Hence, if two flag nodes are facing to each other, they must be neighbors in our configuration. Given this information, there can be only one mapping for hyperplanar grouping function for the input UDG $G = (V,E)$, and only one corresponding coloring of its corresponding hypergprah $H = (X,F)$.

Therefore, an algorithm to solve the problem in polynomial time can also be used to solve 2-coloring of 3-uniform hypergraphs in polynomial time.

at (-14, 15) [fixing lines]{}; (-14,14.5)–(-10,12); (-14,14.5)–(10,12); at (-15,2) [main line]{}; at (-15,1.2) [and support line]{}; =\[draw, fill=white, shape=circle, minimum size=3pt,inner sep=0pt\]; (-16,0) – (13,0); (-16,0.5) – (13,0.5); (-15,0)–(-16.5,0.5)–(-15.5,0.5) – (-14,0) – (-16.5,0.5); (-15,0)–(-15.5,0.5); at (-16.5,0.5) ; at (-15.5,0.5) ;

iin [8.5, 6.5, 4.5]{} [ (-12,i) – (12,i); (-11, i) – (-10, i+ .5) – (-9, i) – (-10, i- .5) – (-11, i); (11, i) – (10, i+ .5) – (9, i) – (10, i- .5) – (11, i); at (-11, i) ; at (-9, i) ; at (11, i) ; at (9, i) ; ]{}

iin [-4.5, -6.5, -8.5]{} [ (-12,i) – (12,i); (-11, i) – (-10, i+ .5) – (-9, i) – (-10, i- .5) – (-11, i); (11, i) – (10, i+ .5) – 