Question about Nested Sets

microsoft.public.sqlserver.programming

19 Jul 2006 5:04 PM

JayDial

When building a BOM using the nested sets model do you need to include
the childrens children in the definition of the assembly like the
example below?

SERVER_RACK
     SERVER_1
          HDD
          MEM
          VID CARD
    SERVER_2
          HDD
          MEM
          VID CARD
    SERVER_3
          HDD
          MEM
          VID CARD

Or can you just list

SERVER_RACK
     SERVER_1
     SERVER_2
     SERVER_3

and have a recursive function build the children of SERVER_1, etc. for
you? How would this affect the LFT,RGT values?

Thanks for your time

-JayDial

19 Jul 2006 5:18 PM

--CELKO--

First, buy a copy of TREES & HIERARCHIES IN SQL. Youn need the
information and I need the royalty money.

Second, here is my usual "cut & paste":

There are many ways to represent a tree or hierarchy in SQL. This is
called an adjacency list model and it looks like this:

CREATE TABLE OrgChart
(emp CHAR(10) NOT NULL PRIMARY KEY,
boss CHAR(10) DEFAULT NULL REFERENCES OrgChart(emp),
salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);

OrgChart
emp       boss      salary
===========================
'Albert' NULL    1000.00
'Bert'    'Albert'   900.00
'Chuck'   'Albert'   900.00
'Donna'   'Chuck'    800.00
'Eddie'   'Chuck'    700.00
'Fred'    'Chuck'    600.00

Another way of representing trees is to show them as nested sets.

Since SQL is a set oriented language, this is a better model than the
usual adjacency list approach you see in most text books. Let us define
a simple OrgChart table like this.

CREATE TABLE OrgChart
(emp CHAR(10) NOT NULL PRIMARY KEY,
lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
CONSTRAINT order_okay CHECK (lft < rgt) );

OrgChart
emp         lft rgt
======================
'Albert'      1   12
'Bert'        2    3
'Chuck'       4   11
'Donna'       5    6
'Eddie'       7    8
'Fred'        9   10

The organizational chart would look like this as a directed graph:

            Albert (1, 12)
            /        \
          /            \
    Bert (2, 3)    Chuck (4, 11)
                   /    |   \
                 /      |     \
               /        |       \
             /          |         \
        Donna (5, 6) Eddie (7, 8) Fred (9, 10)

The adjacency list table is denormalized in several ways. We are
modeling both the Personnel and the organizational chart in one table.
But for the sake of saving space, pretend that the names are job titles
and that we have another table which describes the Personnel that hold
those positions.

Another problem with the adjacency list model is that the boss and
employee columns are the same kind of thing (i.e. names of personnel),
and therefore should be shown in only one column in a normalized table.
To prove that this is not normalized, assume that "Chuck" changes his
name to "Charles"; you have to change his name in both columns and
several places. The defining characteristic of a normalized table is
that you have one fact, one place, one time.

The final problem is that the adjacency list model does not model
subordination. Authority flows downhill in a hierarchy, but If I fire
Chuck, I disconnect all of his subordinates from Albert. There are
situations (i.e. water pipes) where this is true, but that is not the
expected situation in this case.

To show a tree as nested sets, replace the nodes with ovals, and then
nest subordinate ovals inside each other. The root will be the largest
oval and will contain every other node. The leaf nodes will be the
innermost ovals with nothing else inside them and the nesting will show
the hierarchical relationship. The (lft, rgt) columns (I cannot use the
reserved words LEFT and RIGHT in SQL) are what show the nesting. This
is like XML, HTML or parentheses.

At this point, the boss column is both redundant and denormalized, so
it can be dropped. Also, note that the tree structure can be kept in
one table and all the information about a node can be put in a second
table and they can be joined on employee number for queries.

To convert the graph into a nested sets model think of a little worm
crawling along the tree. The worm starts at the top, the root, makes a
complete trip around the tree. When he comes to a node, he puts a
number in the cell on the side that he is visiting and increments his
counter. Each node will get two numbers, one of the right side and one
for the left. Computer Science majors will recognize this as a modified
preorder tree traversal algorithm. Finally, drop the unneeded
OrgChart.boss column which used to represent the edges of a graph.

