Re: Obtaining terms in an expansion

It's not clear to me what you want. Maybe it's something like

my $C;

if (@$ar) {
   $C = 1;
   foreach my $row (@$arr) {
      my $sum = 0;
      foreach my $ele (@$row) {
         $sum += $ele;
      }
      $C *= $sum;
   }
}
[download]

By the way, you shoulnd'y use $a and $b as variable names to avoid conflicts with some special variables.

Update: Or maybe

my $cols = @{$arr->[0]};

my $C;

foreach my $col_idx (0 .. $cols-1) {
   $C->[$col_idx] = 1;
   foreach my $row (@$arr) {
      $C->[$col_idx] *= $row->[$col_idx];
   }
}
[download]

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Re^2: Obtaining terms in an expansion by randyk (Parson) on Jan 05, 2006 at 22:03 UTC
I should have supplied an example of what I was looking for. For example, for `N=2`: `( arr[0][0] + arr[0][1] ) ( arr[1][0] + arr[1][1] )` [download] what I want to find is `C[0] = arr[0][0] arr[1][0] C[1] = arr[0][0] arr[1][1] C[2] = arr[0][1] arr[1][0] C[3] = arr[0][1] arr[1][1]` [download] It doesn't matter if the coefficients come out in this order, of course.	[reply] [d/l] [select]
Re^3: Obtaining terms in an expansion by BrowserUk (Patriarch) on Jan 06, 2006 at 00:26 UTC
If 2^N could become m^n, then you could use a recursive solution: (Updated: cleaned up slightly!) #! perl -slw use strict; sub expand{ return @{ $_[0] } if @_ == 1; map{ my $x = $_; map{ $x * $_ } &expand; } @{ pop @_ }; } my @mat = ( [ -1, 2 ], [ 3, -4 ] ); printf +("%s " x @mat ) . '= ', map{ "[ @$_ ] " } @mat; print join ' ', expand( @mat ); @mat = ( [ 1, 2 ], [ 3, 4 ], [ 5, 6 ], ); printf +("%s " x @mat ) . '= ', map{ "[ @$_ ] " } @mat; print join ' ', expand( @mat ); @mat = ( [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ], ); printf +("%s " x @mat ) . '= ', map{ "[ @$_ ] " } @mat; print join ' ', expand( @mat ); __END__ P:\test>junk [ -1 2 ] [ 3 -4 ] = -3 6 4 -8 [ 1 2 ] [ 3 4 ] [ 5 6 ] = 15 30 20 40 18 36 24 48 [ 1 2 3 ] [ 4 5 6 ] [ 7 8 9 ] = 28 56 84 35 70 105 42 84 126 32 64 96 40 80 120 48 96 144 36 72 108 45 + 90 135 54 108 162 [download] Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error. Lingua non convalesco, consenesco et abolesco. -- Rule 1 has a caveat! -- Who broke the cabal? "Science is about questioning the status quo. Questioning authority". In the absence of evidence, opinion is indistinguishable from prejudice.	[reply] [d/l]
Re^3: Obtaining terms in an expansion by InfiniteSilence (Curate) on Jan 05, 2006 at 22:54 UTC
I don't see an operator between `arr[0][0]` and `arr[1][0]`, so I presume the parenthesis mean you want to multiply. Your description does not adequately explain what you want, particularly the `(a[N-1][0] + a[N-1][1])`. Indices from your example: `0,0 1,0 0,0 1,1 0,1 1,0 0,1 1,1` [download] Don't follow. They should be: `0,0 1,0 0,0 1,1 1,0 1,0 1,1 1,1 2,0 1,0 2,1 1,1 3,0 1,0 3,1 1,1 ...` [download] The wrong element is incrementing -- `a[0][N-1] * a[1][0], a[0][N-1] * a[1][1]`. Furthermore, this makes no sense. Are `a[0][1]` and `a[1][1]` use repeatedly throughout? If so, why not change them into constants? Celebrate Intellectual Diversity	[reply] [d/l] [select]