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# Pascal's triangle or binomial coefficients

• Should this work for all values of :n?
interesting. Padders version works for me (allowing for the fatc that the CONNECT BY trick doesn't work propperly in 9.2)...
```SQL> EXEC :n := 10
PL/SQL procedure successfully completed.
SQL> SELECT * FROM
2  (
3   SELECT EXP (SUM (LN (LEVEL)) OVER ()) / (:n + 1) / (
4              EXP (SUM (LN (GREATEST (LEVEL - 1, 1)))
5                 OVER (ORDER BY LEVEL)) *
6              EXP (SUM (LN (GREATEST (:n - (LEVEL - 1), 1)))
7                 OVER (ORDER BY LEVEL DESC))) n
8    FROM   dual
9    CONNECT BY LEVEL <= CEIL (:n + 1))
10  /
N
----------
1
10
45
120
210
252
210
120
45
10
1
11 rows selected.
SQL> ```
Cheers, APC
• interesting. Padders version works for me (allowing for the fatc that the CONNECT BY trick doesn't work propperly in 9.2)...
There's an easy workaround for that (I'm sure you're well aware, but just for those who don't know yet):
```SELECT EXP (SUM (LN (L)) OVER ()) / (:n + 1) / (
EXP (SUM (LN (GREATEST (L - 1, 1)))
OVER (ORDER BY L)) *
EXP (SUM (LN (GREATEST (:n - (L - 1), 1)))
OVER (ORDER BY L DESC))) n
FROM   ( SELECT LEVEL L
FROM DUAL
CONNECT BY LEVEL <= CEIL (:n + 1)
)
/```
It works perfect on my 9.2 box.

MHE
• I'm using 9.2.0.6.0... so there's a flaw with connect by in this version thats affecting the output of the query sometimes? When I use values 0 through 9 and 11 it appears to return the correct values, but for whatever reason it doesn't work when :n equals 10.

Can anyone explain how the query works, not what each function does individually, but why when you put them together they return pascal's triangle?
• I'm using 9.2.0.6.0
Me likewise
Can anyone explain how the query works
basically the query is calculating the value of each element in the row; it implements the nCr method using analytic functions to generate factorials.

Cheers, APC
• Thanks for your reply APC. I understand the formula (I accept that it works although I have no idea why it works). But how do the functions that padders uses perform the same functionality that a factorial (!) does?
• The following are all the same. This is a (sort of) common trick for implementing factorial.
```select exp(sum(ln(level)))
from dual
connect by level <= N

exp(ln(1) + ln(2) + ... ln(N))

exp(ln(1)) * exp(ln(2)) * ... exp(ln(N))

1*2*...N

N!```
• select x,y,count(1) from (
select x,y,sys_connect_by_path('['||x||','||y||'>',' ') path
from (
select level x from dual
connect by level < 5
), (
select level y from dual
connect by level < 5
) connect by prior x+1 = x and prior y = y
or prior x = x and prior y+1 = y
) group by x,y;

Nice exercise to include into a book.
• Thanks Scott, I must have missed that day in my SQL class.

and my solution is
```SELECT       EXP (SUM (LN (GREATEST (LEVEL - 1, 1))) OVER ())
/ EXP (SUM (LN (GREATEST (LEVEL - 1, 1))) OVER (ORDER BY LEVEL))
/ EXP (SUM (LN (GREATEST (:n - LEVEL, 1))) OVER (ORDER BY LEVEL DESC)
) bin_coef
FROM DUAL
CONNECT BY LEVEL <= :n```
Query Your Dream & Future at
http://www.soqool.com
• Just a couple of variations on the theme using a table function as a row source and a user-defined product aggregate (source available on request).
```Oracle Database 10g Enterprise Edition Release 10.2.0.1.0 - Production
With the Partitioning, OLAP and Data Mining options

SQL> VARIABLE r NUMBER;
SQL> EXEC :r := 5;

PL/SQL procedure successfully completed.

SQL> SELECT product (GREATEST (n, 1)) OVER () /
2         product (GREATEST (n, 1)) OVER (ORDER BY n) /
3         product (GREATEST (:r - n, 1)) OVER (ORDER BY n DESC) n
4  FROM   TABLE (many (0, :r));

N
----------
1
5
10
10
5
1

6 rows selected.

SQL> SELECT SUBSTR (SYS_CONNECT_BY_PATH (
2            r, ','), 2) pascals_triangle
3  FROM  (SELECT a.n an, b.n bn,
4                product (GREATEST (b.n, 1)) OVER (
5                   PARTITION BY a.n) /
6                product (GREATEST (b.n, 1)) OVER (
7                   PARTITION BY a.n ORDER BY b.n) /
8                product (GREATEST (a.n - b.n, 1)) OVER (
9                   PARTITION BY a.n ORDER BY b.n DESC) r
10         FROM   TABLE (many (0, :r)) a,
11                TABLE (many (0, a.n)) b)
12  WHERE  CONNECT_BY_ISLEAF = 1
14  CONNECT BY an = PRIOR an AND bn = PRIOR bn + 1;

PASCALS_TRIANGLE
--------------------------------------------------------------------------------
1
1,1
1,2,1
1,3,3,1
1,4,6,4,1
1,5,10,10,5,1

6 rows selected.

SQL> ```
This discussion has been closed.