1. Igor Rivin, Temple University andIAS School of Mathematics

2. Cant say it better than the Wikipedia: Macromolecular docking is the computationalmodeling of the structure of complexes formed by twoor more interacting biological macromolecules. Protein-protein complexes are the most commonly attemptedtargets of such modeling, followed by protein-nucleicacid complexes. The ultimate goal of docking is the prediction of thethree dimensional structure of the macromolecularcomplex of interest as it would occur in a livingorganism. Docking itself only produces plausiblecandidate structures. These candidates must beranked to predict what would occur in nature. 3. For protein-protein docking, can modelmolecules by three-dimensionalbodies, but since the interaction happensalong the surface, the question can beasked in the context of the geometry ofsurfaces. What surface do we look at? 4. There are many kinds, two favorites are: Van der Waals surface 5. And my favorite, solvent accessible surface (an neighborhood of the van der Waals surface, where is the van der Waals radius of a water molecule. 6. Given two surfaces S1 and S2, we want tofind a coarse quasi-isometry between asbig a subsurface of S1 and S2 as possible. Notice that since the realization in E3 of thesurfaces is floppy (the molecules are notrigid), we are really trying to solve theintrinsic problem. 7. Sowe solve a simpler problem first (the drunk under a streetlight paradigm). 8. Weforget the subsurface confusion, andassume that both our surfaces arehomeomorphic to the sphere S2. And this quasi-isometry thing is a little toopuzzling, so will go to the conformalcategory, which is a little easier to workwith (and clearly is closely related to themetric category). 9. AsBoris Springborn had said, conformalityoccurs for mysterious reasons (probablybecause conformal maps are aestheticallypleasing) In some biological settings, conformality isnatural (regions all grow at the same rate). 10. Given two metrics on S2, how do we findthe conformal map between them? Note that the metrics are still coarse, sowe can ignore the small-scale details. Here is our plan of attack: 11. 1. First, find a conformal map between the surface S1 and the round metric on S2.2. Second, find a conformal map between S2 and S2.3. Match them up somehow. 12. First, find a nice triangulation of the surface S1. To do this, sample a bunch of points from S1. Compute the Delaunay triangulation (dual of Voronoi tesselation), with respect to that point set so, crudely speaking, every point is connected to points close to it. 13. Then,construct a circle packing on the round sphere S2, combinatorially equivalent to the Delaunay triangulation we have just constructed. 14. Givena topological triangulation T of S2,we can draw it in such a way, that a diskcan be centered at every vertex, and twodisks are tangent precisely when thevertices are joined by an edge of T. The packing is then unique (up to Mobiustransformation). 15. The circle packing theorem as stated on the last slide was noticed by W. P. Thurston in the late seventies, though it follows from the work of Koebe on the uniformization of circle domains (done many years before the birth of Thurston, around 1916). 16. Thurston had observed that the circle packing theorem was an immediate corollary of the Andreev theorem on non- obtuse angled polyhedra in three- dimensional hyperbolic space: the circle packing of a triangulation and its dual together constitute a right-angled ideal polyhedron in three-dimensional hyperbolic space, the existence of which followed from Andreevs theorem. 17. Idealpolyhedra are polyhedra with all vertices on the sphere of infinity (ideal boundary) of three-dimensional hyperbolic space. 18. The previous page has the logos of the various versions of Mathematica, the ideal icosahedron logo was created by Henry Cejtin and Igor Rivin, and later modified by Michael Trott. 19. Andreevs argument is non-algorithmic. Thurston gave a procedure to construct a circle packing, but did not give any indication of convergence speed, or indeed a proof that it converged (this was supplied several years later by Al Marden and Burt Rodin [1989, appeared in 1992]) 20. IR(1994) showed that the construction of a convex ideal polyhedron with prescribed dihedral angles is a convex optimization problem, and thus admits a very fast algorithm (quadratic in practice). The algorithm is particularly fast for the special case of circle packing. 