Implementation of recursion in different languages: what happens inthe background ?

[=?ISO-8859-1?Q?S=E9bastien_de_Mapias?=]

Hello,
Hopefully I'm asking my question on the proper forum, but it's some
kind of low-level computer language issue and I guess here, many
people are likely to give me fine enlightenments (and I suppose -?-
that the interpreters I talk about below are coded in C)... I've found
on the page
http://antoniocangiano.com/2007/11/2...s-python-away/
a discussion about the performance differences between several
languages (Ruby, Perl, Python...) to execute the same operation,
using recursive function calls.

Could someone explain me how such differences can be explained ?
Where does it come from ?

Thanks a lot...
Regards,
Seb

[jbroquefere]

Hello,
To my mind, it depends more about the compiler than the language.
But, because im new here, and a young student, my intepretation might
be wrong.

On 12 fév, 12:13, "Sébastien de Mapias" <[e-mail]:

Hello,
Hopefully I'm asking my question on the proper forum, but it's some
kind of low-level computer language issue and I guess here, many
people are likely to give me fine enlightenments (and I suppose -?-
that the interpreters I talk about below are coded in C)... I've found
on the pagehttp://antoniocangiano.com/2007/11/28/holy-shmoly-ruby-19-smokes-pyth...
a discussion about the performance differences between several
languages (Ruby, Perl, Python...) to execute the same operation,
using recursive function calls.

Could someone explain me how such differences can be explained ?
Where does it come from ?

Thanks a lot...
Regards,
Seb

[jacob navia]

Sébastien de Mapias wrote:

Hello,
Hopefully I'm asking my question on the proper forum, but it's some
kind of low-level computer language issue and I guess here, many
people are likely to give me fine enlightenments (and I suppose -?-
that the interpreters I talk about below are coded in C)... I've found
on the page
http://antoniocangiano.com/2007/11/2...s-python-away/
a discussion about the performance differences between several
languages (Ruby, Perl, Python...) to execute the same operation,
using recursive function calls.

Could someone explain me how such differences can be explained ?
Where does it come from ?

Thanks a lot...
Regards,
Seb

Python 2.5.1: 31.507s
Ruby 1.9.0: 11.934s

I just added "C"

not even 1 second...

:-)
#include <stdio.h>
int fib(int n)
{
if (n < 2)
return n;
else
return fib(n-1)+fib(n-2);
}

int main(void)
{
for(int i = 1; i<36;i++) {
printf("fib(%d)=%d\n",i,fib(i));
}
}
fib(1)=1
fib(2)=1
fib(3)=2
fib(4)=3
fib(5)=5
fib(6)=8
fib(7)=13
fib(8)=21
fib(9)=34
fib(10)=55
fib(11)=89
fib(12)=144
fib(13)=233
fib(14)=377
fib(15)=610
fib(16)=987
fib(17)=1597
fib(18)=2584
fib(19)=4181
fib(20)=6765
fib(21)=10946
fib(22)=17711
fib(23)=28657
fib(24)=46368
fib(25)=75025
fib(26)=121393
fib(27)=196418
fib(28)=317811
fib(29)=514229
fib(30)=832040
fib(31)=1346269
fib(32)=2178309
fib(33)=3524578
fib(34)=5702887
fib(35)=9227465
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[Chris Dollin]

jacob navia wrote:

Python 2.5.1: 31.507s
Ruby 1.9.0: 11.934s

I just added "C"

not even 1 second...

Since one expects a factor of between 10 and 100 for interpretive
implementations, that's hardly surprising. How does your implementation
perform for fib(100)?

--
"Well begun is half done." - Proverb

Hewlett-Packard Limited Cain Road, Bracknell, registered no:
registered office: Berks RG12 1HN 690597 England

[jacob navia]

Chris Dollin wrote:

jacob navia wrote:

>Python 2.5.1: 31.507s
Ruby 1.9.0: 11.934s

I just added "C"

not even 1 second...

Since one expects a factor of between 10 and 100 for interpretive
implementations, that's hardly surprising. How does your implementation
perform for fib(100)?

Using long long instead of int and the 64 bit lcc-win compiler
instead of the 32 bit lcc-win32 I obtain:
fib(1)=1
fib(2)=1
fib(3)=2
fib(4)=3
fib(5)=5
fib(6)=8
fib(7)=13
fib(8)=21
fib(9)=34
fib(10)=55
fib(11)=89
fib(12)=144
fib(13)=233
fib(14)=377
fib(15)=610
fib(16)=987
fib(17)=1597
fib(18)=2584
fib(19)=4181
fib(20)=6765
fib(21)=10946
fib(22)=17711
fib(23)=28657
fib(24)=46368
fib(25)=75025
fib(26)=121393
fib(27)=196418
fib(28)=317811
fib(29)=514229
fib(30)=832040
fib(31)=1346269
fib(32)=2178309
fib(33)=3524578
fib(34)=5702887
fib(35)=9227465
fib(36)=14930352
fib(37)=24157817
fib(38)=39088169
fib(39)=63245986
fib(40)=102334155
fib(41)=165580141
fib(42)=267914296
fib(43)=433494437
fib(44)=701408733
fib(45)=1134903170
fib(46)=1836311903
fib(47)=2971215073

