Re: double linked list

--CELKO-- wrote:

> >> When you are ready to demonstrate a superior implementation in
> another SQL
> RDBMS please let us know. And I don't mean superior in terms of some
> theoretical concept like building a database in fifth normal form. I
> mean superior in terms of performance and scalability in the real
> world. <<
>
> Funny you should ask :)!! I have a book coming out later this year on
> TREES & HIERARCHIES IN SQL (Morgan-Kaufmann), which gives several
> different models in Standard SQL-92 for tree structures which are
> superior in terms of performance and scalability compared to the
> CONNECT BY.
>
> A few years back, we ran a test on a huge tree with the following
> method versus cursors and CONNECT BY; it was 20 to 100 times faster
> for 75,000 nodes and 12 levels deep. It could also do things like
> compare the structures of sub-trees in one SELECT statement.
>
> 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
> = (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 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.
>
> -- 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);
>
> 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
> 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 with
> more code and details.

Thanks for your valiant attempt to sell your book. As I have you on my recommended reading list of authors for my students at the University of Washington no doubt I'll buy a copy if you don't get too terribly offensive.

But unless you are going to state that the above can NOT be done in Oracle ... the fact that CONNECT BY exists is not license to call the product, as you did, 'horrible'.

So are you complaining because there is a less superior method available for those not as technically proficient as you, are you complaining because that less superior implementation doesn't exist in Transact SQL, or are you complaining just because you don't personally work in Oracle? Seems to me the posting you did, that started this thread, was 99% hyperbole and that perhaps a restatement would be appropriate without editorializing would be appropriate.

Daniel Morgan Received on Mon Feb 03 2003 - 16:19:08 CST