
Short Title of the Article
Lemma 2.1. For the partition defined above, if the height of the triangle 𝐷is divided into 𝑑number of equal parts,
then, there are
i) 2𝑑2− 2𝑑+ 3 subtriangles and
ii) 𝑑2+𝑑+ 1 vertices,
in the partition.
Proof. i) According to the partition defined, the height of the triangle is divided into 𝑑number of equal parts,
resulting 𝑑+ 1 ′𝑦′values, where 𝑦𝑑+1 is the 𝑦coordinate of the top vertex of 𝐷. Along each horizontal line
𝑦=𝑦𝑗, 𝑗 = 1,2, ..., 𝑑 the line is divided into 𝑑number of equal parts. The coordinates of the newly obtained
points along the line 𝑦=𝑦𝑗are denoted by (𝑥1, 𝑦𝑗),(𝑥2, 𝑦𝑗), ..., (𝑥𝑑+1, 𝑦𝑗).It is observed from the partition that
between two consecutive points on the line 𝑦=𝑦𝑗,two subtriangles are obtained (a normal and an inverted
triangle). Hence, along each line 𝑦=𝑦𝑗,there are 2𝑑subtriangles for 𝑗= 2, ..., 𝑑. Considering the top most three
subtriangles, which are fixed irrespective of 𝑑, there are (2𝑑)(𝑑− 1) + 3 subtriangles in the partition. 𝑖.𝑒, the total
number of subtriangles in the partition is 2𝑑2− 2𝑑+ 3.Hence the proof.
ii) Following the partition, since each horizontal line 𝑦=𝑦𝑗is divided into 𝑑equal parts, there are 𝑑+ 1 new
vertices along each line 𝑦=𝑦𝑗, 𝑗 = 1,2, ..., 𝑑. Now, including the top most vertex 𝐶, of the triangle 𝐷, there are
[(𝑑+ 1)𝑑] + 1 = 𝑑2+𝑑+ 1 vertices in the partition. Hence the proof.
Lemma 2.2. If 𝑑= 3𝑛+ 1, 𝑛 ∈𝑁, then the graph of the partition defined above has chromatic number 3.
Proof. According to the partition, each horizontal line 𝑦=𝑦𝑗is divided into 𝑑equal parts, generating 𝑑+ 1 points
along that line. Let the points be denoted by (𝑥1, 𝑦𝑗),(𝑥2, 𝑦𝑗), ..., (𝑥𝑑+1, 𝑦𝑗).Among these points, (𝑥1, 𝑦𝑗)and (𝑥𝑑+1, 𝑦𝑗)
are points on the two sides of the triangle 𝐷. The remaining points are intermediate points and there 𝑑− 1 such points
along that line, where 𝑑− 1 is a multiple of 3. Now, considering the coloring of theses points with the least number
of colors, since the points (𝑥1, 𝑦1)and (𝑥𝑑+1, 𝑦1)are adjacent with respect to the triangle 𝐷, they have to be colored
differently. Without loss of generality, let (𝑥1, 𝑦1)and (𝑥𝑑+1, 𝑦1)be colored with colors ’1’ and ’2’ respectively. Now,
since the point (𝑥2, 𝑦1)is adjacent to (𝑥1, 𝑦1),it should be colored ’2.’ Similarly, (𝑥3, 𝑦1)is colored ’1’. Proceeding
in this manner, the point (𝑥𝑑−1, 𝑦1)will be colored ’1.’ Then, the point (𝑥𝑑, 𝑦1)has to colored with a different color
other than ’1’ and ’2’, since it is adjacent to both (𝑥𝑑−1, 𝑦1)and (𝑥𝑑+1, 𝑦1).Hence, a minimum of 3 colors are needed
to color the points along this line. The same reasoning can be applied along each of the horizontal lines 𝑦=𝑦𝑗and
along the slanting sides of 𝐷. Thus, it can be established that atleast 3 different colors are needed to properly color the
graph. Therefore, the chromatic number of the graph is 3.
2.2. Construction
Consider a triangular domain 𝐷with vertices (𝑥1, 𝑦1),(𝑥2, 𝑦2),(𝑥3, 𝑦3),colored ’1’, ’2’ and ’3’ respectively. Let
the triangle be partitioned into 𝑁number of subtriangles 𝐷1, 𝐷2, ..., 𝐷𝑁such that 𝐷= ∪𝑁
𝑛=1𝐷𝑛and ∩𝑁
𝑛=1𝐷𝑛=𝜙. The
partitioning is done such that the chromatic number of their corresponding graph is 3. Each subtriangle be numbered
from 1 to 𝑁as shown in Figure 1. Set 𝑃= {(𝑥𝑛𝑗 , 𝑦𝑛𝑗 ) ∶ 𝑗= 1,2,3, 𝑛 = 1,2, ..., 𝑁}to be the set of all vertices of the
subtriangles 𝐷𝑛, 𝑛 = 1,2, ..., 𝑁 . Let 𝑧𝑛𝑗 =𝑓(𝑥𝑛𝑗 , 𝑦𝑛𝑗 )be the corresponding function values.
Let 𝑅= {(𝑥𝑛𝑗 , 𝑦𝑛𝑗 , 𝑧𝑛𝑗 ) ∶ 𝑗= 1,2,3, 𝑛 = 1,2, ..., 𝑁}be the data set. Without loss of generality, let (𝑥𝑛1, 𝑦𝑛1)denotes
the vertex colored ’1’, (𝑥𝑛2, 𝑦𝑛2)be the vertex with color ’2’, and (𝑥𝑛3, 𝑦𝑛3)be the vertex colored ’3’ in 𝐷𝑛.
Consider an invertible, affine map 𝐿𝑛∶𝐷→𝐷𝑛such that
𝐿𝑛(𝑥𝑗, 𝑦𝑗)=(𝑥𝑛𝑗 , 𝑦𝑛𝑗 ),for 𝑗= 1,2,3.(1)
𝑖.𝑒, 𝐿𝑛maps (𝑥𝑗, 𝑦𝑗)to the vertex in 𝐷𝑛, which is colored 𝑗, for 𝑗= 1,2,3.The map 𝐿𝑛used here is given by,
𝐿𝑛(𝑥, 𝑦) = [𝛼𝑛1𝛼𝑛2
𝛼𝑛3𝛼𝑛4][𝑥
𝑦]+[𝛽𝑛1
𝛽𝑛2](2)
First Author et al.: Preprint submitted to Elsevier Page 3 of 14