2.1. Locating-Chromatic Number of Origami Graphs
Theorem 1. Let  be an origami graph for . Then, the locating-chromatic number of ,
 Proof.  Let  with . An origami graph  is a graph with  and ∪∪. Next, to prove the theorem, we consider the following two cases:
 First, we determine the lower bound of . In the origami graphs  for , there are three adjacent vertices (complete graph with three vertices, denoted by ); we then need at least 3-locating coloring. Without loss of generality, we assign three colors for any  in  for , and then the three vertices are dominant vertices. As a result, if we use three colors it is not enough because there are more than one  in  for . Therefore, .
Next, we determine the upper bound of . To show that 4 is an upper bound for the locating-chromatic number for the origami graph  we describe a locating coloring c using four colors as follows:
The coloring c will create the partition  on . We shall show that the color codes of all vertices in  are different. We have: ; ; ; ; ; ; ; ; . Since the color codes of all vertices  are different, c is a locating-chromatic coloring. Thus, .
Case 2.  , for 
 To determine the lower bound, we will show that four colors are not enough. For a contradiction, assume that there exists a 4-locating coloring c on  for . We assign , where  because , . Observe that, on  for , there are n vertices  whose degree is 5. As a result, at least two vertices ,  have the same color codes, which is a contradiction. Therefore, , for .
To show the upper bound for the locating-chromatic number of origami graphs  for , let us differentiate some subcases. 
Subcase 1.  (Odd n), for  odd, 
 Let c be a coloring of origami graph ,  odd, and ; we make the partition  of :
- ; 
-  for odd ∪ for even ; 
-  for even ∪ for even ∪ for odd ; 
- ; 
- . 
For  odd, the color codes of all the vertices of  are:
For 
 we have:
		  
For 
 we have:
 for odd ∪ for even .
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
, we have:
For 
i even, 
 we have:
 for even ∪ for even ∪ for odd 
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
For 
, we have:
For 
, we have:
Since for n odd all vertices have different color codes, c is a locating coloring of origami graphs , so that , for  odd, .
Subcase 2.  (Odd n), for  even, .
 Let c be a coloring of origami graph ,  even, and ; we make the partition  of  as follows:
- ; 
-  for odd ∪ for even ; 
-  for even ∪ for even ∪ for odd ; 
- ; 
- . 
For  even, the color codes of all the vertices of  are:
For 
 we have:
For 
 we have:
 for odd ∪ for even .
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
 for even ∪ for even ∪ for odd 
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
, we have:
Since for n odd all vertices have different color codes, c is a locating coloring of origami graphs , so that , for  even, .
Subcase 3.  (even n), for  odd, .
 Let c be a coloring of origami graph ,  odd, and ; we make the partition  of :
- ; 
-  for odd ∪ for even ; 
-  for even ∪ for odd ; 
- ; 
- . 
For 
 odd, the color codes of all the vertices of 
 are:
For 
 we have:
For 
 we have:
 for odd ∪ for even 
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
 for even ∪ for odd 
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
For 
, we have:
For 
, we have:
Since for n even all vertices have different color codes, c is a locating coloring of origami graphs , so that , for  odd, .
Subcase 4.  (even n), for  even, .
 Let c be a coloring of origami graph ,  even, and ; we make the partition  of  as follows:
- ; 
-  for odd ∪ for even ; 
-  for even ∪ for odd ; 
- ; 
- . 
For  even, the color codes of all the vertices of  are:
For 
 we have:
For 
 we have:
 for odd ∪ for even 
For 
i odd, 
 we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
 for even ∪ for odd 
For 
i odd, 
 we have:
For 
, we have:
For 
i odd, 
 we have:
For 
i even, 
 we have:
For 
i even, 
 we have:
For 
, we have:
For 
, we have:
Since for n even all vertices have different color codes, c is a locating coloring of origami graphs , so that , for  even, . this completes the proof of Theorem 1.    □
Note that 
Figure 1 is an example locating coloring for origami graph 
.
  2.2. Locating-Chromatic Number for Subdivision Outer Edge of Origami Graphs
Theorem 2. Let  be a subdivision outer edge of origami graphs for . Then the locating-chromatic number of , 
 Proof.  Let ,  be a subdivision of an origami graph;  is a graph with  and ∪∪}. Next, to prove the theorem, we consider the following two cases: 
 Case A.   First, we determine the lower bound of .
Without loss of generality, we assign . Then, there are three dominant vertices in A, while we still have vertices on other A that must be colored. As a result, there will be two vertices with the same color codes. Thus, .
Next, we determine the upper bound of . To show that 4 is an upper bound for the locating-chromatic number for a subdivision outer edge of origami graph , we describe a locating coloring c using four colors as follows:
The coloring c will create the partition  on . We shall show that the color codes of all vertices in  are different. We have: ; ; ; ; ; ; ; ; . ; ; . Since the color codes of all vertices  are different, c is a locating-chromatic coloring. Thus, . 
Case B.   for 
 Since a subdivision of origami graphs  for  is obtained by origami graph  with one added vertex in edge , we have  for . The addition of one vertex on the outside does not reduce the number of colors needed because the number of the sets  is the same.
To show the upper bound for the locating-chromatic number for a subdivision outer edge of origami graph  for , let us consider different subcases. 
Subcase a.  (odd n), for  odd, .
 Let c be a coloring for a subdivision outer edge of origami graph , for  odd, and ; we make the partition  of :
- ; 
-  for odd ∪ for even ∪ for odd ; 
-  for even ∪ for even ∪ for odd ∪ for even ; 
- ; 
- . 
For for  odd the color codes of all the vertices of  are:
Since for n odd all vertices have different color codes, c is a locating coloring for subdivision of origami graph , so that , for  odd, .  
Subcase b.  (odd n), for  even, .
 Let c be a coloring for a subdivision outer edge of origami graph , for  even, and ; we make the partition  of :
- ; 
-  for odd ∪ for even ∪ for odd ; 
-  for even ∪ for even ∪ for odd ∪ for even ; 
- ; 
- . 
For  even, the color codes of all the vertices of  are:
Since for n odd all vertices have different color codes, c is a locating coloring for a subdivision of the outer edge of origami graph , so that , for  even, .
Subcase c.  (even n), for  odd, .
 Let c be a coloring for a subdivision outer edge of origami graph , for  odd, and ; we make the partition  of :
- ; 
-  for odd ∪ for even ∪ for odd ; 
-  for even ∪ for odd ∪ for even ; 
- ; 
- .  
For  odd, the color codes of all the vertices of  are:
Since for n even all vertices have different color codes, c is a locating coloring for a subdivision of the outer edge of origami graph , so that , for  odd, . 
Subcase d.  (even n), for  even, .
 Let c be a coloring of subdivision origami graph , for  even, and ; we make the partition  of :
- ; 
-  for odd ∪ for even ∪ for odd ; 
-  for even ∪ for odd ∪ for even ; 
- ; 
- .  
For  even the color codes of all the vertices of  are:
Since for n even all vertices have different color codes, c is a locating coloring for a subdivision outer edge of origami graph , so that , for  even, . This completes the proof of Theorem 2.    □
Note that 
Figure 2 is an example locating coloring for a subdivision of the outer edge of origami graph 
.