Tokyo headlined the last week (problems, results on the left, open round results, my screencastanalysis). The last three problems turned out too difficult to solve during the contest, so it all came down to speed on the first three. yutaka1999 was the fastest, but his five incorrect attempts gave the competitors 25 minutes to overtake him, and apiad did just that and won the first ever AtCoder World Tour. Congratulations!

I was pretty quick with solving A and C1, but then tried to solve B, C2 and E in parallel, switching often from one to another, instead of focusing on just one of them as it was not clear at that point that solving just one more would be enough for a good result. By the time I managed to come up with a solution for B, it was already too late to catch apiad and yutaka1999.

I found the problems C1 and C2 the most exciting. Consider an infinite 2D grid, where each cell can be either black or white. Initially, exactly one cell was black, and then we repeatedly applied the following operation: take some integers x and y, and invert the color of three cells: (x, y), (x+1, y) and (x, y+1). You are given the set of black cells in the final state of the grid. There are at most 105 black cells in C1 and at most 104 in C2, and each black cell has coordinates not exceeding 1017 by absolute value. Your goal is to find the coordinates of the only cell that was black originally. In problem C1 you know that its y-coordinate was 0, and in C2 there are no further constraints. Can you see a way to solve at least C1?

TopCoder SRM 751 followed on Friday (problems, results, top 5 on the left, analysis). Only rng_58 and pashka submitted all three problems, but the challenge phase did not leave them unscathed, and IH19980412 emerged in the first place thanks to three successful challenges, including one on rng_58 himself. Well done!

Open Cup 2018-19 Grand Prix of Bytedance presented problems from the team Moscow SU: Red Panda on Sunday (results, top 5 on the left, analysis). There were several nice problems in this contest, and problem C was the one I've enjoyed solving the most.

You have some amount x of money between 0 and 1. You're playing a betting game where in one turn, you bet some amount y, and with probability p (p<0.5) your amount of money becomes x+y, and with probability 1-p it becomes x-y. Your bet must not exceed your current amount of money. Your goal is to reach amount 1. So far this setup is somewhat standard, but here comes the twist: your bets must be non-decreasing, in other words at each turn you must bet at least the amount you bet in the previous turn. In case you don't have enough money for that, you lose. What is the probability of winning if you play optimally? More precisely, what is the supremum of the set of probabilities of winning of all possible strategies? Both x and p are given as fractions with numerator and denominator not exceeding 106, and you need to return the answer using division modulo 998244353.

Finally, Codeforces Round 542 wrapped up the week on Sunday evening (problems, results, top 5 on the left, analysis). Three contestants solved all problems correctly, and there wasn't much challenge activity, so everything was decided by the problem solving order and speed. mnbvmar was the fastest to solve everything, and did (the slightly faster to solve) problem E before problem D unlike the others. Congratulations on the victory!

Last week I have mentioned another Open Cup problem: there's a hidden not necessarily convex polygon with n vertices (n<=200). Your goal is to find its area, but the only thing you can do is to pick a subset of its vertices by their numbers (the vertices are numbered in the order they appear along the polygon), and the system will tell you the area of the convex hull of the chosen points. You can retrieve the convex hull areas for at most n*(n-1)/2 subsets before you need to give back the area of the hidden polygon.

During the contest we tried to invent a solution based on representing the area of the polygon as the sum of signed areas of triangles using one of its vertices as the base point. We could not figure out a way to deal with "signed" part: we need to determine the orientation of each triangle, and while in most cases we can determine orientation of triangle ACD given the orientation of triangle ABC and ability to ask convex hull area queries, we could not see a way to make it work in all cases. Is there one?

The approach that works involves a completely different idea: first, let's find the area of the convex hull of all vertices. Since our polygon is not necessarily convex, then we need to subtract something from it.

For each particular vertex, we can find whether it lies on the boundary of the convex hull or not by checking if the area of the convex hull of all vertices except this one is smaller. Now we know which vertices do not lie on the convex hull of everything.

Now let's take segments of consecutive vertices that do not lie on the convex hull, together with one vertex of convex hull before and after such segment. We claim that those are precisely the polygons whose areas we need to subtract from the area of the big convex hull to find the answer.

The only remaining step is to recursively apply the same algorithm to find the areas of those smaller polygons.

Thanks for reading, and check back next week!

## Monday, February 18, 2019

### A snack week

CodeChef SnackDown 2019 onsite finals early on Saturday was the main event of the week (problems, results, top 5 on the left). Team ovuvuevuevue enyetuenwuevue ugbemugbem osas looked to have pretty good winning chances when they were the first to solve 8 problems with a couple of hours still left in the contest, but they could not make further progress in the remaining time. Team Dandelion solved the ninth problem with about five minutes remaining to go on top, but team pastry was working on the same problem and could still overtake them on penalty time. It turned out that they were comparing their solution locally against a slower but simpler one, and there were still cases of disagreement as the end of the contest was approaching. With nothing left to lose, they submitted whatever they had 30 seconds before the end of the round — and it passed the system test. Congratulations to team pastry on the super narrow victory!

Later that day, Codeforces hosted its Round 539 (problems, results, top 5 on the left, analysis). The participants of the Codechef finals occupied most of the top spots in this round as well. wxhtxdy was the only contestant to solve all problems, but as his solution to E turned out to be incorrect, Um_nik emerged in the first place. Congratulations to Um_nik on the win!

Finally, the Open Cup 2018-19 Grand Prix of Belarus wrapped up the week (results, top 5 on the left). Team MIT THREE won the second round in a row, this time solving three in the last hour including two hardest ones in the last fifteen minutes. Amazing persistence once again, well done!

Problem A was a very nice interactive one: there's a hidden not necessarily convex polygon with n vertices (n<=200). Your goal is to find its area, but the only thing you can do is to pick a subset of its vertices by their numbers (the vertices are numbered in the order they appear along the polygon), and the system will tell you the area of the convex hull of the chosen points. You can retrieve the convex hull areas for at most n*(n-1)/2 subsets before you need to give back the area of the hidden polygon.

Thanks for reading, and check back next week!

I will also try to post something here and/or on Twitter about the first ever AtCoder World Tour finals in Tokyo on Thursday — already looking forward to the event!

## Sunday, February 10, 2019

### A tourist week

Codeforces hosted its Global Round 1 on Thursday (problems, results, top 5 on the left, analysis). tourist and Um_nik were quite close, both finishing the problem-solving part around 1:20 mark and having some challenge fun thereafter. However, in the end the challenges did not affect the standings, and tourist stayed in first place. Congratulations!

TopCoder SRM 750 followed a day later (problems, results, top 5 on the left, analysis). This time tourist managed to get a commanding lead thanks to solving the 1000-pointer in just 8 minutes, while rng_58 needed 22 minutes and others even more. Well done!

Finally, the Open Cup returned on Sunday with the Grand Prix of Gomel (results, top 5 on the left). This was the first of seven consecutive Open Cup Sundays stretching right up to the ICPC World Finals, and that will provide a good preview as many top-rated ICPC teams are competing. The team from MIT earned the first place with just four minutes remaining, solving two of the hardest problems in the last hour. Amazing performance!

In the previous summary, I have mentioned a TopCoder problem: you are given a 9x47 grid of zeroes and ones. In one step, you can take any cell that contains a one, and another cell that is either in the same row but to the right or in the same column but to the bottom from the first cell, and flip both of them (so one changes to zero, and zero changes to one). You are given the initial and the final state of the grid, and need to produce any (not necessarily the shortest) sequence of operations that transforms one to the other with at most 947 operations.

When we apply this operation to a one and a zero, we're effectively moving the one to the bottom or to the right. By applying this operation several times, the one can move to the bottom and to the right. Moreover, the opposite is true: for any position to the bottom and to the right, we can move the one there in exactly two operations. If the cell in the middle (in the same row or column as both source and target cells) contains a zero, then we just move the one twice in a straightforward manner. If the cell in the middle contains a one, then we can first move that one to the target, and then move the one from the source to the middle cell, restoring the one there.

Finally, the other option of applying the operation to a one and a one, or moving a one onto a one in two operations as described above, results in both ones disappearing. Now we're in a very nice position: we understand the full structure of the problem, and have described everything that is possible in a very concise manner, which allows to see the solution.

More precisely, we need find a "source" one in the initial grid to the left and/or to the top for each one from the final grid, which might also leave some ones in the initial grid unused. Note that if the number of unused ones is odd, there's no solution since all operations preserve parity, and if the number of unused ones is even, we can always get rid of them by moving each to the bottom-right corner in at most two operations.

This observation leaves us with a matching problem which can actually be solved greedily because of the special structure of the graph. If we traverse the ones from the final grid in row-major order, we can simply always pick the rightmost yet-unused one in the initial grid to the left and/or to the top from the current position. This can be proven by a typical exchange argument: let's look at the first time in this row-major traversal when the optimal solution uses a different one to cover the final one in the current position, and uses the greedy choice to cover something else. We can swap the assignments of those two ones and still obtain a valid solution.

Thanks for reading, and check back next week!

## Sunday, February 3, 2019

### A mumbling week

TopCoder SRM 749 was the main event of this week (problems, results, top 5 on the left, my screencast with mumbling). All three problems were quite tricky, and passing the sample cases did not really mean that much. As he often does, rng_58 excelled in such circumstances and claimed the first place with a sizable margin — congratulations!

The only other contestant to solve the hardest problem, pashka, was actually unable to figure out the proper algorithmic solution in time; so he decided to take his chances and treat the problem as a general hamiltonian cycle problem and solve it with a simple but powerful heuristic, just like I did in 2014. It seems that participating in IOI 2002 has finally paid off for both of us :)

The medium problem was quite nice in this round. After removing some unnecessary complications, the statement boiled down to: you are given a 9x47 grid of zeroes and ones. In one step, you can take any cell that contains a one, and another cell that is either in the same row but to the right or in the same column but to the bottom from the first cell, and flip both of them (so one changes to zero, and zero changes to one). You are given the initial and the final state of the grid, and need to produce any (not necessarily the shortest) sequence of operations that transforms one to the other with at most 947 operations.

Thanks for reading, and check back next week!