Polygons on either side of a common edge need to be colored differently, so 2 colors are needed, as a minimum. Polygons of the same shape do not share an edge, so all can be colored the same color.
There should be at least 8 different colors available for coloring the sections. The one color is used to color all the small triangles on the upper most and lower most lines, then there will be required another color so that the edges does not matches with the previous color. For the bigger hexagon shapes in the center we will require different colors for all of them because all of the hexagon shapes touches a line and an edge with each other.
its 2 trust me
its two cause if you think about it and color in the hexagons and triangles two different colors it works
You would assume that in this figure, the number of colored sections with which are not colored with respect to a " touching " colored section, would be half of the total colored sections. However that is not the case, the sections are not alternating as they still meet at a common point. After all, it notes no two touching sections, not adjacent sections. Their is no equation to calculate this requirement with respect to the total number of sections.
Let's say that we take one triangle as the starting. This triangle will be the start of a chain of other triangles that have no two touching sections, specifically 7 triangles. If a square were to be this starting shape, there are 5 shapes that have no touching sections, 3 being a square, the other two triangles. This is presumably a lower value as a square occupies two times as much space, but it also depends on the positioning. Therefore, the least number of colored sections you can color in the sections meeting the given requirement, is 5 sections for this first figure.
Respectively the solution for this second figure is 5 sections as well.