W przypadku stosunkowo małych wartości n
(20 w tym przykładzie) można użyć metody wykorzystującej fakt, że naturalne liczby całkowite są kombinacjami bitów.
Rozwiązanie T-SQL
Przykładowe dane:
DECLARE @Sample AS TABLE
(
item_id tinyint IDENTITY(1,1) PRIMARY KEY NONCLUSTERED,
item nvarchar(500) NOT NULL,
bit_value AS
CONVERT
(
integer,
POWER(2, item_id - 1)
)
PERSISTED UNIQUE CLUSTERED
);
INSERT @Sample
(item)
VALUES
(N'Ann'),
(N'Bob'),
(N'Charles'),
(N'Darren'),
(N'Eric'),
(N'Fred'),
(N'George'),
(N'Harry'),
(N'Ian'),
(N'John'),
(N'Keith'),
(N'Larry'),
(N'Mark'),
(N'Nathan'),
(N'Owen'),
(N'Paul'),
(N'Quentin'),
(N'Ryan'),
(N'Steve'),
(N'Terry');
Rozwiązanie:
-- Maximum integer we need
-- for all combinations of 'n' bits
DECLARE
@max integer =
POWER(2,
(
SELECT COUNT(*)
FROM @Sample AS s
)
) - 1;
SELECT
combination =
STUFF
(
(
-- Choose items where the bit is set
-- and concatenate all matches
SELECT ',' + s.item
FROM @Sample AS s
WHERE
n.n & s.bit_value = s.bit_value
ORDER BY
s.bit_value
FOR XML
PATH (''),
TYPE
).value('(./text())[1]', 'varchar(8000)'), 1, 1, ''
)
-- A standard numbers table
-- (single column, integers from 1 to 1048576, indexed)
FROM dbo.Numbers AS N
WHERE
N.n BETWEEN 1 AND @max;
Próbka wyjściowa:
╔════════════════════════╗
║ combination ║
╠════════════════════════╣
║ Ann ║
║ Bob ║
║ Ann,Bob ║
║ Charles ║
║ Ann,Charles ║
║ Bob,Charles ║
║ Ann,Bob,Charles ║
║ Darren ║
║ Ann,Darren ║
║ Bob,Darren ║
║ Ann,Bob,Darren ║
║ Charles,Darren ║
║ Ann,Charles,Darren ║
║ Bob,Charles,Darren ║
║ Ann,Bob,Charles,Darren ║
║ ... ║
╚════════════════════════╝
Plan wykonania:
To trwa 41 sekund , aby napisać 1,048,576 kombinacje zmiennej na moim laptopie. Dzięki wymuszonemu paralelizmowi mogłem skrócić czas wykonania DOP 8
do 13 sekund.
Jeśli potrzebujesz tabeli Numbers, jest to szybki sposób na jej utworzenie:
SELECT TOP (1048576)
n = ISNULL(CONVERT(integer, ROW_NUMBER() OVER (ORDER BY (SELECT NULL))), 0)
INTO dbo.Numbers
FROM sys.columns AS c
CROSS JOIN sys.columns AS c2
CROSS JOIN sys.columns AS c3;
CREATE UNIQUE CLUSTERED INDEX cuq
ON dbo.Numbers (n)
WITH (MAXDOP = 1, SORT_IN_TEMPDB = ON);
Rozwiązanie SQLCLR
Znacznie bardziej wydajna implementacja jest możliwa w SQLCLR (SQL Server 2005 i nowsze wersje):
CREATE ASSEMBLY [Demo]
AUTHORIZATION [dbo]
FROM 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GO
CREATE FUNCTION dbo.Combinations
(
@ElementsCSV nvarchar (4000)
)
RETURNS TABLE
(
Combination nvarchar (4000) NULL
)
AS EXTERNAL NAME Demo.UserDefinedFunctions.Permute;
Przykładowe użycie:
SELECT
f.Combination
FROM dbo.Combinations('A,B,C,D') AS f;
Wydajność:
╔═════════════╗
║ Combination ║
╠═════════════╣
║ A ║
║ B ║
║ A,B ║
║ C ║
║ A,C ║
║ B,C ║
║ A,B,C ║
║ D ║
║ A,D ║
║ B,D ║
║ A,B,D ║
║ C,D ║
║ A,C,D ║
║ B,C,D ║
║ A,B,C,D ║
╚═════════════╝
Używając 20-elementowego zestawu wcześniejszego:
DECLARE @Sample AS TABLE
(
item_id tinyint IDENTITY(1,1) PRIMARY KEY CLUSTERED,
item nvarchar(50) NOT NULL
);
INSERT @Sample
(item)
VALUES
(N'Ann'),
(N'Bob'),
(N'Charles'),
(N'Darren'),
(N'Eric'),
(N'Fred'),
(N'George'),
(N'Harry'),
(N'Ian'),
(N'John'),
(N'Keith'),
(N'Larry'),
(N'Mark'),
(N'Nathan'),
(N'Owen'),
(N'Paul'),
(N'Quentin'),
(N'Ryan'),
(N'Steve'),
(N'Terry');
Rozwiązanie SQLCLR:
-- Create CSV input
DECLARE
@Elements nvarchar(4000) =
STUFF
(
(
SELECT ',' + s.item
FROM @Sample AS s
ORDER BY
s.item_id
FOR XML
PATH (''),
TYPE
).value('(./text())[1]', 'varchar(8000)'), 1, 1, ''
);
DECLARE
@bitbucket nvarchar(4000);
SELECT
@bitbucket = combination
FROM dbo.Combinations(@Elements);
Czas wykonania wynosi 2,5 sekundy dla 1 048 576 kombinacji na moim laptopie o godzinie DOP 1
.
Tworzenie wejścia CSV:
Znajdowanie kombinacji:
Kod źródłowy C #:
using System;
using System.Collections;
using Microsoft.SqlServer.Server;
public partial class UserDefinedFunctions
{
[SqlFunction
(
DataAccess=DataAccessKind.None,
SystemDataAccess=SystemDataAccessKind.None,
FillRowMethodName="FillRow",
TableDefinition="Permutation nvarchar(4000)"
)
]
public static IEnumerable Permute(string ElementsCSV)
{
// Split CSV
string[] elements = ElementsCSV.Split(new char[] { ',' });
// Highest integer needed
int count = (int)Math.Pow(2, elements.Length) - 1;
// Pre-computed array of 2^n values
int[] powers = new int[elements.Length];
for (int i = 0; i < powers.Length; i++)
{
powers[i] = (int)Math.Pow(2, i);
}
// Test integers
for (int i = 1; i <= count; i++)
{
// Reset output
string s = string.Empty;
// Test each bit that could be set
for (int bit = 0; bit < powers.Length && i >= powers[bit]; bit++)
{
if ((i & powers[bit]) == powers[bit])
{
// Add the element corresponding to the set bit
s += elements[bit] + ',';
}
}
// Return a row via enumeration
yield return s.Substring(0, s.Length - 1);
}
}
// Called by SQL Server to fetch a row
public static void FillRow(object o, out string Permutation)
{
Permutation = (string)o;
}
}