September 2011 - mathfun@mailman.xmission.com

[math-fun] Computer gamers solve protein-folding problem
by Henry Baker 19 Sep '11

19 Sep '11

FYI -- Perhaps some math problems could be converted into games? http://blogs.discovermagazine.com/notrocketscience/2011/09/18/computer-game… Computer gamers solve problem in AIDS research that puzzled scientists for years When scientists struggle with a problem for over a decade, few of them think, ÂI know! IÂll ask computer gamers to help.Â That, however, exactly what Firas Khatib from the University of Washington did. The result: he and his legion of gaming co-authors have cracked a longstanding problem in AIDS research that scientists have puzzled over for years. It took them three weeks. KhatibÂs recruits played Foldit, a programme that reframes fiendish scientific challenges as a competitive multiplayer computer game. It taps into the collective problem-solving skills of tens of thousands of people, most of whom have little or no background in science. HereÂs what I wrote about Foldit last year: The goal of the game is to work out the three-dimensional structures of different proteins. Proteins are feats of biological origami; they consist of long chains of amino acids that fold into very specific and complicated shapes. These shapes can reveal how proteins work, but solving them is fiendishly challenging. To do it, scientists typically need to grow crystals of purified protein before bouncing X-rays off them. Foldit takes a different approach, using the collective efforts of causal gamers to do the hard work. And its best players can outperform software designed to do the same job. Best of all, you donÂt need a PhD to play Foldit. Barely an eighth of the players work in science, and two-thirds of the top scorers have no biochemistry experience beyond high school. The controls are intuitive; tutorial levels introduce the gameÂs mechanics; colourful visuals provide hints; and the interface is explained in simple language. While protein scientists concern themselves with Ârotating alpha-helicesÂ and Âfixing degrees of freedomÂ, Foldit players simply ÂtweakÂ, ÂfreezeÂ, ÂwiggleÂ and ÂshakeÂ their on-screen shapes. FolditÂs success relies on the fact that it doesnÂt shallowly flirt with interactivity Â itÂs a true game. Its creator Seth Cooper designed it to Âattract the widest possible audienceÂ and encourage prolonged engagementÂ. ItÂs competitive: players are scored based on the stability of the structures they end up with and a leader board shows how they rank against other gamers. ThereÂs also a social side: gamers can chat on online forums, work in groups to solve puzzles and share solutions on a wiki. And just like real game development, everything was tuned according to feedback from the players. Tools were added and refined, the difficulty of the tutorials was tweaked to stop frustrated beginners from leaving, and puzzles were matched to the skills of the players. ThereÂs the thrill of contributing to genuine scientific research, but that motivates less than half of the community. The rest do it for the achievement, the social aspects and largely, because the game was fun and immersive. FolditÂs origins lie within Rosetta, a piece of software designed to solve protein structures by simulating and testing thousands of different folds. Rosetta is an example of Â distributed computingÂ, where volunteers run the program on their home computers when they donÂt need it. They effectively donate their computing power to speed up the laborious task of solving protein structures. But the volunteers wanted to use their biological computers Â their brains Â as well as their man-made ones. They suggested an interactive version of the programme and in May 2008, they got their wish with Foldit. Last year, Cooper showed that FolditÂs gamers were better than the Rosetta programme at solving many protein structures. They used a wide range of strategies, they could pick the best places to begin, and they were better at long-term planning. Human intuition trumped mechanical number-crunching. This year, Khatib wanted to see if the Foldit community could solve fresh problems. He entered the players into a twice-yearly contest called CASP (Critical Assessment of Techniques for Protein Structure Prediction), where structural biologists from all over the world compete to predict the structures of proteins that have almost been solved. They get the best predictions from Rosetta to begin with. Then, theyÂre on their own. KhatibÂs gamers, bearing names such as Foldit Contenders Group and Foldit Void Crushers Group, had varying success in the contest. In many of the categories, they did reasonably well but they couldnÂt match the best groups. They werenÂt as good at using the structures of similar proteins to tweak the ones they were working on. They could also head down dead ends if they started at the wrong place. In one case, their strategy of refining their starting structures to the best possible degree led to one of the Âmost spectacular successesÂ in the contest. But mostly, they focused too heavily on tweaking already imperfect solutions that other teams achieved better results by making large-scale changes. Learning from that lesson, Khatib stepped in himself. He agitated the initial protein structures in many random ways, to create a wide variety of terrible answers that the gamers could then refine. In their attempts, they came up with the best-ranked answer to the most difficult challenge in the competition. It was a success, and more would follow. After the competition, the players solved an even more important problem. They discovered the structure of a protein belonging to the Mason-Pfizer monkey virus (M-PMV), a close relative of HIV that causes AIDS in monkeys. These viruses create many of their proteins in one big block. They need to be cut apart, and the viruses use a scissor enzyme Âa protease Â to do that. Many scientists are trying to find drugs that disable the proteases. If they donÂt work, the virus is hobbled Â itÂs like a mechanic that cannot remove any of her tools from their box. To disable M-PMVÂs protease, we need to know exactly what it looks like. Like real scissors, the proteases come in two halves that need to lock together in order to work. If we knew where the halves joined together, we could create drugs that prevented them from uniting. But until now, scientists have only been able to discern the structure of the two halves together. They have spent more than ten years trying to solve structure of a single isolated half, without any success. The Foldit players had no such problems. They came up with several answers, one of which was almost close to perfect. In a few days, Khatib had refined their solution to deduce the proteinÂs final structure, and he has already spotted features that could make attractive targets for new drugs. ÂThis is the first instance that we are aware of in which online gamers solved a longstanding scientific problem,Â writes Khatib. ÂThese results indiÂcate the potential for integrating video games into the real-world scientific process: the ingenuity of game players is a formidable force that, if properly directed, can be used to solve a wide range of scientific problems.Â Reference: Khatib, DiMaio, Foldit Contenders Group, Foldit Void Crushers Group, Cooper, Kazmierczyk, Gilski, Krzywda, Zabranska, Pichova, Thompson, Popovic', Jaskolski & Baker. 2011. Crystal structure of a monomeric retroviral protease solved by protein folding game players. Nature Structural and Molecular Biology http://dx.doi.org/10.1038/nsmb.2119 More on Foldit: Foldit Â tapping the wisdom of computer gamers to solve tough scientific puzzles September 18th, 2011 by Ed Yong in Molecular biology, Select, Technology, Video games | 9 comments | RSS feed | Trackback >

