pg.ddx.io pgsql-sql@postgresql.org mailing list archive
help / color / mirror / Atom feeddouble linked list
11+ messages / 7 participants
[nested] [flat]
* double linked list
@ 2003-01-28 15:44 Juergen <jasche@gmx.de>
2003-01-28 16:45 ` Re: double linked list DA Morgan <damorgan@exesolutions.com>
2003-01-28 23:54 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
0 siblings, 2 replies; 11+ messages in thread
From: jasche@gmx.de @ 2003-01-28 15:44 UTC (permalink / raw)
To: pgsql-sql
Hi folks!,
I've got a table called 'link_t' containing a collection of seller -
buyer relations between two parties.
sql> select * from link_t
S B
- -
C X
A B
B C
C D
D E
5 rows selected.
I am looking for a select statement that returns the concatenation of
seller - buyer relations between the first seller 'A' and the last
buyer 'B'.
the result should be
S B
- -
A B
B C
C D
D E
Currently I fumbling around with self joins but haven't figured out
yet.
Any suggestions
Many tanks in advance
Cheers
Juergen
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
@ 2003-01-28 16:45 ` DA Morgan <damorgan@exesolutions.com>
1 sibling, 0 replies; 11+ messages in thread
From: DA Morgan @ 2003-01-28 16:45 UTC (permalink / raw)
To: pgsql-sql
Juergen wrote:
> Hi folks!,
>
> I've got a table called 'link_t' containing a collection of seller -
> buyer relations between two parties.
>
> sql> select * from link_t
>
> S B
> - -
> C X
> A B
> B C
> C D
> D E
>
> 5 rows selected.
>
> I am looking for a select statement that returns the concatenation of
> seller - buyer relations between the first seller 'A' and the last
> buyer 'B'.
>
> the result should be
>
> S B
> - -
> A B
> B C
> C D
> D E
>
> Currently I fumbling around with self joins but haven't figured out
> yet.
> Any suggestions
>
> Many tanks in advance
>
> Cheers
>
> Juergen
Look at the CONNECT BY built-in function.
Daniel Morgan
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
@ 2003-01-28 23:54 ` --CELKO-- <71062.1056@compuserve.com>
2003-01-29 12:12 ` Re: double linked list Juergen <jasche@gmx.de>
1 sibling, 1 reply; 11+ messages in thread
From: 71062.1056@compuserve.com @ 2003-01-28 23:54 UTC (permalink / raw)
To: pgsql-sql
>> I've got a table called 'link_t' containing a collection of seller
-
buyer relations between two parties. <<
That is not a real linked list, but let's ignore bad terminology. One
way to do this is with cursors, but they will take time and trend to
be proprietary.
Anohter way is to build a tree, with the first seller as the root and
the final buyer as a leaf node.
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, but in
the meantime, you can look at:
http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci537290,00.html
http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci801943,00.html
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
2003-01-28 23:54 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
@ 2003-01-29 12:12 ` Juergen <jasche@gmx.de>
2003-01-29 13:48 ` Re: double linked list Jan Hidders <hidders@REMOVE.THIS.uia.ua.ac.be>
2003-01-29 21:15 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
0 siblings, 2 replies; 11+ messages in thread
From: jasche@gmx.de @ 2003-01-29 12:12 UTC (permalink / raw)
To: pgsql-sql
heavy stuff Celko. I would lie if I would pretend I fully understand
Your answer. I'll let sink it in.
However, I dont store a consistent tree structure. The table at hand
is more a kind of a collection of graphs where I want to find all
possible paths between a given starting point and a given end point
cheers
Juergen
71062.1056@compuserve.com (--CELKO--) wrote in message news:<c0d87ec0.0301281554.581797ee@posting.google.com>...
> >> I've got a table called 'link_t' containing a collection of seller
> -
> buyer relations between two parties. <<
>
> That is not a real linked list, but let's ignore bad terminology. One
> way to do this is with cursors, but they will take time and trend to
> be proprietary.
>
> Anohter way is to build a tree, with the first seller as the root and
> the final buyer as a leaf node.
>
> 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, but in
> the meantime, you can look at:
>
> http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci537290,00.html
>
> http://searchdatabase.techtarget.com/tip/1,289483,sid13_gci801943,00.html
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
2003-01-28 23:54 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
2003-01-29 12:12 ` Re: double linked list Juergen <jasche@gmx.de>
@ 2003-01-29 13:48 ` Jan Hidders <hidders@REMOVE.THIS.uia.ua.ac.be>
1 sibling, 0 replies; 11+ messages in thread
From: hidders@REMOVE.THIS.uia.ua.ac.be @ 2003-01-29 13:48 UTC (permalink / raw)
To: pgsql-sql
Juergen wrote:
>
>However, I dont store a consistent tree structure. The table at hand
>is more a kind of a collection of graphs where I want to find all
>possible paths between a given starting point and a given end point
A collection of graphs? As you presented the problem it was simply a single
graph.
That's not possible in a single SQL statement without using some form of
recursion such as the CONNECT BY in Oracle that was already mentioned or the
recursive queries as are possible in RDB. Another "poor man's solution" could for
example be to add a table Reachable(node, from_a, to_b) with 'from_a' and
'from_b' boolean field that indicate that the node is reachable from a and
that b is reachable from this node. You could compute this relation by
repeating a certain SQL update statement that:
1. sets from_a of node n to true if there is a node n' that is reachable
from a and there is an edge from n' to n, and
2. set to_b of node n to tur if there is a node n' that leads to b and there
is an edge from n to n'.
You repeat that until no more flags are changed. Then you select only those
edges for which the begin and node have both flags set to true.
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
2003-01-28 23:54 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
2003-01-29 12:12 ` Re: double linked list Juergen <jasche@gmx.de>
@ 2003-01-29 21:15 ` --CELKO-- <71062.1056@compuserve.com>
2003-01-29 21:30 ` Re: double linked list Ryan <rgaffuri@cox.net>
1 sibling, 1 reply; 11+ messages in thread
From: 71062.1056@compuserve.com @ 2003-01-29 21:15 UTC (permalink / raw)
To: pgsql-sql
>> The table at hand is more a kind of a collection of graphs where I
want to find all possible paths between a given starting point and a
given end point. <<
For the reachabiity index of a general graph, you need Warshal's
algorithm.
Let V = number of nodes in the graph
Let A[i,j] be the adjacency matrix for the undirected graph
FOR j:= 1 TO V
DO FOR i:= 1 TO V
DO IF A[i,j] = 1
THEN FOR k := 1 TO V
DO IF A[j,k]] = 1
THEN A[i,k]] := 1;
You can also do a summation to get the length of the path from i to j.
You can concatenate names of the nodes into a string that gives the
path, etc.
Her is a first attempt at some SQL; I am sure it can be done better
CREATE TABLE Graph
(i CHAR(2) NOT NULL,
j CHAR(2) NOT NULL,
flag CHAR(1) NOT NULL DEFAULT 'n'
CHECK (flag IN ('n', 'y')),
PRIMARY KEY (i,j));
INSERT INTO Graph (i, j, flag)
SELECT DISTINCT G1.i, G2.j, 'y'
FROM Graph AS G1, Graph AS G1
WHERE G1.i <> G2.j
AND EXISTS
(SELECT *
FROM Graph AS G3
WHERE G3.i = G1.j
AND G3.j = G2.i)
AND NOT EXISTS
(SELECT *
FROM Graph AS G3
WHERE (G3.i = G1.i AND G3.j = G1.j))
OR (G3.i = G2.i AND G3.j = G2.j));
You wll have to run this statement until the size of Graph does not
change -- no new rows are being added.
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
2003-01-28 23:54 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
2003-01-29 12:12 ` Re: double linked list Juergen <jasche@gmx.de>
2003-01-29 21:15 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
@ 2003-01-29 21:30 ` Ryan <rgaffuri@cox.net>
2003-01-30 19:13 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
0 siblings, 1 reply; 11+ messages in thread
From: Ryan @ 2003-01-29 21:30 UTC (permalink / raw)
To: pgsql-sql
are you joe celko, guy who wrote those sql books?
"--CELKO--" <71062.1056@compuserve.com> wrote in message
news:c0d87ec0.0301291315.7ae946eb@posting.google.com...
> >> The table at hand is more a kind of a collection of graphs where I
> want to find all possible paths between a given starting point and a
> given end point. <<
>
> For the reachabiity index of a general graph, you need Warshal's
> algorithm.
>
> Let V = number of nodes in the graph
> Let A[i,j] be the adjacency matrix for the undirected graph
>
> FOR j:= 1 TO V
> DO FOR i:= 1 TO V
> DO IF A[i,j] = 1
> THEN FOR k := 1 TO V
> DO IF A[j,k]] = 1
> THEN A[i,k]] := 1;
>
> You can also do a summation to get the length of the path from i to j.
> You can concatenate names of the nodes into a string that gives the
> path, etc.
>
> Her is a first attempt at some SQL; I am sure it can be done better
>
> CREATE TABLE Graph
> (i CHAR(2) NOT NULL,
> j CHAR(2) NOT NULL,
> flag CHAR(1) NOT NULL DEFAULT 'n'
> CHECK (flag IN ('n', 'y')),
> PRIMARY KEY (i,j));
>
> INSERT INTO Graph (i, j, flag)
> SELECT DISTINCT G1.i, G2.j, 'y'
> FROM Graph AS G1, Graph AS G1
> WHERE G1.i <> G2.j
> AND EXISTS
> (SELECT *
> FROM Graph AS G3
> WHERE G3.i = G1.j
> AND G3.j = G2.i)
> AND NOT EXISTS
> (SELECT *
> FROM Graph AS G3
> WHERE (G3.i = G1.i AND G3.j = G1.j))
> OR (G3.i = G2.i AND G3.j = G2.j));
>
> You wll have to run this statement until the size of Graph does not
> change -- no new rows are being added.
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
2003-01-28 23:54 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
2003-01-29 12:12 ` Re: double linked list Juergen <jasche@gmx.de>
2003-01-29 21:15 ` Re: double linked list --CELKO-- <71062.1056@compuserve.com>
2003-01-29 21:30 ` Re: double linked list Ryan <rgaffuri@cox.net>
@ 2003-01-30 19:13 ` --CELKO-- <71062.1056@compuserve.com>
0 siblings, 0 replies; 11+ messages in thread
From: 71062.1056@compuserve.com @ 2003-01-30 19:13 UTC (permalink / raw)
To: pgsql-sql
>> are you joe celko, guy who wrote those sql books? <<
Yes.
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
@ 2003-01-30 13:10 Christoph Haller <ch@rodos.fzk.de>
2003-01-30 14:30 ` Re: double linked list Alan Gutierrez <ajglist@izzy.net>
0 siblings, 1 reply; 11+ messages in thread
From: Christoph Haller @ 2003-01-30 13:10 UTC (permalink / raw)
To: pgsql-sql; +Cc: 71062.1056@compuserve.com
I've seen CELKO's reply and find it very useful.
But I cannot find anything about
> BEGIN ATOMIC
> DECLARE rightmost_spread INTEGER;
>
> SET rightmost_spread
> = (SELECT rgt
> FROM Frammis
> WHERE part = 'G');
> ...
Is this PostgreSQL at all? Any hints welcome.
Regards, Christoph
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
2003-01-30 13:10 Re: double linked list Christoph Haller <ch@rodos.fzk.de>
@ 2003-01-30 14:30 ` Alan Gutierrez <ajglist@izzy.net>
0 siblings, 0 replies; 11+ messages in thread
From: Alan Gutierrez @ 2003-01-30 14:30 UTC (permalink / raw)
To: Christoph Haller <ch@rodos.fzk.de>; pgsql-sql
On Thursday 30 January 2003 07:10, Christoph Haller wrote:
> I've seen CELKO's reply and find it very useful.
