Team:USTC Software/project

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(Difference between revisions)
(Chain Sorting and Weighting Scheme)
(Chain Sorting and Weighting Scheme)
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Sorting is always performed under some rules. Since there are no natural options, we design one to generate the unique order for a list of chains. The rule we developed is something like that for comparing two strings: comparing first the '''partRef''' and '''partType''' defined in database of the first part of two chains in alphabet order and if equal, then comparing the next until differences or the end of either chain. For convenience, we generate a '''unicode''' (a string) for each chain by putting their '''partRef''' and '''partType''' attributes one by one, for example:
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Sorting is always performed under some rules. Since there are no natural options, we design one to generate the unique order for a list of chains. The rule we developed is something like that for comparing two strings: comparing first the '''partRef''' and '''partType''' defined in database of the first part of two chains in alphabet order and if equal, then comparing the next until differences or the end of either chain. For convenience, we generate a '''unicode''' (a string) for each chain by splicing their '''partReference''' and '''partType''' attribute values in order, for example:
  <font color='Yellow'>[placI185:ForwardDNA][RBS147:ForwardDNA][lacI153:ForwardDNA][TE145:ForwardDNA]</font>
  <font color='Yellow'>[placI185:ForwardDNA][RBS147:ForwardDNA][lacI153:ForwardDNA][TE145:ForwardDNA]</font>
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Back to the question above, if two chains have the same '''unicode''', which should be arranged with preference of the other? The answer lies in the exchange of them. The exchange will change the parts' weight on both chains and thus the weight of some nodes of the trees. The tree structure (with virtual root) is also possible to change. If the tree structure is not changed, the two chains are equivalent meaning that it makes no difference to arrange either chain in front of the other. If the tree structure changes, we must find some ways to determine which tree structure is preferred and based on this to arrange the two chains with equal '''unicode'''. We know each tree could be uniquely converted to a binary tree by taking the first child of each node as the left child and its siblings as its right child in the binary tree. The Huffman algorithm could be used to generate the Huffman code of each node in the binary tree. Since the Huffman code is actually a representation of the path from current node to the root, we can restore the tree structure from the set of Huffman codes of all leaf nodes. Hence, the Huffman codes set of leaf nodes could uniquely represent a tree structure.
+
Back to the question above, if two chains have the same '''unicode''', which should be arranged with preference of the other? The answer lies in the exchange of them. The exchange will change the parts' weight on both chains and thus the weight of some nodes of the trees. The tree structure (with virtual root) is also possible to change. If the tree structure is not changed, the two chains are equivalent meaning that it makes no difference to arrange either chain in front of the other. If the tree structure changes, we must find some ways to determine which tree structure is preferred and based on this to arrange the two chains with equal '''unicode'''. We know each tree could be uniquely converted to a binary tree by taking the first child of each node as the left child and its siblings as its right child in the binary tree. The Huffman algorithm could be used to generate the Huffman code of each node in the binary tree. Since the Huffman code is actually a representation of the path from current node to the root, we can restore the tree structure from the set of Huffman codes of all leaf nodes. Hence, the Huffman codes set of leaf nodes could uniquely represent a tree structure. We could generate a similar '''unicode''' for tree structure by splicing the Huffman codes of all leaf nodes sorted by their weights and adding a asterisk between two adjacent codes.

Revision as of 05:15, 11 October 2010

Chain Sorting and Weighting Scheme

liaochen@mail.ustc.edu.cn

Chains in our Chain-Node Model should not have any order since there are no natural ways to sort them. One chain is not preferred to another, since in a complex of multiple chains, which are connected via binding interaction of binding bites on their own, all chains are equivalent. However, a chain-sorting program is a prerequisite for implementations of many other functions, such as comparison of two Chain-Node structures, or pattern matching of template species to those with definite structure.


Sorting is always performed under some rules. Since there are no natural options, we design one to generate the unique order for a list of chains. The rule we developed is something like that for comparing two strings: comparing first the partRef and partType defined in database of the first part of two chains in alphabet order and if equal, then comparing the next until differences or the end of either chain. For convenience, we generate a unicode (a string) for each chain by splicing their partReference and partType attribute values in order, for example:

[placI185:ForwardDNA][RBS147:ForwardDNA][lacI153:ForwardDNA][TE145:ForwardDNA]

If one part is binded with another part (a binding site), we will mark it with *, the example above becoms (if placI185 is binded with some part, say lacI153 protein):

[placI185:ForwardDNA:*][RBS147:ForwardDNA][lacI153:ForwardDNA][TE145:ForwardDNA]

If unicode of each chain is different with others, then we could sort the chain list by comparing their unicode (just the same with comparison of two strings). However, it is not always so: chains with same unicode may be equivalent or not. It is due to the mark asterisk *. Unicode knows where are the binding sites, but it does not know the binding details of that part. In other words, unicode could not soly determines the order of the chains since it lacks of information of the trees defined in our Chain-Node model. The forest (tree set) in Chain-Node Model describes the binding details of parts in chains. We need to find something like unicode to represent the essence of the associated trees. Our weighting scheme will be first introduced below.


Assume the chains are sorted already. Each part will be assigned a weight based on its position in the chain list. The weight of the jth part on the ith chain is (i,j). For example, the first part on the first chain posses a weight (0,0) (in C style, counting number from zero). After assignment of parts' weight, we turn to weights of the nodes in trees. Since leaf nodes in trees are actually the representation of parts in binded state on chains, the weight of each leaf node could be inherited from the weight of its corresponding part. For parent node, its weight equals to the maximum of weights of all its children nodes (we define a weight (i, j) is greater than another (i',j') if i < i' or i = i' while j < j'). The recursive definition of tree nodes' weight make it possible to sort the forest. For convenience, we add a virtual node ROOT, whose weight value could also be obtained under the general rule, as the root of all trees in the forest and make a conversion from forest to tree. The tree structure could be uniquely determined by recursively sorting all children of each nodes by their weights.


Back to the question above, if two chains have the same unicode, which should be arranged with preference of the other? The answer lies in the exchange of them. The exchange will change the parts' weight on both chains and thus the weight of some nodes of the trees. The tree structure (with virtual root) is also possible to change. If the tree structure is not changed, the two chains are equivalent meaning that it makes no difference to arrange either chain in front of the other. If the tree structure changes, we must find some ways to determine which tree structure is preferred and based on this to arrange the two chains with equal unicode. We know each tree could be uniquely converted to a binary tree by taking the first child of each node as the left child and its siblings as its right child in the binary tree. The Huffman algorithm could be used to generate the Huffman code of each node in the binary tree. Since the Huffman code is actually a representation of the path from current node to the root, we can restore the tree structure from the set of Huffman codes of all leaf nodes. Hence, the Huffman codes set of leaf nodes could uniquely represent a tree structure. We could generate a similar unicode for tree structure by splicing the Huffman codes of all leaf nodes sorted by their weights and adding a asterisk between two adjacent codes.