Proof of Bicker's Theorem in Graph Theory

Pick's theorem is the content of the Olympiad in the fourth grade of elementary school. I accidentally read it from a textbook and found the theorem quite interesting and related to my work. I found some proof materials on the Internet, combined with my own thinking, and a little Excavated the following and chatted for records.

Pick's theorem refers to a formula for calculating the area of a polygon whose vertices are on the grid point in the lattice. This formula can be expressed as S=N+L÷2－1, where N represents the number of points inside the polygon, and L represents the polygon falling on the grid point. The number of points on the boundary, S represents the area of the polygon.

The formula defaults to a small square with a side length of 1, that is, an area of 1. If a grid square has a side length of 2 (with an area of 4), it needs to be multiplied by 4 on the basis of the original formula.

1. General description of the theorem:

Given a grid, each grid consists of unit squares with side length 1. There is a polygon in the grid, and the vertices of the polygon are all at the intersection of the grid, that is to say, none of the vertices fall on the side of the unit square or inside the unit square. Record the area of the polygon as S, and the points inside the polygon The number of is I, and the number of points on the side of the polygon is A, then:

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

Proof, take the following figure as an example, the red points are interior points, and the points distributed on the edge lines are edge points.

First of all, use cutting and supplementing to supplement it into a rectangle. Assuming that the area of the rectangle is S, and the area of the edge triangle is shown in the figure, the area of the irregular polygon is:

$S_{poly}=S-S1-S2-S3-S4-S5$

1. Proof steps

(1) First, prove that it is true for rectangles;

(2) Next, prove that it is true for right triangles;

(3) Then, continue to prove that it is also true for any triangle;

(4) Finally, the proof is still true for the combination of two graphics.

First prove (4)

Suppose the area of any polygon has:

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

For polygon 1:

$S_1=I_1+\frac{A_1}{2} -1$

For polygon 2:

$S_2=I_2+\frac{A_2}{2} -1$

If polygons 1 and 2 share a side, and this side has n side points, then according to the formula, the figure formed by the combination of 1 and 2 is:

$\\S=(I_1+I_2+(n-2)) + \frac{A_1 -n +A_2-n +2}{2}-1 = I_1+I_2-2+\frac{A_1+A_2+2}{2}-1=I_1+I_2+\frac{A_1+A_2}{2}-2=S_1+S_2$

Proven.

Next prove that it holds for rectangles:

The length and width of a rectangle are x and y, respectively.

$\\I=(x-1)(y-1) \\A=2(x+y) \\S=I+\frac{A}{2}-1=(x-1)(y-1)+\frac{2(x+y)}{2}-1 = 1-(x+y)+xy +(x+y)-1 = xy$

established.

For the calculation method of A, you can refer to Euler's theorem, the number of faces F=1, so the number of fixed points is equal to the number of line segments. The number of fixed points held is equal to the number of line segments.

The same is true for triangles, first look at the direct triangle, and make the red right triangle into a rectangle:

Assume that the point through which the diagonal line passes is n. Then for the red triangle:

$I=\frac{(x-1)(y-1)-(n-2)}{2}$

$A=x+y+n-1$

but:

$\\ S=I+\frac{A}{2}-1 =\frac{(x-1)(y-1)-(n-2)}{2} +\frac{x+y+n-1}{2} -1=\frac{xy-x-y+1-n+2+x+y+n-1 - 2}{2}=\frac{xy}{2}$

established.

Any triangle can be spliced by 1 rectangle = several right triangles + this triangle, which can be proved by the above dismantling method.

$x_1,x_2,x_3,x_4,x_5,x_6$ is the length, $n_1,n_2,n_3$ is the number of points passed by the three sides of the triangle, and $I_1,I_2,I_3$ is the number of interior points of the three right triangles, then:

$S_{\Delta}= S-S_1-S_2-S_3$

$S=x\cdot y$

$S_1=I_1+\frac{x_1+x_2+n_1-1}{2}-1$

$S_2=I_2+\frac{x_3+x_4+n_2-1}{2}-1$

$S_3=I_3+\frac{x_5+x_6+n_3-1}{2}-1$

Use the geometric method and Biketing respectively to find the area of the irregular triangle:

Geometry:

$\\S_{\Delta}= S-S_1-S_2-S_3=xy-(I_1+I_2+I_3)-\frac{(x_1+x_2+n_1-1)+(x_3+x_4+n_2-1)+(x_5+x_6+n_3-1)}{2} -3$

Pick's theorem method:

