Here's the same thing for a 2×2 grid. The triangle in the middle has to go one way or the other, so you have to break the vertical line of symmetry

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Consider the algorithm "M(x): if x<0 return -x, else return M(x-M(x-1))/2". This algorithm terminates for all real x, though this is not so easy to prove. In fact, Peano Arithmetic cannot prove the statement "M(x) terminates for all natural x". Paper to come! Joint work with @jeffgerickson and @alreadydone

Mine does.

http://somethingorotherwhatever.com/nice-calculator

@christianp Nice. I wish there were a way to do the same at arXiv, but things are a bit too dense there.

What I use most often at arXiv does work at MathWorld though: http://mathworld.wolfram.com/search/?query=%22open+problem%22

http://mathworld.wolfram.com/search/?query=surprisingly&x=0&y=0

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