Abstract
In this study, from a tree with a quasi-spanning face, the algorithm will route Hamiltonian cycles. Goodey pioneered the idea of holding facing 4 to 6 sides of a graph concurrently. Similarly, in the three connected cubic planar graphs with two-colored faces, the vertex is incident to one blue and two red faces. As a result, all red-colored faces must gain 4 to 6 sides, while all obscure-colored faces must consume 3 to 5 sides. The proposed routing approach reduces the constriction of all vertex colors and the suitable quasi-spanning tree of faces. The presented algorithm demonstrates that the spanning tree parity will determine the arbitrary face based on an even degree. As a result, when the Lemmas 1 and 2 theorems are compared, the greedy routing method of Hamiltonian cycle faces generates valuable output from a quasi-spanning tree. In graph idea, a dominating set for a graph is a subset of . The range of vertices in the smallest dominating set for is the domination number (). Vizing’s conjecture from 1968 proves that the Cartesian fabricated from graphs domination variety is at least as big as their domination numbers production. Proceeding this work, the Vizing’s conjecture states that for each pair of graphs .
1. Introduction
Finite integral multipliers are used in the greedy routing algorithm. For the maximal tree, subgraphs are used, in which subgraph is denoted as . If the edges are suitably labeled, the two trees are distributed among them. Here the variation of a tree’s maximum number is established on the vertices and it is known as the Cayley. In general, a graph is drawn from the spanning tree vertices. The spanning tree is evaluated by using a single edge . To define the system’s vertex, a diagonal matrix is introduced. The variation between the adjacency matrix and the incidence matrix is determined by the spanning tree. Thus, the subgraph contains all the vertices, and the diameter for any single tree graph is denoted as
A spinning graph diameter is determined by one of the two trees, or , and is denoted as
The graph tree operation is denoted as The subsequent matrix is to combine rows and columns. Next to defining the spanning tree estimation, the product value is obtained. Cauchy-Binet present the estimation. This entire calculation is followed by vertices and adjacency calculations. The edges of the cycle are counted by the sign, and an insertion form appears [1]. To classify subgraphs and establish paths from subgraphs, the edge is connected to the spanning tree. The geometric cycle is not equal to the value of the subgraph.
2. Hamiltonian Cycle from Quasi-Spanning Tree of Faces
There is precisely one Hamiltonian cycle along with no cubic graphs, which is a unique Hamiltonian graph because a minimum of three Hamiltonian cycles are there in a Hamiltonian cubic graph. In 1978, Thomason showed that in a graph with the vertices of the odd degree, in an even number of Hamiltonian cycles, all edges are confined, proving Smith’s result [2]. Hence, uniquely Hamiltonian graphs unless even degree vertices and especially regular exclusively. Odd does not have Hamiltonian graphs. What is even ? Thomason showed that by Lovász local lemma, -regular exclusively even does not have Hamiltonian graphs [3], by a cautious option of parameters, theirs statements provide 73 rather of 300. That was modified by Haxell, Seamone, and Verstraete to . No 4-regular exclusively Hamiltonian graphs existed assumed by Sheehan. The fact of this assumption would indicate that cycles are the only regular exclusively Hamiltonian graphs, as according to Petersen’s 2-factor theorem.
According to Thomason’s result, an exclusively Hamiltonian graph has a necessity of minimum of two even degree vertices. This connection between the degree of the graph and either or not it is exclusively Hamiltonian increases several ordinary enquiries, such as either there are any exclusively Hamiltonian graphs of degree 3. Swart and Entringer gave a positive answer to that question by relating in closely cubic graphs an infinite family, that is, graphs along precisely two degrees 4 vertices and all of the other cubic vertices. Fleischer lately demonstrated that there are graphs with every vertex having a degree of 4 or 14 that are uniquely Hamiltonian [4].
Jackson and Bondy examined that an individually Hamiltonian graph of order consumes minimum one-degree vertex maximum which means the minimum degree is smaller than this number, here . Jamshed and Abbasi modified that to , here . Jackson and Bondy were especially attentive to planar exclusively Hamiltonian graphs in their article. A graph necessity has a minimum of two vertices of degree 2 or 3 that are displayed by them and assumed that all planar individually Hamiltonian graphs must have a minimum of two vertices of degree 2.
