Computational Aspects of Carbon and Boron Nanotubes
Abstract
:1. Introduction
2. Basic Properties of Nanotubes of Armchair Model
- (1)
- A carbon hexagonal nanotube of order n × m has nm vertices and m(3n–2)/2 edges.
- (2)
- A boron triangular nanotube of order n × m has 3nm/2 vertices and 3m(3n–2)/2 edges.
- (3)
- A boron α-nanotube of order n × m has 4 nm/3vertices and m(7n–4)/2 edges when n is a multiple of 3.
3. Independent Set of Three Nanotubes
4. Perfect Matching and Matching Ratio
5. Broadcasting Problem of Carbon and Boron Nanotubes
5.1. Broadcasting Algorithm for Carbon Hexagonal Nanotubes
- Step 1:
- Let O be the source node with the message and E denote the eccentric node of O. Unfold the carbon hexagonal nanotube into a rectangular sheet in such a way that the eccentric vertex E lies on the perimeter of the rectangular sheet. See Figure 6.
- Step 2:
- Draw lines TOZ (at angle 30°), ROX (vertical), and SOY (at angle 150°). These lines create six zones, namely, zones ROS, SOT, TOX, XOY, YOZ and ZOR. See Figure 7.
- Step 3:
- Delete all the edges of zone ROS which are perpendicular to OS. Similarly, delete all the edges perpendicular to OT in zone SOT, edges perpendicular to OX in zone TOX, edges perpendicular to OY in zone XOY, edges perpendicular to OZ in zone YOZ, edges perpendicular to OR in zone ZOR. The resulting tree is the broadcasting tree of the carbon hexagonal nanotubes. See Figure 8.
- Step 4:
- Message is disseminated from source O based on farthest-distance-first protocol where a node with the message chooses an uninformed adjacent node which leads to longest path in the tree. If a node has label i, it means that the node receives the message from its neighbor at ith time unit. See Figure 9.
5.2. Broadcasting Problem of Boron Triangular Nanotubes
- Step 1:
- Let O be the source node with the message and E denote the eccentric node of O. Unfold the boron triangular nanotube into a rectangular sheet in such a way that the eccentric vertex E lies on the perimeter of the rectangular sheet. See Figure 10.
- Step 2:
- Draw lines SOY (horizontal), TOZ (at angle 60°), and XOR (at angle 120°). These lines create six zones, namely, zones ROS, SOT, TOX, XOY, YOZ and ZOR. See Figure 11.
- Step 3:
- Delete all the edges of zone ROS except the edges parallel to OS. Similarly retain only the edges parallel to OT in zone SOT, edges parallel to OX in zone TOX, edges parallel to OY in zone XOY, edges parallel to OZ in zone YOZ, and edges parallel to OR in zone ZOR. The resulting tree is the broadcasting tree of the boron triangular nanotube. See Figure 12.
- Step 4:
- The message is disseminated from source O based on farthest-distance-first protocol where a node with the message chooses an uninformed adjacent node which leads to longest path in the tree. If a node has label i, it means that the node receives the message from its neighbor at ith time unit. See Figure 13.
Conclusions
Acknowledgements
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Manuel, P. Computational Aspects of Carbon and Boron Nanotubes. Molecules 2010, 15, 8709-8722. https://doi.org/10.3390/molecules15128709
Manuel P. Computational Aspects of Carbon and Boron Nanotubes. Molecules. 2010; 15(12):8709-8722. https://doi.org/10.3390/molecules15128709
Chicago/Turabian StyleManuel, Paul. 2010. "Computational Aspects of Carbon and Boron Nanotubes" Molecules 15, no. 12: 8709-8722. https://doi.org/10.3390/molecules15128709
APA StyleManuel, P. (2010). Computational Aspects of Carbon and Boron Nanotubes. Molecules, 15(12), 8709-8722. https://doi.org/10.3390/molecules15128709