Formulas and/or algorithms for FFT's are available a zillion places
on the net although I wouldn't bother with the one at numerical
recipes. Formulas for MESA are also available on net although they
are a little harder to find as I think someone who is probably the
best known for this technique although not the originator of the
algorithm or even its use with price data has done his best to
squelch a lot of what otherwise would be available.
Writing
efficient AFL for either is fairly straight forward with no
need for
extra hardware ( FFT Chips ?! ) or DLL's ...
--- In [EMAIL PROTECTED]ps.com,
"Ton Sieverding"
<ton.sieverding@...>
wrote:
>
> Thanks Tomasz. First I do not understand why it
involves 3.000 * (
Bars# ) * ( Bars# ). Why not just 3.000 * ( Bars#
) ? I did the same
test in Excel with 3.000 * ( Bars# ) and got no
delay. After your
email I did it again with 3.000 * ( Bars# ) * (
Bars# ) and got a
delay of several minutes. But why two times 1.000
?
> Secondly I just copied the code I got from Fred and tried to
speed
it up by removing the second and third harmonic calculation.
Of
course it was faster but still to slow for me. Frankly I do not
understand how you can remove Cum() and LastValue() as the results
are being used for the calculation of the harmonic.
> Finally
I agree that Log calculations will speed up the process and
will
look for FFT code that does use Log's. I only have used full
Fourier
Analysis in electronics and have no experience with the Fast
Fourier
Transform. Although I agree with Ara that a FFT card is
probably
needed for the real good performance. Do you have any
experience
with these goodies ?
>
> Ton.
>
> -----
Original Message -----
> From: Tomasz Janeczko
> To: [EMAIL PROTECTED]ps.com
> Sent: Thursday, September 07, 2006 1:32 PM
> Subject: Re:
[amibroker] Re: Cycles and Mesa
>
>
>
>
Hello,
>
> You are doing 24 * 16 * 8 * (NUMBER OF BARS)
iterations per bar
(in the most inner loop you are using Cum()
function which is
cummulative
> sum over all bars). So entire
execution involves 24 * 16 * 8 *
(Number of bars) * (Number of
bars). If you say have 1000 bars you
end up with
> 3 billion
operations.
> To speed it up you would need to REMOVE Cum() and
LastValue()
from inner loop (they are not needed in fact).
>
> The code below is a sample of very inefficient coding. Properly
coded FFT requires only N*logN operations
>
> Best
regards,
> Tomasz Janeczko
> amibroker.com
> -----
Original Message -----
> From: Ton Sieverding
> To: [EMAIL PROTECTED]ps.com
> Sent: Wednesday, September 06, 2006 12:16 PM
> Subject:
Re: [amibroker] Re: Cycles and Mesa
>
>
> Rakesh
when talking about the Fourier code ( Fred -> Ehler ? )
you sent
me and looking for what you want to get, why not just taking
the
first harmonic and forgetting the rest of the code. In this way
you
have your current cycle length - at least this is what I think
you
want to get - and this will also solve an important part of the
speed problem. So take Y01 and take out the linear Y and you should
get the first harmonic. But as I already told you, this is not what
I
have in mind for the first harmonic ...
>
> Also I do
not understand why this code is still so slow. When
looking to the
For Loops I am getting 24 * 16 * 8 being 3.072
iterations per Bar.
So why is the code so slow ? Thomasz am I to
optimistic ? Dimitri do
you have any idea ? You can speedup the
calculation by setting n =
6, g01=1.0 ( starting g=1 ) and stp0 = 200
( starting stp0 = 400 ).
I do not see any difference in the graph and
the speed should be of
course 12 * 8 * 4 = 384 or 8x the old
speed ...
>
>
Ton.