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[math-fun] 1750, Big and Small, iterate
by Eric Angelini 19 Sep '11

19 Sep '11

Hello Math-fun,, a light (iterating) idea (too light, I'm afraid)... Take an integer N (N = 1750 here) Make the sum of N's digits and call it D (D = 13 here) Now build the integers B (big) and S (small) like this: - B is the biggest integer you can write using one digit of N and all the digits of D (here 7 and 1,3 --> B = 731) - S is the smallest integer you can write using one digit of N and all the digits of D (here 0 and 1,3 --> S = 103) Now iterate, computing B, then S, then B, then S, etc. Let's see what happens, starting with 1 (the line under the sequence shows D (digit sum) then B (big), then D, then S (small), then D, then B, then D, then S, ... 1---11---12---32---25---75---125---85---135---95---145 1 B 2 S 3 B 5 S 7 B 12 S 8 B 13 S 9 B 14 S 145---510---16---76---136---610---17---87 ... S 10 B 6 S 7 B 13 S 10 B 7 S 8 B ... Where do we get? (loops, fixed points, flights, etc.) Best, É. (and what about iterating only B's? Or S's?)

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[math-fun] Scalenicity?
by Fred lunnon 18 Sep '11

18 Sep '11

On 9/16/11, Stanley Rabinowitz <stan.rabinowitz(a)comcast.net> wrote: > ... > So in the future, it is good to look at scalene triangles when > testing hypotheses. > Which brings up an interesting question. What is the best triangle to > use for a test? > What does it mean to be "very scalene"? > > Of all triangles, which one is the most scalene (whatever that means)? Here's a nice rational example: vertices [0,0], [0,12], [15,20]; side lengths {12,17,25} . The side lengths above ascend with ratios roughly 3/2; clearly this ratio must be neither large nor close to unity, so (for test purposes) what might its optimal value be? WFL

