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Re: Fibonacci numbers
by suaveant (Parson) on Oct 02, 2001 at 00:59 UTC

$i = 1;
print "$i,";
print($i++,',') for(;;);
This will generate the fibonacci numbers... (and a few others), the professor can sort them out :)
 Ant
 Some of my best work  Fish Dinner
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Re: Fibonacci numbers
by jeroenes (Priest) on Oct 02, 2001 at 00:54 UTC

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Yes, it is. To reduce the problem size, the following program is likely to eventually print out any Fibonacci numbers between 0 and N1, where N is the command line argument. In order, even. However, you must find them buried in all the other numbers, and it will take an indeterminate amount of time. It will not stop upon hitting a sequence of fibonacci numbers. (Hint: The output is random.)
#!/usr/bin/perl w
use strict;my$N=$_[0];$\=' ';for (;;){print int rand$N}
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Re: Fibonacci numbers
by runrig (Abbot) on Oct 02, 2001 at 01:18 UTC

Can anybody make a program to generate Fibonacci numbers
Well, no I don't think anybody can do it. I had trouble getting my daughter to write just a 'hello world' program. But you might be able to do it by just writing a program that generates a sequence that follows these rules:
i_{0}=0, i_{1}=1, i_{n, n>=2}=i_{n1}+i_{n2}  [reply] 
(jeffa) Re: Fibonacci numbers
by jeffa (Bishop) on Oct 02, 2001 at 00:58 UTC

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Re: Fibonacci numbers
by demerphq (Chancellor) on Oct 02, 2001 at 01:07 UTC

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Re: Fibonacci numbers
by petdance (Parson) on Oct 02, 2001 at 05:25 UTC

Based on your description of the problem, I think this
should work just fine:
print "1,1,2,3,5,8,13,21,...\n";
If you want something that "does more", try this:
print join( ",", qw( 1 1 2 3 5 8 13 21 ... ) );
xoxo,
Andy

<megaphone>
Throw down the gun and tiara and come out of the float!
</megaphone>
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Re (tilly) 1: Fibonacci numbers
by tilly (Archbishop) on Oct 02, 2001 at 01:48 UTC

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Re: Fibonacci numbers
by chromatic (Archbishop) on Oct 02, 2001 at 03:16 UTC

Do you know LISP? It's pretty easy to translate into Perl:
#!/usr/bin/perl w
use strict;
*prev = sub {
my $a = $_[0]  0;
sub { $a };
};
*fib = sub {
($a, $b) = ($_[0] + $_[1]>(), $_[0]);
printf "%d$/", $a;
$_ = sub { fib( $a, prev($b) ) };
};
print "$_$/" for ++$;
fib(1, prev(0));
for my $var (1 .. shift  10) {
$_>();
}
Alright, that's not necessary *good* LISP, but you should have no trouble deciphering the algorithm without extraneous parenthesis.
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Re: Fibonacci numbers
by miyagawa (Chaplain) on Oct 02, 2001 at 05:46 UTC

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Re: Fibonacci numbers
by saiful (Novice) on Oct 02, 2001 at 14:22 UTC

Well i've tried this but is not giving the desired result...
$a=1;
$b=1;
$c=$a+$b;
while($a <=100){
print("$a\n");
print("$b\n");
print("$c\n");
$a=$b+$c;
$b=$a+$c;
}
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OK, if you're going to make an attempt at it, then we can help
Lets see....at the start you set $a and $b to 1, and $c to $a+$b = 2
Then, on each iteration round the loop, you print out all three variables, then add 2 ($c from your initialisation) to $a and $b, then repeat.
The assignment of $c=$a+$b simply sets $c according to the values of $a and $b at the time of assignment, and does not 'recalculate' on each trip around the loop.
You will also have to rethink the calculation, as it doesn't quite work.
p.s. try using strict and warnings.
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Re: Fibonacci numbers
by Sweeper (Pilgrim) on Oct 05, 2001 at 00:09 UTC

foreach (1..10) { print fibo($_), "\n" }
sub fibo {
$_ = '1'.'2'x($_[0]).'1';
while (/[1568]$/)
{s=([2589]?[01358])=int(abs((67*$11467)*$1+1423)/140)=eg}
y/02678//d;
return length
}
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Whoa, thats wild... I'm still picking it apart, but I've got a slight obfu change, if that's what you're going for.
Replace the 'return length' line with:
s;.;+1;g;eval;
now, back to figuring out exactly what the rest of that sub
is doing....
Blake
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This is one of my two submissions to OPC5.
And now, here is the solution file:
Solution

This program computes the Fibonacci numbers, using regexps.
First, I'll prove that $_ contains only 0's, 1's and 2's, and
that all other digits are there for obfuscation purpose only.
Then, I'll prove that the numbers of 0's and 1's are
Fibonacci numbers.
Lastly, I'll explain the result.
1 Which digits?

