Comments on: OR aggregate

By: Laurent Schneider

Laurent Schneider — Thu, 06 Dec 2007 11:15:22 +0000

wow! this is exactly the same, I like this

arigato!

By: Aketi Jyuuzou

Aketi Jyuuzou — Thu, 06 Dec 2007 10:59:09 +0000

http://oraclesqlpuzzle.hp.infoseek.co.jp/8-20.html
My site mentions these solutions (written by Japanese language)
Thank you

By: Laurent Schneider

Laurent Schneider — Mon, 03 Dec 2007 09:34:02 +0000

for AGGREGATE_BITXOR, I would then use :

select
mod(sum(sign(bitand(32,n))),2)*32+
mod(sum(sign(bitand(16,n))),2)*16+
mod(sum(sign(bitand( 8,n))),2)* 8+
mod(sum(sign(bitand( 4,n))),2)* 4+
mod(sum(sign(bitand( 2,n))),2)* 2+
mod(sum(sign(bitand( 1,n))),2) n
from t;

By: Laurent Schneider

Laurent Schneider — Mon, 03 Dec 2007 09:27:42 +0000

and for AGGREGATE_BITAND, we could use MIN then!

By: Laurent Schneider

Laurent Schneider — Mon, 03 Dec 2007 09:20:38 +0000

this makes sense!

By: Aketi Jyuuzou

Aketi Jyuuzou — Mon, 03 Dec 2007 08:54:53 +0000

oops
We can use solution which is more simple.

with t as (
select 10 n from dual union all
select 12 from dual union all
select 6 from dual)
select
max(bitand(32,n))+
max(bitand(16,n))+
max(bitand( 8,n))+
max(bitand( 4,n))+
max(bitand( 2,n))+
max(bitand( 1,n)) as N
from t;

By: Laurent Schneider

Laurent Schneider — Mon, 03 Dec 2007 08:14:54 +0000

Nice approach, I wished I found it myself

By: Aketi Jyuuzou

Aketi Jyuuzou — Mon, 03 Dec 2007 07:04:28 +0000

Oh
Good solution!

I post homeMadeBitOr

with t as (
select 10 n from dual union all
select 12 from dual union all
select 6 from dual)
select
sum(distinct(bitand(32,n)))+
sum(distinct(bitand(16,n)))+
sum(distinct(bitand( 8,n)))+
sum(distinct(bitand( 4,n)))+
sum(distinct(bitand( 2,n)))+
sum(distinct(bitand( 1,n))) as N
from t;

By: Rob van Wijk

Rob van Wijk — Sat, 24 Nov 2007 07:44:54 +0000

Good points.

Although the preconditions were not mentioned, your solution handles them and is therefore more robust.

Regards,
Rob.

By: Laurent Schneider

Laurent Schneider — Fri, 23 Nov 2007 16:06:35 +0000

well, if the table contains only one row, it will return null and if the third row is a NULL, it will exit too early (therefore UNTIL PRESENTV)


with mytable as (
  select 1 p, 63 n from dual union all
  select 2,1 from dual union all
  select 2,1 from dual union all
  select 2,null from dual union all
  select 2,1 from dual)
select p,n
from mytable
model
return updated rows
partition by (p)
dimension by (row_number() over (partition by p order by null) rn)
measures (n)
rules iterate (10000)
until (presentv(n[ITERATION_NUMBER+3],1,2)=2)
(
  n[1] = n[1] + nvl(n[ITERATION_NUMBER+2],0) -
  bitand(n[1],nvl(n[ITERATION_NUMBER+2],0))
)
/

         P          N
---------- ----------
         1         63
         2          1