OTN Discussion Forums : SQL puzzle :) ...

antu

Posts: 285
Registered: 09/16/07

SQL puzzle :)
Posted: Jun 8, 2008 3:47 AM

I want to find all possible X up to the most highest number possbile with a single SQL.

X - Y = T + Z

where

X as a positive number,
Y is the reverse of X,
T is the sum of each number which X has
Z is the product of each number which X has

like in this example;

63 - 36 = 9 + 18

- Which is the most efficient way of producing for example numbers between 1 and 1000000000000, CONNECT BY from DUAL?
- Is there an alternative way to reverse a number other than undocumented REVERSE function to_number(reverse(to_char(anumber)) this way there will be three function calls for each number,
- What is the best way to break a number into pieces to sum or make product in SQL.

Thank you very much

wateenmooiedag

Posts: 476
Registered: 01/28/06

Re: SQL puzzle :)
Posted: Jun 8, 2008 4:13 AM

in response to: antu

Reply

Is calling user defined PL/SQL functions allowed or not?

antu

Posts: 285
Registered: 09/16/07

Re: SQL puzzle :)
Posted: Jun 8, 2008 4:28 AM

in response to: wateenmooiedag

Reply

wateenmooiedag the problem will be the performance most probably, but if it can not be handled within already SQL suppllied functionality then yes of course

By the way I will be using an 10.2 instance.

Asif Momen

Posts: 1,906
Registered: 02/10/07

Re: SQL puzzle :)
Posted: Jun 8, 2008 4:31 AM

in response to: antu

Reply

SQL> create or replace function sum_prod(p_val in number) return number is

  2   l_val Varchar2(500) := p_val;

  3   l_sum number := 0;

  4   l_n1  number;

  5   l_n2  number;

  6   l_product number := 1;

  7   l_first number;

  8  begin

  9    for i in 1..length(p_val)

 10    loop

 11      l_first := substr(l_val, 1, 1);

 12      l_sum := l_sum + l_first;

 13      l_product := l_product * l_first;

 14      l_val := substr(l_val, 2);

 15    end loop;

 16    return(l_sum + l_product);

 17  end;

 18  /



Function created.



SQL> select rno from ( select rownum rno

  2                      from dual

  3                    connect by level <= 100)

  4  where rno - reverse(to_char(rno)) = sum_prod(rno);



  RNO


   63



SQL> select rno from ( select rownum rno

  2                      from dual

  3                    connect by level <= 1000)

  4  where rno - reverse(to_char(rno)) = sum_prod(rno);



  RNO


   63

  726



SQL> select rno from ( select rownum rno

  2                      from dual

  3                    connect by level <= 10000)

  4  where rno - reverse(to_char(rno)) = sum_prod(rno);



  RNO


   63

  726

 8937

I think SQL experts can write it more elegantly.

antu

Posts: 285
Registered: 09/16/07

Re: SQL puzzle :)
Posted: Jun 8, 2008 5:01 AM

in response to: Asif Momen

Reply

Citrus thank you for your time, but let me rephrase my concerns below, if I use 1000000000000 with this method even on a strong machine it takes half a day to finish.

- Which is the most efficient way of producing for example numbers between 1 and 1000000000000, CONNECT BY from DUAL?
- Is there an alternative way to reverse a number other than undocumented REVERSE function to_number(reverse(to_char(anumber)) this way there will be three function calls for each number,
- What is the best way to break a number into pieces to sum or make product in SQL.

N. Gasparotto

Posts: 18,604
Registered: 08/22/02

Re: SQL puzzle :)
Posted: Jun 8, 2008 11:40 AM

in response to: antu

Reply

Hi,

The brute force won't help too much in your case.
You may want to exclude at least the number where the last digit is higher than the first.
And still, with such big number, I'm not sure SQL and PL/SQL will be very efficient.
Lastly, even if that takes half ad day to finish, I'm sure it's not a scheduled job, after running it once, it's finish.

Anyway, here below my first try on my laptop :

SQL> create or replace type nb_obj as object (n1 number, n2 number, n3 number, n4 number);

  2  /



Type created.



Elapsed: 00:00:00.00

SQL>

SQL> create or replace type nb_list as table of nb_obj;

  2  /



Type created.