$\\ S_{\Delta} = (x-1)(y-1)-(I_1+I_2+I_3)-(n_1-2+n_2-2+n_3-2) + \frac{n_1+n_2+n_3-3}{2}-1 \\ =xy-(x+y)-(I_1+I_2+I_3)-(n_1-2+n_2-2+n_3-2) + \frac{n_1+n_2+n_3-3}{2} \\ =xy-(x+y)-(I_1+I_2+I_3)-\frac{n_1+n_2+n_3-9}{2}$

Next simplify the geometry:

because:

$\\ \frac{(x_1+x_2+n_1-1)+(x_3+x_4+n_2-1)+(x_5+x_6+n_3-1)}{2}=\frac{(x_1+x_2+x_3+x_4+x_5+x_6)+(n_1+n_2+n_3-3)}{2}=\frac{2(x+y)+(n_1+n_2+n_3-3)}{2}=(x+y)+\frac{(n_1+n_2+n_3-3)}{2}$

so:

$\\ S_{\Delta}= S-S_1-S_2-S_3=xy-(I_1+I_2+I_3)-\frac{(x_1+x_2+n_1-1)+(x_3+x_4+n_2-1)+(x_5+x_6+n_3-1)}{2} -3 \\ =xy-(I_1+I_2+I3)-(x+y)-\frac{(n_1+n_2+n_3-3)}{2}-3 \\ = xy-(x+y)-(I_1+I_2+I3)-\frac{(n_1+n_2+n_3-9)}{2}$

It can be seen that the two methods are finally transformed into the form:

$xy-(x+y)-(I_1+I_2+I3)-\frac{(n_1+n_2+n_3-9)}{2}$

Therefore, the proposition is proved, and Pick's theorem applies to irregular triangles as well.

In Bicker's theorem, it is divided into prismatic grids and triangular grids. We will prove that it is also true for prismatic networks.

Proving that Prism Networks Conform to Bicker's Theorem

We know from elementary geometry that a quadrilateral is not a stable shape, and the area and shape of a quadrilateral can be changed by fixing one side and translating the other side. In linear algebra, this change is called a linear transformation, because it does not change the properties of parallelism and 0-point immobility. We can consider this transformation for the following graph:

You will see that a triangle is still a triangle after this transformation, and there is no change in the number and relative positions of its interior points, exterior points, and edge points, but the graphics area is reduced in the same proportion. Comparing the triangle below, there is no change in the number of interior and edge points before and after the transformation.

Because it is a linear transformation, the area of any part is reduced in the same proportion, so the grid point polygon counted as a small prism still satisfies the Bicker's theorem, which is a natural deduction of the linear transformation.

Therefore, for the grid composed of prisms, Pick's theorem can also be applied.

holds for the triangle

For convenience, take an equilateral triangle as an example (any triangle is fine, in fact, an equilateral triangle does not bring additional information to the proof, it is just convenient for drawing).

It is proved that any triangle conforms to Picker's theorem, and based on the addition of new sides on any triangle to form a new shape, each step also proves that it conforms to Picker's theorem, so no matter how complicated the surface composed of triangles is, the final figure It will also conform to Pick's theorem, and the proof is over.

Or using the previous conclusion, two regular triangles form a prism, which conforms to Bicker's theorem, and the prism is decomposed into a regular triangle, without bringing additional changes in the interior and exterior points, but the smallest area unit changes from the prism to the original Half, the area of a regular triangle.

Therefore, if it is calculated as a regular triangle, compared with the formula of the original prism, the final result should be multiplied by 2. The area of a prism is twice that of a regular triangle.

After the proof is completed, look at the figure below to find the area of a large triangle, isn't it very simple?

As long as the graph is decomposed into prisms or regular triangles, this problem can be solved by applying Pick's theorem, so I won't go into details.

think:

Since any irregular figure can be decomposed into triangles and rectangles, and rectangles can be further decomposed into triangles, in fact, this problem only needs to consider triangles. Therefore, in fact, as long as the previous conclusions 3 and 4 are proved, the Pick's theorem can be proved.

Lenovo:

In fact, since any irregular figure can be decomposed into triangles, the condition of rectangles can be made weaker. We only need to prove that Picker's theorem is applicable to irregular triangles and conclusion 4 to deduce the universality of Picker's theorem in conclusion. Because of the design of the GPU, any primitive can be decomposed into a combination of triangular planes. So naturally all lattice geometric figures are also applicable to the conclusion of Bicker's theorem.

The simplest surface in two-dimensional space is a triangle surface. During the vertex processing and rasterization stages of the GPU pipeline, the geometric primitives are decomposed into triangles for processing, and the PIPELINE algorithm will also be specially optimized for triangles.

No matter how complicated the geometric shape is, it can be seen as a splicing of infinitely many tiny triangles.

References

"Graphics Fundamentals" Note 2. Graphics Processing Unit GPU_gpu Why are graphics drawn in triangular shapes_Zhou Jie's Blog-CSDN Blog