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[math-fun] The Solvable Octic x^8-x^7+29x^2+29 = 0
by Bill Gosper 18 Sep '11

18 Sep '11

Really From: tpiezas(a)gmail.com Hello all, The octic, x^8-x^7+29x^2+29 = 0 (eq.1) (by Igor Schein) is solvable, but not as easy as merely factoring over a square root extension. Rather, this can be solvable by the 29th root of unity. My solution to (eq.1) is, [tweaked (with permission) by rwg on the basis of numerical evidence: {8 x} = {1-a-b+c+d-e-f-g, 1+a+b+c-d-e-f+g, 1-a+b-c-d+e-f-g, 1+a-b-c+d+e-f+g, 1-a-b-c-d-e+f+g, 1+a+b-c+d-e+f-g, 1+a-b+c-d+e+f-g, 1-a+b+c+d+e+f+g} ] where each {a,b,c,d,e,f,g} = Sqrt[4v_i+1] and the v_i are the 7 roots of the septic, 8903+47647v+39672v^2+7192v^3-522v^4-174v^5+v^7 = 0 (eq.2) The solution of which was given by Peter Montgomery as, v_i = 2(w^11+w^13+w^16+w^18)-2(w+w^12+w^17+w^28)-(w^2+w^5+w^24+w^27)+ (w^3+w^7+w^22+w^26)+(w^4+w^10+w^19+w^25)-(w^8+w^9+w^20+w^21) and one can set w_i = {t, t^7, t^23, t^25, t^16, t^20, t^24}, and t = exp(2Pi*I/29). P.S. Similarly, a solvable 32-deg equation in x can be solved by a 31-deg Lagrange resolvent in z in the form, x = z1^(1/2) +/- z2^(1/2) +/- ... +/- z31^(1/2) though, unfortunately, no explicit examples are yet known. - Tito Piezas III http://sites.google.com/site/tpiezas/ (Check this out! --rwg) [rwg: the septic roots are all real: v->-2 (2 cos((2 π)/29)+2 cos((3 π)/29)+cos((4 π)/29)-2 cos((5 π)/29)-cos((6 π)/29)+2 cos((7 π)/29)-sin(π/58)-sin((3 π)/58)-sin((7 π)/58)+sin((9 π)/58)+sin((11 π)/58)-sin((13 π)/58)) v->2 (cos(π/29)+cos((2 π)/29)-cos((4 π)/29)-cos((5 π)/29)-2 cos((6 π)/29)-2 sin(π/58)-sin((3 π)/58)-sin((5 π)/58)-sin((7 π)/58)-sin((9 π)/58)-2 sin((11 π)/58)+2 sin((13 π)/58)) v->2 (2 cos(π/29)-cos((2 π)/29)-cos((3 π)/29)+cos((4 π)/29)+cos((5 π)/29)-cos((7 π)/29)-2 sin((3 π)/58)-2 sin((5 π)/58)-2 sin((7 π)/58)+sin((9 π)/58)+sin((11 π)/58)-sin((13 π)/58)) v->-2 (2 cos(π/29)-cos((2 π)/29)+cos((3 π)/29)+cos((5 π)/29)+cos((6 π)/29)+cos((7 π)/29)+sin(π/58)-sin((3 π)/58)-2 sin((5 π)/58)-sin((7 π)/58)-2 sin((11 π)/58)+2 sin((13 π)/58)) v->-2 (cos(π/29)+cos((2 π)/29)-2 cos((3 π)/29)-2 cos((4 π)/29)-cos((5 π)/29)+cos((6 π)/29)-2 cos((7 π)/29)+sin(π/58)-sin((5 π)/58)-2 sin((9 π)/58)+sin((11 π)/58)-sin((13 π)/58)) v->2 (cos(π/29)+2 cos((2 π)/29)+cos((3 π)/29)+cos((4 π)/29)-2 cos((5 π)/29)+cos((6 π)/29)+cos((7 π)/29)+sin(π/58)+2 sin((3 π)/58)-sin((5 π)/58)+2 sin((7 π)/58)+sin((9 π)/58)) v->-2 (cos(π/29)-cos((3 π)/29)+2 cos((4 π)/29)-2 cos((6 π)/29)-cos((7 π)/29)-2 sin(π/58)+sin((3 π)/58)-sin((5 π)/58)+sin((7 π)/58)+2 sin((9 π)/58)-sin((11 π)/58)+sin((13 π)/58))]

