Comments on: Fibonacci Sequence … in PowerShell http://huddledmasses.org/fibbonaci-sequence-in-powershell/ You can do more than breathe for free... Mon, 15 Mar 2010 20:36:59 +0000 http://wordpress.org/?v=2.9.2 hourly 1 By: Joel 'Jaykul' Bennett http://huddledmasses.org/fibbonaci-sequence-in-powershell/comment-page-1/#comment-140826 Joel 'Jaykul' Bennett Sun, 03 Feb 2008 05:35:47 +0000 http://HuddledMasses.org/fibbonaci-sequence-in-powershell/#comment-140826 Yeah, it's true, I didn't think of that when I read it at first, but the example that Hanselman gave in ruby is almost literally PowerShell, except it lacks dollar signs, and has those pesky extra words ;-) (not sure why his example for ruby assumes the target number is in a variable @size@ already, but let's go with that, for the sake of equal comparisons): h5. Ruby <code lang="ruby"> x1,x2 = 0,1; 0.upto(size){puts x1; x1 += x2; x1,x2 = x2,x1} </code> h5. PowerShell <code lang="posh"> $x1,$x2 = 0,1; 0..$size|%{$x1; $x1,$x2 = ($x1+$x2),$x1} </code> Of course, although we all know that these solutions perform *way* better than the recursive algorithms, the only reason anyone bothers with fibonacci as an algorithm is to teach recursion, right? It's a simple elegant recursive problem ... and a great example of why recursion isn't always the right answer, even when it's elegant. :-| Oh, and just for the record, if you wanted to get the correct output to match the output of the "fib" function above, you'd need to change it just a tad (I think the Ruby solution is wrong too, but I don't have an interpreter handy, and I'm in a hurry): <code lang="posh"> $x1,$x2 = 0,1; 0..$size|%{$x1,$x2 = ($x1+$x2),$x1}; $x1 </code> See also "Recursion in PowerShell ... Slow":http://huddledmasses.org/recursion-in-powershell-slow/ Yeah, it’s true, I didn’t think of that when I read it at first, but the example that Hanselman gave in ruby is almost literally PowerShell, except it lacks dollar signs, and has those pesky extra words ;-) (not sure why his example for ruby assumes the target number is in a variable size already, but let’s go with that, for the sake of equal comparisons):

Ruby
x1,x2 = 0,1; 0.upto(size){puts x1; x1 += x2; x1,x2 = x2,x1}
 
PowerShell
$x1,$x2 = 0,1; 0..$size|%{$x1; $x1,$x2 = ($x1+$x2),$x1}
 

Of course, although we all know that these solutions perform way better than the recursive algorithms, the only reason anyone bothers with fibonacci as an algorithm is to teach recursion, right? It’s a simple elegant recursive problem … and a great example of why recursion isn’t always the right answer, even when it’s elegant. :-|

Oh, and just for the record, if you wanted to get the correct output to match the output of the “fib” function above, you’d need to change it just a tad (I think the Ruby solution is wrong too, but I don’t have an interpreter handy, and I’m in a hurry):

$x1,$x2 = 0,1; 0..$size|%{$x1,$x2 = ($x1+$x2),$x1}; $x1
 

See also Recursion in PowerShell … Slow

]]> By: Jeffery Hicks http://huddledmasses.org/fibbonaci-sequence-in-powershell/comment-page-1/#comment-140376 Jeffery Hicks Thu, 31 Jan 2008 19:31:40 +0000 http://HuddledMasses.org/fibbonaci-sequence-in-powershell/#comment-140376 I just found this mentioned on a related blog from Bruce's book (I must have not gotten that far yet) which is even better. This will display the sequence up to 1000 $c=$p=1; while ($c -lt 1000) { $c; $c,$p = ($c+$p), $c } I just found this mentioned on a related blog from Bruce’s book (I must have not gotten that far yet) which is even better. This will display the sequence up to 1000

$c=$p=1; while ($c -lt 1000) { $c; $c,$p = ($c+$p), $c }

]]> By: Jeffery Hicks http://huddledmasses.org/fibbonaci-sequence-in-powershell/comment-page-1/#comment-140375 Jeffery Hicks Thu, 31 Jan 2008 19:26:52 +0000 http://HuddledMasses.org/fibbonaci-sequence-in-powershell/#comment-140375 This only gives you the last number in the sequence. Here's a solution I came up with awhile ago (I'm very geeky for stuff like this) that uses recursion so you get the full sequence: Function Get-Fibonacci { Param([int]$n=0) if ($n -eq 0) {$result = 0} elseif ($n -eq 1) { $result =1} else { $result=(Get-Fibonacci($n-1)) + (Get-Fibonacci ($n-2)) } write $result } for ($x=0;$x -le 11;$x++) { Get-Fibonacci $x } This only gives you the last number in the sequence. Here’s a solution I came up with awhile ago (I’m very geeky for stuff like this) that uses recursion so you get the full sequence:

Function Get-Fibonacci {
Param([int]$n=0)

if ($n -eq 0) {$result = 0}
elseif ($n -eq 1) { $result =1}
else {

$result=(Get-Fibonacci($n-1)) + (Get-Fibonacci ($n-2))
}

write $result

}

for ($x=0;$x -le 11;$x++) {
Get-Fibonacci $x
}

]]> By: Andy http://huddledmasses.org/fibbonaci-sequence-in-powershell/comment-page-1/#comment-140353 Andy Thu, 31 Jan 2008 15:55:07 +0000 http://HuddledMasses.org/fibbonaci-sequence-in-powershell/#comment-140353 I like the pipeline method. Very cool. You can also use double variable assignment. http://www.get-powershell.com/ http://getpowershell.wordpress.com/2008/01/24/fibonacci-series-in-powershell/ I like the pipeline method. Very cool. You can also use double variable assignment.

http://www.get-powershell.com/
http://getpowershell.wordpress.com/2008/01/24/fibonacci-series-in-powershell/

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