OpenEuphoria

1. which is faster?

• Posted by don cole <doncole at pacbell.net> May 08, 2007
• 683 views

Hello everybody,

   Has anybody benchmarked this:

function won_lost(sequence a,sequence b)
      if equal(a,b)  then
         return "Won"
      end if
  return "Lost"
end function

-------------------------or-----------------------

function won_lost(sequence a,sequence b)
      if equal(a,b)  then
         return "Won"
      else
         return "Lost"
      end if
end function

Thanks Bernie,

Don Cole

3. Re: which is faster?

• Posted by egg <egege at rambler.ru> May 08, 2007
• 650 views

You may use this simple speed comparer:

include sort.e
include machine.e
include misc.e
include get.e
include graphics.e

atom start,finish
integer cycles,x,y
sequence s,t,a

s = "Äà"
t = "Íåò"

cycles = 10000000

puts(1,"Start...\n")

start = time()

function won_lost1(sequence a,sequence b)
      if equal(a,b)  then
         return "Won"
      end if
  return "Lost"
end function

-------------------------or-----------------------

function won_lost2(sequence a,sequence b)
      if equal(a,b)  then
         return "Won"
      else
         return "Lost"
      end if
end function

for i=1 to cycles do
	--construction1
	a = won_lost1(s,t)
end for

finish = time() - start

position(23,2)
printf(1,"time1 = %f sec\n ",finish)

start = time()

for i=1 to cycles do
	--construction2
	a = won_lost2(s,t)
end for

finish = time() - start

printf(1,"time2 = %f sec\n",finish)

x = wait_key()

5. Re: which is faster?

• Posted by don cole <doncole at pacbell.net> May 09, 2007
• 658 views

Hello egg,

Thanks for the trouble.

Don Cole

7. Re: which is faster?

• Posted by Jules <jdavy at dsl.p?pex.co?> Nov 08, 2007
• 644 views

this is what I want to optimise:

global function st_dev(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    s -= m
    for i = 1 to length(s) do 
	sum += power(s[i], 2)
    end for
    sd = sqrt(sum/(length(s)-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

can this be improved?

9. Re: which is faster?

• Posted by CChris <christian.cuvier at agricu?ture.gouv.?r> Nov 08, 2007
• 630 views

I didn't quote original message, because the formatting was very messy and
unreadable, I don't know why for sure.

Instead of substracting m to every element, just do

m = mean(s)
for i = 1 to l do -- sum of squared deviations
    sum += power(s[i], 2)                                                    
   end for
   sum -= m*m -- same result, but avoids the s-=m statement

I didn't check whether this was better:

atom si
--... 
   m = mean(s)
for i = 1 to l do -- sum of squared deviations
      a = s[i] 
      sum += a*a   
   end for
   sum -= m*m

but I'd bet it's faster than calling power().

Also, computing the sum of elements of s inside the for loop will shave the call
to mean() and speed things even more.
CChris

11. Re: which is faster?

• Posted by Michael J. Sabal <m_sabal at ya?oo.com> Nov 08, 2007
• 630 views

Jules wrote:
> 
> I'm squaring values in a loop, is it faster to use (s[i]-a)*(s[i]-a) or
> power((s[i]-a),
> 2)?
> Thanks.

The interpreter optimizes power(x,2) to x*x, so at runtime they are the same.

13. Re: which is faster?

• Posted by Jules <jdavy at dsl.pipe?.?om> Nov 08, 2007
• 664 views

wow thanks! I'm amazed at the comparative slowness of variant #0...

15. Re: which is faster?

• Posted by Matt Lewis <matthewwalkerlewis at gmail.??m> Nov 08, 2007
• 630 views
• Last edited Nov 09, 2007

CChris wrote:
> 
> 
> I thought that precomputing s[i] would speed things up, but this didn't occur.
> Not sure why.

I think you didn't use s[i] enough to make the difference.  Also, the 
interpreter is smart enough to reuse the temp variable that it creates.
Here's the IL that's generated when you don't try to cache s[i]:

40: 058 119          # STARTLINE: string.exw(119)<<sum += power(s[i], 
                     #     2)>>
42: 025 171 178 179  # RHS_SUBS: [s:171] sub [i:178] => [_temp_:179]
46: 013 179 179 179  # MULTIPLY: [_temp_:179], [_temp_:179] => 
                     #     [_temp_:179]
50: 011 174 179 174  # PLUS: [sum:174], [_temp_:179] => [sum:174]

Here's the IL when you *do* cache s[i]:
52: 058 138          # STARTLINE: string.exw(138)<<sd = s[i]>>
54: 025 181 189 185  # RHS_SUBS: [s:181] sub [i:189] => [sd:185]
58: 101 185          # ATOM_CHECK: [sd:185]
60: 087 185          # DISPLAY_VAR: [sd:185] 
62: 058 139          # STARTLINE: string.exw(139)<<sum += sd*sd>>
64: 013 185 185 190  # MULTIPLY: [sd:185], [sd:185] => [_temp_:190]
68: 011 184 190 184  # PLUS: [sum:184], [_temp_:190] => [sum:184]

The STARTLINE and DISPLAY_VAR codes are due to with trace (which I put in
so we could see the actual lines for clarity) and don't really affect
the algorithm (assuming that a production release would make sure that
without trace was in effect).

