1. optimization....help
- Posted by Euman <euman at BELLSOUTH.NET>
Oct 20, 2000
-
Last edited Oct 21, 2000
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Hello all,
Can the below loop be optimized further?
I know that takeing the multiplies out should do it
but am at a lose on this.......
----- begin -----------
count =3D 1=20
across =3D 262
down =3D 236
seq =3D repeat(0,down) =20
=20
for x =3D 1 to down do =20
seq[x] =3D r[count..across * x]=20
count =3D across * x + 1
end for
--- end -------------------
euman at bellsouth.net
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<DIV><FONT face=3DArial size=3D2>Hello all,</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>Can the below loop be optimized=20
further?</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>I know that takeing the multiplies out =
should do=20
it</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>but am at a lose on =
this.......</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>----- begin -----------</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>count =3D 1 </FONT></DIV>
<DIV><FONT face=3DArial size=3D2>across =3D 262</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>down =3D 236</FONT></DIV>
<DIV><FONT face=3DArial size=3D2> </DIV></FONT>
<DIV><FONT face=3DArial size=3D2>seq =3D=20
<DIV><FONT face=3DArial size=3D2>for x =3D 1 to down =
do =20
<BR> seq[x] =3D r[count..across * x] =
<BR> =20
count =3D across * x + 1</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>end for</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2>--- end =
-------------------</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV><FONT face=3DArial size=3D2><A=20
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2. Re: optimization....help
- Posted by Bernie <xotron at PCOM.NET>
Oct 20, 2000
-
Last edited Oct 21, 2000
On Fri, 20 Oct 2000 22:59:33 -0400, Euman <euman at BELLSOUTH.NET> wrote:
>Can the below loop be optimized further?
How can anyone optimize an example when they
don't know what sequence r is ?
what type of variables you are using, or
what you are trying to accomplish, etc.
3. optimization....help
- Posted by Euman <euman at BELLSOUTH.NET>
Oct 20, 2000
-
Last edited Oct 21, 2000
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Found on the net..........
Programmers tend to over-estimate the usefulness of the programs they =
write. The approximate value of an optimization is:=20
number of runs =D7 number of users =D7 time savings =D7 user's =
salary - time spent optimizing =D7 programmer's salary
even if the program will be run hundreds of times by thousands of users, =
an extra day spent saving 40 milliseconds probably isn't going to help.=20
euman at bellsouth.net
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<P>Found on the net..........
<P>Programmers tend to over-estimate the usefulness of the programs they =
write.=20
The approximate value of an optimization is: <PRE> number of runs =D7 =
number of users =D7 time savings =D7 user's salary - time spent =
optimizing =D7 programmer's salary
</PRE>
<P>even if the program will be run hundreds of times by thousands of =
users, an=20
extra day spent saving 40 milliseconds probably isn't going to help. =
</P>
<P> </P>
<P><A href=3D"mailto:euman at bellsouth.net">euman at bellsouth.net</A></P>
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4. optimization....help
- Posted by Euman <euman at BELLSOUTH.NET>
Oct 20, 2000
-
Last edited Oct 21, 2000
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Bernie wrote:
How can anyone optimize an example when they
don't know what sequence r is ?
what type of variables you are using, or
what you are trying to accomplish, etc.
well, r is a sequence obviously across * down
of 0's or 1's or 61,832 bytes long.
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<DIV><FONT face=3DArial size=3D2>Bernie wrote:</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
<DIV>How can anyone optimize an example when they<BR> don't know =
what=20
sequence r is ?<BR> what type of variables you are using, =
or<BR> =20
what you are trying to accomplish, etc.</DIV>
<DIV> </DIV>
<DIV><FONT face=3DArial size=3D2>well, r is a sequence obviously across =
*=20
down</FONT></DIV>
<DIV><FONT face=3DArial size=3D2>of 0's or 1's or 61,832 bytes =
long.</FONT></DIV>
<DIV><FONT face=3DArial size=3D2></FONT> </DIV>
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5. Re: optimization....help
Euman, this will be somewhat faster:
for x = 1 to down do
seq[x] = r[count..across]
count = across + 1
across += 262
end for
jiri
>From: Euman <euman at BELLSOUTH.NET>
>Reply-To: Euphoria Programming for MS-DOS <EUPHORIA at LISTSERV.MUOHIO.EDU>
>To: EUPHORIA at LISTSERV.MUOHIO.EDU
>Subject: optimization....help
>Date: Fri, 20 Oct 2000 22:59:33 -0400
>
>Hello all,
>
>Can the below loop be optimized further?
>
>I know that takeing the multiplies out should do it
>but am at a lose on this.......
>
>----- begin -----------
>
>count = 1
>across = 262
>down = 236
>
>seq = repeat(0,down)
>
>for x = 1 to down do
> seq[x] = r[count..across * x]
> count = across * x + 1
>end for
>
>--- end -------------------
>
>euman at bellsouth.net
>
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6. Re: optimization....help
Thanks Jiri
my mistake was @ across += 262 where I added
across again where 262 is now...........
< \Kick myself>
----- Original Message -----
From: "jiri babor" <jiri_babor at HOTMAIL.COM>
To: <EUPHORIA at LISTSERV.MUOHIO.EDU>
Sent: Saturday, October 21, 2000 2:09 AM
Subject: Re: optimization....help
> Euman, this will be somewhat faster:
>
> for x = 1 to down do
> seq[x] = r[count..across]
> count = across + 1
> across += 262
> end for
>
> jiri
>
>
> >From: Euman <euman at BELLSOUTH.NET>
> >Reply-To: Euphoria Programming for MS-DOS <EUPHORIA at LISTSERV.MUOHIO.EDU>
> >To: EUPHORIA at LISTSERV.MUOHIO.EDU
> >Subject: optimization....help
> >Date: Fri, 20 Oct 2000 22:59:33 -0400
> >
> >Hello all,
> >
> >Can the below loop be optimized further?
> >
> >I know that takeing the multiplies out should do it
> >but am at a lose on this.......
> >
> >----- begin -----------
> >
> >count = 1
> >across = 262
> >down = 236
> >
> >seq = repeat(0,down)
> >
> >for x = 1 to down do
> > seq[x] = r[count..across * x]
> > count = across * x + 1
> >end for
> >
> >--- end -------------------
> >
> >euman at bellsouth.net
> >
>
> _________________________________________________________________________
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> Share information about yourself, create your own public profile at
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