OpenEuphoria

Re: modified reverse

• Posted by Igor Kachan <kinz at peterlink.ru> Aug 06, 2001
• 480 views

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Robert Craig writes:

> Derek Parnell writes:
> > Strange but I found rforno's version slightly slower (96%) than the
> > standard. 
> 
> Yes, I searched the mailing list archives, and found a comparison
> I did between the current reverse() and a version identical to
> rforno's 3 years ago. I was using my old Pentium-150 machine,
> and found the current reverse() was faster. That's why I chose it.
> Now I have a Pentium II with a different cache size etc.

I have some results under extremal conditions of swapping on Dos-32
platform.
Timing on the complex sequence with length 80000 was :

reverse () -- 52.9      92.93 -- RDS
reverse1() -- 58.05     31.75 -- rforno
reverse2() -- 124.07    165.49
reverse3() -- 107.93    102.49
reverse4() -- 107.93    180.87
reverse2() -- 53.94     66.19 -- Derec

On Windows in Dos console on the simple sequence with length
194169 without swapping results were :

RDS    2.75   2.80    2.97   2.96   2.75
r1     3.73   4.34    4.07   3.95   5.21
r2     3.73   3.84    3.84   3.79   3.85
r3     4.56   4.56    4.51   4.56   4.62
r4     4.51   4.61    4.5    4.5    4.56
Derec  2.86   2.96    3.02   2.91   3.08

But on plain Dos Derec's function seems the fastest,
not always.
I notice a problem -- with tick_rate(100), swapping
on plain Dos immediately hungs up and swap-file gets
boo-boo, scandisk helps. With tick_rate(300), machine
hungs up. My machine is simplest -- 386   

I have attached my test program, try and choose
reverse which you like.  Again  

Regards,
Igor Kachan
kinz at peterlink.ru

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Content-Type: application/octet-stream; name="Reverse.ex"
Content-Transfer-Encoding: quoted-printable
Content-Description: Reverse.ex (EX )
Content-Disposition: attachment; filename="Reverse.ex"

--rforno
--Rob:
--The following modified reverse routine is simpler and, most of the =
times,
--slightly faster than the standard one. However, I noticed that the
--comparison between the two is affected by the type of data (integer or =
not,
--simple or compound sequence), and even it seems that the way of =
generating
--the data (by means of a single sentence or by a loop) affects the =
resulting
--timings. Any explanation for such a behavior?

sequence S S=3D{}
sequence R R=3D{}
sequence n n=3D{}
atom T

global function reverse(sequence s)
-- reverse the top-level elements of a sequence.
-- Thanks to Hawke' for helping to make this run faster.
    integer lower, n
    sequence t
   =20
    n =3D length(s)
    t =3D repeat(0, n)
    lower =3D 1
    for upper =3D n to floor(n/2)+1 by -1 do
	t[upper] =3D s[lower]
	t[lower] =3D s[upper]
	lower +=3D 1
    end for
    return t
end function

global function reverse1(sequence s)
    integer n
    sequence t
    n =3D length(s)
    t =3D repeat(0,n)
    for i =3D 1 to n do
	t[n] =3D s[i]
	n -=3D 1
    end for
    return t
end function

global function reverse2(sequence s)
     integer n
    sequence t
    t =3D s
    n =3D length(s)
    for i =3D 1 to n do
     t[n] =3D s[i]
       n -=3D 1
    end for
   return t
end function

global function reverse3(sequence s, integer n)
  sequence t
   t =3D s
    for i =3D 1 to n do
     t[n] =3D s[i]
       n -=3D 1
    end for
  return t
end function

global function reverse4(sequence s, integer n)
   R =3D s
    for i =3D 1 to n do
     R[n] =3D s[i]
       n -=3D 1
    end for
   return R
end function

global function reverse5(sequence s)
 integer n,m
   n =3D length(s)
   R =3D s m =3D n
    for i =3D 1 to n do
     R[m] =3D s[i]
       m -=3D 1
    end for
   return R
end function

global function reverse6(sequence s)
    integer lower, n, max
    sequence t
   =20
    n =3D length(s)
    t =3D repeat(0, n)
    lower =3D 1

    -- Force an integer result.
    max =3D floor(n/2)+1
   =20
    -- Now 'upper' knows that it can safely do integer comparisions
    -- because both 'n' and 'max' are integers.
    for upper =3D n to max by -1 do
	t[upper] =3D s[lower]
	t[lower] =3D s[upper]
	lower +=3D 1
    end for
    return t
end function

procedure init(integer N)
for i=3D1 to N  do
    for j=3D1 to rand(50) do
	  n &=3D rand(100000)
    end for
 S  =3D append(S,n)
 S &=3D rand(100000)
   n=3D{}
end for
? length(S)
end procedure

procedure test()
 T=3Dtime()
R=3Dreverse(S)
puts(1,"rds ")?time()-T
R=3D{}

T=3Dtime()
R=3Dreverse1(S)
puts(1,"r1  ")?time()-T
R=3D{}

T=3Dtime()
R=3Dreverse2(S)
puts(1,"r2  ")?time()-T
R=3D{}

T=3Dtime()
R=3Dreverse3(S, length(S))
puts(1,"r3  ")?time()-T
R=3D{}

T=3Dtime()
R=3Dreverse4(S, length(S))
puts(1,"r4  ")?time()-T
R=3D{}

T=3Dtime()
R=3Dreverse5(S)
puts(1,"r5  ")?time()-T
R=3D{}

T=3Dtime()
R=3Dreverse6(S)
puts(1,"r6  ")?time()-T
R=3D{}
end procedure


init(18000)
test()


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