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Pascal's triangle or binomial coefficients
Comments

interesting. Padders version works for me (allowing for the fatc that the CONNECT BY trick doesn't work propperly in 9.2)...Should this work for all values of :n?
SQL> EXEC :n := 10
Cheers, APC
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> 
There's an easy workaround for that (I'm sure you're well aware, but just for those who don't know yet):interesting. Padders version works for me (allowing for the fatc that the CONNECT BY trick doesn't work propperly in 9.2)...
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? 
Me likewiseI'm using 9.2.0.6.0basically the query is calculating the value of each element in the row; it implements the nCr method using analytic functions to generate factorials.Can anyone explain how the query works
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
start with x = 1 and y = 1
) group by x,y;
Nice exercise to include into a book. 
Thanks Scott, I must have missed that day in my SQL class.

good job, padders.
and my solution isSELECT 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 userdefined 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 13 START WITH bn = 0 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.