Cheers

I finally approved your comment. And yep, that’s what the Hungarian mathematician who first told me this puzzle would call the “Ah-haa!” solution. The perimeter never decreases as the infection spreads so to infect an NxN board you need at least N initially infected squares.

On a personal note: I found this puzzle because one of my co-founders saw http://appaholic.com/ and your interest in my previous company, iLike, and said “we’ve got to bring this guy in for an interview”. Check out http://vadver.com/jobs.php and let me know when you can come in to learn more about our plans.

To illustrate the line of thinking I want to use:
Assume we can fill an m-by-m grid with k cells initially coloured in. Then we can fill an (m+1)-by-(m+1) grid with k+1 cells by adding a row to the bottom and column to the right side of our m-by-m grid and colouring the bottom right cell. (We colour the first k cells as in our m-by-m set up). Clearly the first k cells will still do their thing and fill the m-by-m square in the top left corner. Then, or before then, the bottom right cell will cause the bottom row and right-most column to fill.

Since we can fill a 1-by-1 square with 1 cell initially it follows inductively that we can fill an n-by-n with n cells filled initially.

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To prove a lower bound along this line of reasoning we need to show that if you can’t fill an m-by-m with k cells then you can’t fill an (m+1)-by-(m+1) with k+1 cells. If we can’t fill it with k+1 cells then clearly we can’t fill it with less then k+1 either.

If we can show this then the result follows fairly simply via induction: we can’t fill a 1-by-1 grid with 0 cells coloured initially.

We know that the infection can’t spread further right/left/up/down then the right/left/up/down-most square initially coloured in. To see this for the down-most case:

Assume we have some down-most cell A (and no ties for this title) which isn’t in the bottom row. Since we can’t generate coloured squares diagonally, if a coloured square gets generated further down then A it must be generated directly below A. But the way generation works means there needs to be another square touching the square directly below A. Necessarily this other square must be further down then A – our downmost square – contradiction. Hence we can’t generate a square lower then A.

It follows then that in any completely filled grid there were initially coloured cells in the bottom and top rows and in the left and right edge columns. Note this can be achieved with two cells by placing them in diagonally opposite corners.

Out of time unfortunately, perhaps I’ll get back to this later.

I’ve never seen this before, and I agree with Eimantas. Obviously we have an upper bound of n by marking everything on a diagonal. But it’s not particularly hard to prove that it’s the lower bound. In order for a square to turn black, it must already be in a row or column that another black one is in. And it takes n squares to put a mark in every row and column. So the lower bound is n.

-E