1Molecular Model Discovery Laboratory, Department of Chemistry and Biotechnology, Faculty of Science, Engineering and Technology, Swinburne University of Technology, Hawthorn Campus, Hawthorn, Victoria, Australia
2Graduate School of Sciences, Information Technology and Engineering and Centre of Informatics and Applied Optimization, Faculty of Science and Technology, The Federation University Australia, Mount Helen Campus, Mount Helen, Ballarat, Victoria, Australia
Received date: January 14, 2016; Accepted date: January 21, 2016; Published date: January 23, 2016
Citation: Zhang J (2016) Packing Rectangular-Box Cells Optimally for the Crystal Structure. Med chem 6:039. doi:10.4172/2161-0444.1000320
Copyright: © 2016 Zhang J. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.
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In crystal structures, if each crystal cell is a special rectangular box (length × width × height=23 × 11.5 × 11.5 angstroms3), then how to pack these cells onto a enough hard base of a (46 × 46 angstroms2) square in order to make the crystal reach the stablest/optimal structure as high as the crystal it is? This is a practical problem that we are always thinking in our mind. This short article is to present an algorithm to solve this problem.
We will give an algorithm to solve the problem:
Step 1: Put 2 boxes on one of the four boundaries (denoted as B1) of the base, with their lengths occupying B1.
Step 2: Put 1 box on each of the neighbored boundaries (denoted as B2 and B3) respectively, with each length side occupying each boundary respectively.
Step 3: Put 2 boxes on the remaining boundary (denoted as B4) of the base, with their lengths occupying B4.
Step 4: Now a 23 × 23 angstroms2 square is inner (or in the center of) the base. Put 2 boxes onto the inner square, with their length sides paralleling to B1 and/or B4. Now we finished the layout of the 1st layer.
Step 5: Determine how many layers that you will lay the boxes, according to your own constraints. Once the number of layers was determined, we will repeat Step 6 to finish the layout onto the inner 23 × 23 angstroms2 square.
Step 6: Put 2 boxes in the even number of the layer, with their length sides paralleling to B2 and/or B3; put 2 boxes in the odd number of the layer, with their length sides paralleling to B1 and/or B4.
Step 7: Now we lay boxes onto the outside of the inner 23 × 23 angstroms2 square: repeating Steps 8 ~ 9 until reaching the maximum number of layers.
Step 8: In the even number of the layer, put 1 box on both sides paralleling to B1 and B4 respectively; in the odd number of the layer, put 1 box on both sides paralleling to B2 and B3 respectively.
Step 9: In the even number of the layer, put 2 boxes on both sides paralleling to B2 and B3 respectively; in the odd number of the layer, put 2 boxes on both sides paralleling to B1 and B4 respectively.
The above algorithm is illuminated in Figure 1 in a workshop. If the maximum number of layers goes to infinity, and each layer produced by the above algorithm is looked as one solution of the algorithm, then the above algorithm is convergent in the strict algorithm convergence theory [1-3].
This research was supported by a Victorian Life Sciences Computation Initiative (VLSCI) grant numbered VR0063 on its Peak Computing Facility at the University of Melbourne, an initiative of the Victorian Government, Australia.