Question: I'm struggling with a tricky SQL query that I'm trying to write. Have a look at the following table:

```
+---+---+
| A | B |
+---+---+
| 1 | 2 |
| 1 | 3 |
| 2 | 2 |
| 2 | 3 |
| 2 | 4 |
| 3 | 2 |
| 3 | 3 |
| 4 | 2 |
| 4 | 3 |
| 4 | 4 |
+---+---+

```

Now, from this table, I essentially want a list of all As which have the exact same set of Bs and give each set an incrementing ID.

Hence, the output set for the above would be:

```
+---+----+
| A | ID |
+---+----+
| 1 |  1 |
| 3 |  1 |
| 2 |  2 |
| 4 |  2 |
+---+----+

```

Thanks.

Edit: If it helps, I have a list of all distinct values of B that are possible in another table.

Edit: Thank you so much for all the innovative answers. Was able to learn a lot indeed.

Answer: 
Here is mathematical trick to solve your tricky select:

```
with pow as(select *, b * power(10, row_number() 
              over(partition by a order by b)) as rn from t)
select a, dense_rank() over( order by sum(rn)) as rn 
from pow
group by a
order by rn, a

```

Fiddle <http://sqlfiddle.com/#!3/6b98d/11>

This of course will work only for limited distinct count as you will get overflow. Here is more general solution with strings:

```
select a, 
dense_rank() over(order by (select  '.' + cast(b as varchar(max))
                            from t t2 where t1.a = t2.a
                            order by b
                            for xml path(''))) rn
from t t1
group by a
order by rn, a

```

Fiddle <http://sqlfiddle.com/#!3/6b98d/29>
Here is a very simple, fast, but approximate solution.
It is possible that `CHECKSUM_AGG` returns the same checksum for different sets of B.

```
DECLARE @T TABLE (A int, B int);

INSERT INTO @T VALUES
(1, 2),(1, 3),(2, 2),(2, 3),(2, 4),(3, 2),(3, 3),(4, 2),(4, 3),(4, 4);

SELECT
    A
    ,CHECKSUM_AGG(B) AS CheckSumB
    ,ROW_NUMBER() OVER (PARTITION BY CHECKSUM_AGG(B) ORDER BY A) AS GroupNumber
FROM @T
GROUP BY A
ORDER BY A, GroupNumber;

```

**Result set**

```
A    CheckSumB    GroupNumber
-----------------------------
1    1            1
2    5            1
3    1            2
4    5            2

```

For exact solution group by `A` and concatenate all `B` values into a long (binary) string using either FOR XML, CLR, or T-SQL function. Then you can partition `ROW_NUMBER` by that concatenated string to assign numbers to the groups. As shown in other answers.