This has some predictable results that we can use for building queries.
The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM
TreeTable)); leaf nodes always have (left + 1 = right); subtrees are
defined by the BETWEEN predicate; etc. Here are two common queries
which can be used to build others:

1. An employee and all their Supervisors, no matter how deep the tree.

SELECT O2.*
   FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
    AND O1.emp = :myemployee;

2. The employee and all their subordinates. There is a nice symmetry
here.

SELECT O1.*
   FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
    AND O2.emp = :myemployee;

3. Add a GROUP BY and aggregate functions to these basic queries and
you have hierarchical reports. For example, the total salaries which
each employee controls:

SELECT O2.emp, SUM(S1.salary)
   FROM OrgChart AS O1, OrgChart AS O2,
        Salaries AS S1
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
    AND O1.emp = S1.emp
GROUP BY O2.emp;

4. To find the level of each emp, so you can print the tree as an
indented listing. Technically, you should declare a cursor to go with
the ORDER BY clause.

SELECT COUNT(O2.emp) AS indentation, O1.emp
   FROM OrgChart AS O1, OrgChart AS O2
WHERE O1.lft BETWEEN O2.lft AND O2.rgt
GROUP BY O1.lft, O1.emp
ORDER BY O1.lft;

5. The nested set model has an implied ordering of siblings which the
adjacency list model does not. To insert a new node, G1, under part G.
We can insert one node at a time like this:

BEGIN ATOMIC
DECLARE rightmost_spread INTEGER;

SET rightmost_spread
    = (SELECT rgt
         FROM Frammis
        WHERE part = 'G');
UPDATE Frammis
   SET lft = CASE WHEN lft > rightmost_spread
                  THEN lft + 2
                  ELSE lft END,
       rgt = CASE WHEN rgt >= rightmost_spread
                  THEN rgt + 2
                  ELSE rgt END
WHERE rgt >= rightmost_spread;

INSERT INTO Frammis (part, lft, rgt)
VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
COMMIT WORK;
END;

The idea is to spread the (lft, rgt) numbers after the youngest child
of the parent, G in this case, over by two to make room for the new
addition, G1. This procedure will add the new node to the rightmost
child position, which helps to preserve the idea of an age order among
the siblings.

6. To convert a nested sets model into an adjacency list model:

SELECT B.emp AS boss, E.emp
FROM OrgChart AS E
       LEFT OUTER JOIN
       OrgChart AS B
       ON B.lft
          = (SELECT MAX(lft)
               FROM OrgChart AS S
              WHERE E.lft > S.lft
                AND E.lft < S.rgt);

7. To find the immediate parent of a node:

SELECT MAX(P2.lft), MIN(P2.rgt)
FROM Personnel AS P!, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
   AND P1.emp = :my_emp ;

8. To convert an adjacency list to a nested set model, use a push down
stack. Here is version with a stack in SQL/PSM.

-- Tree holds the adjacency model
CREATE TABLE Tree
(node CHAR(10) NOT NULL,
parent CHAR(10));

-- Stack starts empty, will holds the nested set model
CREATE TABLE Stack
(stack_top INTEGER NOT NULL,
node CHAR(10) NOT NULL,
lft INTEGER,
rgt INTEGER);

CREATE PROCEDURE TreeTraversal ()
LANGUAGE SQL
DETERMINISTIC
BEGIN ATOMIC
DECLARE counter INTEGER;
DECLARE max_counter INTEGER;
DECLARE current_top INTEGER;

SET counter = 2;
SET max_counter = 2 * (SELECT COUNT(*) FROM Tree);
SET current_top = 1;

--clear the stack
DELETE FROM Stack;

-- push the root
INSERT INTO Stack
SELECT 1, node, 1, max_counter
FROM Tree
WHERE parent IS NULL;

-- delete rows from tree as they are used
DELETE FROM Tree WHERE parent IS NULL;

WHILE counter <= max_counter- 1
DO IF EXISTS (SELECT *
              FROM Stack AS S1, Tree AS T1
             WHERE S1.node = T1.parent
               AND S1.stack_top = current_top)
   THEN BEGIN -- push when top has subordinates and set lft value
        INSERT INTO Stack
        SELECT (current_top + 1), MIN(T1.node), counter, NULL
          FROM Stack AS S1, Tree AS T1
         WHERE S1.node = T1.parent
           AND S1.stack_top = current_top;

        -- delete rows from tree as they are used
        DELETE FROM Tree
         WHERE node = (SELECT node
                         FROM Stack
                        WHERE stack_top = current_top + 1);
        -- housekeeping of stack pointers and counter
        SET counter = counter + 1;
        SET current_top = current_top + 1;
     END;
     ELSE
     BEGIN -- pop the stack and set rgt value
       UPDATE Stack
          SET rgt = counter,
              stack_top = -stack_top -- pops the stack
        WHERE stack_top = current_top;
       SET counter = counter + 1;
       SET current_top = current_top - 1;
     END;
END IF;
END WHILE;
-- SELECT node, lft, rgt FROM Stack;
-- the top column is not needed in the final answer
-- move stack contents to new tree table
END;

I have a book on TREES & HIERARCHIES IN SQL which you can get at
Amazon.com right now.

