This SQL will compute the permutations without repetitions:
WITH recurse(Result, Depth) AS
(
SELECT CAST(Value AS VarChar(100)), 1
FROM MyTable
UNION ALL
SELECT CAST(r.Result + '+' + a.Value AS VarChar(100)), r.Depth + 1
FROM MyTable a
INNER JOIN recurse r
ON CHARINDEX(a.Value, r.Result) = 0
)
SELECT Result
FROM recurse
WHERE Depth = (SELECT COUNT(*) FROM MyTable)
ORDER BY Result
If MyTable
contains 9 rows, it will take some time to compute, but it will return 362,880 rows.
Update with explanation:
The WITH
statement is used to define a recursive common table expression. In effect, the WITH
statement is looping multiple times performing a UNION
until the recursion is finished.
The first part of SQL sets the starting records. Assuming 3 rows named 'A', 'B', and 'C' in MyTable
, this will generate these rows:
Result Depth
------ -----
A 1
B 1
C 1
Then the next block of SQL performs the first level of recursion:
SELECT CAST(r.Result + '+' + a.Value AS VarChar(100)), r.Depth + 1
FROM MyTable a
INNER JOIN recurse r
ON CHARINDEX(a.Value, r.Result) = 0
This takes all of the records generated so far (which will be in the recurse
table) and joins them to all of the records in MyTable
again. The ON
clause filters the list of records in MyTable
to only return the ones that do not exist already in this row's permutation. This would result in these rows:
Result Depth
------ -----
A 1
B 1
C 1
A+B 2
A+C 2
B+A 2
B+C 2
C+A 2
C+B 2
Then the recursion loops again giving these rows:
Result Depth
------ -----
A 1
B 1
C 1
A+B 2
A+C 2
B+A 2
B+C 2
C+A 2
C+B 2
A+B+C 3
A+C+B 3
B+A+C 3
B+C+A 3
C+A+B 3
C+B+A 3
At this point, the recursion stops because the UNION
does not create any more rows because the CHARINDEX
will always be 0
.
The last SQL filters all of the resulting rows where the computed Depth
column matches the # of records in MyTable
. This throws out all of the rows except for the ones generated by the last depth of recursion. So the final result will be these rows:
Result
------
A+B+C
A+C+B
B+A+C
B+C+A
C+A+B
C+B+A