Home  >  Article  >  Database  >  How to Find All Connected Subgraphs in an Undirected Graph Using a Recursive CTE?

How to Find All Connected Subgraphs in an Undirected Graph Using a Recursive CTE?

Patricia Arquette
Patricia ArquetteOriginal
2024-10-31 06:38:30506browse

How to Find All Connected Subgraphs in an Undirected Graph Using a Recursive CTE?

How to Find All Connected Subgraphs of an Undirected Graph

Problem:

Given a table with two columns containing identifiers, find all groups of identifiers that are connected to each other.

Example Table:

ID Identifier1 Identifier2
1 a c
2 b f
3 a g
4 c h
5 b j
6 d f
7 e k
8 i
9 l h

Desired Output:

Identifier Gr_ID Gr.Members
a 1 (a,c,g,h,l)
b 2 (b,d,f,j)
c 1 (a,c,g,h,l)
d 2 (b,d,f,j)
e 3 (e,k)
f 2 (b,d,f,j)
g 1 (a,c,g,h,l)
h 1 (a,c,g,h,l)
j 2 (b,d,f,j)
k 3 (e,k)
l 1 (a,c,g,h,l)
i 4 (i)

Solution:

The following query uses a single recursive query to find all connected subgraphs:

<code class="sql">WITH
CTE_Idents
AS
(
    SELECT Ident1 AS Ident
    FROM @T

    UNION

    SELECT Ident2 AS Ident
    FROM @T
)
,CTE_Pairs
AS
(
    SELECT Ident1, Ident2
    FROM @T
    WHERE Ident1 <> Ident2

    UNION

    SELECT Ident2 AS Ident1, Ident1 AS Ident2
    FROM @T
    WHERE Ident1 <> Ident2
)
,CTE_Recursive
AS
(
    SELECT
        CAST(CTE_Idents.Ident AS varchar(8000)) AS AnchorIdent 
        , Ident1
        , Ident2
        , CAST(',' + Ident1 + ',' + Ident2 + ',' AS varchar(8000)) AS IdentPath
        , 1 AS Lvl
    FROM 
        CTE_Pairs
        INNER JOIN CTE_Idents ON CTE_Idents.Ident = CTE_Pairs.Ident1

    UNION ALL

    SELECT 
        CTE_Recursive.AnchorIdent 
        , CTE_Pairs.Ident1
        , CTE_Pairs.Ident2
        , CAST(CTE_Recursive.IdentPath + CTE_Pairs.Ident2 + ',' AS varchar(8000)) AS IdentPath
        , CTE_Recursive.Lvl + 1 AS Lvl
    FROM
        CTE_Pairs
        INNER JOIN CTE_Recursive ON CTE_Recursive.Ident2 = CTE_Pairs.Ident1
    WHERE
        CTE_Recursive.IdentPath NOT LIKE CAST('%,' + CTE_Pairs.Ident2 + ',%' AS varchar(8000))
)
,CTE_RecursionResult
AS
(
    SELECT AnchorIdent, Ident1, Ident2
    FROM CTE_Recursive
)
,CTE_CleanResult
AS
(
    SELECT AnchorIdent, Ident1 AS Ident
    FROM CTE_RecursionResult

    UNION

    SELECT AnchorIdent, Ident2 AS Ident
    FROM CTE_RecursionResult
)
SELECT
    CTE_Idents.Ident
    ,CASE WHEN CA_Data.XML_Value IS NULL 
    THEN CTE_Idents.Ident ELSE CA_Data.XML_Value END AS GroupMembers
    ,DENSE_RANK() OVER(ORDER BY 
        CASE WHEN CA_Data.XML_Value IS NULL 
        THEN CTE_Idents.Ident ELSE CA_Data.XML_Value END
    ) AS GroupID
FROM
    CTE_Idents
    CROSS APPLY
    (
        SELECT CTE_CleanResult.Ident+','
        FROM CTE_CleanResult
        WHERE CTE_CleanResult.AnchorIdent = CTE_Idents.Ident
        ORDER BY CTE_CleanResult.Ident FOR XML PATH(''), TYPE
    ) AS CA_XML(XML_Value)
    CROSS APPLY
    (
        SELECT CA_XML.XML_Value.value('.', 'NVARCHAR(MAX)')
    ) AS CA_Data(XML_Value)
WHERE
    CTE_Idents.Ident IS NOT NULL
ORDER BY Ident;</code>

Sample Output:

Identifier Gr_ID Gr.Members
a 1 (a,c,g,h,l)
b 2 (b,d,f,j)
c 1 (a,c,g,h,l)
d 2 (b,d,f,j)
e 3 (e,k)
f 2 (b,d,f,j)
g 1 (a,c,g,h,l)
h 1 (a,c,g,h,l)
i 4 (i)
j 2 (b,d,f,j)
k 3 (e,k)
l 1 (a,c,g,h,l)
z 5 (z)

Explanation:

  • The query uses a recursive CTE to find all paths in the graph that follow the edges defined in the CTE_Pairs table.
  • The CTE_Idents table contains all the unique identifiers in the graph.
  • The CTE_CleanResult table extracts the connected identifiers for each anchor identifier.
  • The final SELECT statement uses a combination of FOR XML PATH and CROSS APPLY to concatenate the connected identifiers for each group.
  • DENSE_RANK() is used to assign unique Group IDs to each group.

The above is the detailed content of How to Find All Connected Subgraphs in an Undirected Graph Using a Recursive CTE?. For more information, please follow other related articles on the PHP Chinese website!

Statement:
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn