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Herman,
We may have another approach through while() statement.
[For the simplicity, the //next decimals are in a loop form ]
//Sqrt(X) approximation
X=2;Precision=7;
a=1;b=2;
st0=0.1;//the initial step
z=0;
//1st decimal
i=a;
while(i^2<X)
{
z=i;
i=i+st0;
}
//next decimals
for(n=1;n<Precision;n++)
{
st0=0.1*st0;
i=z;
while(i^2<x)
{
z=i;
i=i+st0;
}
}
Title="Sqrt("+WriteVal(X,1.0)+")="+WriteVal(z,1+0.1*n);
When we have the Sqrt(2)=1.41, the next loop will search 1.411,
1.412, 1.413, 1.414, 1.415 and will stop there, since 1.415 gives
false output.
In this way, the 3rd loop needs 5 steps instead of 10. Since decimals
will be equally disributed below and above 5 [I hope you dont always
search an 1.99999 !!] we gain many steps.
Without this, one should begin from 1 and search every 0.00001 up to
1.41421.
This is done by the full code
//Sqrt(X) full approximation
X=2;Precision=5;
a=1;b=2;
st0=10^(-Precision);//the initial step
z=0;
//decimals
i=a;
while(i^2<X)
{
z=i;
i=i+st0;
}
Title="Sqrt("+WriteVal(X,1.0)+")="+WriteVal(z,1+0.1*Precision);
and it is not the best choice !!
A simple criterion to see the difference, is your endless loop
detection threshold.
If it is set, for example, to 100000 iterations, the first method
will permit Precision=7 whereas the 2nd[full] method will not go more
than Precision=5.
The steps of the 1st method depend on the result : You will not be
that lucky with Sqrt(3.56). It is 1.886796, all the decimal digits
are >5 and the steps per digit will be 9+9+7+8+10+7 respectively !!
It is much better to search Sqrt(2.2811), it is 1.510331 and will
take 6+2+1+4+4+2 steps per digit.
You may also put a small counter to count the total steps of the
approximation, it is interesting.
Dimitris Tsokakis
--- In amibroker@xxxxxxxxxxxxxxx, "Herman vandenBergen" <psytek@xxxx>
wrote:
> Thank you DT, as usual your reply is not only informative but
entertaining
> :-) an 8x improvement in speed would be just fine. If my own
solution is
> general I will post it.
>
> I think I received enough ideas to try a few things, solving a small
> challenge with your owninput is half the fun and a better way to
learn.
>
> Thanks everybody and have a great day!
> herman
> -----Original Message-----
> From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> Sent: January 29, 2004 2:16 AM
> To: amibroker@xxxxxxxxxxxxxxx
> Subject: [amibroker] Re: Successive Approximation in afl
>
>
> Herman,
> my method is a bit different.
> I begin with the 1st digit accuracy [10 steps maximum, 1.0 to 1.9]
> and localize
> 1.4<sqrt(2)<1.5
> Then, for the 2nd digit, 10 steps [maximum] again to obtain
> 1.41<sqrt(2)<1.42
> and so on.
> In the average, I need 5 steps per digit and, for a 3-decimal
> accuracy it will take about
> 5*5*5=125 steps instead of the normal 1000.
> Archimedes, the copyright of the method, was not working with
> decimals.
> For some reason [not explained anywhere], his first choice was
the
> sevenths of the unity 1/7, 2/7, 3/7 etc .
> So, in his famous "Measurement of a Circle", he proves that
> 3+1/7>pi>3+10/71
> As you see, he slightly increases the denominator [he does not
> increase the numerator].
> There are some obscure statements, he never explained, for
example,
> how did he came to the [useful approximation]
> 265/153<sqrt(3)<1351/780 [!!!].
> but, we can not always have what we want...
> Dimitris Tsokakis
> --- In amibroker@xxxxxxxxxxxxxxx, "Herman vandenBergen"
<psytek@xxxx>
> wrote:
> > thanks DT, I have to study your code but I think you have the
> general idea.
> > I have an impossible formula to transform (it contains HHV and
LLVs,
> > stochastic mutation) and want to find the x that would give me
the
> given y.
> > Right now I linearly increment x untill I hit my y-target,
this is
> awfully
> > slow.
> >
> > like y = function(x); // y ranges 0-100 and I want 2 decimal
places
> for x
> > that gives me a given y
> >
> > I thought cutting the range in half, see whether it is greater
or
> less, cut
> > the result in half again, etc. I am not a math guy but I have
used
> > AD-converters that worked like that and what we can do in
hardware
> we can do
> > in software :-)
> >
> > thanks for the starter DT,
> > herman
> >
> >
> >
> >
> > -----Original Message-----
> > From: DIMITRIS TSOKAKIS [mailto:TSOKAKIS@x...]
> > Sent: January 28, 2004 9:18 PM
> > To: amibroker@xxxxxxxxxxxxxxx
> > Subject: [amibroker] Re: Successive Approximation in afl
> >
> >
> > Herman,
> > You mean a procedure like this
> >
> > a=1;b=2;z=0;
> > st0=0.1;st1=0.01;st2=0.01;
> > //1st decimal
> > for(i=a;i<b;i=i+st0)
> > {
> > if(i^2<2 AND (i+st0)^2>2)
> > z=i;
> > }
> > //2nd decimal
> > for(i=z;i<z+st0;i=i+st1)
> > {
> > if(i^2<2 AND (i+st1)^2>2)
> > z=i;
> > }
> > //...etc
> > Plot(z,"sqrt(2)",1,1);
> >
> > to find sqrt(2) without using all the values from 1 to 2 ?
> > Dimitris Tsokakis
> > --- In amibroker@xxxxxxxxxxxxxxx, "Herman vandenBergen"
> <psytek@xxxx>
> > wrote:
> > > Hello,
> > >
> > > has anybody come accross a successive approximation routine
in
> afl?
> > Or
> > > perhaps js?
> > >
> > > thanks,
> > > herman
> >
> >
> >
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