Not an alternative, but introductory work

This was not intended to be an alternative, but for explanation on how to begin such a query. This one only returns all required friendships including those already existing.

Then better delete it completely.

documents/musterloesung-db-2012-09-24/d2c1.2.sql

@@ -1,5 +0,0 @@
-SELECT f1.person2, f2.person2 
-    FROM FriendshipSymmetric f1
-    JOIN FriendshipSymmetric f2 ON f1.person1 = f2.person1
-    WHERE f1.person2 != f2.person2
-        AND f1.person1 = <id>;