If solution is thought of as symmetric matrix with all-unique elements (i.e. "edge labels") per row (and therefore, per column) and crossed-out dummy diagonal, then quite formal algorithm to build it could be as follows. Code is not pretty, at least I hope I streamlined it enough after couple of, even uglier, attempts.

use strict;
use warnings;
use feature 'say';

use constant LETTERS => join '', 'A'..'Z', 'a'..'z';    # template
use constant N => 51;                                   # number of "e
+dges"

die if not N % 2        # (1) prohibit odd nodes number
    or N > 51;          # (2) "template" limitation

say '_' . substr LETTERS, 0, N;
my $last = substr LETTERS, N - 1, 1;    # "collect" last line here

for ( 1 .. N - 1 ) {
    my $s   = substr LETTERS, 0, N;
    my $sfx = substr $s, 0, $_ - 1, '';
    $s  .= $sfx;                        # rotate left       
    $sfx = substr $s, $_, 1, '_';    
    $s  .= $sfx;                        # ...& append replaced diagona
+l item

    say $s;
    $last .= substr $s, -1;
}

say $last . '_';

__END__
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Extended 2:53 2/20/22: now here's formal generation code for both odd/even number of "edges", at expense of introducing additional "edge label" (a star) in latter case:

use strict;
use warnings;
use feature 'say';

use constant LETTERS => join '', 'A'..'Z', 'a'..'z';
use constant N => 12;

say '_' . substr LETTERS, 0, N;
my $last = substr LETTERS, N - 1, 1;

for ( 1 .. N - 1 ) {
    my $s   = substr LETTERS, 0, N;
    my $sfx = substr $s, 0, $_ - 1, '';
    $s  .= '*' unless N % 2;
    $s  .= $sfx;
    $sfx = substr $s, $_, 1, '_';
    $s  .= $sfx if N % 2;

    say $s;
    $last .= substr $s, -1;
}

say $last . '_';

__END__

_ABCDEFGHIJKL
A_CDEFGHIJKL*
BC_EFGHIJKL*A
CDE_GHIJKL*AB
DEFG_IJKL*ABC
EFGHI_KL*ABCD
FGHIJK_*ABCDE
GHIJKL*_BCDEF
HIJKL*AB_DEFG
IJKL*ABCD_FGH
JKL*ABCDEF_HI
KL*ABCDEFGH_J
L*ABCDEFGHIJ_
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