But the bottom line is that you've added an extra ATOM_CHECK op that 
doesn't happen when you use it straight.  I think the interpreter is able
to do this because it's on the same line (though I'll admit that the
logic of when to use temps and when to save them is still opaque to
me, so Rob is probably the only one who could explain this without pouring
over the code).

Matt

17. Re: which is faster?

• Posted by CChris <christian.cuvier at ?gri?ulture.gouv.fr> Nov 09, 2007
• 642 views

Matt Lewis wrote:
> 
> CChris wrote:
> > 
> > 
> > I thought that precomputing s[i] would speed things up, but this didn't
> > occur.
> > Not sure why.
> 
> I think you didn't use s[i] enough to make the difference.  Also, the 
> interpreter is smart enough to reuse the temp variable that it creates.
> Here's the IL that's generated when you don't try to cache s[i]:
> 
> 40: 058 119          # STARTLINE: string.exw(119)<<sum += power(s[i], 
>                      #     2)>>
> 42: 025 171 178 179  # RHS_SUBS: [s:171] sub [i:178] => [_temp_:179]
> 46: 013 179 179 179  # MULTIPLY: [_temp_:179], [_temp_:179] => 
>                      #     [_temp_:179]
> 50: 011 174 179 174  # PLUS: [sum:174], [_temp_:179] => [sum:174]
> 
> Here's the IL when you *do* cache s[i]:
> 52: 058 138          # STARTLINE: string.exw(138)<<sd = s[i]>>
> 54: 025 181 189 185  # RHS_SUBS: [s:181] sub [i:189] => [sd:185]
> 58: 101 185          # ATOM_CHECK: [sd:185]
> 60: 087 185          # DISPLAY_VAR: [sd:185] 
> 62: 058 139          # STARTLINE: string.exw(139)<<sum += sd*sd>>
> 64: 013 185 185 190  # MULTIPLY: [sd:185], [sd:185] => [_temp_:190]
> 68: 011 184 190 184  # PLUS: [sum:184], [_temp_:190] => [sum:184]
> 
> The STARTLINE and DISPLAY_VAR codes are due to with trace (which I put in
> so we could see the actual lines for clarity) and don't really affect
> the algorithm (assuming that a production release would make sure that
> without trace was in effect).
> 
> But the bottom line is that you've added an extra ATOM_CHECK op that 
> doesn't happen when you use it straight.  I think the interpreter is able
> to do this because it's on the same line (though I'll admit that the
> logic of when to use temps and when to save them is still opaque to
> me, so Rob is probably the only one who could explain this without pouring
> over the code).
> 
> Matt

I see... using an object sd instead of atom sd would remove the check....

CChris

19. Re: which is faster?

• Posted by CChris <christian.cuvier at a?riculture.gouv.?r> Nov 09, 2007
• 632 views

Chris Bensler wrote:
> 
> CChris wrote:
> > 
> > Here is a benchmark to prove and disprove stuff....
> 
> }}}
<eucode>
> Try adding this bit of test code to verify that the variations are accurate.
> 
> sequence test
> test = {
>   st_dev0(s[1],0)
>  ,st_dev1(s[1],0)
>  ,st_dev2(s[1],0)
>  ,st_dev3(s[1],0)
>  ,st_dev4(s[1],0)
>  ,st_dev5(s[1],0)
> }
> test = (test = test[1])
> test[1] = 0
> for i = 2 to length(test) do
>   if test[i] = 0 then
>     test[1] += 1
>     printf(1,"st_dev%d() != st_dev0()\n",{i-1})
>   end if
> end for
> if test[1] then
>   machine_proc(26,{})
>   abort(1)
> end if
> </eucode>
{{{

> 
> Chris Bensler
> Code is Alchemy

I had tried this, and all values differ, though by very tiny amounts. This is
known as roundoff error accumulating.

Adding N numbers with some fuzz in a ]-eps..+eps[ range will result in a total
error whose magnitude is in the order of eps*sqrt(N). With 53 bit precision on fp
numbers and N=10000, you can easily expect that the few last bits are different
when you change the method of computation.

But there is hardly a point in estimating a deviation estimate with many
significant digits, is it?