For a good article on using CTEs and recursion, see:
http://www.sqlservercentral.com/columnists/fBROUARD/recursivequeriesinsql1999andsqlserver2005.asp

19 Jul 2006 5:33 PM

JayDial

Mr Celko,

I own your book, and also your programming style book. They are both
great. I am still confused on how to build the nested set model
correctly in SQL Server. I could see building an assembly with every
tier, but is this necessary if one of levels of the tier is defined as
a root elsewhere in the assemblies table?

If this is defined in your book, which section is it defined in?

Thanks

JayDial

--CELKO-- wrote:
Show quote

> First, buy a copy of TREES & HIERARCHIES IN SQL. Youn need the
> information and I need the royalty money.
>
> Second, here is my usual "cut & paste":
>
> There are many ways to represent a tree or hierarchy in SQL. This is
> called an adjacency list model and it looks like this:
>
> CREATE TABLE OrgChart
> (emp CHAR(10) NOT NULL PRIMARY KEY,
>   boss CHAR(10) DEFAULT NULL REFERENCES OrgChart(emp),
>   salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);
>
> OrgChart
> emp       boss      salary
> ===========================
> 'Albert' NULL    1000.00
> 'Bert'    'Albert'   900.00
> 'Chuck'   'Albert'   900.00
> 'Donna'   'Chuck'    800.00
> 'Eddie'   'Chuck'    700.00
> 'Fred'    'Chuck'    600.00
>
> Another way of representing trees is to show them as nested sets.
>
> Since SQL is a set oriented language, this is a better model than the
> usual adjacency list approach you see in most text books. Let us define
> a simple OrgChart table like this.
>
> CREATE TABLE OrgChart
> (emp CHAR(10) NOT NULL PRIMARY KEY,
>   lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
>   rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
>   CONSTRAINT order_okay CHECK (lft < rgt) );
>
> OrgChart
> emp         lft rgt
> ======================
> 'Albert'      1   12
> 'Bert'        2    3
> 'Chuck'       4   11
> 'Donna'       5    6
> 'Eddie'       7    8
> 'Fred'        9   10
>
> The organizational chart would look like this as a directed graph:
>
>             Albert (1, 12)
>             /        \
>           /            \
>     Bert (2, 3)    Chuck (4, 11)
>                    /    |   \
>                  /      |     \
>                /        |       \
>              /          |         \
>         Donna (5, 6) Eddie (7, 8) Fred (9, 10)
>
> The adjacency list table is denormalized in several ways. We are
> modeling both the Personnel and the organizational chart in one table.
> But for the sake of saving space, pretend that the names are job titles
> and that we have another table which describes the Personnel that hold
> those positions.
>
> Another problem with the adjacency list model is that the boss and
> employee columns are the same kind of thing (i.e. names of personnel),
> and therefore should be shown in only one column in a normalized table.
> To prove that this is not normalized, assume that "Chuck" changes his
> name to "Charles"; you have to change his name in both columns and
> several places. The defining characteristic of a normalized table is
> that you have one fact, one place, one time.
>
> The final problem is that the adjacency list model does not model
> subordination. Authority flows downhill in a hierarchy, but If I fire
> Chuck, I disconnect all of his subordinates from Albert. There are
> situations (i.e. water pipes) where this is true, but that is not the
> expected situation in this case.
>
> To show a tree as nested sets, replace the nodes with ovals, and then
> nest subordinate ovals inside each other. The root will be the largest
> oval and will contain every other node. The leaf nodes will be the
> innermost ovals with nothing else inside them and the nesting will show
> the hierarchical relationship. The (lft, rgt) columns (I cannot use the
> reserved words LEFT and RIGHT in SQL) are what show the nesting. This
> is like XML, HTML or parentheses.
>
> At this point, the boss column is both redundant and denormalized, so
> it can be dropped. Also, note that the tree structure can be kept in
> one table and all the information about a node can be put in a second
> table and they can be joined on employee number for queries.
>
> To convert the graph into a nested sets model think of a little worm
> crawling along the tree. The worm starts at the top, the root, makes a
> complete trip around the tree. When he comes to a node, he puts a
> number in the cell on the side that he is visiting and increments his
> counter. Each node will get two numbers, one of the right side and one
> for the left. Computer Science majors will recognize this as a modified
> preorder tree traversal algorithm. Finally, drop the unneeded
> OrgChart.boss column which used to represent the edges of a graph.
>
> This has some predictable results that we can use for building queries.
> The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM
> TreeTable)); leaf nodes always have (left + 1 = right); subtrees are
> defined by the BETWEEN predicate; etc. Here are two common queries
> which can be used to build others:
>
> 1. An employee and all their Supervisors, no matter how deep the tree.
>
> SELECT O2.*
>    FROM OrgChart AS O1, OrgChart AS O2
>   WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>     AND O1.emp = :myemployee;
>
> 2. The employee and all their subordinates. There is a nice symmetry
> here.
>
> SELECT O1.*
>    FROM OrgChart AS O1, OrgChart AS O2
>   WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>     AND O2.emp = :myemployee;
>
> 3. Add a GROUP BY and aggregate functions to these basic queries and
> you have hierarchical reports. For example, the total salaries which
> each employee controls:
>
> SELECT O2.emp, SUM(S1.salary)
>    FROM OrgChart AS O1, OrgChart AS O2,
>         Salaries AS S1
>   WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>     AND O1.emp = S1.emp
>   GROUP BY O2.emp;
>
> 4. To find the level of each emp, so you can print the tree as an
> indented listing. Technically, you should declare a cursor to go with
> the ORDER BY clause.
>
> SELECT COUNT(O2.emp) AS indentation, O1.emp
>    FROM OrgChart AS O1, OrgChart AS O2
>   WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>   GROUP BY O1.lft, O1.emp
>   ORDER BY O1.lft;
>
> 5. The nested set model has an implied ordering of siblings which the
> adjacency list model does not. To insert a new node, G1, under part G.
> We can insert one node at a time like this:
>
> BEGIN ATOMIC
> DECLARE rightmost_spread INTEGER;
>
> SET rightmost_spread
>     = (SELECT rgt
>          FROM Frammis
>         WHERE part = 'G');
> UPDATE Frammis
>    SET lft = CASE WHEN lft > rightmost_spread
>                   THEN lft + 2
>                   ELSE lft END,
>        rgt = CASE WHEN rgt >= rightmost_spread
>                   THEN rgt + 2
>                   ELSE rgt END
> WHERE rgt >= rightmost_spread;
>
> INSERT INTO Frammis (part, lft, rgt)
> VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
> COMMIT WORK;
> END;
>
> The idea is to spread the (lft, rgt) numbers after the youngest child
> of the parent, G in this case, over by two to make room for the new
> addition, G1. This procedure will add the new node to the rightmost
> child position, which helps to preserve the idea of an age order among
> the siblings.
>
> 6. To convert a nested sets model into an adjacency list model:
>
> SELECT B.emp AS boss, E.emp
>   FROM OrgChart AS E
>        LEFT OUTER JOIN
>        OrgChart AS B
>        ON B.lft
>           = (SELECT MAX(lft)
>                FROM OrgChart AS S
>               WHERE E.lft > S.lft
>                 AND E.lft < S.rgt);
>
> 7. To find the immediate parent of a node:
>
> SELECT MAX(P2.lft), MIN(P2.rgt)
>   FROM Personnel AS P!, Personnel AS P2
> WHERE P1.lft BETWEEN P2.lft AND P2.rgt
>    AND P1.emp = :my_emp ;
>
> 8. To convert an adjacency list to a nested set model, use a push down
> stack. Here is version with a stack in SQL/PSM.
>
> -- Tree holds the adjacency model
> CREATE TABLE Tree
> (node CHAR(10) NOT NULL,
> parent CHAR(10));
>
> -- Stack starts empty, will holds the nested set model
> CREATE TABLE Stack
> (stack_top INTEGER NOT NULL,
> node CHAR(10) NOT NULL,
> lft INTEGER,
> rgt INTEGER);
>
> CREATE PROCEDURE TreeTraversal ()
> LANGUAGE SQL
> DETERMINISTIC
> BEGIN ATOMIC
> DECLARE counter INTEGER;
> DECLARE max_counter INTEGER;
> DECLARE current_top INTEGER;
>
> SET counter = 2;
> SET max_counter = 2 * (SELECT COUNT(*) FROM Tree);
> SET current_top = 1;
>
> --clear the stack
> DELETE FROM Stack;
>
> -- push the root
> INSERT INTO Stack
> SELECT 1, node, 1, max_counter
>   FROM Tree
> WHERE parent IS NULL;
>
> -- delete rows from tree as they are used
> DELETE FROM Tree WHERE parent IS NULL;
>
> WHILE counter <= max_counter- 1
> DO IF EXISTS (SELECT *
>               FROM Stack AS S1, Tree AS T1
>              WHERE S1.node = T1.parent
>                AND S1.stack_top = current_top)
>    THEN BEGIN -- push when top has subordinates and set lft value
>         INSERT INTO Stack
>         SELECT (current_top + 1), MIN(T1.node), counter, NULL
>           FROM Stack AS S1, Tree AS T1
>          WHERE S1.node = T1.parent
>            AND S1.stack_top = current_top;
>
>         -- delete rows from tree as they are used
>         DELETE FROM Tree
>          WHERE node = (SELECT node
>                          FROM Stack
>                         WHERE stack_top = current_top + 1);
>         -- housekeeping of stack pointers and counter
>         SET counter = counter + 1;
>         SET current_top = current_top + 1;
>      END;
>      ELSE
>      BEGIN -- pop the stack and set rgt value
>        UPDATE Stack
>           SET rgt = counter,
>               stack_top = -stack_top -- pops the stack
>         WHERE stack_top = current_top;
>        SET counter = counter + 1;
>        SET current_top = current_top - 1;
>      END;
> END IF;
> END WHILE;
> -- SELECT node, lft, rgt FROM Stack;
> -- the top column is not needed in the final answer
> -- move stack contents to new tree table
> END;
>
> I have a book on TREES & HIERARCHIES IN SQL which you can get at
> Amazon.com right now.
>
> For a good article on using CTEs and recursion, see:
> http://www.sqlservercentral.com/columnists/fBROUARD/recursivequeriesinsql1999andsqlserver2005.asp