Proposition 1. Given consumes a Hamiltonian cycle through the exterior red face outdoor, all blue face within, and an edge is shared by no two red faces are together within, then the reduced graph H consumes a face’s spanning tree through in D does not contain the external face.
Proof. consumes a Hamiltonian cycle through the exterior red face external, every blue faces consistent to vertices in within, and an edge is shared by no two red faces are together insides, if and only if the reduced graph consumes a face’s quasi-spanning tree through in D does not contain the external face.
Theorem 1. Assume that all red faces have 6 sides or 4 sides, whereas blue faces have 3 or 5 sides, and that blue faces through 3 sides or 5 sides are adjacent to a minimum of one red face along with 4 sides (no conjecture is created for blue faces by 4, 6, 7, 8, 9, … sides). The reduced graph , which is obtained by crumbling blue faces, then has a correct quasi-spanning tree of faces, prove a Hamiltonian cycle.
2.1. We Now Prove the Main Result of This Section
Assume that all of ’s red faces have 4 sides or 6 sides, whereas the faces of blue are chance. Assume that the reduced graph contains a triangle with a minimum one vertex within, and no triangle in is none a face (i.e., includes minimum a vertex inside), and that no digon within is not a face (i.e., includes minimum one vertex within). We shall simplify the inside of the triangle one step at a time while preserving the property that which is no digon inside of that is not a face but authorizing the presence of triangles inside of that are not faced, subject to the succeeding conditions. Handle entire sets of parallel edges like a single edge. Assume and are different triangles within of , along including and perhaps like as , where is not a face, and so that there is no triangle differ from and like that includes and includes . Then, we assume that is a child of . We will need that no triangle consumes three different children , , and , any steps in explanation of the inside of the triangle .
The invariant property of is that no digon within is not a face, and no triangle within consumes three different children.
Lemma 1. Assume consumes a minimum of two vertices within and fulfills the invariant property, proving that it is feasible to choose a triangle ′ that is to say a face within and crimple ′ into an only vertex so that even gratify the invariant property [4].
Proof. Assume that triangle within includes a minimum of two vertices and that not any triangle within is not a face. Take = , in , we declare that consumes a minimum of two different neighbors , . Else, if consumes no neighbors, so goes to a triangle within with an edge parallel to the side of , which is a contradiction to the hypothesis that has no digon within of it is nonface, and while has unique like a neighbor within of , therefore is a triangle within of it is not a face, which is also a contradiction to the hypothesis. Then we can select and to , , are successive neighbors of , and crumble the triangle . There will be no digons that are not faced as a result of this because like a digon gets here before the crumpling from a triangle which is not a face within , a contradiction to the hypothesis. Inside , though, triangles that do not face may appear. Similar triangles derived from quadrilaterals , , and . The quadrilaterals can be one of two types: they can contain or they cannot contain , but they cannot have diagonal edges or , because then either a triangle this isn't a face was within the quadrilateral, otherwise crumpling the side is does not provide the quadrilateral a triangle which is non a face. Such indicates that all like quadrilaterals including are couple included in all other, and every such quadrilateral that do not comprise are pairwise included together [9]. The analogous properties prove that for the quadrilaterals , However, there is only one kind of these, namely those that contain . Else, and we consume the diagonal edge . The quadrilaterals contain analogous properties, but they are of a unique kind [10, 11], specifically do not comprise , then they are included in the triangle = . A quadrilateral including essential also include at all quadrilateral that does not contain and any quadrilateral that does not contain , and any quadrilateral that does not contain must also contain any quadrilateral that contains [12]. These assurances that these quadrilaterals do not take a main, next crumpling , inside , three triangles are not faces and do not conclude together, therefore conserving the property that three children are not taken by triangles [13‐15].
