Re: Bicubic Interpolation Math Function

[BrowserUk]

Do you have a feel for what result you expect for your example? Does 0.75515 sound right?

This won't win any awards for speed, but it might serve as a starting point for something quicker (assuming it is correct):

#! perl -slw
use strict;
use Data::Dump qw[ pp ];

sub cubicInterpolate {
    my $x = shift;
    return $_[1] + 0.5 * $x*( $_[2] - $_[0] + $x*( 2*$_[0] - 5*$_[1] +
+ 4*$_[2] - $_[3] + $x*( 3*( $_[1]-$_[2] ) + $_[3] - $_[0] ) ) );
}

sub bicubicInterpolate {
    my( $x, $y, $aref ) = @_;
    return cubicInterpolate( $x, map cubicInterpolate( $y, @{ $aref->[
+ $_] } ), 0 .. $#$aref );
}

sub readTable {
    my $fh = shift;
    my @table;
    my @xs = split ' ', <DATA>;
    my( @ys, @vals );

    while( <$fh> ) {
        my( $y, @vals ) = split;
        @{ $table[ $y ] }[ @xs ] = @vals;
    }
    return \@table;
}

sub get4x4 {
    my( $x, $y, $table ) = @_;
    my @m;
    $x = int( $x );
    $y = int( $y );
    for my $yi ( $y-1 .. $y+2 ) {
        my $yt = $yi < 0 ? 0 : $yi > $#{ $table } ? $#{ $table } : $yi
+;
        push @m, [];
        for my $xi ( $x-1 .. $x+2 ) {
            my $xt = $xi < 0 ? 0 : $xi > $#{ $table->[ $yt ] } ? $#{ $
+table->[ $yt ] } : $xi;
            push @{ $m[ $#m ] }, $table->[ $yt ][ $xt ];
        }
    }
    return \@m;
}

my $table = readTable( *DATA ); pp $table;
my $m4x4 = get4x4( 2.5, 5.8, $table ); pp $m4x4;

my $interpolated = bicubicInterpolate( 0.5, 0.8, $m4x4 );
print $interpolated;


__DATA__
    1   2   3
4   0.1 0.2 0.3
5   0.4 0.5 0.6
6   0.7 0.8 0.9

Output:

C:\test>1162861.pl
[
  undef,
  undef,
  undef,
  undef,
  [undef, 0.1, 0.2, 0.3],
  [undef, 0.4, 0.5, 0.6],
  [undef, 0.7, 0.8, 0.9],
]

[
  [0.1, 0.2, 0.3, 0.3],
  [0.4, 0.5, 0.6, 0.6],
  [0.7, 0.8, 0.9, 0.9],
  [0.7, 0.8, 0.9, 0.9],
]

0.75515

It's based upon this;

[anirbanphys]

Yes, it should work the way you had written the code. Very sorry for the delayed response. I was checking with the real/actual numbers and from the actual calculation the code result deviating 5%. Now I have to work more to get the more accurate result. But this is a very good reference to work with. I am really thankful for providing such guidance.

[BrowserUk]

Now I have to work more to get the more accurate result.

Take note of the section entitled:"The first and the last interval".

The code I posted uses the first method; perhaps if you switched it to the second method it would be closer to what you are looking for.

[anirbanphys]

Sure, thank you so much for your valuable feedback :)