You can compute a mean with more accuracy by doing:

m += ((s[i] - m)/i)
w/eucode>
starting with m = s[1]. The roundoff error buildup is way slower. But because of
the many integer divisions, the algorithm is way slower too.

CChris }}}

21. Re: which is faster?

• Posted by Fernando Bauer <fmbauer at h?tm?il.com> Nov 10, 2007
• 642 views

Hi,
I think the reason of relative low speed of variant 0 is because, commonly, the
computed mean is not an Euphoria integer. Then, the subtraction needs to use
floating point which is slower. Also, in squaring the deviations, floating point
operations again occur.
But if the sequence (data) contains mainly Euphoria integers (as in the showed
case), it's possible to round the mean and obtain more rapidly the integer
deviations and their squares. In the end, the effect of using the rounded mean is
compensated.
Below, variants 6, 7, 8 and 9 use this trick and are compared with the other
implementations. The magnitude of the mean (offset) can influence the speed of
the variants (1,2,3,4,5) that don't subtract the mean.
Finally, in the case of non-integers, the simulations show that it's better not
use the trick.

function mean(sequence s)
-- adds verything up in s, returns average
atom result
result=s[1]
for i=2 to length(s) do result+=s[i] end for
return result/length(s)
end function

global function st_dev0(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    s -= m
    for i = 1 to length(s) do
	sum += power(s[i], 2)
    end for
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev1(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    for i = 1 to length(s) do
	sum += power(s[i], 2)
    end for
    sum -= m*m*l
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev2(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    for i = 1 to length(s) do
	    sd = s[i]
        sum += sd*sd
    end for
    sum -= m*m*l
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev3(sequence s, atom X)
    atom m, sum, sd , m2
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    m2 = m*m
    for i = 1 to length(s) do
	    sd = s[i]
        sum += sd*sd
    end for
    sum -= m2
    sd = sqrt((sum/(l-1)) - m2)
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev4(sequence s, atom X)
    atom m, sum, sum_sq, sd , m2
    integer l
    sum_sq = 0.0
    sum = 0.0
    l = length(s)
    for i = 1 to length(s) do
	    sd = s[i]
	    sum += sd
        sum_sq += sd*sd
    end for
    m = sum/l
    m2 = m*m
    sum_sq -= m2
    sd = sqrt((sum_sq/(l-1)) - m2)
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev5(sequence s, atom X)
    atom m, sum, sum_sq, sd , m2
    integer l
    object x

    sum_sq = 0.0
    sum = 0.0
    l = length(s)
    for i = 1 to length(s) do
	    x = s[i]
        sum += x
        sum_sq += x*x
    end for
    m = sum/l
    m2 = m*m
    sum_sq -= m2
    sd = sqrt((sum_sq/(l-1)) - m2)
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev6(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    s -= floor(m)
    for i = 1 to length(s) do
    sum += power(s[i], 2)
    end for
    sum -= l*power(m-floor(m),2)
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev7(sequence s, atom X)
    atom m, sum, sd, m1
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    m1 = floor(m)
    for i = 1 to length(s) do
	    sum += power(s[i]-m1, 2)
    end for
    sum -= l*power(m-m1,2)
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev8(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    s -= floor(m)
    s *= s
    for i = 1 to length(s) do
	    sum += s[i]
    end for
    sum -= l*power(m-floor(m),2)
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

global function st_dev9(sequence s, atom X)
    atom m, sum, sd
    integer l
    sum = 0
    l = length(s)
    m = mean(s)
    s = power(s-floor(m),2)
    for i = 1 to length(s) do
	    sum += s[i]
    end for
    sum -= l*power(m-floor(m),2)
    sd = sqrt(sum/(l-1))
    if X = 0 then
	return sd
    else -- return the z-score
	return (X - m)/sd
    end if
end function

atom t, t0
procedure show(integer i)
printf(1,"Variant form #%d took %5.2fs. (%5.1f%%)\n", {i, t, 100*t/t0})
end procedure

procedure test(sequence title, sequence s)
atom junk
printf(1,"%s\n", {title})
t=time()
for i=1 to length(s) do
   junk=st_dev0(s[i],0.0)
end for
t = time()-t
t0 = t
show(0)

t=time()
for i=1 to length(s) do
   junk=st_dev1(s[i],0.0)
end for
t = time()-t
show(1)

t=time()
for i=1 to length(s) do
   junk=st_dev2(s[i],0.0)
end for
t = time()-t
show(2)

t=time()
for i=1 to length(s) do
   junk=st_dev3(s[i],0.0)
end for
t = time()-t
show(3)

t=time()
for i=1 to length(s) do
   junk=st_dev4(s[i],0.0)
end for
t = time()-t
show(4)