>
>
>
> _SECTION_BEGIN("Fourier
Analysis Elementary");
> //
=============================================================
> // Elementary Fourier analysis, by D. Tsokakis, May 2004
> //
=============================================================
> t=Cum(1)-1;
> A=Param("Rsi",50,1,100,1);
> B=Param("smooth",100,1,120,1);
>
C1=MA(RSI(A),B);
> start=Cum(IsTrue(C1))==1;
>
t1=ValueWhen(start,t);
>
> //
PlotShapes(shapeDownTriangle*start,colorYellow);
>
> //
C10=ValueWhen(start,C1);Plot(C1,"C1",colorBlack,8);
> GraphXSpace=2;
> x = Cum(1);
> lastx =
LastValue( x );
> Daysback =
LastValue(Cum(IsTrue(C1)));
> aa = LastValue(
LinRegIntercept( C1, Daysback) );
> bb = LastValue( LinRegSlope(
C1, Daysback ) );
> yy = Aa + bb * ( x - (Lastx - DaysBack) );
> yy=IIf( x >= (lastx - Daysback), yy, -1e10 );
>
> // Plot( yy, "yy", colorRed );
> detrend=C1-yy;
>
new1=detrend;
>
Hor=LastValue(Cum(new1)/Cum(IsTrue(C1)));
>
pi=4*atan(1);
> n=12;
>
> //
===============================================
> // Fundamental period, crude approximation
> //
===============================================
>
> error00=10000;
> per01=0;
> g01=0;
>
phi01=0;
> stg0=0.5;
> stp0=100;
>
>
for(phi=0;phi<2*pi;phi=phi+pi/n)
> {
>
for(g=0.5;g<=8;g=g+stg0)
> {
>
for(per=300;per<=1000;per=per+stp0)
> {
>
f=1/per;
> y=Hor+g*sin(2*pi*f*(t-t1)+phi);
>
error=LastValue(Cum(abs(y-new1)));
>
if(error<error00)
>
{error00=error;per01=per;g01=g;phi01=phi;}
> }
> }
> }
>
> f01=1/per01;
>
y01=Hor+g01*sin(2*pi*f01*(t-t1)+phi01);
>
>
Plot(y01,"y01",colorRed,4);
>
>
_SECTION_END();
>
>
> ----- Original Message
-----
> From: Ara Kaloustian
> To: [EMAIL PROTECTED]ps.com
> Sent: Wednesday, September 06, 2006 6:12 AM
> Subject:
Re: [amibroker] Re: Cycles and Mesa
>
>
>
>
Rakesh,
>
> John Ehler's code is quite usable with AB... it
produces the
dominant cycle value. It is a bit CPU intensive but I
tried using it
at the start of a new bar (instead of using it with
every trade) and
that made the CPU load quite acceptable.
>
> My issue with both Ehler's code anf FFT is that I was not
satisfied with either.
>
> Since you were getting good
results with FFT, you might try
Ehler's code.
>
>
Theoretically ehler's code should be better since it looks at
the
most recent data, while FFT would need much longer data to
produce
usefuk cycle lengths.
>
> Good luck
>
>
Ara
> ----- Original Message -----
> From: Rakesh Sahgal
> To: [EMAIL PROTECTED]ps.com
> Sent: Tuesday, September 05, 2006 5:33 PM
> Subject: Re:
[amibroker] Re: Cycles and Mesa
>
>
> Ton
>
> Back in the old MetaStock days I had fiddled around with
using the packaged FFT in MS. I had used it to extract the current
dominant cycle length in a market and then used it to compute
studies. The results were quite satisfactory. Then I changed
platforms to AB and the whole idea got shelved. Subsequently I have
tried to find a way of extracting current dominant cycle length in
an
issue/market in AB but have not seen any way of using it which my
non-
engineering/mathmetician brain could comprehend. It was in
this
context I tried DT's code. Unfortunately (a) it was computing
power
intensive and (b) the results were beyond my comprehension so
I gave
up on it. I still would like to find a way of ascertaining
what the
current cycle length is in an issue but have not been able
to make
much progress. Perhaps someone on the list could throw up
some ideas
which are implementable on the AB platform.
>
>
> R
>
>
> On 9/6/06, Ton Sieverding
<ton.sieverding@...> wrote:
> Thanks Rakesh. I've
tried underneath mentioned AFL code
for Fourier analysis. It does
something but I have some questions :
>
> 1. I have the
feeling that the code uses a lot of
computer power. When modifying
the parameters it takes several
seconds ( about 5 sec ) before I
have a result on the graph. I am
using a Ghz 2.6 CPU with 1GB
internal and have never seen my computer
so slow. Do you have the
same experience ?
