Codeforces Round 383 was the main event of the last week (problems, results, top 5 on the left, analysis). TooDifficult has regained the second place in the overall rankings thanks to his victory; his 2017 ACM ICPC World Finals rival mnbvmar is now ranked sixth thanks to his second place in this round. Well done!

In my previous summary, I have mentioned an unsolved NEERC 2016 problem where your program explores a maze with at most 20 rooms which look exactly the same. Each room of the maze has the same amount

The solution is explained with pictures in the published analysis PDF (problem I), but let me repeat the outline here. We will implement a depth-first traversal of our graph. However, depth-first traversal sometimes needs to go back, and we can't do that since the passages are one-way. However, since the graph is strongly connected (it looks like I forgot to mention this property in the problem statement summary), there's always some way to get back. The main idea is: when leaving the vertex for the last time, mark the passage that leads as high up as possible in the depth first search tree. This way when we find ourselves in an already processed vertex, we can just follow the marked passages to get back to the current depth first search path. And inside that path, we will mark the passages that lead down this path, so that when we return to this path, we can then get down to the vertex that's being currently processed by the dfs. We can use left/right positioning of the rock to distinguish the vertices on the path ("grey") and the already processed vertices ("black"). There are some more technical details to the solution, but all ideas are above.

I find this problem very appealing for multiple reasons, one of them being that the solution is "tight": the amount of information we can store in the visited rooms turns out to be barely enough to perform the traversal. How can one come up with such tight problems?

In my previous summary, I have mentioned an unsolved NEERC 2016 problem where your program explores a maze with at most 20 rooms which look exactly the same. Each room of the maze has the same amount

*m*(also at most 20) of outgoing one-way passages which also look exactly like one another, arranged in a circle. The only way not to get completely lost in this maze is to use the fact that each room also has a movable rock. Initially there's a rock is in the center of each room. When you leave a room, you can put the rock next to any outgoing passage, and moreover, you can choose whether to put it to the left or to the right of this passage (that means that there are 2*m*ways to put it). If you ever arrive in this room again, you will see whether the rock is placed to the left or to the right of a passage - but since all passages look the same, you won't have any other information! So now you will be able express the number of the passage to take, and the number of the passage to move the rock to by the number in clockwise order starting from the one marked with the rock when you arrive. Your goal is to visit all rooms in at most 20000 steps.The solution is explained with pictures in the published analysis PDF (problem I), but let me repeat the outline here. We will implement a depth-first traversal of our graph. However, depth-first traversal sometimes needs to go back, and we can't do that since the passages are one-way. However, since the graph is strongly connected (it looks like I forgot to mention this property in the problem statement summary), there's always some way to get back. The main idea is: when leaving the vertex for the last time, mark the passage that leads as high up as possible in the depth first search tree. This way when we find ourselves in an already processed vertex, we can just follow the marked passages to get back to the current depth first search path. And inside that path, we will mark the passages that lead down this path, so that when we return to this path, we can then get down to the vertex that's being currently processed by the dfs. We can use left/right positioning of the rock to distinguish the vertices on the path ("grey") and the already processed vertices ("black"). There are some more technical details to the solution, but all ideas are above.

I find this problem very appealing for multiple reasons, one of them being that the solution is "tight": the amount of information we can store in the visited rooms turns out to be barely enough to perform the traversal. How can one come up with such tight problems?

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