24 Jul 2006 4:26 AM

--CELKO--

>> I am still confused on how to build the nested set model correctly in SQL Server. I could see building an assembly with every tier, but is this necessary if one of levels of the tier is defined as a root elsewhere in the assemblies table? <<

You have two choices with nested sets. The first is to put all of the
"atomic parts" (screws, nuts, bolts and things we buy as a unit from
outside suppliers like a motor) in an inventory table, then have the
full assembly as THE root of the whole thing. At each non-leaf node,
you have to give the subassembly a name.

The second way is a bit more complicated, and think this is what you
want. Instead of a tree (i.e. one root), you have a "forest" of many
trees. For example, if you actually build your own motors and use them
in many pplace in one product, then one tree in the forest is for that
motor, something like this:

CREATE TABLE Forest
(tree_name CHAR(15) NOT NULL,
component_name CHAR(15) NOT NULL,
component_type CHAR(1) DEFAULT 'A' NOT NULL,
CHECK (component_type IN 'A', 'T')),
lft INTEGER NOT NULL CHECK (lft > 0),
rgt INTEGER NOT NULL,
CHECK (lft < rgt),
etc.);

So 'Motor' is a tree when we make it in-house and atomic when we buy it
as a unit.

The integrity checks are a bit ugly because SQL Server does not support
CREATE ASSERTION statements, so you have rto do it with VIEWs or
TRIGGGERs.

19 Jul 2006 7:53 PM

J. M. De Moor

>>I need the royalty money<<

OK, wait a sec...

First you say you consult at 4 figures a day plus expenses and are overrun
by cleanup jobs that take of the mess the rest of us make. And you STILL
need royalty $$ on top of that?!? I would love to see your mortgage and
alimony commitments. Hahaha!

Joe

19 Jul 2006 8:07 PM

JayDial

Ah, cmon not even a comment on the topic at hand?

