UPD: Google confirmed that there will be no DCJ in 2020. If you try real hard, I believe that you can think of another established rule of the competition that was broken because of covid-19. If I understand correctly, they're fixing only(!) Then, counting by rows, we must have exactly i*(N-k) free slots in that i x j submatrix. For every other cell, make an edge between a cell vertex and a number vertex if that number can be put into that cell without breaking the Latin square properties. In the end, I used bipartite matching between x and y with edges which were made by empty positions (x,y). Is the following part of the editorial correct? It's OK to do this for the qualification round. Google Code Jam 2020 -- Round 3 By TimonKnigge , history , 5 months ago , Just a reminder that GCJ Round 3 is about an hour away. As we all know Google Code Jam 2020 Qualification Round Ended. However, I ran into the following case. After that, I forcibly place every number if some instance of that number appears on the diagonal with bipartite matching. I found this one https://vstrimaitis.github.io/google_codejam_stats. I couldn't come up with any idea to determine if the permutation was correct in https://codingcompetitions.withgoogle.com/codejam/round/0000000000051679/0000000000146183 (which was the simplest problem in the round as per no. I also got the "not qualified" notification and also got the participation certificate but now both are gone and its showing "We are Reviewing". No It will be updated within a day. I get now why Halls applies here at last <3. It's simple and requires a few observations on how the state of bits change. Good news is all these numbers can be represented by picking $$$2$$$ numbers, one appearing $$$2$$$ times and other appearing $$$n-2$$$ times. The claim I'm asserting consequently is that inserting the numbers row by row as opposed to number by number are isomorphic operations. But I think the idea is: suppose you've already placed numbers 1..k, so each row/column has N-k slots free. In the lower-left (n-2)*2 subgrid add two horizontal patterns 3,4,...,n starting from the first column and second last column. See you in Round 2, and best of luck! So I simply query the pairs and save the rsut, and then at 11th, 21th, ... queries I recheck one of the "equal" pairs if it has changed (1 query), and if it has, I update all the known "equal" pairs. Also, thank you sister_of_contestfucker for this nice, polite and inspiring comment. Google has many special features to help you find exactly what you're looking for. My blog. First try each random pen once until you hit empty one. 4) Weird Monte Carlo (e.g. After that, I place the remaining numbers in increasing order. I wrote a solution to this problem which gave correct answers for the sample cases. You cannot prove that bipartite matching is going to work for those since the 4s and 5s are already partially filled, right? An hour of struggle to try submit it. 2) is just swapping first row and second row,then we can swap 2 digits to get intended diagonal. Since 2003, Google Code Jam has brought together professional and student programmers from all over the world to solve tough algorithmic puzzles. In the upper-right 2*(n-2) subgrid add two vertical patterns 3,4,...,n starting from the two last rows. - The Code Jam Team". EDIT: If you haven't yet scored 30 (i.e. So the script retrieves the data in windows of size 200 until all the results are downloaded. We're still working out the details, but we'll be in touch again soon. Please Help! EDIT: OK, seeing the explanation of xiaowuc1 I don't think that caseology was intended solution :P. I had the same construction for cases 1-3. 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