> 2. What I would like to see is a couple of
sine waves
being the harmonics of the original time series. So more
or less the
same picture as Fred's Cycles. But that's not what I
get. Also the
calculations for the Fourier analysis does not look
familiar to me.
Where can I find the logical background behind these
formulas ?
>
> Ton.
>
>
> -----
Original Message -----
> From: Rakesh Sahgal
> To: [EMAIL PROTECTED]ps.com
> Sent: Tuesday, September 05, 2006 2:12 PM
> Subject: Re:
[amibroker] Re: Cycles and Mesa
>
>
> If you are
interested in Fourier Analysis in AB
environment you should refer to
the work of Dmitris Tsokasis who
shared his work on Fourier Analysis
with the group. Am pasting below
his code. I have never used it and
would not know how to apply it in
a meaningful manner. Hope you find
it useful.
>
>
> R
>
>
>
===============================
> // Elementary
Fourier analysis, by D. Tsokakis, May 2004
>
>
t=Cum(1)-1;
>
>
A=Param("Rsi",50,1,100,1);
>
>
B=Param("smooth",100,1,120,1);
>
>
C1=MA(RSI(A),B);
>
>
start=Cum(IsTrue(C1))==1;
>
>
t1=ValueWhen(start,t);
>
>
PlotShapes(shapeDownTriangle*start,colorYellow);
>
>
C10=ValueWhen(start,C1);Plot(C1,"C1",colorBlack,8);
>
> GraphXSpace=2;
>
> x = Cum(1);
>
> lastx = LastValue( x );
>
> Daysback =
LastValue(Cum(IsTrue(C1)));
>
> aa = LastValue(
LinRegIntercept( C1, Daysback) );
>
> bb = LastValue(
LinRegSlope( C1, Daysback ) );
>
> yy = Aa + bb * ( x -
(Lastx - DaysBack) );
>
> yy=IIf( x >= (lastx -
Daysback), yy, -1e10 );
>
> Plot( yy, "yy", colorRed
);
>
> detrend=C1-yy;
>
>
new1=detrend;Hor=LastValue(Cum(new1)/Cum(IsTrue(C1)));
>
> pi=4*atan(1);n=12;
>
> // Fundamental period,
crude approximation
>
>
error00=10000;per01=0;g01=0;phi01=0;stg0=0.5;stp0=100;
>
> for(phi=0;phi<2*pi;phi=phi+pi/n)
>
>
{
>
> for(g=0.5;g<=8;g=g+stg0)
>
>
{
>
>
for(per=300;per<=1000;per=per+stp0)
>
>
{f=1/per;
>
>
y=Hor+g*sin(2*pi*f*(t-t1)+phi);
>
>
error=LastValue(Cum(abs(y-new1)));
>
>
if(error<error00)
>
>
{error00=error;per01=per;g01=g;phi01=phi;}
>
> }}}
>
>
f01=1/per01;y01=Hor+g01*sin(2*pi*f01*(t-t1)+phi01);
>
> Plot(y01+yy,"y01",colorSkyblue,4);
>
> Title=Name()+" [
Sample="+WriteVal(Daysback,1.0)+"
bars
]"+"\nyS0="+WriteVal(Hor,1.2)+
>
>
"\nyS01="+
>
>
WriteVal(g01,1.1)+"*sin(2*pi*(1/"+
>
>
WriteVal(per01,1.0)+")*(t-t1)+"+
>
>
WriteVal(12*phi01/pi,1.0)+"*pi/"+WriteVal(n,
1.0)+"),