J. M. De Moor wrote:
Show quote

> >>I need the royalty money<<
>
> OK, wait a sec...
>
> First you say you consult at 4 figures a day plus expenses and are overrun
> by cleanup jobs that take of the mess the rest of us make. And you STILL
> need royalty $$ on top of that?!? I would love to see your mortgage and
> alimony commitments. Hahaha!
>
> Joe

19 Jul 2006 9:34 PM

Frank

On 19 Jul 2006 13:07:13 -0700, "JayDial" <JayD***@gmail.com> wrote:

in <1153339633.560961.170***@m73g2000cwd.googlegroups.com>

Show quote

>Ah, cmon not even a comment on the topic at hand?
>
>
>J. M. De Moor wrote:
>> >>I need the royalty money<<
>>
>> OK, wait a sec...
>>
>> First you say you consult at 4 figures a day plus expenses and are overrun
>> by cleanup jobs that take of the mess the rest of us make. And you STILL
>> need royalty $$ on top of that?!? I would love to see your mortgage and
>> alimony commitments. Hahaha!
>>
>> Joe

What makes you think that even deserves a comment, insect?

Presumably you command at least one or two hundred a day?

24 Jul 2006 4:34 AM

--CELKO--

>> I would love to see your mortgage and alimony commitments. <<

No alimony yet and my wife is labor-intensity instead of pocketbook
intensity -- the advantage of marrying a Zen Monk who has been
practicing for 6+ years instead of a 20-year old stripper with a desire
for a red sports car.

What you don't see is how I got fried back in the "dot com disasters"
of 2000+ when we were all going to be zillionaires on our stock
options. A 20+ year old kid can recover, but a 60+ year man needs his
book royalties -- the geek equivalent of a "Union Pension Plan" -- or
those &%^@! stock options.

19 Jul 2006 9:57 PM

Paul Nielsen (MVP)

BOM typically needs to allow for many parts to be used in many assemblies,
which is different than an org chart because each part has mulitple paths up
and down.

Joe, How exactly do you intend to model this many to many relationship in a
Nested Set pattern? Maybe there's something about the Nested Set pattern I
don't undertand the same way you understand.

Jay, I've always solved one hierachical problem with a BOM associative
reflective table.

CREATE TABLE Part (
PartID
...
)

CREATE TABLE BOM (
MasterPartID,
ComponentPartID
Quantity INT
)

If you need more for the example please email me.

