Is there an elegant way to store a dual relationship (i.e. user 1 and user 2 are friends)

问题

I've run into the same problem in two different pieces of work this month:

Version 1: User 1 & User 2 are friends
Version 2: Axis 1 & Axis 2 when graphed should have the quadrants colored...

The problem is, I don't see an elegant way, using a RDBMS, to store and query this information.

There are two obvious approaches:

Approach 1:

store the information twice (i.e. two db rows rows per relationship):
u1, u2, true 
u2, u1, true
u..n, u..i, true
u..i, u..n, true

have rules to always look for the inverse on updates: 
on read, no management needed
on create, create inverse
on delete, delete inverse
on update, update inverse

Advantage:    management logic is always the same.
Disadvantage: possibility of race conditions, extra storage (which is admittedly cheap, but feels wrong)

Approach 2:

store the information once (i.e. one db row per relationship)
u1, u2, true
u..n, u..i, true

have rules to check for corollaries:
on read, if u1, u2 fails, check for u2, u1 
on create u1, u2: check for u2, u1, if it doesn't exist, create u1, u2
on delete, no management needed
on update, optionally redo same check as create

Advantage: Only store once
Disadvantage: Management requires different set of cleanup depending on the operation

I'm wondering if there's a 3rd approach that goes along the lines of "key using f(x,y) where f(x,y) is unique for every x,y combination and where f(x,y) === f(y,x)"

My gut tells me that there should be some combination of bitwise operations that can fulfill these requirements. Something like a two-column:

key1 = x && y key2 = x + y

I'm hoping that people who spent more time in the math department, and less time in the sociology department have seen a proof of the possibility or impossibility of this and can provide a quick "[You moron,] its easily proven (im)possible, see this link" (name calling optional)

Any other elegant approach would also be very welcome.

Thanks

回答1:

There is also a way to use the 2nd approach by adding an extra constraint. Check that u1 < u2:

CREATE TABLE User
( Name VARCHAR(10) NOT NULL
, PRIMARY KEY (Name)
) ;

CREATE TABLE MutualFriendship
( u1 VARCHAR(10) NOT NULL
, u2 VARCHAR(10) NOT NULL
, PRIMARY KEY (u1, u2)
, FOREIGN KEY (u1) 
    REFERENCES User(Name)
, FOREIGN KEY (u2) 
    REFERENCES User(Name)
, CHECK (u1 < u2) 
) ;

The rules to read, create, insert or update will have to use the (LEAST(u1,u2), GREATEST(u1,u2)).

回答2:

In SQL it's easy to implement the constraints to support your first approach:

CREATE TABLE MutualFriendship
(u1 VARCHAR(10) NOT NULL,
 u2 VARCHAR(10) NOT NULL,
 PRIMARY KEY (u1,u2),
 FOREIGN KEY (u2,u1) REFERENCES MutualFriendship (u1,u2));

INSERT INTO MutualFriendship VALUES
('Alice','Bob'),
('Bob','Alice');

回答3:

For anyone that's interested, I played around with a few bitwise operations and found that the following seems to fulfill the criteria for f(x,y):

#Python, returns 3 tuple
def get_hash(x, y):
  return (x & y, x | y, x * y)

I can't prove it, though.

回答4:

"x is a friend of y".

Define a table of (x,y) pairs and enforce a canonical form, e.g. x<y. This will ensure that you cannot have both (p,q) and (q,p) in your database, thus it will ensure "store once".

Create a view as SELECT x,y FROM FRIENDS UNION SELECT x as y, y as x FROM FRIENDS.

Do your updates against the base table (downside : updaters must be aware of the enforced canonical form), do your queries against the view.

回答5:

You seem to limit the number of friends to 1. If this is the case then I would use something like u1,u2 u2,u1 u3,null u4,u5 u5,u4

u3 does not have a friend.

来源：https://stackoverflow.com/questions/8808994/is-there-an-elegant-way-to-store-a-dual-relationship-i-e-user-1-and-user-2-are