# recursion problem

• 06-20-2002, 06:47 PM
Steven
recursion problem

I really need some help here.

I have a table [family]:parent_name, my_name

and I want to pass 'adam' as a parameter and get the number of family members
down 3 generations

the data would be:
Michael, Lucifer
John, Matthew
Matthew, David
David, Peter
Peter, Paul
Paul, Wendy

so 'Adam' has 5 members (3 levels deep, john&michael&Lucifer&Matthew&David),
David and his offspring are ignored.

and cannot understand 'books online' so am pretty much completly stuck.
If anyone can help I would really appreciate it
Stephen
• 06-21-2002, 09:16 AM
David Satz
Re: recursion problem
I assume you read "Expanding Hierarchies" in SQL Server BOL.

this is what I call the Joe Celko model:

The usual example of a tree structure in SQL books is called an adjacency
list model and it looks like this:

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

Personnel
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 Personnel
table like this, ignoring the left (lft) and right (rgt) columns for now.
This problem is always given with a column for the employee and one for his
boss in the textbooks. This table without the lft and rgt columns is called
the adjacency list model, after the graph theory technique of the same name;
the pairs of nodes are adjacent to each other.

CREATE TABLE Personnel
(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) );

Personnel
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 first 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, 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 rgt and lft columns (I cannot use the reserved words LEFT
and RIGHT in SQL) are what shows the nesting.

If that mental model does not work, then imagine a little worm crawling
anti-clockwise along the tree. Every time he gets to the left or right side
of a node, he numbers it. The worm stops when he gets all the way around
the tree and back to the top.

This is a natural way to model a parts explosion, since a final assembly is
made of physically nested assemblies that final break down into separate
parts.

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 Personnel.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 P2.*
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P1.emp = :myemployee;

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

SELECT P2.*
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P2.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 P2.emp, SUM(S1.salary)
FROM Personnel AS P1, Personnel AS P2,
Salaries AS S1
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
AND P1.emp = S1.emp
GROUP BY P2.emp;

4. To find the level of each node, so you can print the tree as an indented
listing.

DECLARE Out_Tree CURSOR FOR
SELECT P1.lft, COUNT(P2.emp) AS indentation, P1.emp
FROM Personnel AS P1, Personnel AS P2
WHERE P1.lft BETWEEN P2.lft AND P2.rgt
GROUP BY P1.emp
ORDER BY P1.lft;

5. The nested set model has an implied ordering of siblings which the
adjacency list model does not. To insert a new node as the rightmost
sibling.

BEGIN
DECLARE right_most_sibling INTEGER;

SET right_most_sibling
= (SELECT rgt
FROM Personnel
WHERE emp = :your_boss);

UPDATE Personnel
SET lft = CASE WHEN lft > right_most_sibling
THEN lft + 2
ELSE lft END,
rgt = CASE WHEN rgt >= right_most_sibling
THEN rgt + 2
ELSE rgt END
WHERE rgt >= right_most_sibling;

INSERT INTO Personnel (emp, lft, rgt)
VALUES ('New Guy', right_most_sibling, (right_most_sibling + 1))
END;

6. To convert an adjacency list model into a nested set model, use a push
down stack algorithm. Assume that we have these tables:

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

INSERT INTO Tree
SELECT emp, boss FROM Personnel;

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

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;

INSERT INTO Stack
SELECT 1, emp, 1, NULL
FROM Tree
WHERE boss IS NULL;

DELETE FROM Tree
WHERE boss IS NULL;

WHILE counter <= (max_counter - 2)
LOOP IF EXISTS (SELECT *
FROM Stack AS S1, Tree AS T1
WHERE S1.emp = T1.boss
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.emp), counter, NULL
FROM Stack AS S1, Tree AS T1
WHERE S1.emp = T1.boss
AND S1.stack_top = current_top;

DELETE FROM Tree
WHERE emp = (SELECT emp
FROM Stack
WHERE stack_top = current_top + 1);

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 IF;
END LOOP;
END;

This approach will be two to three orders of magnitude faster than the
adjacency list model for subtree and aggregate operations.

For details, see the chapter in my book JOE CELKO'S SQL FOR SMARTIES
(Morgan-Kaufmann, 1999, second edition)

--CELKO--
===========================
Please post DDL, so that people do not have to guess what the keys,
constraints, Declarative Referential Integrity, datatypes, etc. in your
schema are.

--
HTH,
David Satz
Principal Web Engineer
Hyperion Solutions
{ SQL Server 2000 SP2/6.5 SP5a } { Cold Fusion 5 SP1 } { VSS }
-----------------------------------------------------------------
"Steven" <sgebbie@sglan.freeserve.co.ul> wrote in message
news:3d124d87\$1@10.1.10.29...
>
> I really need some help here.
>
> I have a table [family]:parent_name, my_name
>
> and I want to pass 'adam' as a parameter and get the number of family

members
> down 3 generations
>
> the data would be:
> Michael, Lucifer
> John, Matthew
> Matthew, David
> David, Peter
> Peter, Paul
> Paul, Wendy
>
> so 'Adam' has 5 members (3 levels deep,

john&michael&Lucifer&Matthew&David),
> David and his offspring are ignored.
>
> and cannot understand 'books online' so am pretty much completly stuck.
> If anyone can help I would really appreciate it