20200108, 08:23  #45 
"Hugo"
Jul 2019
Germany
31_{10} Posts 
Since we are not getting a solution from IBM for the time being, we could at least try to find out together what a record solution could have looked like. That would be a little more fun than complaining about the inability of the puzzle team to update their website. I can only assume that IBM's management stopped supporting the puzzle team because the puzzle solvers are not the company's commercial target group.
I can only contribute what I found out together with Hermann Jurksch, with whom I work from time to time. We found a solution that had approximately the record MSE. f(x) = c3 + x * (x / (c0 / (c1  x*x) + c2) + x / (x*x / (d0 / (d1  x*x) + d2) + d3 )) c0= 1.3555986881256104 c1= 0.51428204774856567 c2= 2.8334267139434814 c3= 4.7942781820893288E003 d0= 0.33841833472251892 d1= 0.51365238428115845 d2= 0.70969653129577637 d3= 2.6548692956566811E002 MSE ~= 1.8215E7. Unfortunately, this solution required one operation too much, so 16. I have not found a trick how I could change this to get rid of the operation. If I omit the constant term c3 and refit the coefficients, the deviation will increase approximately to 2 * (record value), which is not bad, but would not have been awarded the "*". Does anyone of you have an idea how to improve it? 
20200108, 08:36  #46 
Oct 2017
143_{8} Posts 
„I can only assume that IBM's management stopped supporting the puzzle team because the puzzle solvers are not the company's commercial target group.“
But the puzzlemaster reacts quickly to wrong solutions. I have submissed such one at 2.1. and he has responded a few hours later — so I was able to correct my code. 
20200108, 09:36  #47 
Jun 2003
4844_{10} Posts 
A much simpler explanation is that the puzzlemasters screwed up while publishing the solution (maybe they published under a typoed url), and the server doesn't know how to reach it. This is evidenced by the fact that someone actually enabled the solution link (which doesn't show up automatically, AFAICT).

20200108, 09:37  #48 
Jun 2003
2^{2}×7×173 Posts 
Can you inform them that the December solution link is dead ? I am sure they will fix it if they are aware of the issue.

20200108, 18:03  #49 
Oct 2017
3^{2}·11 Posts 

20200108, 22:02  #50 
Sep 2017
2^{2}×23 Posts 

20200108, 22:30  #51 
Feb 2017
Nowhere
2^{2}·7·149 Posts 
I didn't try to optimize my answer; I merely sought an answer good enough to beat the bound with 15 operators.
It occurred to me to formulate the MSE integral for the approximating function f(x) on [1,1] separately in [1,0] where x = x, and [0,1] where x = x. This leads to an expression of the MSE as an integral from 0 to 1, with integrand (x  e(x))^2 + o(x)^2, where e(x) = (f(x) + f(x))/2 is the even part of f(x), and o(x) = (f(x)  f(x))/2 is the odd part of f(x). Thus, if f is neither even nor odd, taking the even part of f makes the MSE smaller. For example, taking the even part 1/2 of the example (1+x)/2, gives an MSE of 1/12. I was able to optimize polynomial approximations. Doing the formal integration of (x  e(x))^2 with e(x) a even polynomial gives a formula which is a quadratic polynomial in the coefficients. Equating the partial derivatives to 0 gives a system of linear equations which can always be solved. Using PariGP to bludgeon out "best" even polynomial approximations to x over [0,1] yielded f(x) = 1/2, MSE = 1/12; f(x) = 3/16 + 15/16*x^2, MSE = 1/192, and f(x) = 15/128+105/64*x^2105/128*x^4, MSE = 1/768. These polynomials aren't good enough approximating functions to get a formula that beats the bound with 15 operators. However, polynomials aren't the only game in town. Another definition of x is the (positive) square root of x^2. This suggests using Newton's method to improve a polynomial approximation with a rational function. We refine f(x) to g(x) = (f(x) + x^2/f(x))/2. Trying this with the constant and quadratic polynomials yields rational functions g(x) with MSE values that exceed the given bound of 0.0001. However, the quartic polynomial gives a g(x) with an MSE of approximately 0.0000867, which is less than 0.0001. And g(x) = (0.05859375 + x*x*(0.8203125  x*x*0.41015625)) + x*x/(0.234375 + x*x*(3.28125  x*x*1.640625)) [coefficients are exact terminating decimals] uses precisely 15 operators. Problem solved. However, there are better rationalfunction approximations. Alas, with rational functions, optimizing the coefficients is much more complicated  the formal integrals can involve inverse tangents and/or logarithms with the coefficients buried inside. I looked up Pade approximants, but didn't pursue them. I also found a result called "Newman's Theorem" which gave a formula for an approximating rational function to x which is quite good for large degrees. I worked out some lowdegree cases, and did obtain a much better answer than the one I submitted. 
20200108, 22:52  #52 
Jan 2020
1_{2} Posts 
I used the following Mathematica code to attain a MSE of 2.92237E6
Code:
f[x_, a_, b_, c_, d_, m_, v_, g_] := 1/2*(Abs[x]((x*x*(x*x*(a*x*x+b)+c)+g)/(x*x*(x*x+v)+m)))^2; NMinimize[{NIntegrate[f[x, a, b, c, d, m, v, g], {x, 1, 1}]}, {a, b, c, d, m, v, g}, Method > {"RandomSearch"}] 
20200111, 06:00  #53 
Dec 2019
Kansas
2^{4} Posts 
Wow! After 11 days a solution still has not been published, and the names haven't been updated
Last fiddled with by what on 20200111 at 06:01 
20200111, 07:43  #54 
Oct 2017
3^{2}·11 Posts 
I have pointed out this to the puzzlemaster : 8.1.2020
Last fiddled with by Dieter on 20200111 at 07:44 
20200112, 03:03  #55 
Oct 2017
1100011_{2} Posts 
He has answered. They have indeed technical problems, but they will update.
The December solution is deferred to another month — so it is still possible to get a „*“ for December. 
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