44

@hexomino's answer is correct and well-reasoned, as always. Here's another approach, which to me feels much more.. "axe-to-the-head" is what I'd call it in my native language, so I thought it might be interesting enough to warrant posting. Lower bound: (This is what @trolley813's encrypted comment is saying.) Upper bound:


12

I think the answer is Reasoning As trolley813 mentions in the comments


6

Short way to guess at the answer: Proving that this works: Now you asked about strategy:


5

Surely must be enough, and also required. After white has two unavoidable captures coming up: on white's next move, the Black can prevent only one of those moves, so the first objective is completed. Continuing from there, white will play next. It threatens black's F-pawn, and also the queen's rook or some other piece in front of it. All those pieces ...


5

EDIT: All solutions up to mirror symmetry, rotation, colour permutation found by brute force: End of EDIT. One possibility (I write X,O,+ for 0,1,2): A bit on how this was found:


4

Using a modified version of Albert Lang's method on the previous question, the best I've managed so far is As follows


4

Here's (what I think is) a simpler proof of Albert's lemma than the one in loopy_wall's answer. We'll find either a king-path of 0-squares connecting N and S sides, or a king-path of 1-squares connecting W and E sides. The basic idea is to walk along the boundary between 0-squares and 1-squares until we reach an edge of the board. So here's an example board; ...


3

Here is a solution with six lines: It's difficult to tell how I found this, except that I already knew a solution for a 3 by 3 grid with 4 lines, which can be found e.g. here on our sister site Mathematics Stack Exchange. It's also possible that a solution with 5 lines exists. (By the way, the puzzle is missing the requirement that the lines must be ...


3

Proof of Albert's lemma (which solves the bonus question; please note that some repetitive details are omitted to keep down overall length to something reasonable):


3

Another proof of Albert's lemma (and one that I believe is much more elegant than the others presented): I will prove a stronger lemma instead. Namely: Proof: (A proof of the reduced statement follows. This proof is similar to Gareth's answer to the same question, but does not rely on an arbitrary choice of "leftmost".)


2

It feels like this can be improved but the best I've been able to do so far is As follows


2

I would like to present an argument that shows that In fact Proof


1

I used integer linear programming (and, sorry, a computer):


1

I don't know if this satisfies the intended problem but it follows the rules. I approached it by looking at the minimum multiple choice answer, and trying to exclude solutions of that size.


1

I finally wrote a heuristic solver that finds good solutions. I found many solutions that achieve the longest chain length of Here are some example solutions


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