pa***@sqlserverbible.com

Show quote

"--CELKO--" <jcelko***@earthlink.net> wrote in message
news:1153329526.144931.90280@p79g2000cwp.googlegroups.com...
> First, buy a copy of TREES & HIERARCHIES IN SQL. Youn need the
> information and I need the royalty money.
>
> Second, here is my usual "cut & paste":
>
> There are many ways to represent a tree or hierarchy in SQL. This is
> called an adjacency list model and it looks like this:
>
> CREATE TABLE OrgChart
> (emp CHAR(10) NOT NULL PRIMARY KEY,
> boss CHAR(10) DEFAULT NULL REFERENCES OrgChart(emp),
> salary DECIMAL(6,2) NOT NULL DEFAULT 100.00);
>
> OrgChart
> emp       boss      salary
> ===========================
> 'Albert' NULL    1000.00
> 'Bert'    'Albert'   900.00
> 'Chuck'   'Albert'   900.00
> 'Donna'   'Chuck'    800.00
> 'Eddie'   'Chuck'    700.00
> 'Fred'    'Chuck'    600.00
>
> Another way of representing trees is to show them as nested sets.
>
> Since SQL is a set oriented language, this is a better model than the
> usual adjacency list approach you see in most text books. Let us define
> a simple OrgChart table like this.
>
> CREATE TABLE OrgChart
> (emp CHAR(10) NOT NULL PRIMARY KEY,
> lft INTEGER NOT NULL UNIQUE CHECK (lft > 0),
> rgt INTEGER NOT NULL UNIQUE CHECK (rgt > 1),
> CONSTRAINT order_okay CHECK (lft < rgt) );
>
> OrgChart
> emp         lft rgt
> ======================
> 'Albert'      1   12
> 'Bert'        2    3
> 'Chuck'       4   11
> 'Donna'       5    6
> 'Eddie'       7    8
> 'Fred'        9   10
>
> The organizational chart would look like this as a directed graph:
>
>            Albert (1, 12)
>            /        \
>          /            \
>    Bert (2, 3)    Chuck (4, 11)
>                   /    |   \
>                 /      |     \
>               /        |       \
>             /          |         \
>        Donna (5, 6) Eddie (7, 8) Fred (9, 10)
>
> The adjacency list table is denormalized in several ways. We are
> modeling both the Personnel and the organizational chart in one table.
> But for the sake of saving space, pretend that the names are job titles
> and that we have another table which describes the Personnel that hold
> those positions.
>
> Another problem with the adjacency list model is that the boss and
> employee columns are the same kind of thing (i.e. names of personnel),
> and therefore should be shown in only one column in a normalized table.
> To prove that this is not normalized, assume that "Chuck" changes his
> name to "Charles"; you have to change his name in both columns and
> several places. The defining characteristic of a normalized table is
> that you have one fact, one place, one time.
>
> The final problem is that the adjacency list model does not model
> subordination. Authority flows downhill in a hierarchy, but If I fire
> Chuck, I disconnect all of his subordinates from Albert. There are
> situations (i.e. water pipes) where this is true, but that is not the
> expected situation in this case.
>
> To show a tree as nested sets, replace the nodes with ovals, and then
> nest subordinate ovals inside each other. The root will be the largest
> oval and will contain every other node. The leaf nodes will be the
> innermost ovals with nothing else inside them and the nesting will show
> the hierarchical relationship. The (lft, rgt) columns (I cannot use the
> reserved words LEFT and RIGHT in SQL) are what show the nesting. This
> is like XML, HTML or parentheses.
>
> At this point, the boss column is both redundant and denormalized, so
> it can be dropped. Also, note that the tree structure can be kept in
> one table and all the information about a node can be put in a second
> table and they can be joined on employee number for queries.
>
> To convert the graph into a nested sets model think of a little worm
> crawling along the tree. The worm starts at the top, the root, makes a
> complete trip around the tree. When he comes to a node, he puts a
> number in the cell on the side that he is visiting and increments his
> counter. Each node will get two numbers, one of the right side and one
> for the left. Computer Science majors will recognize this as a modified
> preorder tree traversal algorithm. Finally, drop the unneeded
> OrgChart.boss column which used to represent the edges of a graph.
>
> This has some predictable results that we can use for building queries.
> The root is always (left = 1, right = 2 * (SELECT COUNT(*) FROM
> TreeTable)); leaf nodes always have (left + 1 = right); subtrees are
> defined by the BETWEEN predicate; etc. Here are two common queries
> which can be used to build others:
>
> 1. An employee and all their Supervisors, no matter how deep the tree.
>
> SELECT O2.*
>   FROM OrgChart AS O1, OrgChart AS O2
> WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>    AND O1.emp = :myemployee;
>
> 2. The employee and all their subordinates. There is a nice symmetry
> here.
>
> SELECT O1.*
>   FROM OrgChart AS O1, OrgChart AS O2
> WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>    AND O2.emp = :myemployee;
>
> 3. Add a GROUP BY and aggregate functions to these basic queries and
