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.
@@ -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>;