For residual case in crumpling a triangle, here is a triangle which consumes any one child or two children and , here together and consume precisely one vertex within. Assume shares no sides through any or . We must take the quadrilaterals , , and once more when writing = . Quadrilaterals including , , including , including , and not including may not exist at the same time. For if , then is not a face and therefore equals , thus is a vertex of and the quadrilateral cannot include while , is , and the similar argument applies, and while , then and , so the triangle is , this is not possible because the quadrilateral would be inside the triangle , it is called a face. As a result of symmetry, we can assume that after identifying and , there is either no quadrilateral including , otherwise no quadrilateral not including , which will provide an increase to a fresh triangle which is not a face. Crumpling the triangle identifies and and creates unique triangles through pairwise confinement introduce the new vertex , except the triangle , so conserving the property that three children are not taken by triangles. Assume = shares a single side by , it is a side , then one of the other two sides is not shared by , say the side , and the quadrilaterals unable to include , thus repeatedly we were able to crumple the triangle by within , producing unique triangles through pairwise confinement introducing the new vertex , except the triangle , so conserving the property that no triangle consumes three children. While and share aside , then every quadrilateral that includes also includes [16]. As a result, crumpling with inside provides two families of triangles through pairwise confinements concerning , one including and the another including, conserving the property that three children are not taken by triangles.
The succeeding proposition incorporates Herbert Fleischner’s result [17].
Proposition 2. Let us consider blue faces remain random and ’s red faces get 4 to 6 sides. The reduced graph has only one triangle which is in the outer layer and it does not have any faces, other than that the graph has no triangles. also incorporated no diagonal direction which is not even considered to face. has a spanning tree face which is triangles and is said to be Hamiltonian when contains odd number vertices.
Proof. While saving the invariant property, collaborate triangle faces into single vertices and redo Lemma 1. The total of vertices stays odd until the outer face remains by reducing the vertices by two. Eventually, a spanning tree is formed by the collaborated triangle. The main observation that results to this result is as follows:
Lemma 2. .In Theorem 1, take as same. If the graph has triangle with only one vertex, there is no other triangle inside , which is not considered to be a face also as it does not have any digons. To find out the acceptable quasi-spanning tree of faces for the graph ′, identifying the appropriate quasi-spanning tree face is reduced. By separating all inside vertices () and incident edges, it tends to incorporate the look-alike edge inside to every edge of obtained from the reduction graph .
Proof. As shown in the previous Lemma, by collapsing the triangle faces repeatedly we can wind up a in or else make nothing inside . In a quasi-spanning tree of faces, choose one of the three triangles which imply , which corresponds to the one in three diagonal directions for the sides of in . And we might either choose triangle in in a face of quasi-spanning trees. When the time T holds an off vertex and which is inside of , in this scenario the vertex which is in the is obtained, and then when the moment has an even number of vertices and which is in , in this case, we reached which has no vertices.
The parity inside the is represented by the two cases. Initially, if there is a digon named , has one endpoint which is in , and to frame a triangle we need to collaborate , the framed triangle does not have any faces out of the quadrilaterals such as , again there are a family of two quadrilaterals, consisting of two triangles as and . Quadrilaterals have and . The quadrilaterals provide triangles with pairwise boundaries of each family, which assures the property invariant that does not have , which is equivalent to or them have no children. Theorem allows as connected cubic bipartite planar graph of three nodes. Let us assume, be the subgraph of and reduced graph is ,
Here we got the results by removing all the possible edges with successive side by side edges. If the graph has one and two and three connected elements since has face’s spanning trees, then contains a Hamiltonian cycle. In the occurrence of a single element for , all faces among three colors of classes are considered.
We demonstrate vertex inside of only when there are no digons of . Which pertained to one of four triangles that share with side . After that, an appropriate quasi-spanning tree is built, two triangles and are included in the suitable spanning tree of faces and it does not share its edge. The collaborated triangles which are to remove , identify upon also identifies with and convert 5 vertices to only 2 vertices, also change the number of vertices. Hence the complete proof of Lemma is derived.
We can write = when there are no digons inside initially. There need to be 2 vertices inside of is present, if not the single vertex which inside have a degree and it does not have any digons, assuming that the blue face with 3 sides is needed to be adjacent to one red face with at least 4 sides. This indicates need to have at least two distinct neighbors inside , if not the case, is considered to be only one vertex of , since there are no such triangles as faces. If we calculate and , then contains a degree of 4. Similarly, edges holds at least a degree of 4. Further, there is no such vertex of that has degree 3 or 5, as all of the blue faces with 3 to 5 sides are close to one red face with 4 sides, as a result, a vertex is considered to be incident to digon. As per Euler’s formula, there must be three vertices of degree 4 in , on the other hand, there are 6 vertices of degree 4, is present. Let us assume which is inside contains four consecutive neighbors: . The quadrilateral share one edge with = , as we know indicates triangle. As and are getting shared, has only one adjacent neighbor which is in and it has degree 3 and not 4. Let us say might be shared with . In this scenario, make to an appropriate vertex of quasi and choose the two triangles such as and . Here, recognizing and detaching identifying and lessens the total number of vertices by 3. The quadrilaterals have the edge of which produces fresh triangles that contain of quadrilaterals which also gives new triangles that contain of edges. The quadrilaterals have edges which gives triangles that are newly created and those triangles having quadrilaterals of does not have the edge . By recognizing and , we tend to attain two families of newly created triangles with every family giving containment which is considered as pairwise that occurs among its triangles.
This gives that the property does not have triangle and equal to or else inside of having three children. Before proceeding to minimize the number of vertices by , which increases to two till a single vertex is not inside of and thus finishes off the proof with variation in parity of numbers inside of . As recently expressed, this decreases the issue of tracking down an appropriate semitraversing the tree of countenances for to the assignment of erasing the vertices inside and interfacing equal edges to the sides of to acquire ′.
Theorem 1 produces Lemma as a digon is considered as the outer face or else a triangle which has vertices inside of it. There is a triangle that has vertices inside and it does not have any triangles or two vertices of diagon inside or else the diagon contains vertices inside and in the same manner it does not have any triangles or diagons vertices inside. By removing the vertices and adding the same parallel edges to the side of , this has vertices inside and it does not have any triangle either. It can be clarified as per Lemma 1. The digon have a triangle with vertices inside, when a digon has vertices inside but it does not have any triangle and it has a of a single vertex. Among and , either one considered as a digon; only had the degree. For instance, it happens when is a digon. After removing the vertex , we can moreover choose the digon otherwise the triangle , that represent also not selecting or choosing the digon which has developed a face. When the outer face has no vertices in it and that the graph is simplified. In such a case, what is considered to complete this process is while selecting the face involved all the vertices in the quasi-spanning tree faces of and hence Theorem 1 is proved.
Coming up next is a rundown of corollary is an uncommon instance of Theorem 1 that sums up Goodey’s outcome to diagrams with just 4 sides or 6 sides.
Corollary 1. Assume be a 3-connected cubic planar bipartite graph, while the faces are three colored, through all vertex incident to a face of all color, since two of the three color classes include only that have 4 sides or 6 sides. The reduced graph , which is acquired by crumpling the class of the third color, thus includes a correct face’s quasi-spanning tree, and hence is a Hamiltonian cycle.
2.2. NP Complete and Polynomial Problems
The following result is for a face’s spanning tree where the majority of the faces are digons.
Theorem 2. Consider stay a 3-connected cubic planar bipartite graph. Assume the reduced graph for , and the subgraph of found by eliminating each edge with consecutive parallel edges. has a face’s spanning tree if it includes one or two or three connected components, and contains a Hamiltonian cycle. In one of the three color classes, all the faces are squares in the case of a single component for .
Proof. We can take be a spanning tree that corresponds to a spanning tree of digons in , while is a single linked component.
We can take a face of which takes vertices from together components if has two connected components. For the two components of , we assume two spanning trees of digons, starting with this face , and enhance that digons are unique at the same time show they do not create a cycle including . The single face and the added digons desire eventually span .
Although consumes three connected components, it is possible that has a face that touches each three, and we can move from to two components by examining for the two components, the three spanning trees of digons. Otherwise, we consider the first component, which has faces that contact it, as well as the second and third components, which also contain faces that contact it and the third component. We can select a face contact the first component and second component, and a face contact the first component and third component, so that those two faces do not divide each vertex, thus a cut of has a minimum of four edges because of 3-connectivity and the reality that at all cut consumes an edge’s even number. Initial through those two faces from the three spanning trees we can enhance digons for the three components thus far, a face’s spanning tree for is found since they do not form a cycle.
The result for three connected components applies to four connected components as well, but the result is not valid for five connected components.
Following that, we show how to decide in polynomial time that the reduced graph consumes a face’s spanning tree that is digons or triangles. Simply expands of the result, the case of a face’s spanning tree where all but a face’s constant number are digons or triangles.
3. Domination in Graphs
Consider = be a graph through the vertex set and the boundary set . If each vertex in is adjacent to the vertex in , it is a dominant set of . The domain number of , mentioned by () that is called the minimum cardinality of a dominant set of .
