The problem is, that although the counts at each position will be a maximum of depth, meaning that you can get away with using a single byte per position upto depth=255, the size of the matrix grows (hyper?) geometrically. That means that even using a single byte per position, with no further overhead, by the time you reach depth=8, you already require 4GB of storage.
By the time you reached level 16, the storage requirement is 16e18 bytes, which according to this table is roughly 3 times the 'All words ever spoken by human beings':
depth: X x Y required m
+emory:
1: 4 x 4 == 16 _b
2: 16 x 16 == 256 _b
3: 64 x 64 == 4 Kb
4: 256 x 256 == 64 Kb
5: 1024 x 1024 == 1 Mb
6: 4096 x 4096 == 16 Mb
7: 16384 x 16384 == 256 Mb
8: 65536 x 65536 == 4 Gb
9: 262144 x 262144 == 64 Gb
10: 1048576 x 1048576 == 1 Tb
11: 4194304 x 4194304 == 16 Tb
12: 16777216 x 16777216 == 256 Tb
13: 67108864 x 67108864 == 4 Pb
14: 268435456 x 268435456 == 64 Pb
15: 1073741824 x 1073741824 == 1 Eb
16: 4294967296 x 4294967296 == 16 Eb
17: 17179869184 x 17179869184 == 256 Eb
18: 68719476736 x 68719476736 == 4 Zb
19: 274877906944 x 274877906944 == 64 Zb
20: 1099511627776 x 1099511627776 == 1 Yb
21: 4398046511104 x 4398046511104 == 16 Yb
22: 17592186044416 x 17592186044416 == 256 Yb
23: 70368744177664 x 70368744177664 == 4 2^90b
24: 281474976710656 x 281474976710656 == 64 2^90b
25: 1125899906842624 x 1125899906842624 == 1 2^100b
26: 4503599627370496 x 4503599627370496 == 16 2^100b
27: 18014398509481984 x 18014398509481984 == 256 2^100b
28: 72057594037927936 x 72057594037927936 == 4 2^110b
29: 288230376151711740 x 288230376151711740 == 64 2^110b
30: 1152921504606847000 x 1152921504606847000 == 1 2^120b
31: 4611686018427387900 x 4611686018427387900 == 16 2^120b
32: 18446744073709552000 x 18446744073709552000 == 256 2^120b
But if you think about it, even if you could calculate the data at say depth 8, then you would need a screen 70 feet square to view it.
As with all fractals, as you 'zoom' in, you only want to calculate the data required to produce some small subset of the total picture: say 1280x 1024. Because you want the ordered positions in each row, you would have to calculate a 'slice' of the data, but still far less than doing the whole thing.
At depth 1, the value for any given X,Y position is simply: $p = X == Y ? $depth : $depth - 1;
At level 2, the value for any given X,Y position is slightly more complicated:
if( X == Y ) {
return depth;
}
elsif ( ( X,Y ) is in a 4x4 block on the major diagonal ) {
return $depth - 1;
}
else {
## Move the 4x4 block to the origin and treat as for depth -1
X %= 4;
Y %= 4;
depth--;
recurse;
}
This is a recursive data structure, and calculating the value for any given (X,Y,depth) combinaton can be done recursively, quickly and without requiring a lot of memory. The following routine is close (but no cigar!):
#! perl -slw
use strict;
$|++;
our $DEPTH ||= 3;
sub fract2D {
use integer;
my( $x, $y, $depth ) = @_;
if( $x < 16 and $y < ) { ## A block on the major diagonal
return $x == $y ? $depth : $depth -1;
}
elsif( $depth > 1 ) {
return fract2D( $x % 4**$depth, $y % 4**$depth, $depth -1 );
}
else {
return fract2D( $x % 4**$depth, $y % 4**$depth, $depth );
}
}
for my $depth ( $DEPTH ) {
for my $y ( 0 .. 4**$depth-1 ) {
my $c = 0;
for my $x ( 0 .. 4 ** $depth -1 ) {
printf +( ( ++$c % 4 ) == 0 ? " %2d " : " %2d" ),
fract2D( $x, $y, $depth );
}
print +( $y % 4 ) == 3 ? "\n" : '';
}
print '';
}
But I seem to have left my tuits in my other head, and its washday.
Examine what is said, not who speaks -- Silence betokens consent -- Love the truth but pardon error.
"Science is about questioning the status quo. Questioning authority".
In the absence of evidence, opinion is indistinguishable from prejudice.
|