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<FONT face=Arial color=#0000ff
size=2>Thank you so much DT, regretfully I'll be enroute back to Canada
for the next week or so, so I won't be able to hit the keys as often as I would
like :-(
<FONT face=Arial color=#0000ff
size=2>
<FONT face=Arial color=#0000ff
size=2>I will surely look at your code later and try to get to read the AB posts
now and then :-)
<FONT face=Arial color=#0000ff
size=2>
<FONT face=Arial color=#0000ff
size=2>take care,
<FONT face=Arial color=#0000ff
size=2>herman.
<FONT face=Tahoma
size=2>-----Original Message-----From: DIMITRIS TSOKAKIS
[mailto:TSOKAKIS@xxxxxxxxx]Sent: January 29, 2004 7:57
PMTo: amibroker@xxxxxxxxxxxxxxxSubject: [amibroker] Re:
Successive Approximation in aflHerman,We may have
another approach through while() statement.[For the simplicity, the //next
decimals are in a loop form ]//Sqrt(X)
approximationX=2;Precision=7;a=1;b=2;st0=0.1;//the initial
stepz=0;//1st
decimali=a;while(i^2<X){z=i;i=i+st0;}//next
decimalsfor(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
approximationX=2;Precision=5;a=1;b=2;st0=10^(-Precision);//the
initial
stepz=0;//decimalsi=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@xxxx]> 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@xxxx]> > 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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