Reputation: 115
I have written a code to find determinant of a 10X10 matrix. This code gives the proper result till 9X9 matrix. But for 10X10 matrix gives the following error
"Use of Uninitialized value in multiplication <*> at line 23
Illegal division by zero at line 21"
I tried for 11X11 matrix also, but it is giving the wrong answer.
Why this code is giving such error...
Following is the code:
#!/usr/bin/perl
use strict;
use warnings;
my @x1=(
[5, 6, 3, 2, 4, 9, 3, 5, 4, 2],
[12, 9, 8, 3, 3, 0, 6, 9, 3, 4],
[8, 6, 5, 8, 9, 3, 9, 3, 9, 5],
[6, 4, 3, 0, 6, 4, 8, 2, 22, 8],
[8, 3, 2, 5, 2, 12, 7, 1, 6, 9],
[5, 9, 3, 9, 5, 1, 3, 8, 4, 2],
[3, 10, 4, 16, 4, 7, 2, 12, 9, 6],
[2, 12, 9, 13, 8, 3, 1, 16, 0, 6],
[3, 6, 8, 5, 12, 8, 4, 19, 8, 5],
[2, 5, 6, 4, 9, 10, 3, 11, 7, 3]
);
# Matrix of nxn
for (my $i=0;$i le 9;$i++) {
for (my $j=0;$j le 9;$j++) {
if($j>$i) {
my $ratio = $x1[$j][$i]/$x1[$i][$i];
for(my $k = 0; $k le 9; $k++){
$x1[$j][$k] -= $ratio * $x1[$i][$k];
}
}
}
}
my $det1 = 1;
for(my $i = 0; $i le 9; $i++){
$det1 *= $x1[$i][$i];
}
printf $det1," ";
Upvotes: 3
Views: 612
Reputation: 53508
le
doesn't do what you think it does. http://perldoc.perl.org/perlop.html
Binary "le" returns true if the left argument is stringwise less than or equal to the right argument.
print 10 le 9,"\n";
print 10 <= 9,"\n";
It's a stringwise comparison not a numeric one.
So "10" le "9"
is true, because alphabetically 10
is before 9
.
But this would work fine for a smaller matrix, because 9 le 8
is a valid comparison and works the 'right way'.
You should use <=
instead:
Binary "<=" returns true if the left argument is numerically less than or equal to the right argument.
You can also probably auto-scale by using $#x1
for your comparison, which is the value of the last array index. In your example above, $#x1
is 9
, because your array is 0-9
Upvotes: 6