, Protein representation Group: Core::CSB / Package: Protein representation
, classes Pack: Protein representation / Class: Polypeptide chain representation Gives access to a number of high level accessors and iterators to manipulate a polypeptide chain. Allows access to the three structures described in the previous sub-section in a unique class. Pack: Protein representation / Class: Protein representation Gives access to a number of specified polypeptide chains from a protein quaternary structure, The SBL provides many applications which rely on different structures tied to a polypeptide chain: ? topological information i.e. the covalent bondsGroup: Core::CSB / Package: Molecular covalent structure
, Molecular distances / Class: SBL::CSB::RMSD comb for motifs We provide a new class for the Molecular distances package. Given a set of structural motifs, this class builds the motif graph
, Group: SBL::Applications / Package: Molecular distances flexible We provide an application which, given a set of polypeptide chains as well as "subdomain" definitions (labeled residue ranges), computes the RMSD Comb.. The specification of labels is provided from SBL::Models::MolecularSystemLabelTraits. Example specification files can be found in the documentation. The application provides three executables: ? sbl-flexible-rmsd-proteins, Molecular distances / Class: SBL::Modules::RMSD comb for motifs module We provide the module enabling the use of the previous class in a workflow
, ? sbl-flexible-rmsd-conformations.exe is used to compare conformations of an identical protein ? sbl-flexible-rmsd-motifs.exe is used to compute the RMSD Comb. of two chains with user specified structural motifs
, Pre-requisites Following the contributions from ADDREF, we provide a novel package in the SBL. Given two polypeptide chains, the goal of this package is to identify structural motifs using any of the four methods from ADDREF, SBL::Applications / Package: Structural motifs Bibliography
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, 2 Statistical significance of our motifs, when compared against random motifs with two non parametric two-sample tests. Second column: p-value for the Wilcoxon Mann-Whitney U test
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, When qualified by the suffix iter
, Method SBL executable Option Align-Apurva-SFD sbl-structural-motifs-chains-apurva.exe Align-Apurva-CD sbl-structural-motifs-chains-apurva.exe-use-cd-filtration Align-Kpax-SFD sbl-structural-motifs-chains-kpax.exe Align-Kpax-CD sbl-structural-motifs-chains-kpax.exe-use-cd-filtration Align-Identity-SFD sbl-structural-motifs-conformations
, This section is devoted to the proof of Theorem 6.1. For the sake of readability, we splitted this proof into three parts: Theorems D.2, D.5 and D.6. Notice that the last two proofs are quite similar
, We say that ? L-reduces to ? is there are two polynomial-time algorithms f , g and constants ?, ? > 0 such that for each instance I of ?: 1. Algorithm f produces an instance I = f (I) of ? such that the optima of I and I
, Given any solution of I with cost c , algorithm g produces a solution of I with cost c such that OP T ? (I) ? c ? ?(OP T ? (I ) ? c )
, It is known that if ? is AP X-hard and L-reduces to ? , then ? is AP X-hard as well, that case, ? does not admit a P T AS (Polynomial Time Approximation Scheme) unless P = N P
, For any D ? 2, the D-family-matching problem is AP X-hard even if the maximum degree ? is at most 4 and the weights are 2 and 5. In our reduction, we use a special case of set packing problem
, an integer k ? 1, set packing problem consists in determining whether there exists a packing C of size |C| = k. Set packing problem is NP-complete even if |Y i | = 3 for every i ? {1
, By Theorem 6.2, given D ? 1, there is an O(D 2 n)-time complexity algorithm for the Dfamily-matching problem because ? = 2. We prove in Lemma D.3 a better time complexity algorithm for the D-family-matching problem
, Let E = {{v j , v j+1 } | 1 ? j ? n ? 1}. We define the function ? D as follows. For every t ? {1,. .. , n} and every i ? {max(1, t ? D),. .. , t + 1}, then ? D (v t , i) is the score of an optimal solution S of the D-family-matching problem, for the sub-path induced by the set of nodes {v 1, there exists an O(Dn)-time complexity algorithm for the D-family-matching problem for G. Proof of Lemma D.3. Let V = {v 1
