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

[Sébastien 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/2007/11/28/holy-shmoly-ruby-19-smokes-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.

 

[jacob navia]

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/2007/11/28/holy-shmoly-ruby-19-smokes-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

 

[Chris Dollin]

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)?

 

[jacob navia]

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...

 

[Sébastien de Mapias]

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

 

[Chris Dollin]

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.

 

[Spiros Bousbouras]

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]

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.

 

[jacob navia]

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.

In ONLY 0.05 seconds you can do it in lcc-win:

fib( 0)= 0
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
fib( 48)= 4807526976
fib( 49)= 7778742049
fib( 50)= 12586269025
fib( 51)= 20365011074
fib( 52)= 32951280099
fib( 53)= 53316291173
fib( 54)= 86267571272
fib( 55)= 139583862445
fib( 56)= 225851433717
fib( 57)= 365435296162
fib( 58)= 591286729879
fib( 59)= 956722026041
fib( 60)= 1548008755920
fib( 61)= 2504730781961
fib( 62)= 4052739537881
fib( 63)= 6557470319842
fib( 64)= 10610209857723
fib( 65)= 17167680177565
fib( 66)= 27777890035288
fib( 67)= 44945570212853
fib( 68)= 72723460248141
fib( 69)= 117669030460994
fib( 70)= 190392490709135
fib( 71)= 308061521170129
fib( 72)= 498454011879264
fib( 73)= 806515533049393
fib( 74)= 1304969544928657
fib( 75)= 2111485077978050
fib( 76)= 3416454622906707
fib( 77)= 5527939700884757
fib( 78)= 8944394323791464
fib( 79)= 14472334024676221
fib( 80)= 23416728348467685
fib( 81)= 37889062373143906
fib( 82)= 61305790721611591
fib( 83)= 99194853094755497
fib( 84)= 160500643816367088
fib( 85)= 259695496911122585
fib( 86)= 420196140727489673
fib( 87)= 679891637638612258
fib( 88)= 1100087778366101931
fib( 89)= 1779979416004714189
fib( 90)= 2880067194370816120
fib( 91)= 4660046610375530309
fib( 92)= 7540113804746346429
fib( 93)= 12200160415121876738
fib( 94)= 19740274219868223167
fib( 95)= 31940434634990099905
fib( 96)= 51680708854858323072
fib( 97)= 83621143489848422977
fib( 98)= 135301852344706746049
fib( 99)= 218922995834555169026
fib(100)= 354224848179261915075
TimeThis : Command Line : tfib-bigq
TimeThis : Start Time : Tue Feb 12 22:14:56 2008


TimeThis : Command Line : tfib-bigq
TimeThis : Start Time : Tue Feb 12 22:14:56 2008
TimeThis : End Time : Tue Feb 12 22:14:56 2008
TimeThis : Elapsed Time : 00:00:00.054

Note that we arrive to the same result for fib(100).
That is somehow reassuring

Program:

#include <math.h>
#include <stdio.h>
#include <qfloat.h>
int main(void)
{
qfloat GoldenRatio = (1+sqrt(5.0Q))/2.0q;

for (int i=0; i<=100;i++) {
qfloat phiN = pow(GoldenRatio,(qfloat)i);
qfloat s = pow((1-GoldenRatio),(qfloat)i);
qfloat fib = (phiN-s)/sqrt(5.0q);

printf("fib(%3d)=%29.25qg\n",i,fib);
}
}

 

[Bartc]

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)

 

[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]

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.

As for wider integers, maybe, but as for memoization, what's so ugly with:
#include <stdio.h>
#include <errno.h>
#include <limits.h>
#define MAX 47 /* assuming 32-bit long */

long int fib(int n)
{
static long int values[MAX] = { 0, 1 };
static int i = 1;
if (n < 0)
return (n % 2 ? -1 : 1) * fib(-n);
if (n >= MAX) {
errno = ERANGE;
return LONG_MAX;
}
while (i < n) {
i++;
values = values[i - 1] + values[i - 2];
}
return values[n];
}

int main(void)
{
int i = 0;
while (1) {
long r = fib(i);
if (errno == ERANGE)
break;
printf("%2d %10ld\n", i, r);
i++;
}
return 0;
}

(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...)

 

[jacob navia]

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...

 

[Spoon]

(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]

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/Successione_di_Fibonacci#Versione_a_complessit.C3.A0_logaritmica

 

[Army1987]

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}.)