t=time()
for i=1 to length(s) do
   junk=st_dev5(s[i],0.0)
end for
t = time()-t
show(5)

t=time()
for i=1 to length(s) do
   junk=st_dev6(s[i],0.0)
end for
t = time()-t
show(6)

t=time()
for i=1 to length(s) do
   junk=st_dev7(s[i],0.0)
end for
t = time()-t
show(7)

t=time()
for i=1 to length(s) do
   junk=st_dev8(s[i],0.0)
end for
t = time()-t
show(8)

t=time()
for i=1 to length(s) do
   junk=st_dev9(s[i],0.0)
end for
t = time()-t
show(9)
puts(1,'\n')
end procedure

sequence s,s1  -- data
constant num_runs=5000, len_data=6000
s=repeat(0,num_runs)
s1=repeat(0,len_data)

-- generate random data
for k=1 to num_runs do
   for i=1 to len_data do
      s1[i]=rand(12345+k)
   end for
   s[k]=s1
end for
puts(1,"Data initialised, starting benchmark...\n")

test("Original random data",s)
test("Data 50%=0 50%=1",repeat((repeat(0,len_data/2)&repeat(1,len_data/2)),
num_runs))
test("Data 50%=1000
50%=1001",repeat((repeat(1000,len_data/2)&repeat(1001,len_data/2)), num_runs))
test("Data 50%=10000
50%=10001",repeat((repeat(10000,len_data/2)&repeat(10001,len_data/2)), num_runs))
test("Data 50%=0.1
50%=1.1",repeat((repeat(0.1,len_data/2)&repeat(1.1,len_data/2)), num_runs))
?machine_func(26,0)

My results:
Data initialised, starting benchmark...
Original random data
Variant form #0 took  4.47s. (100.0%)
Variant form #1 took  2.24s. ( 50.1%)
Variant form #2 took  2.84s. ( 63.5%)
Variant form #3 took  2.86s. ( 64.0%)
Variant form #4 took  2.84s. ( 63.5%)
Variant form #5 took  2.41s. ( 53.9%)
Variant form #6 took  2.51s. ( 56.2%)
Variant form #7 took  2.35s. ( 52.6%)
Variant form #8 took  2.83s. ( 63.3%)
Variant form #9 took  2.84s. ( 63.5%)

Data 50%=0 50%=1
Variant form #0 took  4.47s. (100.0%)
Variant form #1 took  1.48s. ( 33.1%)
Variant form #2 took  2.10s. ( 47.0%)
Variant form #3 took  2.09s. ( 46.8%)
Variant form #4 took  2.00s. ( 44.7%)
Variant form #5 took  1.64s. ( 36.7%)
Variant form #6 took  1.59s. ( 35.6%)
Variant form #7 took  1.49s. ( 33.3%)
Variant form #8 took  1.98s. ( 44.3%)
Variant form #9 took  2.00s. ( 44.7%)

Data 50%=1000 50%=1001
Variant form #0 took  4.46s. (100.0%)
Variant form #1 took  2.06s. ( 46.2%)
Variant form #2 took  2.69s. ( 60.3%)
Variant form #3 took  2.70s. ( 60.5%)
Variant form #4 took  2.64s. ( 59.2%)
Variant form #5 took  2.27s. ( 50.9%)
Variant form #6 took  1.59s. ( 35.7%)
Variant form #7 took  1.50s. ( 33.6%)
Variant form #8 took  1.97s. ( 44.2%)
Variant form #9 took  1.98s. ( 44.4%)

Data 50%=10000 50%=10001
Variant form #0 took  4.49s. (100.0%)
Variant form #1 took  2.18s. ( 48.6%)
Variant form #2 took  2.83s. ( 63.0%)
Variant form #3 took  2.82s. ( 62.8%)
Variant form #4 took  2.79s. ( 62.1%)
Variant form #5 took  2.46s. ( 54.8%)
Variant form #6 took  1.62s. ( 36.1%)
Variant form #7 took  1.52s. ( 33.9%)
Variant form #8 took  1.98s. ( 44.1%)
Variant form #9 took  2.00s. ( 44.5%)

Data 50%=0.1 50%=1.1
Variant form #0 took  5.34s. (100.0%)
Variant form #1 took  4.30s. ( 80.5%)
Variant form #2 took  4.66s. ( 87.3%)
Variant form #3 took  4.68s. ( 87.6%)
Variant form #4 took  4.52s. ( 84.6%)
Variant form #5 took  4.16s. ( 77.9%)
Variant form #6 took  5.31s. ( 99.4%)
Variant form #7 took  5.45s. (102.1%)
Variant form #8 took  5.14s. ( 96.3%)
Variant form #9 took  5.32s. ( 99.6%)

Regards,
   Fernando