Largest Common Point Set (LCP) problem Given e>0 and two point sets A and B find a transformation T and equally sized subsets A’ (a subset of A) and B’ (a subset of B) of maximal cardinality such that dist(A’,T(B’)) ≤ e. Bottleneck metric: optimal solution in O(n 32.5 ) C. Ambuhl et al RMSD metric: open problem

“alignment” technique 1. Generate 3D transformations. For example: for each pair of triplets, one from each molecule which define e-congruent triangles compute the rigid transformation that superimposes them. 2. For each transformation compute the maximal matching (for the bottleneck metric apply the bipartite matching). Accuracy of the result depends on

Approximation Algorithm Assume that L is the size of the LCP (bottleneck) with error ε. Definition: A β-approximation algorithm for the LCP problem guarantees to find the LCP of size at least L with error (ε+β).

R 2 Construct transformations between each point pair (a,b) and (p,q) in the following way: a b q’ Let T(p)=p*, T(q)=q*. Let T opt be the optimal transformation and T opt (p)=p’, T opt (q)=q’. p* p’ Select the grid size to be: β/3 Select: γ= √2 β/3 -> d(p’,p*) ≤ SQRT((β/3) 2 + (β/3) 2 )/2 = β/(3√2) -> min T d(q’,T(q)) ≤ SQRT( (β/(3√2)) 2 + (γ/2) 2 ) ≤ β/3 q* γ Circle segment < γ

Estimate the error, i.e. d(T(v), T opt (v)) < ? d(T(v), T opt (v)) ≤ d(R◦I◦T(v), T(v)) where R and I are defined as follows: a p* b q’ q* p’ I(T(p))=p’ R(I(T(q))=q’->R◦I◦T = T opt d(R◦I◦T(v), T(v)) ≤ d(R◦I◦T(v), I◦T(v)) + d(I◦T(v), T(v)) ≤ d(R◦I◦T(v), I◦T(v)) + β/3 d(R◦I◦T(v), I◦T(v)) ≤ d(R◦I◦T(q), I◦T(q)) why? d(R◦I◦T(q), I◦T(q)) ≤ d(q’,q*) + d(q*, I◦T(q)) ≤ β/3 + β/3 d(R◦I◦T(v), T(v)) ≤ β

Time Complexity for R 2 : O(n 2 n 2 (2ε/(β/3)) 2 (2ε/(√2 β/3)) ) O(n 4 (2ε/(β/3)) 2 (2ε/(√2 β/3)) ) O(n 4 (ε/β) 3 ) with large constant ~153. For example if ε=3 and we want β=0.5 -> ~ n 4 On the other hand, the optimal algorithm for R 2 is O(n 8.5 ) For n = 300 -> 1.5*10 11 n 4