Elapsed: 00:00:00.04

SQL>

SQL> create or replace function f_maths (p_max number) return nb_list pipelined is

  2      x number:=0;

  3      y number;

  4      t number;

  5      z number;

  6

  7  begin

  8      loop

  9          --X is a positive number

 10          x := x+1;

 11          if    substr(x,-1,1) > substr(x,1,1)

 12          then  x := x+11-substr(x,-1,1);

 13          end if;

 14          exit when x > p_max;

 15

 16          y := null;

 17          t := 0;

 18          z := 1;

 19          for j in 1..length(x) loop

 20              --Y is the reverse of X

 21              y:=y||to_char(substr(x,-j,1));

 22              --T is the sum of each number which X has

 23              t:=t+substr(x,-j,1);

 24              --Z is the product of each number which X has

 25              z:=z*substr(x,-j,1);

 26              exit when y > substr(x,1,j);

 27          end loop;

 28          --X - Y = T + Z

 29          if x - y = t + z then

 30              pipe row (nb_obj(x,y,t,z));

 31          end if;

 32      end loop;

 33  end;

 34  /



Function created.



Elapsed: 00:00:00.04

SQL> show err

No errors.

SQL> select *

  2  from   table(f_maths(10000));



        N1         N2         N3         N4


----------
----------

        63         36          9         18

       726        627         15         84

      8937       7398         27       1512



Elapsed: 00:00:00.06

SQL>

SQL> select *

  2  from   table(f_maths(100000));



        N1         N2         N3         N4


----------
----------

        63         36          9         18

       726        627         15         84

      8937       7398         27       1512



Elapsed: 00:00:00.60

SQL>

SQL> select *

  2  from   table(f_maths(1000000));



        N1         N2         N3         N4


----------
----------

        63         36          9         18

       726        627         15         84

      8937       7398         27       1512



Elapsed: 00:00:07.01

SQL>

SQL> select *

  2  from   table(f_maths(10000000));



        N1         N2         N3         N4


----------
----------

        63         36          9         18

       726        627         15         84

      8937       7398         27       1512



Elapsed: 00:01:21.81

SQL>

SQL> select *

  2  from   table(f_maths(100000000));



        N1         N2         N3         N4


----------
----------

        63         36          9         18

       726        627         15         84

      8937       7398         27       1512



Elapsed: 00:15:36.81

Nicolas.

antu

Posts: 285
Registered: 09/16/07

Re: SQL puzzle :)
Posted: Jun 8, 2008 12:44 PM

in response to: N. Gasparotto

Reply

N. Gasparotto thank you for your responce.

Related to just using SQL functions I opened another thread here -
http://forums.oracle.com/forums/thread.jspa?forumID=75&threadID=666861

But I guess since there are several function calls those solutions do not help both in performance and easy undertanding compared to a user defined pl/sql like yours. This was an interesting experience

Asif Momen

Posts: 1,906
Registered: 02/10/07

Re: SQL puzzle :)
Posted: Jun 8, 2008 12:44 PM

in response to: antu

Reply

I replaced "SUM_PROD" function with wateenmooiedag's query:

http://forums.oracle.com/forums/message.jspa?messageID=2575576#2575576

SQL> select rno from ( select rownum rno

  2                          from dual

  3                        connect by level <= 1000)

  4      where rno - reverse(to_char(rno)) =

  5                           (select to_number(extractvalue(xmltype(dbms_xmlgen.getxml('select '

  6                                   ||regexp_replace(rno,'(.)','\1+') || '0'||'

  7                                      s from dual')),'/ROWSET/ROW/S'))

  8                          +        to_number(extractvalue(xmltype(dbms_xmlgen.getxml('select '

  9                                   ||regexp_replace(rno,'(.)','\1*') || '1'||'

 10                                   p from dual')),'/ROWSET/ROW/P'))

 11                              from dual);



       RNO


        63

       726



SQL>

N. Gasparotto

Posts: 18,604
Registered: 08/22/02

Re: SQL puzzle :)
Posted: Jun 8, 2008 12:57 PM

in response to: antu

Reply

YEah, XML and REGEXP are slow down. I'm pretty sure you can find some more rules to add in your function to exclude more and more numbers.

Nicolas.