, For every i ? {max
, v t } is a set of this solution. We then modify this solution by adding node v t+1 in the last set, and we obtain the optimal solution for the D-family-matching problem, for the sub-path induced by the set of nodes {v i, p.1
,
, Any solution must contain the set {v t+1 }. Thus, we have to consider an optimal solution for the D-family-matching problem for the sub-path induced by the set of nodes {v i ,. .. , v t }. We now prove the result for ? D (v t+1 , t + 2), p.1
,
, Let D ? N +. Consider any intersection graph G = (V, E, w) that is an even cycle. Then, there exists an O(D 2 n)-time complexity algorithm for the Dfamily-matching problem for G
, Consider any instance of the D-family-matching problem such that: ? for every i ? {1,. .. , r}, there exist j 1 , j 2 ? {1,. .. , r } such that F i ? F j = ? for any j ? {1
, there exist i 1 , i 2 ? {1,. .. , r} such that F j ?F i = ? for any i ? {1
, D 2 )-time complexity algorithm for the D-family-matching problem. Say otherwise, Corollary D.2 shows that there is a polynomial time algorithm for the D-family-matching problem if any set in F ? F has a non-empty intersection with at most two other sets of F ? F, Then, there exists an O((r + r )
, Let T r be any spanning tree of G rooted at node r ? V. For every v ? V , we define H(G, T r , v) as the set of all H ? H(G, v) such that the graph induced by the set of nodes V (H) ? V (T v ) is a (connected) sub-tree rooted at v. Let H(G, T r ) = ? v?V H(G, T r , v), D.6 Appendix-Generic approach based on spanning trees Let us first introduce some notations. For every v ? V , let H(G, v) be the set of all different sub-graphs of G that contain v and of diameter at most D. Let H(G) = ? v?V H(G, v)
, Let N (v) = {v 1 ,. .. , v q } be the set of q ? 1 neighbors of v in T v. Suppose we have computed ? D (v j , H) for every j ? {1, A leaf is a node of degree one and different than the root r
, Algorithms based on spanning trees Proof of Lemma 6.5. For some k ? 1, consider an optimal solution S = {S 1 ,. .. , S k } for the D-familymatching problem for G. For every i ? {1
, By construction of T , S is an admissible solution for the D-family-matching problem for G, Let T be any rooted spanning tree of G such that E(T i ) ? E(T ) for every i ?
, Algorithm 1 returns ? D (G), that is an optimal solution for the D-family-matching problem for G, Given any positive integer D ? 1 and any intersection graph G
, Furthermore, the time complexity of Algorithm 1 is O(|T (G)| max Tr?T (G) h(G, T r ) ? n)
, Let G be any intersection graph. Then, there exists a rooted spanning tree T of G
, For some k ? 1, consider an optimal solution S = {S 1 ,. .. , S k } for the 2-familymatching problem for G. For every i ? {1
, Indeed, since D = 2, G[S i ] is necessarily a complete bipartite graph and its number of nodes is at most 2?. It is sufficient to select the maximum star as T, Let T be any rooted spanning tree of G such that E(T i )
, D) returns ? D (T ? ) (Theorem 6.2). prove the result by induction. Clearly, ? 1,?1(G),?1(G) (1) = 0. Assume that we have computed ? y,x ? ,x + (D) for every D ?, Given any intersection graph G, Algorithm 1 returns a 2?-approximation for the 2-familymatching problem for G if: ? ?(M)
, We necessarily have ? y,x ? ,x + (D +1) = ? y,x ? ,x + (D) because we cannot start a new plateau since x ? < ? D+1 (G) < x +
, We cannot start a new plateau because x ? < x +. Thus we have to find the best y plateaus such that the lower bound is at least ? D+1 (G) = x ? and at most x +. We get that ? y,x ? ,x + (D + 1) = min x?P D+1
,
, In the second case, the score is minimum score among all the optimal solutions composed of y ?1 plateaus
, D + 1) is the minimum among these two scores
, D+1 (G) < x ? or ? D+1 (G) > x + , then there is no admissible solution and, by convention, ? If ?
, There are O(D 4 G ) such computations. All the cases (but the fourth), can be calculated in O(D G ) time. Thus, we get the O(D 5 G )-time complexity. Now consider the fourth case in which x ? = x +. Thus, for every D ?