20201231, 08:10  #1 
Jul 2015
24_{10} Posts 
January 2021

20201231, 09:57  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{4}·11·53 Posts 
Quote:
What a year! What a problem! 

20210102, 16:12  #3 
Romulan Interpreter
Jun 2011
Thailand
10010000111011_{2} Posts 
Nice one.
And not difficult (search space only about \(\pi\) millions ). 
20210103, 01:17  #4 
"Ed Hall"
Dec 2009
Adirondack Mtns
111000101110_{2} Posts 
Is this one easy or am I doing something wrong? I appear to be able to get 188 pairs of coordinates rather easily for N=50. I think I'll try for the first bonus. . .

20210103, 05:07  #5 
Romulan Interpreter
Jun 2011
Thailand
22073_{8} Posts 

20210103, 12:24  #6 
"Composite as Heck"
Oct 2017
7·109 Posts 
It might be easy for a computer to brute force but I'm going to run out of pen ink :P

20210103, 13:22  #7 
"Ed Hall"
Dec 2009
Adirondack Mtns
2·3·5·11^{2} Posts 

20210104, 03:25  #8 
Jan 2017
3·29 Posts 
I wonder whether an answer to the ** bonus question is actually known by the problem setters or guaranteed to exist?

20210104, 06:39  #9 
Romulan Interpreter
Jun 2011
Thailand
5^{2}×7×53 Posts 
It is known.
Hint: the ranges are not contiguous, for example the only square grids that can be filled with no bot are n=1 (trivial), 2, and 5. For N=3, 4, 6, 7, 8, 9, 10, .... you need 1 bot (yep, I don't know why they talk about N=6, to confuse us, because, if I understand the problem right, for N=4 you also need a bot, so they should just stay at the example they provided, with N=4). Then you need 2 bots "sometime in the future" as N increases, but again, after it, you may suffice with 1 bot only, for a while, and so on, until 1 bot is not possible anymore. In a certain point further you will need 3 bots. But then, some other higher grids can use only 2 bots for a while (only a guess, by seeing how the grid is filled for low values, I do not have solutions for * and **). This is shitty in the sense that you can't do a binary search by N, you need to take N up, one by one, and do all combinations for it. You can't rely on symmetry either (the grid is not symmetric, due to initial orientation). You can however make some optimizations, for example, you can skip the squares not touched without any bot, when you put the first bot, and you can skip the squares not touched with one bot when you place the second bot, this reduces the cases to about a half (and doubles the speed), but the downside is that you need to keep evidence (you need anyhow, because you need to know when a walk finishes, i.e. it filled all the board, or it cycles  filling all the board is easy, just a sum of all vaccinations, but cycling, well... you need to keep the direction for every cell that reaches a 2, and if you reach it again with the same direction, then you are cycling). This is only after playing a bit in Excel. I do not have a solution for the * and **, but I have a nice animation of what the vaccination bot is doing (that bot follows a quite inefficient path, actually, haha, to paraphrase Serge, "what a waste of resources!"), and I have some "gut heuristic" how to set the "anti" bots, which works most of the time (I may post that on YT after the solution is published, the walking of the bot is quite "funny"), and I could find a solution to N=50 "by hand" with it, after few trials in Excel. Also, due to the modularity of the board, it doesn't matter where you start. You can start anywhere (and shift it accordingly when you send the solution). If we are starting in the middle of the board, the animation looks much better (and it is easier to guess the "heuristic" about placing the Xes (bots). Last fiddled with by LaurV on 20210104 at 06:47 
20210104, 14:16  #10 
Jan 2017
1010111_{2} Posts 

20210120, 21:46  #11 
Jun 2016
2^{3} Posts 
Has anyone made any progress on the second bonus question?
My initial attempts have been largely computational, without really doing any clever analysis of the problem. I found a heuristic which has worked pretty well so far for finding 2bot placements while searching a very small section of the problem space. That is giving me values of N up to several hundred with 2bot solutions (assuming my code is correct  I've submitted a solution but it hasn't been verified yet). But if there is a value of N for which 2 bots won't work, I don't see any purely computationbased way to verify it other than exhaustively testing every possible pair of placements, which would take days for N that large with my current code (which I think is reasonably optimized). That all leads me to the conclusion that there must be some clever way to analyse the problem mathematically rather than just brute forcing it with compute, but I don't really see any way to get to grips with the problem that way. Has anyone had any luck with a more analytical approach? 
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