Re: Modeling Friendship Relationships

On 11/11/2014 02:38 PM, Robert DiFalco wrote:
> I have a question about modeling a mutual relationship. It seems basic
> but I can't decide, maybe it is 6 of one a half dozen of the other.
>
> In my system any user might be friends with another user, that means
> they have a reciprocal friend relationship.
>
> It seems I have two choices for modeling it.
>
> 1. I have a table with two columns userOne and userTwo. If John is
> friends with Jane there will be one row for both of them.
> 2. I have a table with two columns owner and friend. If John is
> friends with Jane there will be two rows, one that is {John, Jane} and
> another {Jane, John}.
>
> The first option has the advantage of saving table size. But queries
> are more complex because to get John's friends I have to JOIN friends
> f ON f.userA = "John" OR f.userB = "John" (not the real query, these
> would be id's but you get the idea).
>
> In the second option the table rows would be 2x but the queries would
> be simpler -- JOIN friends f ON f.owner = "John".
>
> There could be >1M users. Each user would have <200 friends.
>
> Thoughts? Do I just choose one or is there a clear winner? TIA!

What you are describing is basically an adjacency-list without any
hierarchy information, i.e. there isn't a John reports to Dick reports
to Jane type of tree.

One-million-users at 200 friends each would (order-of-magnitudeish) be
200-million rows which tends to argue for saving space. It also reduces
the number of rows impacted by deletes and avoids the risk of ending up
with John,Jane without a corresponding Jane,John.

Getting John's friends isn't too complicated but I suspect the form of
the query you gave won't lead to optimal query plans. For a two-column
format I would imagine that ...userB as friend where userA='John' union
userA as friend where userB='John'... would yield a more optimal plan
assuming an index on each column.

I'm guessing that you will also want to study common-table-expressions
and recursive queries (description and examples are available in the
PostgreSQL docs) so you can start to write single queries to answer
things like "list everyone who is a friend of a friend of John."

Cheers,
Steve