Pick’s Holes

When building a shape with a hole in it, there must be a quadrilateral added in the process that acts as a bridge. For example, in the picture below, the dark gray square is bridging the gap on the light gray polygon to form one hole.

According to Pick’s Theorem, the two areas are $A = \frac{1}{2}B_1 + I_1 - 1$ and $A = \frac{1}{2}B_2 + I_2 - 1$ for a combined area of $A = \frac{1}{2}(B_1 + B_2) + I_1 + I_2 - 2$ , where $B_1$ and $I_1$ are the number of border and interior points on the dark gray square, and $B_2$ and $I_2$ are the number of border and interior points on the light gray polygon.

However, when viewing the two shapes as one combined shape, the number of $B$ border points and $I$ interior points is not simply $B = B_1 + B_2$ and $I = I_1 + I_2$ , because some of the original border points change to interior points and some of the border points are counted twice. Letting $S_1$ and $S_2$ be the number of border points on the two shared sides (including the endpoints), the resulting shape has the same number of border and interior points as the original two polygons except that $2(S_1 - 2)$ border points change to $S_1 - 2$ interior points, the $2$ endpoints of $S_1$ are repeated, $2(S_2 - 2)$ border points change to $S_2 - 2$ interior points, and the $2$ endpoints of $S_2$ are repeated. Therefore, $B = B_1 + B_2 - 2(S_1 - 2) - 2 - 2(S_2 - 2) - 2$ and $I = I_1 + I_2 + (S_1 - 2) + (S_2 - 2)$ .

So the area of the combined shape in terms of Pick’s Theorem for polygons (without holes) would be $\frac{1}{2}(B_1 + B_2 - 2(S_1 - 2) - 2 - 2(S_2 - 2) - 2) + (I_1 + I_2 + (S_1 - 2) + (S_2 - 2)) - 1$ which simplifies to $(\frac{1}{2}B_1 + I_1 - 1) + (\frac{1}{2}B_2 + I_2 - 1) - 1$ (the area of the dark gray square plus the area of the light gray square minus $1$ ), but the expected value is $(\frac{1}{2}B_1 + I_1 - 1) + (\frac{1}{2}B_2 + I_2 - 1)$ (the area of the dark gray square plus the area of the light gray square), for a discrepancy of $1$ . For every hole that is created, an extra $1$ is subtracted, so the equation can be balanced by adding the number of holes in the shape. Therefore, Pick’s Theorem for holes is the same as Pick’s Theorem without holes except for adding $h$ , which is

$A = \frac{1}{2}B + I + H - 1$

and so $a = \frac{1}{2}$ , $b = 1$ , $c = 1$ , and $d = -1$ , which means that $|a| + |b| + |c| + |d|$ $=$ $|\frac{1}{2}| + |1| + |1| + |-1|$ $=$ $\boxed{3.5}$ .