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Minimize the perimeter



A child's game uses twenty-one unique shapes that comprise from one to five squares. This puzzle asks how tightly can the shapes be packed without overlap so as to achieve a figure with the smallest perimeter? The individual shapes may be rotated and turned over (reflected) as desired.



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1 + 2+ 2*3+4*5+12*5 = 89.

For real numbers, given an area A and a MxN rectangle of that area, the minimum perimeter 2*(M+N) is equal to 2*sqrt(2*M*N) and is achieved for M=N (since (a+b)^2 >= 2*a*b for any a, b). 

In this case, minimum perimeter is achieved for a rectangle MXN if M = N or M = N+1. For integers, that means a 10 x 9 rectangle covering area 90, since 89 is prime.

It doesn't matter where the extra cell is placed, even if it is on the border, the perimeter is still 2*(10+9)=38.


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Thinking about this ... One might make the case for three different values of perimeter for three different placements of the "hole," yielding a distinct best answer.


I wanted to say the minimum perimeter is the same but somehow we must be having a different definition, which your "three" reference points to.

Re-running the argument I had in my previous post - a 9x10 board is minimum as a rectangle. I did not talk about circle-type figures. With real coordinates, the circle is the solution with minimum perimeter for a given area. With squares I built on the definition of perimeter as movements along the x/y axis only. 

If one goes along with this definition, then the perimeter is 38 when the hole is inside or in a corner and 40 when the hole is border (but not corner). 

If one takes into account the definition of a perimeter as a convex hull then the minimum such hull is different if one is allowed to use in the perimeter something different than the edges of the squares. In this case, the minimum is achieved for boards when the hole is in a corner =38-2+sqrt(2) which is approx 37.83. I suspect in this case, a more circle-ish figure may have a smaller perimeter.

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