In the investigated branch of the diagram concept, supremacy in diagrams was used. The superiority of the diagrams was utilized in the examined division of the diagram idea. Blending problems with optimal problems, classical problems, and combinatorial problems is a growing principle. It has several applications in a range of fields, including body sciences, engineering, life sciences and society, and so on. The research interest in the graph concept these days is centered on dominance. This is essentially a list of new parameters that may be improved from basic dominance definitions. The NP completeness of elementary domination problems and investigate the relation to another NP completeness by them and action growth in the domination principle.
When in a graph every vertex is incident on at least one edge in , the set of edges is said to cover . The edge covering a set of a graph is said to be an edge covering or a cover subgraph or simply a cover (e.g., a spanning tree in a linked graph is a cover). The example of a computer network over the relation minimum vertex coverage is shown in Figure 1 [5].
3.1. Applications of Domination in Graph
The graph applications of domination have been applied in a variety of fields. The dominion comes from structural challenges in which there is a constant type of centers (e.g., hearth stations, hospitals) and space must be kept to a minimum. To diminish the number of locations where a surveyor needs to commit to taking peak measurements for a whole area, surveyors use standards of domination.
3.2. Domination Path
A graph containing a dominating path is one where each vertex exterior of includes a neighbor on . Let represents the vertex and represents a graph’s edge set. represents a vertex’s neighborhood in and denotes its degree. For D, represented by letting = and letting = and = ∣∣. Likewise, () represents the minimum vertex degree and () represents the maximum vertex degree.
Theorem 3. Since , each connected -vertex graph along contains a dominating path and proves the inequality is acute.
Proof. The sharpness structure is declared for . In general, assume for the structure be a clique over vertices, . Then the three cliques jointly moreover include vertices, . Now presume that is a connected graph of n-vertex over which include no dominating path; find that , here . Assume first that is 2-connected. Dirac showed that essential since containing a cycle over at least min vertices. A path over minimum is a dominating path vertex, so we can connect . Once is 2-connected, we consume , and contain a cycle of length minimum . While dominating path does not have the vertex set, further few vertex and on its neighbors are not. Since is connected, here is the shortest path start at and end at . Summing to a path along with at one end and ’s alternative neighbor at the other end (existing then ) gains a path along minimum vertices. While it is not that a dominating path, therefore neglects its neighborhood and some other vertex, it needs , an inconsistency. Therefore, must include a cut-vertex . Every component of has minimum vertices, so has a maximum of three components. Since has maximum vertices, consuming three components along minimum vertices needs one through precisely vertices. So, a component shall be a complete graph through all vertices adjacent to . Further two components take order maximum , therefore a vertex in like a component is nonadjacent to maximum one other vertex of , while is nonadjacent to . As a result, contains a dominating path that include and in this case two vertices each from the two majors components. In the leftover case, consumes two components, though also contains a cut-vertex , therefore include three closely complete components flexible in a dominating path as shown in the above section. While every component of is 2-connected, since all contain a cycle which is spanning or consumes minimum vertices, then removing depart a minimum degree at least . All component contains at most vertices because it includes minimum vertices. We get a path across that is dominating and neglects very few vertices. On other hand provides a brief proof of nearly the optimal threshold for 2-connected graphs. Dirac’s theorem proves too that suppose , therefore is Hamiltonian-connected, sense that some two vertices are a spanning path’s terminus. In a path start at and end at and , consider represent the instant successors set vertices of with , and consider represents the set of instant predecessors. We know that once includes no terminus of .
3.3. Vizing’s Conjecture in Domination
The comparison of the dimensions of minimal dominant sets and Vizing conjectures in the and graphs, in the Cartesian product graph that is called a dominant set. The proof of the Vizing theorem with the use of some colors, every simple nonoriented graph can be multicolored.
Let be determinate. In vertex subsets, dominates though , that is, while each vertex is in , otherwise is adjacent to a vertex. dominates outwardly , when , are separate and dominate . The domain number is the lowest represented cardinal dominating . Although dominates , further, dominates and this is a ’s dominant set.
Each closed quarter in must span any dominant set of . Hence, the domain number of is minimum similar to the cardinality of whatever set consuming the characteristics that for different , in , and . So, a set is known as 2-packing and is represented the maximum cardinality of a 2-packing in and is named the 2-packing number of . The independence number of a vertex in is the maximum cardinality of an independent set of vertices in , and the smallest cardinality of a dominant set that is likewise independent is represented .