TimeThis : Command Line : tfib
TimeThis : Start Time : Tue Feb 12 13:14:25 2008
TimeThis : End Time : Tue Feb 12 13:16:51 2008
TimeThis : Elapsed Time : 00:02:25.941

I think no C compiler will go (using exactly the same program as given
of course) beyond fib(50) in a reasonable amount of time.

fib(100) would take more than the age of the Universe...
--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[=?ISO-8859-1?Q?S=E9bastien_de_Mapias?=]

Could you please tell me how you 'timed' your
execution ? Thanks.

[Chris Dollin]

jacob navia wrote:

Chris Dollin wrote:

>jacob navia wrote:

>>Python 2.5.1: 31.507s
Ruby 1.9.0: 11.934s

I just added "C"

not even 1 second...

Since one expects a factor of between 10 and 100 for interpretive
implementations, that's hardly surprising. How does your implementation
perform for fib(100)?

Using long long instead of int and the 64 bit lcc-win compiler
instead of the 32 bit lcc-win32 I obtain:

Sorry, Jacob, I was confusing `fib` and `fact` and thinking of the
size of the numbers rather than the time taken. The fix for the
recursion time is easy.

I'd say that you were comparing apples and oranges, except that's
easy.

--
"Ashes are burning the way." - Renaissance, /Ashes Are Burning/

Hewlett-Packard Limited registered no:
registered office: Cain Road, Bracknell, Berks RG12 1HN 690597 England

[Spiros Bousbouras]

On Feb 12, 11:13*am, "Sébastien de Mapias" <[e-mail]>
wrote:

Hello,
Hopefully I'm asking my question on the proper forum, but it's some
kind of low-level computer language issue and I guess here, many
people are likely to give me fine enlightenments (and I suppose -?-
that the interpreters I talk about below are coded in C)... I've found
on the pagehttp://antoniocangiano.com/2007/11/28/holy-shmoly-ruby-19-smokes-pyth...
a discussion about the performance differences between several
languages (Ruby, Perl, Python...) to execute the same operation,
using recursive function calls.

Could someone explain me how such differences can be explained ?
Where does it come from ?

I doubt that there is a simple answer to your
question. One would have to dig deep in the
internals of the interpreters/compilers of the
languages under question to see how they
implement function calls, arithmetic, etc.
and how that might affect performance. So a
group or mailing list for the developers of
interpreters/compilers of the languages would
be a better place.

The only general comment I can make is that
some languages optimize tail recursive calls
by turning them into loops. Scheme for example
is required to do that by its standard. I
don't know what the situation is with Ruby and
Python and it's not relevant to the Fibonacci
benchmark since the recursive call there is
not tail recursive.

Crossposting and setting follow-ups for
comp.programming where I think this is more
on topic.

[Kaz Kylheku]

On Feb 12, 4:19*am, jacob navia <[e-mail]:

I think no C compiler will go (using exactly the same program as given
of course) beyond fib(50) in a reasonable amount of time.

fib(100) would take more than the age of the Universe...

fib(100) cannot be done using that program because the answer is the
69 bit integer 354224848179261915075.

In a high level language, you can memoize your function with some
trivial spelling change.

In Python, simple memoization can be done with a custom function
decorator, which turns into a trivial blurb that you add in front of
the function definition.

In CLISP, with my own memoization macro:

[1](define-memoized-function fib (n) (if (< n 2) n (+ (fib (1- n))
(fib (- n 2)))))
[2](compile 'fib)
FIB ;
NIL ;
NIL
[3](time (fib 100))
Real time: 1.0E-6 sec.
Run time: 0.0 sec.
Space: 9944 Bytes
354224848179261915075

You can do memoization and wider integers in C, but the program will
be so ugly that the Fibonacci part of it won't even be recognizeable
any longer, except for the name.

[Bartc]

"jacob navia" <ja***@nospam.comwrote in message
news:fo**********@aioe.org...

Sébastien de Mapias wrote:

>Hello,
Hopefully I'm asking my question on the proper forum, but it's some
kind of low-level computer language issue and I guess here, many
people are likely to give me fine enlightenments (and I suppose -?-
that the interpreters I talk about below are coded in C)... I've found
on the page
http://antoniocangiano.com/2007/11/2...s-python-away/
a discussion about the performance differences between several
languages (Ruby, Perl, Python...) to execute the same operation,
using recursive function calls.

Could someone explain me how such differences can be explained ?
Where does it come from ?