We start with a string which looks like "12222221" (with six 2's,
because we want to compute fib(6)).
This string contains substrings belonging to the following set:
"0", "1", "2", "00", "01", "10", "11", "02", "12", "22" and "21".
Of these substrings, only "0", "1", and "21" can match the regexp.
For these three substrings, the formula gives:
0 > 10
1 > 0
21 > 1
The resulting string will therefore contain substrings belonging to
the same set. And the deobfuscated lines would read:
while (/1$/){s=(2?[01])=int(abs((67*$11467)*$1+1423)/140)=eg}
y/02//d;
2 How Fibonacci is computed

Theorem: at the end of the ith iteration, the string contains
fib(i1) "1"s
mixed with fib(i) "0"s
then (ni) "2"s
then either a "1" or a "0"
Demonstration: at the start of the program, the string contains a
single "1", n "2"s and another "1". After the first iteration,
the string contains a single "0", (n1) "2"s and a "1".
That is, the first part of the string contains 0 "1"s (and
fib(11) = 0) and only one "0" (fib(1) = 1).
Suppose this is true for i, we show it remains true for j=i+1.
 In the first part of the resulting string, all "1"s come
from the substitution 0 > 10. There were fib(i) "0"s in the
original string, therefore there are fib(i) = fib(j1) "1"s in
the resulting string.
 In the first part of the resulting string, all "0"s come
either from the substitution 1 > 0 (fib(i1) such substitutions),
or from 0 >10 (fib(i) such substitutions). Therefore, in the
resulting string, there are fib(i1) + fib(i) "0"s, that is
fib(i+1) or fib(j).
 In the second part of the original string, only the last "2"
match the regexp. Actually, this "2" and the last "1" match
"21". Both digits are replaced with a single "1", in effect
deleting this "2". The number ni becomes ni1, that is, nj.
3 How many iterations?

We start with n "2"s. Each iteration removes one "2". So, after
n iterations, we have a string consisting of:
fib(n1) "1"s
mixed with fib(n) "0"s
and then a single "1", no longer prefixed by any "2"s.
After this iteration, the loop condition (end with "1") is still
true. So there is a (n+1)th iteration, which produces:
fib(n) "1"s
mixed with fib(n+1) "0"s
and then a single "0", because of the 1 > 0 substitution (used
for a different purpose).
Then, we remove the (fib(n+1)+1) "0"s, with y/02//d, and the last
string now contains only fib(n) "1"s.
By the way, y/02//d could even be replaced by y/0//d, because
there are no "2"s anymore.
4 Comments

Where did the formula come from?
I needed a formula producing:
0 > 10
1 > 0
21 > 1
I tried a polynom y = a x**2 + b x + c, and I found
y = (67 x**2  1467 x + 1400)/140
or, with the Horner formula
y = ((67 x  1467) x + 1400)/140
The problem is, only the three values give an positive integer
result, which would be an indication that these three values
are the important ones, and the other values can be
discarded. So I introduced the functions "abs" and "int".
Since I had done that, I could alter a bit the
polynom, and therefore I changed it to
y = ((67 x  1467) x + 1423)/140
Duration of the program.
This program is very inefficient. During the ith iteration,
you have:
fib(i1) substitutions "1 > 0"
fib(i) substitutions "0 > 10" (which extends the string)
and 1 substitution "21 > 1".
Therefore, the ith iteration has fib(i+1)+1 substitutions, that is
0.723 x 1.618 ** i + small quantity.
The total cost is 0.723 x sum(1.618 ** i + small quantities),
that is O(1.618**n). Not very good, but much fun.
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Re: Fibonacci numbers
by davorg (Chancellor) on Oct 02, 2001 at 14:53 UTC

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Re: Fibonacci numbers(Again)
by saiful (Novice) on Oct 02, 2001 at 16:40 UTC

Here's the most simple program to do it.
You can test it.
$a=0;
$b=1;
while($a <=200)
{print("$a\n");
($a,$b)=($b,$a+$b);
}
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($a,$b)=($a,$a+$b);
does the trick? I couldn't understand...can somebody explain?  [reply] [d/l] 

($a,$b)=($b,$a+$b);
It's a simple list assignment. $a gets the
value of $b and $b gets $a+$b.
Incidently, it can be a bad idea to use $a and
$b in Perl scripts as they are "special" variables
because of their use in sorting.
Blessed Be
The Pixel  [reply] [d/l] 

Dear Anonymous,
It's very simple. In every iteration the $a receives the earlier value of $b...that's the trick. Bye the way, u can see my ealier attempt, which had one serious mistake... that can be rewritten to give the desired result. See:
$a=1;
$b=1;
while($a <=200)
{$c=$a+$b;
print("$a\n$b\n$c\n");
$a=$b+$c;
$b=$c+$a;
}
Thanks.  [reply] [d/l] 

Oh, for goodness sake. This is homework, right?
For the answer to this latest question, see the first answer in the thread...
(Alternatively, read what you're quoting: it might make more sense then.)
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