Assume is not a complete graph prove for every vertex pair and which are not adjacent to , it is proved that . While S has the property that for that pair of nonadjacent vertices, since is critical concerning the domain (or critical for brevity).
The graph , which has the domain number . is known as a separable graph if all of its vertices can be enclosed by all of its subgraphs.
Theorem 4. Suppose a decomposable graph have a spanning subgraph , so that = , then holds for each graph ,
Proof. Undirected, finite graphs, coherent, and simple are all considered. Specific, let denote a graph have the edge set and the vertex set . are two vertices and its neighbors, otherwise in case . The ’s open neighborhood belongs to and it is the ’s neighbor set, denoted as , whereas the closed neighborhood . The s open neighborhood contained in and it is the set of all neighbors of vertices in , denoted as , whereas the s closed neighborhood is . But is detached from the context, it perhaps represented by and or and correspondingly. The space among two vertices is in the shortest length path and is represented by . In two graphs, the Cartesian products are and represented by , is a vertex-set graph and edge set .
A subset of vertices is known as a dominant set of half sum, if , and every vertex a vertex occurs, thus . When is a dominant set of half-sums in the induced subgraph of , a vertex set semidominates a vertex set . The semitotal dominance number of , denoted as , is known as a minimum half-sum dominating set size of . A 2-pack is a subset of vertices in which each pair of s vertices are a minimum of 3 separate. The maximum 2-pack size of is known as the 2-pack number [6–8].
Theorem 5. To are all isolate-free graphs. Then,
Proof. Let us take be a max of 2 packages of graph. Consider without restrictions as . Every vertex in the graph is at least three far from the packing of vertices. The closed adjacent are represented as pairwise disjoint and for . Consider is said to be a partition of just like for . Let be an -set. For , . Let . Moreover, consider a minimum set of vertices that dominate completely and include as several vertices as feasible in . Further . Next is not present in when has a vertex, and is considered to be the uniquely determined vertex that entirely dominates for ∈ . Since contains neighbors that pertain to , vertices in dominate that all neighbors, even now is a semisumerally dominant set once is changed to in . Hence, vertices set that almost fully dominate and have further vertices in , therefore, called , is an inconsistency. Since are subsets, and thus is a partial dominance of in persuaded by . Then partially dominance , . Therefore, .
The subtotal must be calculated of the domination number and the results of Vizing’s type based on it. Separating minimum half dominating sets into partially dominating sets which is considered as completely dominate. is considered to be a minimum of a semidominant set of graphs , note that it is suitable for each graph. It might be partitioned into two sets of . Here represents vertices set of that are nearer to anyone vertex of , on the other hand, represented as on . Take as the minimum dominating set of vertices for every graph , also take , and represents partitions in allied and free sets considerably. Therefore, it represents , so as a result, is considered to be the max extent for , a maximum relayed semitotal dominant set of . Maximum of allied partition of the graph is represented as . The set denote a maximum related set of , and the set a minimum free set of . Each maximal related partition of assume = and .
4. Conclusion
Hamiltonian cycle’s quasi-spanning tree of faces is executed in this research. In a cubic bipartite planer graph, a polynomial time technique is utilized to reduce the issues in the minimum quasi spanning tree. For another graph-like products and other domination, numerous researchers have conducted Vizing’s conjecture. But still, this conjecture is not yet demonstrated. To prove Vizing’s conjecture, a graph theory described one or two conjectures which are still considered as wider problems. To separate free graphs, Vizing types are based on the subtotal of domination number and it proved as well. Vizing’s conjecture is said to be true when the polynomial time was positive concerning the particularly build ideal. And here Vizing’s conjecture is designed by the graph theory as an appropriate pair.
Data Availability
No data were used to support this study.
Disclosure
An earlier version of the manuscript is presented as preprint in [4] in the link https://theory.stanford.edu/∼tomas/barnew.pdf.
Conflicts of Interest
The authors declare that there are no conflicts of interest regarding the publication of this article.
Acknowledgments
The authors thank VR Siddhartha Engineering College, Vijayawada, Shree Institute of Technical Education, Tirupati, and Bangabandhu Sheikh Mujibur Rahman Science and Technology University, Gopalganj, Bangladesh.