> you have hierarchical reports. For example, the total salaries which
> each employee controls:
>
> SELECT O2.emp, SUM(S1.salary)
>   FROM OrgChart AS O1, OrgChart AS O2,
>        Salaries AS S1
> WHERE O1.lft BETWEEN O2.lft AND O2.rgt
>    AND O1.emp = S1.emp
> GROUP BY O2.emp;
>
> 4. To find the level of each emp, so you can print the tree as an
> indented listing. Technically, you should declare a cursor to go with
> the ORDER BY clause.
>
> SELECT COUNT(O2.emp) AS indentation, O1.emp
>   FROM OrgChart AS O1, OrgChart AS O2
> WHERE O1.lft BETWEEN O2.lft AND O2.rgt
> GROUP BY O1.lft, O1.emp
> ORDER BY O1.lft;
>
> 5. The nested set model has an implied ordering of siblings which the
> adjacency list model does not. To insert a new node, G1, under part G.
> We can insert one node at a time like this:
>
> BEGIN ATOMIC
> DECLARE rightmost_spread INTEGER;
>
> SET rightmost_spread
>    = (SELECT rgt
>         FROM Frammis
>        WHERE part = 'G');
> UPDATE Frammis
>   SET lft = CASE WHEN lft > rightmost_spread
>                  THEN lft + 2
>                  ELSE lft END,
>       rgt = CASE WHEN rgt >= rightmost_spread
>                  THEN rgt + 2
>                  ELSE rgt END
> WHERE rgt >= rightmost_spread;
>
> INSERT INTO Frammis (part, lft, rgt)
> VALUES ('G1', rightmost_spread, (rightmost_spread + 1));
> COMMIT WORK;
> END;
>
> The idea is to spread the (lft, rgt) numbers after the youngest child
> of the parent, G in this case, over by two to make room for the new
> addition, G1. This procedure will add the new node to the rightmost
> child position, which helps to preserve the idea of an age order among
> the siblings.
>
> 6. To convert a nested sets model into an adjacency list model:
>
> SELECT B.emp AS boss, E.emp
> FROM OrgChart AS E
>       LEFT OUTER JOIN
>       OrgChart AS B
>       ON B.lft
>          = (SELECT MAX(lft)
>               FROM OrgChart AS S
>              WHERE E.lft > S.lft
>                AND E.lft < S.rgt);
>
> 7. To find the immediate parent of a node:
>
> SELECT MAX(P2.lft), MIN(P2.rgt)
> FROM Personnel AS P!, Personnel AS P2
> WHERE P1.lft BETWEEN P2.lft AND P2.rgt
>   AND P1.emp = :my_emp ;
>
> 8. To convert an adjacency list to a nested set model, use a push down
> stack. Here is version with a stack in SQL/PSM.
>
> -- Tree holds the adjacency model
> CREATE TABLE Tree
> (node CHAR(10) NOT NULL,
> parent CHAR(10));
>
> -- Stack starts empty, will holds the nested set model
> CREATE TABLE Stack
> (stack_top INTEGER NOT NULL,
> node CHAR(10) NOT NULL,
> lft INTEGER,
> rgt INTEGER);
>
> CREATE PROCEDURE TreeTraversal ()
> LANGUAGE SQL
> DETERMINISTIC
> BEGIN ATOMIC
> DECLARE counter INTEGER;
> DECLARE max_counter INTEGER;
> DECLARE current_top INTEGER;
>
> SET counter = 2;
> SET max_counter = 2 * (SELECT COUNT(*) FROM Tree);
> SET current_top = 1;
>
> --clear the stack
> DELETE FROM Stack;
>
> -- push the root
> INSERT INTO Stack
> SELECT 1, node, 1, max_counter
> FROM Tree
> WHERE parent IS NULL;
>
> -- delete rows from tree as they are used
> DELETE FROM Tree WHERE parent IS NULL;
>
> WHILE counter <= max_counter- 1
> DO IF EXISTS (SELECT *
>              FROM Stack AS S1, Tree AS T1
>             WHERE S1.node = T1.parent
>               AND S1.stack_top = current_top)
>   THEN BEGIN -- push when top has subordinates and set lft value
>        INSERT INTO Stack
>        SELECT (current_top + 1), MIN(T1.node), counter, NULL
>          FROM Stack AS S1, Tree AS T1
>         WHERE S1.node = T1.parent
>           AND S1.stack_top = current_top;
>
>        -- delete rows from tree as they are used
>        DELETE FROM Tree
>         WHERE node = (SELECT node
>                         FROM Stack
>                        WHERE stack_top = current_top + 1);
>        -- housekeeping of stack pointers and counter
>        SET counter = counter + 1;
>        SET current_top = current_top + 1;
>     END;
>     ELSE
>     BEGIN -- pop the stack and set rgt value
>       UPDATE Stack
>          SET rgt = counter,
>              stack_top = -stack_top -- pops the stack
>        WHERE stack_top = current_top;
>       SET counter = counter + 1;
>       SET current_top = current_top - 1;
>     END;
> END IF;
> END WHILE;
> -- SELECT node, lft, rgt FROM Stack;
> -- the top column is not needed in the final answer
> -- move stack contents to new tree table
> END;
>
> I have a book on TREES & HIERARCHIES IN SQL which you can get at
> Amazon.com right now.
>
> For a good article on using CTEs and recursion, see:
> http://www.sqlservercentral.com/columnists/fBROUARD/recursivequeriesinsql1999andsqlserver2005.asp
>

24 Jul 2006 4:39 AM

--CELKO--

>> Joe, How exactly do you intend to model this many to many relationship in a Nested Set pattern? Maybe there's something about the Nested Set pattern I don't undertand the same way you understand. <<

the tree structure (BOM) is one table that rerferences a second table
of nodes (inventory) t the (lft,rgt) pairt is unique, but the node
("#5 machine screw") can be anywhere on the leaf nodes. Then you can
keep a repeated sub-assembly in its own tree in a forest -- see another
post in this thread -- or do a sub-tree copy everywhere it is used.