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Re: [math-fun] Scalenicity?
by Dan Asimov 17 Sep '11

17 Sep '11

The configuration space of triangles up to similarity is naturally a 30-60-90 triangle itself, with its short leg and its hypotenuse forming the locus of all the isosceles (and the vertex between them representing the equilateral). So the "most scalene" would naturally be as far away as possible from that equilateral vertex, along the angle bisector of the two sides respresenting isosceles. That would put it on the line of degenerate triangles having angles (p*pi, (1-p)*pi, 0), to be exact at the one with angles (2*pi/3, pi/3, 0). Since this isn't so interesting, what if we exclude obtuse triangles? In that case the scalenest non-obtuse triangle turns out to be the 30-60-90 one. --Dan Stanley Rabinowitz asked: << Of all triangles, which one is the most scalene (whatever that means)? Sometimes the brain has a mind of its own.

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[math-fun] The radical solution of a solvable 17th deg eqn
by Tito Piezas 17 Sep '11

17 Sep '11

This post is rather long, but the intro is necessary, and the question is at the end. Here's a "natural" solvable 17-th deg eqn with small coefficients: x^17-6 x^16-24 x^15-42 x^14-31 x^13-23 x^12-7 x^11-x^10-4 x^9-11 x^8-7 x^7-13 x^6-x^5+x^3+x^2+x-1 = 0 (eq.1) Its unique real root is exactly given by (in Mathematica) as x = zeta_48 DedekindEta[tau]/(Sqrt[2]DedekindEta[2tau]) = 9.1630942... with the root of unity zeta_48 = exp(2Pi I/48), tau = (1+Sqrt[-d])/2, and d = 383. This d has class number h(-d) = 17. To solve this, depress eq.1 (get rid of its x^(n-1) term), by letting x = (y+6)/17 to get, y^17-11832 y^15-1124346 y^14-55393735 y^13-1784741617 y^12-41171464807 y^11-711423456455 y^10-9455898295636 y^9-99724287747103 y^8-887992943070295 y^7-7665207188897171 y^6-70479807472769473 y^5-592167373130143650 y^4-3496187093606980919 y^3-8695712981307573757 y^2+68265051092799270505 y-427806967360317821039 = 0 (eq.2) Its 16-deg resolvent, a polynomial with INTEGER coefficients, call this R_16, has roots, z_k = ((y1 + w^k y2 + w^(2k) y8 + w^(3k) y7 + w^(4k) y16 + w^(5k) y4 + w^(6k) y12 + w^(7k) y15 + w^(8k) y11 + w^(9k) y10 + w^(10k) y14 + w^(11k) y13 + w^(12k) y5 + w^(13k) y17 + w^(14k) y6 + w^(15k) y9 + w^(16k) y3)/17)^17 for k = {1 to 16} where w is any complex 17th root of unity, and y_n follows the root object Root[poly, n] ordering in Mathematica. Approximately, these are, y1 = 149.7726 {y2, y3} = -27.62 -/+ 18.49i {y4, y5} = -21.61 -/+ 7.52i {y6, y7} = -16.58 -/+ 6.34i {y8, y9} = -10.57 -/+ 15.32i {y10, y11} = -5.02 -/+ 13.71i {y12, y13} = -2.34 -/+ 13.15i {y14, y15} = 2.57 -/+ 2.60i {y16, y17} = 6.31 -/+ 7.04i R_16 has *extremely* large integer coefficients, with the largest being the 248-digit constant term 429534618434587^17 which, naturally enough, is a 17th power. (Note: R_16 can easily be formed using 500-digits precision or more on the y_n, and multiplying the 16 factors together to form the polynomial.) The polynomial R_16 can be factored into two octics over the ext Sqrt[17]. This, in turn, can be factored into 2 quartics over Sqrt[2(17+Sqrt[17])]. This can be factored further into 2 quadratics using an expression involved in the 17th root of unity. (Apparently, to solve R_16 = 0, only square roots of square roots of square roots, etc, are needed.) The real root of eq.2 in radicals is then, y1 = z_1^(1/17) + z_2^(1/17) + z_3^(1/17) + … + z_16^(1/17) = 149.7726… *Problem*: Express the roots of R_16 purely in terms of the complex 17th root of unity. (If anyone knows how to contact the mathematician Peter-Lawrence Montgomery, he probably will know how, since he has done something similar with a septic root and the 29th root of unity.) Sincerely, - Tito

