for odd length of buffers.

My current understanding is that the logic needs to be different for each of 4 possibilities:

1. Odd length left/even length right buffers;
2. Even length right/odd length left buffers;
3. Even length left/even length right buffers;
4. odd length left/odd length right buffers

Also, your approach seems to do considerable more moving/swapping than is optimal; and Perl's convenient-but-slow:

cmpthese -1, { 
    a => q[ 
        my @a = 1 .. 1e5;
        @a[ $_, $_+2 ] = @a[ $_+2, $_ ] for 0 .. $#a-2 
    ],
    b => q[ 
        my( @a, $t ) = 1 .. 1e5;
        $t = $a[ $_ ], $a[ $_ ] = $a[ $_+2 ], $a[ $_+2 ] = $t for 0 ..
+ $#a-2
    ] 
};;
    Rate    a    b
a 9.65/s   -- -13%
b 11.1/s  15%   --
[download]

array-slice to array-slice assignment, hides a lot of sins.

Here is my next try. Based on the know how from which place a particular value needs to replaced. This requires an explicit loop and some temp integers as well. Still in Perl but easy to translate to C and pointer arithmetics.

use strict;
use warnings;

my @buffer = qw[X0 X1 X2 X3 X4 Y0 Y1 Y2 Y3 Y4 Y5 Y6 Y7];

sub swapbuffers {
  my ($n1, $buf) = @_;
  my $n2 = @$buf-$n1;
  my $here = 0;
  my $tmp = $$buf[$here];
  my $istart = $here;
  for (0..@$buf-1) {
    my $from = $here < $n2 ? $here+$n1 : $here-$n2; 
    if( $from == $istart ) { # loop finished
      $$buf[$here] = $tmp;
      $here = ++$istart; 
      $tmp = $$buf[$istart]
    } else {
      $$buf[$here] = $$buf[$from];
      $here = $from;
    }
    print "@$buf\n";
  }
}

print "@buffer\n";
swapbuffers 5, \@buffer;
print "@buffer\n";
[download]

That is a fascinating algorithm! Here's my attempt to grok what's going on (swapped the Xn and Yn for alphas and numbers as X1 & Y1 look far too similar):

C:\test>1118256
        0 1 2 3 4 5 6 7 8 9 0 1 2 3

tmp     a b c d e f g h 1 2 3 4 5 6      ist her frm
a      [1]b c d e f g h   2 3 4 5 6       0   8   8
a       1 b   d e f g h[c]2 3 4 5 6       0   2   2
a       1 b[3]d e f g h c 2   4 5 6       0  10  10
a       1 b 3 d   f g h c 2[e]4 5 6       0   4   4
a       1 b 3 d[5]f g h c 2 e 4   6       0  12  12
a       1 b 3 d 5 f   h c 2 e 4[g]6       0   6   6

b       1   3 d 5 f[a]h c 2 e 4 g 6       1   1   0
b       1[2]3 d 5 f a h c   e 4 g 6       1   9   9
b       1 2 3   5 f a h c[d]e 4 g 6       1   3   3
b       1 2 3[4]5 f a h c d e   g 6       1  11  11
b       1 2 3 4 5   a h c d e[f]g 6       1   5   5
b       1 2 3 4 5[6]a h c d e f g         1  13  13
b       1 2 3 4 5 6 a   c d e f g[h]      1   7   7

3       1 2   4 5 6 a[b]c d e f g h       2   2   1
        1 2[3]4 5 6 a b c d e f g h
[download]

But, it still needs a little work, I Ican't see how to fix it. Here's one of my basket case tests:

C:\test>1118256
        0 1 2 3 4 5 6 7 8
        
tmp     a b c d e f 1 2 3        ist her frm
a      [3]b c d e f 1 2           0   8   8
a       3 b c d e f 1  [2]        0   7   7
a       3 b c d e f  [1]2         0   6   6
a       3 b c d e  [f]1 2         0   5   5
a       3 b c d  [e]f 1 2         0   4   4
a       3 b c  [d]e f 1 2         0   3   3
a       3 b  [c]d e f 1 2         0   2   2
a       3  [b]c d e f 1 2         0   1   1
a       3[a]b c d e f 1 2         1   1   0
        3 a b c d e f 1 2
[download]

---

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