December 2019

[what]

I got mse of 8.27604×10^-6 (12 times better)

[dg211]

I think there was some discussion earlier about whether expressions would be allowed which used brackets in such a way that they have subexpressions that can't be evaluated in a single pass without storing intermediate results.

Could one of the people who has had a solution accepted please confirm whether an expression like (x - 1) * (x + 2) would be allowed for example?

[SmartMersenne]

Originally Posted by dg211

I think there was some discussion earlier about whether expressions would be allowed which used brackets in such a way that they have subexpressions that can't be evaluated in a single pass without storing intermediate results.

Could one of the people who has had a solution accepted please confirm whether an expression like (x - 1) * (x + 2) would be allowed for example?

The point was not to store the results but saving operations by substitution. For example:

x*x*x*x*x*x*x*x requires 7 operations, but you can do it using only 3 operations if you were allowed to save and substitute the results:

y = x*x
z = y*y
z*z

so z would be x*x*x*x and z*z would give the x*x*x*x*x*x*x*x

In your case you can do it in 3 operations in any case, by finding the subexpressions and mutliply them or

x*x+x-2

or

(1+x)*x - 2

In computer language: keeping the intermediate results in registers is allowed, but you can't assign them to variables and reuse them later.

[what]

I wonder how someone got an error of only 1.803E-7. The best I have managed is 2.9E-6, and I can't think of anything to decrease it.

[Dieter]

Originally Posted by SmartMersenne

The point was not to store the results but saving operations by substitution. For example:

x*x*x*x*x*x*x*x requires 7 operations, but you can do it using only 3 operations if you were allowed to save and substitute the results:

y = x*x
z = y*y
z*z

so z would be x*x*x*x and z*z would give the x*x*x*x*x*x*x*x

In your case you can do it in 3 operations in any case, by finding the subexpressions and mutliply them or

x*x+x-2

or

(1+x)*x - 2

In computer language: keeping the intermediate results in registers is allowed, but you can't assign them to variables and reuse them later.

That‘s right. I have asked the puzzlemaster (9.12.):

„Is it allowed to write for example

y=x*x (one operation) and f =(a*y+b)*y+... (the other operations),

or have all 15 operations to be in the same expression?“

The answer was:

You can not define new variables.

[Dieter]

Originally Posted by SmartMersenne

The point was not to store the results but saving operations by substitution. For example:

x*x*x*x*x*x*x*x requires 7 operations, but you can do it using only 3 operations if you were allowed to save and substitute the results:

y = x*x
z = y*y
z*z

so z would be x*x*x*x and z*z would give the x*x*x*x*x*x*x*x

In your case you can do it in 3 operations in any case, by finding the subexpressions and mutliply them or

x*x+x-2

or

(1+x)*x - 2

In computer language: keeping the intermediate results in registers is allowed, but you can't assign them to variables and reuse them later.

In computer language: keeping the intermediate results in registers is allowed, but you can't assign them to variables and reuse them later.

You will say: in ONE Register.

Otherwise I could write:

movsd xmm1,x
mulsd xmm1,x (x*x in xmm1)

movsd xmm0,a
mulsd xmm0,xmm1
addsd xmm0,b (a*x*x+b in xmm0)
mulsd xmm0,xmm1
addsd xmm0,c ((a*x*x+b)*x*x+c in xmm0)

and so on. I think that this is not allowed.

[SmartMersenne]

Originally Posted by Dieter

In computer language: keeping the intermediate results in registers is allowed, but you can't assign them to variables and reuse them later.

You will say: in ONE Register.

Otherwise I could write:

movsd xmm1,x
mulsd xmm1,x (x*x in xmm1)

movsd xmm0,a
mulsd xmm0,xmm1
addsd xmm0,b (a*x*x+b in xmm0)
mulsd xmm0,xmm1
addsd xmm0,c ((a*x*x+b)*x*x+c in xmm0)

and so on. I think that this is not allowed.

Well, this is practically assigning it to a new variable and re-using it.

[SmartMersenne]

Originally Posted by what

I wonder how someone got an error of only 1.803E-7. The best I have managed is 2.9E-6, and I can't think of anything to decrease it.

It may be a bug in their integral computation

[Dr Sardonicus]

I note that they corrected the MSE for their example from .125 to 1/6.

I fault the statement of the problem, in that it did not specify the type(s) of constants that were acceptable. I guessed that decimal fractions were OK since the MSE bound and the original, incorrect MSE value for the example were stated as such.

I note that they also changed the criteria for "bonus" asterisk(s). In the original statement of the problem, they referred to "elegant" solutions. In the current incarnation, they refer to smallest MSE to date.

[what]

Originally Posted by SmartMersenne

It may be a bug in their integral computation

Might just be

[what]

I only managed to get the 1.803E-7 MSE after using 21 operations. I wonder what I'm missing