The second regex doesn't work because you're not capturing anything to $1 or $2. To capture something to $1, etc, use parens.

Further there's no need to escape the space in the replacement.

But, ignoring the issue of "non-English" (do you mean a different character set, like Greek or Japanese by any chance?), you're close. Try this:

#!/usr/bin/perl
use strict;
use warnings;

my $in = "foo bar blivitz done";

$in =~ s/\s+/\n/g;      # replace spaces with \n

print "after first regex: $in \n";

# $in =~ s/[a-zA-Z]\n[a-zA-Z]/$1\ $2/g;
$in =~ s/([a-zA-Z])\n([a-zA-Z])/$1 $2/g;  # remove newlines, restore s
+paces

print "after second regex: $in \n";


=head execution:

ww@GIG:~/pl_test$ perl 675941.pl
after first regex: foo
bar
blivitz
done
after second regex: foo bar blivitz done
ww@GIG:~/pl_test$


=cut
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