![]() ![]() Think of shifting a set of vertical stripes horizontally by one stripe. For example, translational symmetry is present when the pattern can be translated (shifted) some finite distance and appear unchanged. A symmetry of a pattern is, loosely speaking, a way of transforming the pattern so that the pattern looks exactly the same after the transformation. Imagine you want to invent a pattern to make a tessellation. So any tessellation has a fundamental restriction in Euler's formula (in our case, it is 0 = V - E F).Īs for the tessellations in itself, it's not exactly the shape of the polygon that matters, but its simmetry group. Proof: At any vertex, you have 3 or more edges, so $V \leq \frac = A$, which is a contradiction. I remember a very simple of proof of the fact that any polyhedron has at least a face that has 5 sides or less: In any case, however, the function gives conditions on the polygons you are working with. Proofs involving the Euler Characteristic can be extremely simple, but may be really complex too (it is widely used in algebraic topology). So, in some sense, it is a measure of the curvature of the space you are in. You can test the function (Euler Characteristic) with any polyhedron, for example. When the full set of possible graphs is enumerated it is found that only 17 have Euler characteristic 0. a wallpaper group and if it is negative it will have a hyperbolic structure. If the Euler characteristic is positive then the graph has an elliptic (spherical) structure if it is zero then it has a parabolic structure, i.e. The Euler Characteristic is $\chi = V - E F$, where V is the number of corners (vertices), E is the number of edges and F is the number of faces. But you can gain some intuiton using the Euler Characteristic.Ī graph can be viewed as a polygon with face, edges, and vertices, which can be unfolded to form a possibly infinite set of polygons which tile either the sphere, the plane or the hyperbolic plane. The question you are asking is by no means trivial. Fedorov found that there are exactly five 3-dimensional parallelohedra. You may be interested in the question.Įdit 2 : Let us consider $3D$ version. For example, (3,3,3,3,6) means there exist four equilateral triangles and one hexagon at every vertex.Įdit 1 : This is a question which I asked at mathoverflow. Also, you'll find some figures in the same page as above. You can find helpful comments in other's answer. Second, let's see the case we can use more than two distinct polygons and its copies to tessellate the plane. For example, you can divide a hexagon of (4) into two congruent pentagons. (6) Some pentagons with a special condition can tessellate the plane. (5) The fact (4) means that any quadrilateral can tessellate the plane because you can make a hexagon with three pairs of parallel edges using two copies of a quadrilatelral which is not a parallelogram. (4) A hexagon with three pairs of parallel edges can tessellate the plane. (3) The fact (2) means that any triangle can tessellate the plane because you can make a parallelogram using two copies of a triangle. ![]() ![]() (I found some figures here though the language is Japanese.) (2) You'll see that any parallelogram can tessellate the plane. (1) We can easily prove that there are only three regular polygons $(n=3,4,6)$ which tessellate the plane with one polygon. First, let's see the case that we use only one polygon and its copies to tessellate the plane. ![]()
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