Thanks a lot...
Regards,
Seb

Python 2.5.1: 31.507s
Ruby 1.9.0: 11.934s

I just added "C"

not even 1 second...

:-)
#include <stdio.h>
int fib(int n)
{
if (n < 2)
return n;
else
return fib(n-1)+fib(n-2);
}

int main(void)
{
for(int i = 1; i<36;i++) {
printf("fib(%d)=%d\n",i,fib(i));
}
}

That's not quite the same code as Ruby/Python, it should be more like:

#include <stdio.h>
int fib(int n)
{
if (n==0 || n==1) /* test this way */
return n;
else
return fib(n-1)+fib(n-2);
}

int main(void)
{
for(int i = 0; i<36;i++) { /* start from 0 */
printf("fib(%d)=%d\n",i,fib(i));
}
}

Your changes gave you an advantage of some 10% :-)

For the record, my own timings were: Ruby(1.8.6) 90 secs, Python(2.5.1) 30
secs, C(lccwin) 0.52 seconds.

(And one of my own compiled C-like languages, 0.5 seconds, one of my own
interpreted languages, 10 secs. If new Ruby is really that fast then I may
have a bit of work to do to stay ahead)

--
Bart

[Army1987]

jacob navia wrote:

I think no C compiler will go (using exactly the same program as given
of course) beyond fib(50) in a reasonable amount of time.

army1987@army1987-laptop:~$ cat fib.c
#include <stdio.h>
long long fib(long long n)
{
if (n < 2)
return n;
else
return fib(n-1)+fib(n-2);
}

int main(void)
{
for(long long i = 1; i<52;i++) {
printf("fib(%lld)=%lld\n",i,fib(i));
}
}
army1987@army1987-laptop:~$ gcc -std=c99 -O3 fib.c -o fib
army1987@army1987-laptop:~$ time ./fib
fib(1)=1
fib(2)=1
[...]
fib(50)=12586269025
fib(51)=20365011074

real 19m4.878s
user 18m50.311s
sys 0m3.824s

(No, it's not reasonable...)
(BTW, what does it do when n < 0?)
--
Army1987 (Replace "NOSPAM" with "email")

[jacob navia]

Army1987 wrote:

jacob navia wrote:

>I think no C compiler will go (using exactly the same program as given
of course) beyond fib(50) in a reasonable amount of time.

army1987@army1987-laptop:~$ cat fib.c

[snip]

fib(51)=20365011074

real 19m4.878s
user 18m50.311s
sys 0m3.824s

(No, it's not reasonable...)
(BTW, what does it do when n < 0?)

Off by one bug?

:-)

Try fib(55) when your computer has nothing else to do.

In any case, the non recursive function will do this
in MUCH less than a second.

When it is less than zero it returns its argument
unchanged...

--
jacob navia
jacob at jacob point remcomp point fr
logiciels/informatique
http://www.cs.virginia.edu/~lcc-win32

[Spoon]

Army1987 wrote:

(And you can use a floating type instead of long int to have a wider
range, if you don't care about results becoming inexact for large n...)

I offer another fast (and not quite correct) implementation.

#include <stdint.h>
uint64_t fib(int n)
{
uint64_t temp, u = 0, v = 1;
while (--n) { temp = u + v; u = v; v = temp; }
return v;
}

[Army1987]

jacob navia wrote:

Army1987 wrote:

[fibonacci function]

>(BTW, what does it do when n < 0?)

Off by one bug?

No. If you use a signed parameter for fib(), you ought to look up how
fibonacci(-n) is defined. Another possibility is to use an unsigned
parameter.

:-)

Try fib(55) when your computer has nothing else to do.

In any case, the non recursive function will do this in MUCH less than a
second.

Indeed.

<OT>
army1987@army1987-laptop:~$ cat fib.py
#!/usr/bin/python
def fib(n):
a, b = 0, 1
for i in xrange(n):
a, b = b, a + b
return a

if __name__ == "__main__":
for i in xrange(101):
print i, fib(i)

army1987@army1987-laptop:~$ time ./fib.py
0 0
1 1
[...]
99 218922995834555169026
100 354224848179261915075

real 0m0.016s
user 0m0.016s
sys 0m0.000s
</OT>
I know there is an implementation which gives exact results without using
floating point with O(log n) time.
http://it.wikipedia.org/wiki/Success...A0_logaritmica
--
Army1987 (Replace "NOSPAM" with "email")

[Army1987]

Spoon wrote:

Army1987 wrote:

I offer another fast (and not quite correct) implementation.

#include <stdint.h>
uint64_t fib(int n)
{
uint64_t temp, u = 0, v = 1;
while (--n) { temp = u + v; u = v; v = temp; }
return v;
}

It's not correct only when n <= 0.
(When fibonacci(n) is greater than the number of grains of wheat on that
chessboard, the result is not correct in Z, but it *is* correct in Z_{2^64}.)
--
Army1987 (Replace "NOSPAM" with "email")