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x#y=floor$sqrt$min(x+y)\$1+2*min x y


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## Explanation

This answer calculates the following formula:

$$\left\lfloor\sqrt{\min(a+b,2\min(a,b)+1)}\right\rfloor$$

Why does this formula work? Well lets start by noting the following:

Every square of even side length can be tiled by $$\2\times 1\$$ tiles.

and

Every square of odd length can be tiled, spare a single $$\1\times 1\$$ square, by $$\2\times 1\$$ tiles.

Now we note that if we put these $$\2\times 1\$$ tiles on a chessboard each would lay on top of one black square and on white square. So if we make an even chessboard every tile needs to have a pair of the other color, and if we make an odd chessboard every tile but one needs a pair of the other color. This tells us that the answer is never more than $$\\left\lfloor\sqrt{2\min(a,b)+1}\right\rfloor\$$. $$\2\min(a,b)\$$ is the maximum number of pairs we can make and the $$\+1\$$ is for the last square that doesnt' need a pair. The problem with this is that if $$\a=b\$$ we will not have the extra square for the odd case. So we add another condition: Our result cannot be more than $$\\left\lfloor\sqrt{a+b}\right\rfloor\$$. That is we can't make a square which has more tiles than we have available.

So we just take the lesser of the two options.

$$\left\lfloor\sqrt{\min(a+b,2\min(a,b)+1)}\right\rfloor$$