[math-fun] Redundant computations on today's computer architectures
I recently attended a conference on modern parallel computers, and they are one heck of a lot different from what I grew up with. In particular, parallel redundant computations can be free. It occurred to me that the redundant representation of a complex number a+bi by the 2x2 matrix [a -b] [b a] may no longer be wasteful. In particular, multiplication of complex*complex may actually be faster using these matrices than when using pairs. If you have enough parallelism, addition/subtraction won't cost any more. Even norm(a+bi)=aa+bb=determinant(a+bi) can faster than when using pairs. The reason: it may cost more to try to read a & b simultaneously by more than one arithmetic processor, or by more than one part of a single arithmetic processor. Multi-ported memories are more expensive and slower than single-ported memories. Ditto for quaternions (heavily used in graphics processing today) represented by matrices. Since communication of these complex values can be expensive, it may pay to constantly convert back & forth between pair & 2x2 formats. --- I can imagine all sorts of numeric processes in which different formulae/methods are utilized, and some voting mechanism is used to select the "correct" answer. If some of the formulae/methods blow up due to bugs and/or singularities and/or ill-conditioning, hopefully there will be others to pick up the slack. In some of these computers, there may be tens of arithmetic units sitting around idle, so the redundancy may not cost anything (except perhaps some amount of electrical energy). In particular, one could conceive of performing a floating point calculation in many different orders simultaneously to see how divergent the values might be. --- It's a new day in computer land!
Stirling (in 1717) prove that 9 points uniquely define a cubic. McLaurin (in 1720) proved that two cubics intersect in at most nine points. Bezout gets credit for it, due to his incorrect proof which came years later. Around 1750, Euler and Cramer noticed that these seem to contradict, since different cubics are passing through the same nine points. With that set-up, here's my question. Are there some really nice sets of 9 points that have multiple interesting cubics passing through all nine points? In a Mathematica wrapper, here's a set of four cubics that pass through the same 9 points. ContourPlot[{ 41x - 5x^3 == 12y, -41y + 5y^3 == 12x, 365x - 41x^3 == 12y^3, -365y + 41y^3 == 12x^3 }, {x, -4, 4}, {y, -4, 4}, Frame -> False] Cubics in the Triangle Plane http://pagesperso-orange.fr/bernard.gibert/index.html likely has a set of nine triangle centers that leads to a variety of odd cubics, but I haven't figured out a good way to isolate a good case. --Ed
Ed, If you work projectively (i.e. introduce a new variable z and work with the set of monomials in x,y,z which have degree 3), you find that there are precisely 10 cubic monomials in 3 variables. If you have 9 points, then you'll find that if the determinant of the linear system that you get from the 9 points is non-zero this gives you a unique solution (this is essentially what Stirling was probably thinking of). So, for there to be more than one curve passing through the 9 points it is necessary and sufficient that the determinant be 0. This conditions will be some large polynomial in all of the coordinates of all the points which must vanish. Once you have this you'll have a whole pencil of solutions. Victor On Fri, Nov 13, 2009 at 12:40 AM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
Stirling (in 1717) prove that 9 points uniquely define a cubic.
McLaurin (in 1720) proved that two cubics intersect in at most nine points. Bezout gets credit for it, due to his incorrect proof which came years later.
Around 1750, Euler and Cramer noticed that these seem to contradict, since different cubics are passing through the same nine points.
With that set-up, here's my question. Are there some really nice sets of 9 points that have multiple interesting cubics passing through all nine points? In a Mathematica wrapper, here's a set of four cubics that pass through the same 9 points.
ContourPlot[{ 41x - 5x^3 == 12y, -41y + 5y^3 == 12x, 365x - 41x^3 == 12y^3, -365y + 41y^3 == 12x^3 }, {x, -4, 4}, {y, -4, 4}, Frame -> False]
Cubics in the Triangle Plane http://pagesperso-orange.fr/bernard.gibert/index.html likely has a set of nine triangle centers that leads to a variety of odd cubics, but I haven't figured out a good way to isolate a good case.
--Ed _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (3)
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Ed Pegg Jr -
Henry Baker -
victor miller