## Monday, July 29, 2024

### An Eliška week

EGOI 2024 in Veldhoven, the Netherlands was the main event of last week (problems, results, top 5 on the left, analysis). Eliška has won for the second time in a row, this time with an even more commanding margin — huge congratulations! She was one of the four girls (together with Lara, Vivienne and Anja) who has participated in all four EGOIs, but it seems that this was the last year for all of them. Nevertheless, the new stars are already there as well, with so many medalists having several EGOI years ahead of them. It is great to see that this wonderful new community has been built and thrives!

To get the full score on this problem you must always have w<=3, but I find that solving for w<=5 (which would give 95 points out of 100) is more approachable and potentially more fun. As another hint, two of the subtasks in the problem were as follows:
• n=2
• the given permutation is a reverse permutation

Can you see how to move from those subtasks to a general solution with w<=5?

Codeforces ran Pinely Round 4 on Sunday (problems, results, top 5 on the left, analysis). As Um_nik rightly points out, tourist has continued his amazing run of form, and is potentially one more win away from crossing the magical boundary of 4000. Very well done!

I also share Yui's sentiment: it is very cool and quite surprising that problems H and I can in fact be solved. During the contest, I did not manage to make any significant progress in both of them in about 1 hour and 15 minutes I had left after solving the first 7. On the other hand, the (nicely written!) editorial almost makes them look easy :)

If you have not checked out the editorial yet, you can try crack problem I yourself. The statement is quite simple and beautiful: you are given an empty n times m grid (4 <= n, m <= 10). You need to write all integers from 1 to nm exactly once in the grid. You will then play the following game on this grid as the second player and need to always win. The first player takes any cell of the grid for themselves. Then the second player takes any cell that is not taken yet, but is adjacent (shares a side) to a taken cell, for themselves. Then the first player takes any cell that is not taken yet, but is adjacent (shares a side) to a taken cell, for themselves, and so on until all cells are taken. The second player wins if the sum of the numbers of the cells they have in the end is smaller than the sum of the numbers of the cells of the first player.

Given that the first player can choose the starting cell arbitrarily, it seems quite hard to believe that the second player can have any advantage. Can you see the way?