Error1 ="+
>
>
WriteVal(LastValue(Cum(abs(y01-new1))),1.0)+", Error1/bar
="+
>
>
WriteVal(2*LastValue(Cum(abs(y01-new1)))/Daysback,1.2)+"
%";;
>
> // Fundamental period, detailed
approximation
>
>
error0=10000;per1=0;g1=0;phi1=0;stg=0.5;stp=10;
>
> for(phi=0;phi<2*pi;phi=phi+pi/n)
>
>
{
>
> for(g=0.5;g<=8;g=g+stg)
>
>
{
>
>
for(per=per01-stp0;per<=per01+stp0;per=per+stp)
>
> {f=1/per;
>
>
y=Hor+g*sin(2*pi*f*(t-t1)+phi);
>
>
error=LastValue(Cum(abs(y-new1)));
>
>
if(error<error0)
>
>
{error0=error;per1=per;g1=g;phi1=phi;}
>
>
}}}
>
>
f1=1/per1;y1=Hor+g1*sin(2*pi*f1*(t-t1)+phi1);
>
> Plot(y1+yy,"y1",colorBlue,4);
>
> Title=Title+
>
> "\nyS1="+
>
>
WriteVal(g1,1.1)+"*sin(2*pi*(1/"+
>
>
WriteVal(per1,1.0)+")*(t-t1)+"+
>
>
WriteVal(12*phi1/pi, 1.0)+"*pi/"+WriteVal(n,1.0)+"),
Error1 ="+
>
>
WriteVal(LastValue(Cum(abs(y1-new1))),1.0)+", Error1/bar
="+
>
>
WriteVal(2*LastValue(Cum(abs(y1-new1)))/Daysback,1.2)+"
%";;
>
> // 2nd Harmonic
>
>
error0=10000;
>
>
for(phi=0;phi<2*pi;phi=phi+pi/n)
>
> {
>
> for(g=0;g<=8;g=g+0.1)
>
> {
>
> per2=per1/2;f=1/per2;
>
>
y2=y1+g*sin(2*pi*f*(t-t1)+phi);
>
>
error2=LastValue(Cum(abs(y2-new1)));
>
>
if(error2<error0)
>
>
{error0=error2;g2=g;phi2=phi;}
>
> }}
>
>
f2=1/per2;y2=y1+g2*sin(2*pi*f2*(t-t1)+phi2);
>
> Plot(y2+yy,"y1",colorYellow,8);
>
>
Title=Title+
>
> "\nyS2="+
>
>
WriteVal(g2,1.1)+"*sin(2*pi*(1/"+
>
>
WriteVal(per2,1.0)+")*(t-t1)+"+
>
>
WriteVal(12*phi2/pi,1.0)+"*pi/"+WriteVal(n,1.0)+"),
Error2 ="+
>
>
WriteVal(LastValue(Cum(abs(y2-new1))),1.0)+", Error2/bar
="+
>
>
WriteVal(2*LastValue(Cum(abs(y2-new1)))/Daysback,1.2)+"
%";;
>
> // 3rd Harmonic
>
>
error0=10000;
>
>
for(phi=0;phi<2*pi;phi=phi+pi/n)
>
> {
>
> for(g=0;g<=8;g=g+0.1)
>
> {
>
> per3=per2/2;f=1/per3;
>
>
y3=y2+g*sin(2*pi*f*(t-t1)+phi);
>
>
error3=LastValue(Cum(abs(y3-new1)));
>
>
if(error3<error0)
>
>
{error0=error3;g3=g;phi3=phi;}
>
> }}
>
>
f3=1/per3;y3=y2+g3*sin(2*pi*f3*(t-t1)+phi3);
>
> Plot(y3+yy,"y1",colorWhite,8);
>
>
Title=Title+
>
> "\nyS3="+
>
>
WriteVal(g3,1.1)+"*sin(2*pi*(1/"+
>
>
WriteVal(per3,1.0)+")*(t-t1)+"+
>
>
WriteVal(12*phi3/pi,1.0)+"*pi/"+WriteVal(n,1.0)+"),
Error3 ="+
>
>
WriteVal(LastValue(Cum(abs(y3-new1))),1.0)+", Error3/bar
="+
>
>
WriteVal(2*LastValue(Cum(abs(y3-new1)))/Daysback,1.2)+"
%";
>
> /*
>
>
===============================
>
>
>
> On 9/5/06, Ton Sieverding <ton.sieverding@...>
wrote: