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Re: SQL competition: the shortest interval coalesce/pack query

From: Lennart Jonsson <lennart_at_kommunicera.umea.se>
Date: 4 Dec 2004 07:42:15 -0800
Message-ID: <6dae7e65.0412040742.572da937@posting.google.com>


"John Gilson" <jag_at_acm.org> wrote in message news:<ujesd.60193$Vk6.30697_at_twister.nyc.rr.com>...
> "Lennart Jonsson" <lennart_at_kommunicera.umea.se> wrote in message
> news:6dae7e65.0412031821.4df452c6_at_posting.google.com...
> > lennart_at_kommunicera.umea.se (Lennart Jonsson) wrote in message
> news:<6dae7e65.0412031228.3051dff_at_posting.google.com>...
> > > "Mikito Harakiri" <mikharakiri_at_iahu.com> wrote in message
> news:<2uPrd.61$G45.36_at_news.oracle.com>...
> >
> > [...]
> > >

[...]
>
> Lennart, as you've noted, your query does exhibit problems. For example,
> starting with the sample interval set, if you remove (20,21) so that you no
> longer have a gap, that is,
> INSERT INTO Intervals (x,y)
> VALUES (1,3),(2,5),(4,11),(10,12);
> then the query returns
> 1 11
> 2 12
> Also, if we add an interval to the interval set that's contiguous with say
> (20,21), e.g.,
> INSERT INTO Intervals (x,y)
> VALUES (1,3),(2,5),(4,11),(10,12),(20,21),(21,25);
> then the query returns
> 1 12
> 20 21
> 21 25
>
> However, using recursion is an interesting take on this so here's an
> offering:
>
> CREATE TABLE Intervals
> (
> x INT NOT NULL,
> y INT NOT NULL,
> CHECK (y > x),
> PRIMARY KEY (x,y)
> );
>
> WITH Cover (x,y,n)
> AS (SELECT x,y,
> (SELECT COUNT(*) FROM Intervals)
> FROM Intervals
> UNION ALL
> SELECT CASE WHEN C.x <= I.x THEN C.x ELSE I.x END,
> CASE WHEN C.y >= I.y THEN C.y ELSE I.y END,
> C.n - 1
> FROM Intervals AS I, Cover AS C
> WHERE I.x <= C.y AND
> I.y >= C.x AND
> (I.x <> C.x OR I.y <> C.y) AND
> C.n > 1)
> SELECT DISTINCT C1.x, C1.y
> FROM Cover AS C1
> WHERE NOT EXISTS (SELECT *
> FROM Cover AS C2
> WHERE C2.x <= C1.x AND
> C2.y >= C1.y AND
> (C1.x <> C2.x OR C1.y <> C2.y))
> ORDER BY C1.x;

Hi John. Very interesting and I will have a look at it later.

In fact I hadnt started worrying about the correctness of my attempt, but merely the performance. Because of the nature of the problem the cover set grows very fast. I took the liberty to take you code for a run

select * from intervals

X Y
----------- -----------

          1           3
          2           5
          4          11
         10          12
         20          21
          4          16
          5          17
          6          18

WITH Cover (x,y,n)
AS (SELECT x,y,

                      (SELECT COUNT(*) FROM Intervals)
       FROM Intervals
       UNION ALL
       SELECT CASE WHEN C.x <= I.x THEN C.x ELSE I.x END,
                      CASE WHEN C.y >= I.y THEN C.y ELSE I.y END,
                      C.n - 1
       FROM Intervals AS I, Cover AS C
       WHERE I.x <= C.y AND
                      I.y >= C.x AND
                      (I.x <> C.x OR I.y <> C.y) AND
                      C.n > 1)

SELECT count(C1.x)
FROM Cover AS C1;

1



SQL0347W The recursive common table expression "LELLE.COVER" may contain an
infinite loop. SQLSTATE=01605

    2379720

But your query terminates very quickly so perhaps the optimizer is smart enough to push the predicates inside the cover clause?

/Lennart Received on Sat Dec 04 2004 - 09:42:15 CST

Original text of this message

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