> But I cannot find anything about
>
> > BEGIN ATOMIC
> > DECLARE rightmost_spread INTEGER;
> >
> > SET rightmost_spread
> > = (SELECT rgt
> > FROM Frammis
> > WHERE part = 'G');
> > ...
>
> Is this PostgreSQL at all? Any hints welcome.
Mr Haller
No, this is a dialect SQL-92 (SQL-99?) that Mr Celko uses for his
examples since his solutions are vendor nutral. He is big on standards,
so posting using the standard is his way of boosting them.
BEGIN ATOMIC is BEGIN in PG.
I am not sure how to declare a variable in PG in normal SQL. I don't do
it that often, but when I do I do this:
CREATE TEMPORARY TABLE Rightmost_Spread
AS SELECT rightmost_spread
FROM Frammis
WHERE part = 'G';
I wonder what the alterntatives are?
Alan Gutierrez - ajglist@izzy.net
http://khtml-win32.sourceforge.net/ - KHTML on Windows
^ permalink raw reply [nested|flat] 11+ messages in thread
* Re: double linked list
@ 2003-01-30 14:26 Christoph Haller <ch@rodos.fzk.de>
0 siblings, 0 replies; 11+ messages in thread
From: Christoph Haller @ 2003-01-30 14:26 UTC (permalink / raw)
To: pgsql-sql; +Cc: ajglist@izzy.net
>
> On Thursday 30 January 2003 07:10, Christoph Haller wrote:
> > I've seen CELKO's reply and find it very useful.
> > But I cannot find anything about
> >
> > > BEGIN ATOMIC
> > > DECLARE rightmost_spread INTEGER;
> > >
> > > SET rightmost_spread
> > > =3D (SELECT rgt
> > > FROM Frammis
> > > WHERE part = 'G');
> > > ...
> >
> > Is this PostgreSQL at all? Any hints welcome.
>
> Mr Haller
>
> No, this is a dialect SQL-92 (SQL-99?) that Mr Celko uses for his
> examples since his solutions are vendor neutral. He is big on
standards,
> so posting using the standard is his way of boosting them.
>
> BEGIN ATOMIC is BEGIN in PG.
>
> I am not sure how to declare a variable in PG in normal SQL. I don't
do
> it that often, but when I do I do this:
>
> CREATE TEMPORARY TABLE Rightmost_Spread
> AS SELECT rightmost_spread
> FROM Frammis
> WHERE part = 'G';
>
> I wonder what the alternatives are?
>
plpgsql is the best I can think of.
And thanks for the quick reply.
Regards, Christoph
^ permalink raw reply [nested|flat] 11+ messages in thread
end of thread, other threads:[~2003-01-30 19:13 UTC | newest]
Thread overview: 11+ messages (download: mbox mbox.gz follow: Atom feed)
-- links below jump to the message on this page --
2003-01-28 15:44 double linked list Juergen <jasche@gmx.de>
2003-01-28 16:45 ` DA Morgan <damorgan@exesolutions.com>
2003-01-28 23:54 ` --CELKO-- <71062.1056@compuserve.com>
2003-01-29 12:12 ` Juergen <jasche@gmx.de>
2003-01-29 13:48 ` Jan Hidders <hidders@REMOVE.THIS.uia.ua.ac.be>
2003-01-29 21:15 ` --CELKO-- <71062.1056@compuserve.com>
2003-01-29 21:30 ` Ryan <rgaffuri@cox.net>
2003-01-30 19:13 ` --CELKO-- <71062.1056@compuserve.com>
2003-01-30 13:10 Re: double linked list Christoph Haller <ch@rodos.fzk.de>
2003-01-30 14:30 ` Alan Gutierrez <ajglist@izzy.net>
2003-01-30 14:26 Re: double linked list Christoph Haller <ch@rodos.fzk.de>
This inbox is served by DDX for PostgreSQL; see mirroring instructions
for how to clone and mirror all data and code used for this inbox