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[math-fun] Scalenicity?
by Kerry Mitchell 17 Sep '11

17 Sep '11

> Of all triangles, which one is the most scalene (whatever that means)? The least scalene triangle is the equilateral triangle, which has the largest area for its size, so the most scalene triangle is probably a degenerate triangle with zero area. From a standard deviation perspective, the limit is side lengths (0, 1, 1), but that's essentially isosceles. Another perspective on unequal sides is to maximize the differences between the side lengths. Treating (maximum - middle) and (middle - minimum) as two sides of a rectangle, the limits of the sides that maximize the area of the rectangle are in the proportions of (1, 3, 4), but that's also degenerate. Balancing the desire for the triangle to have maximal area and for the side lengths to be maximally unequal (in the rectangular area sense, above), I get side lengths in the proportion of (0.284634809, 0.754329904, 1), or approximately (263, 697, 924). Kerry -- lkmitch(a)gmail.com www.kerrymitchellart.com

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Re: [math-fun] New record for continued frac[t]ion for pi
by Bill Gosper 17 Sep '11

17 Sep '11

I'm not sure who should be announcing this, but Neil noticed that Eric Weisstein has quietly smashed Neil's record, and *finally* my 878M largest term. --rwg ---------- Forwarded message ---------- From: Neil Bickford <techie314(a)gmail.com> Date: Fri, Sep 16, 2011 at 10:12 PM Subject: Re: Looks like Weisstein's got the record again To: Eric Weisstein <eww(a)wolfram.com> Cc: Bill Gosper <billgosper(a)gmail.com> Wow! A new largest term (at less than a trillion) and 2.9B terms? That's amazing! Congratulations! Also, clearly the computer you ran it on has much better specs than mine, and perhaps this new version of Mathematica (M--?) has a new, faster ContinuedFraction! (Or maybe it's just parallelized to multiple cores, that would work too) Anyways, congratulations on trouncing my record (by 6x, no less)! You can certainly count me in for the next version. --Neil On Sep 16, 2011, at 1:40 PM, Eric Weisstein <eww(a)wolfram.com> wrote: > On 9/16/11 3:17 PM, Bill Gosper wrote: >> Eric, what's the new largest term?! > > Unfortunately, still 5408240597; see http://oeis.org/A033089. I didn't do anything special; just ran the computation for 3.75 CPU-hours using 45 GB of RAM with a developmental version of M--. I now have access to a machine with a bit more memory, so I'll try to push it a but further over the weekend. > > Cheers (and congratulations to you, Neil). > -Eric > >> On Fri, Sep 16, 2011 at 8:15 AM, Neil Bickford<techie314(a)gmail.com> wrote: >> >>> See http://mathworld.wolfram.com/PiContinuedFraction.html , bottom row of >>> table. >>> I have no idea how much memory he used to accomplish that! >>> --Neil >

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Re: [math-fun] Scalenicity?
by William Rex Marshall 17 Sep '11

17 Sep '11

From: Fred lunnon On 9/16/11, Stanley Rabinowitz wrote: > > ... > > What does it mean to be "very scalene"? > > > > Of all triangles, which one is the most scalene (whatever that means)? One possible definition of "most scalene": The triangle which is worst at covering its own mirror image. Let A(T) be the maximum possible area of intersection of a triangle T of area 1 and its mirror image T' (also of area 1) in the Euclidean plane. For which T is A(T) minimised?

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[math-fun] New record for continued fracion for pi
by Bernie Cosell 16 Sep '11

16 Sep '11

<http://blog.wolfram.com/2011/09/15/from-pi-to-puzzles/> /b\ -- Bernie Cosell Fantasy Farm Fibers mailto:bernie@fantasyfarm.com Pearisburg, VA --> Too many people, too few sheep <--

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