21. Adirect variational algorithm for circlepacking (no polyhedra) was constructed byYves Colin de Verdiere (1991) using adifferent functional. The two functionals are Legendretransforms of each other as shown by A.Bobenko and B. Springborn (2004). 22. Anothercharacterization/algorithm (IR, around 2000?): Consider hyperideal polyhedra (all vertices beyond infinity) with prescribed combinatorics. The volume is a concave function of the dihedral angles. The function is improper, the maximum is achieved when all of the interboundary distances collapse, which means that it gives you the circle packing. 23. Lots of related work has been done (X.Bao/F. Bonahon, J-M Schlenker,Bobenko/Springborn). Cruel irony: volume is convex on ideal andhyperideal polyhedra, but not for compactpolyhedra (if it were, hyperbolization wouldbe easy). 24. The idea of using circle packings to construct conformal mappings is also due to Thurston, but the first proof that this actually works is due to Burt Rodin and Dennis Sullivan [1987] (without any convergence rate estimate). Later, much work on the subject was done by He and Schramm [1993-1998] 25. I have never seen any comparison between circle packing and the more traditional methods of conformal mapping. No question that CP looks cool, but how good is the convergence speed? 26. We have constructed an approximation to a conformal mapping from S1 onto the round sphere via a circle packing scheme, and similarly from S2. These two mappings give us two densities f1 and f2 on the round sphere S2. We try to find a Mobius transformation of S2, which comes closest to transforming f1 to f2. 27. So first, lets look for the best rotation. 28. So,lets look at the problem in one dimension less: 29. Given two (positive) functions f and g onthe circle S1, find a rotation r, such that of f g r is as small as possible. Where by small, we mean in L2 norm,since that is easier to analyze. 30. Given two(finite) sets S and T (of the same cardinality) of points on the circle, find the rotation which minimizes the distance between them (where the level 0 question is: how do you define distance?) 31. We have the Fourier transform, which is anisometry, So that minimizing the L2 norm of f g r isthe same as maximizing the real part ofthe the scalar product , whichis a trigonometric polynomial (in therotation angle). 32. Dowe really have to sample densely? No! The maximal value of a trig polynomialp(x) is the smallest number Y, such thatq(x)=Y p(x) is non-negative everywhereon S1. A trig polynomial q(x) is non-negative ifthere another trig polynomial w(x) suchthat q(x)=|w(x)|2 (Fejer-Riesztheorem), and that 33. (is a semidefiniteness condition, so findingthe best rotation is a semi-definiteprogram, so convex! And fast (OK, notvery slow). If we are willing to be sleazy, there is avery fast, and very easy to implementalgorithm 34. maxshift2[t1_, t2_] := Block[{trans1 = Fourier[t1], trans2 = Fourier[t2]}, InverseFourier[ trans1 Conjugate[trans2] +Conjugate[Reverse[trans1]]Reverse[trans2]]] bestrot[t1_, t2_] :=Ordering[Abs[maxshift2[t1, t2]], -1] - 1 35. So far, all we know is how to rotate 36. Fora measure on the boundary at infinity of a Gromov-hyperbolic space, we define the conformal barycenter of to be: 37. Whichis (surprise!) (geodesically) convex,so computing the argmin reduces toconvex programming. But not easy convex programming, sinceone needs to work with the Klein model ofhyperbolic space, and then the problem isnot actually convex. Still, this can be dealtwith. 38. There are also dynamics based(identical) algorithms due (independently) to J. Milnor and W. Abikoff/Ye (2002?), but they do not generalized to higher dimensions (at least not obviously), since they use complex analysis. 39. Now, to find the Mobius transformation, we compute the conformal barycenter of our two functions, apply a (hyperbolic) translation to map one to the other, then a rotation, and we are done 40. Sincewe are done in one dimension lowerthan we started. In three dimension everything works, butthe Fourier transformation step nowinvolves spherical harmonics, and WignerD-matrices, and it is possible thatconvexity is lost (but maybe not, work inprogress) 41. Allthis leaves many more questions than we had answers, but thats the way it should be.