Any better Fibonacci series generator using pure Oracle SQL?

问题

I wonder if there is any way to generate Fibonacci numbers that beat in simplicity and efficiency this one I wrote:

WITH d (seq) AS
       (SELECT     LEVEL
        FROM       DUAL
        CONNECT BY LEVEL < 195)
SELECT   seq
        ,fib
FROM     d
MODEL
  DIMENSION BY(seq)
  MEASURES(0 AS fib)
  RULES
    (fib [1] = 0,
    fib [2] = 1,
    fib [seq BETWEEN 3 AND 194] = fib[CV(seq) - 2] + fib[CV(seq) - 1],
    fib [seq > 194] = NULL)
ORDER BY 1
/
Execution Plan
----------------------------------------------------------
Plan hash value: 2245903385

---------------------------------------------------------------------------------------
| Id  | Operation                      | Name | Rows  | Bytes | Cost (%CPU)| Time     |
---------------------------------------------------------------------------------------
|   0 | SELECT STATEMENT               |      |     1 |    13 |     2   (0)| 00:00:01 |
|   1 |  SQL MODEL ORDERED             |      |     1 |    13 |            |          |
|   2 |   VIEW                         |      |     1 |    13 |     2   (0)| 00:00:01 |
|*  3 |    CONNECT BY WITHOUT FILTERING|      |       |       |            |          |
|   4 |     FAST DUAL                  |      |     1 |       |     2   (0)| 00:00:01 |
---------------------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   3 - filter(LEVEL<195)


Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          0  consistent gets
          0  physical reads
          0  redo size
       4798  bytes sent via SQL*Net to client
        500  bytes received via SQL*Net from client
         14  SQL*Net roundtrips to/from client
          1  sorts (memory)
          0  sorts (disk)
        194  rows processed

SQL>

Note: LEVEL < 195 was not arbitrarily chosen, higher values make the algorithm lose precision so I decided not to include them in order to keep correct results only.

回答1:

on the simplicity side of things, the query can rely on the built in features (ITERATE () and ITERATION_NUMBER) of MODEL:

select * from dual
model
  dimension by (0 seq)
  measures (0 val)
  rules iterate (195) 
  (
     val[iteration_number] = val[iteration_number-1] + val[iteration_number-2],
     val[2] = 1, 
     val[1] = 0, 
     val[0] = 0
  )
;

回答2:

You can use a recursive sub-query factoring clause:

WITH fib ( lvl, value, next ) AS (
  SELECT 1, 0, 1
  FROM DUAL
UNION ALL
  SELECT lvl + 1, next, value + next
  FROM fib
  WHERE lvl < 195
)
SELECT lvl, value FROM fib

回答3:

Something like this should be (much?) faster:

with
     constants ( x, y, z ) as (
       select 0.5 * ( 1 + sqrt(5) ),
              0.5 * ( 1 - sqrt(5) ),
              sqrt(5)
       from   dual
     )
select level as seq, round( ( power(x, level - 1) - power(y, level - 1) ) / z ) as fib
from   constants
connect by level < 195
;

The point being, you don't need to use the recursive formula; the terms can be written in closed form. Since computers can't do arithmetic with real numbers, only with rational number approximations, I needed to add a ROUND(...) but even so this should be faster than recursive approaches.

EDIT: At the OP's request I traced the execution of this code. I don't see the recursive calls the OP is referring to in the Comment below.

Execution Plan
----------------------------------------------------------
Plan hash value: 1236776825

-----------------------------------------------------------------------------
| Id  | Operation                    | Name | Rows  | Cost (%CPU)| Time     |
-----------------------------------------------------------------------------
|   0 | SELECT STATEMENT             |      |     1 |     2   (0)| 00:00:01 |
|*  1 |  CONNECT BY WITHOUT FILTERING|      |       |            |          |
|   2 |   FAST DUAL                  |      |     1 |     2   (0)| 00:00:01 |
-----------------------------------------------------------------------------

Predicate Information (identified by operation id):
---------------------------------------------------

   1 - filter(LEVEL<195)


Statistics
----------------------------------------------------------
          0  recursive calls
          0  db block gets
          0  consistent gets
          0  physical reads
          0  redo size
       6306  bytes sent via SQL*Net to client
        684  bytes received via SQL*Net from client
         14  SQL*Net roundtrips to/from client
          1  sorts (memory)
          0  sorts (disk)
        194  rows processed

EDIT #2

I suspect the simple generation of levels in a recursive query may be expensive. It's possible that a cross-join of similarly generated, but smaller sequences of integers may work a bit faster. The code looks more complicated (of course); the only change, though, is the way I generate the powers.

with
     constants ( x, y, z ) as (
       select 0.5 * ( 1 + sqrt(5) ),
              0.5 * ( 1 - sqrt(5) ),
              sqrt(5)
       from   dual
     ),
     powers ( n ) as (
       select 14 * a.p + b.q
       from   (select level - 1 p from dual connect by level <= 14) a
              cross join
              (select level - 1 q from dual connect by level <= 14) b
     )
select n + 1 as seq, round( ( power(x, n) - power(y, n) ) / z ) as fib
from   constants cross join powers
where  n < 195
;

来源：https://stackoverflow.com/questions/44415801/any-better-fibonacci-series-generator-using-pure-oracle-sql

标签

sql

Oracle

algorithm

optimization

fibonacci