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Re: Recursive Query

From: --CELKO-- <71062.1056_at_compuserve.com>
Date: 19 Mar 2003 17:31:42 -0800
Message-ID: <c0d87ec0.0303191731.1401afd9@posting.google.com>


>> I have a table with an Id that references itself recursively and need help building the query. <<

The usual example of a tree structure in SQL books 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, 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 emps are adjacent to each other.

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 first table is denormalized in several ways. We are modeling both the OrgChart 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 OrgChart 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 OrgChart), 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 emps with ovals, then nest subordinate ovals inside each other. The root will be the largest oval and will contain every other emp. The leaf emps 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 emp, 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 emp 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 emp, he puts a number in the cell on the side that he is visiting and increments his counter. Each emp 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 emps 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 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 theadjacency 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

 INSERT INTO Frammis (part, lft, rgt)
 VALUES ('G1', rightmost_spread, (rightmost_spread + 1));  COMMIT WORK;
END; The idea is to spread the lft and 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, P.emp
  FROM OrgChart AS P

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

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

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;

WHILE counter <= max_counter- 1
 IF EXISTS (SELECT *

              FROM Stack AS S1, Tree AS T1
             WHERE S1.node = T1.parent
               AND S1.stack_top = current_top)
     
     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;
-- the top column is not needed in the final answer SELECT node, lft, rgt FROM Stack;
END; I have a book on TREES & HIERARCHIES IN SQL coming out in 2003. Received on Wed Mar 19 2003 - 19:31:42 CST

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