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Two black neighbours implies black

Two black neighbours implies black

Q: Consider a 10 x 10 array of white squares. Colour some initial set black. Now, if a square has two black orthogonal neighbours, colour it black. Repeat as much as possible. Show that if the initial set has 9 squares we cannot end with the whole array black.
[Dima Fon-Der-Flaass, 24 Feb 1997]

Q: More generally, find the minimum initial # blacks required to turn an m x n array black.

A: ceil((m + n) / 2)

Necessity: The colouring process cannot increase the perimeter of the black region [Dima Fon-Der-Flaass]. With k initial blacks the initial perimeter is <= 4k and the final perimeter is 2m + 2n, hence k >= (m + n) / 2, and k is an integer, hence the ceil( ).

Sufficiency: Wlog m (# rows) <= n (# columns). Use m squares to colour the leading diagonal black. Now start at the top right corner and work left along the top row colouring alternate squares black, ending in either column m + 1 or m + 2. The diagonal will cause the left m x m square to be blackened, and then the rest of the array can be blackened in m x 2 strips seeded by the squares in the top row.

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