N. Gasparotto

Posts: 18,604
Registered: 08/22/02

Re: SQL puzzle :)
Posted: Jun 8, 2008 2:05 PM

in response to: N. Gasparotto

Reply

Just a new try, ~3 minutes less for hundred million...

SQL> drop type nb_list;



Type dropped.



Elapsed: 00:00:00.11

SQL> create or replace type nb_obj as object (n1 number, n2 number, n3 number, n4 number);

  2  /



Type created.



Elapsed: 00:00:00.01

SQL>

SQL> create or replace type nb_list as table of nb_obj;

  2  /



Type created.



Elapsed: 00:00:00.15

SQL>

SQL> create or replace function f_maths (p_min number,p_max number) return nb_list pipelined is

  2      x number:=p_min-1;

  3      y number;

  4      t number;

  5      z number;

  6      x1 number;

  7      x2 number;

  8  begin

  9      loop

 10          --X is a positive number

 11          x := x+1;

 12          x1 := substr(x,1,floor(length(x)/2));

 13          x2 := substr(x,-floor(length(x)/2));

 14          --dbms_output.put_line(x||' '||x1||' '||x2);

 15          if    x1 < x2 or length(x2) < length(x1)

 16          then  x := ceil(x/power(10,length(x2)))*power(10,length(x2));

 17          end if;

 18          exit when x > p_max;

 19

 20          y := null;

 21          t := 0;

 22          z := 1;

 23          for j in 1..length(x) loop

 24              --Y is the reverse of X

 25              y:=y||to_char(substr(x,-j,1));

 26              exit when y > substr(x,1,j);

 27              --T is the sum of each number which X has

 28              t:=t+substr(x,-j,1);

 29              --Z is the product of each number which X has

 30              z:=z*substr(x,-j,1);

 31          end loop;

 32          --X - Y = T + Z

 33          if x - y = t + z then

 34              pipe row (nb_obj(x,y,t,z));

 35          end if;

 36      end loop;

 37  end;

 38  /



Function created.



Elapsed: 00:00:00.04

SQL> show err

No errors.

SQL> select *

  2  from   table(f_maths(1,100000000));



        N1         N2         N3         N4


----------
----------

        63         36          9         18

       726        627         15         84

      8937       7398         27       1512



Elapsed: 00:12:43.79

SQL>

Nicolas.

michaels2

Posts: 5,875
Registered: 09/24/06

Re: SQL puzzle :)
Posted: Jun 9, 2008 2:17 AM

in response to: antu

Reply

Probaly just interesting from an academical point of view. Performancewise it is just not acceptable:

SQL> select * from
  2   xmltable('declare function local:reverse($a)
  3             {
  4               if (string-length($a) != 0) then
  5                concat(substring($a,string-length($a),1), local:reverse(substring($a,1,string-length($a)-1)))
  6               else ()
  7             }; (: eof :)
  8             declare function local:sum($a)
  9             {
 10               if (string-length($a) != 0) then
 11                 xs:integer(substring($a,1,1)) + xs:integer(local:sum(substring($a,2)))
 12               else (0)
 13             }; (: eof :)
 14             declare function local:prod($a)
 15             {
 16               if (string-length($a) != 0) then
 17                 xs:integer(substring($a,1,1)) * xs:integer(local:prod(substring($a,2)))
 18               else (1)
 19             }; (: eof :)
 20             for $i in 1 to 10000
 21              where $i - local:reverse(xs:string($i)) = local:sum(xs:string($i)) + local:prod(xs:string($i))
 22                    return $i' columns x integer path '.')
 23  / 
 
         X
----------
        63
       726
      8937
 
Abgelaufen: 00:01:24.51

N. Gasparotto

Posts: 18,604
Registered: 08/22/02

Re: SQL puzzle :)
Posted: Jun 9, 2008 2:40 AM

in response to: michaels2

Reply

Abgelaufen: 00:01:24.51

Around 1400 times slower than function defined earlier above, who said SQL was always faster than PL/SQL ?

Nicolas.

michaels2

Posts: 5,875
Registered: 09/24/06

Re: SQL puzzle :)
Posted: Jun 9, 2008 2:43 AM

in response to: N. Gasparotto

Reply

Even without xml no faster than Nicolas

SQL> select * from (select level x from dual connect by level <= 10000)
  2         connect by nocycle level <= length (x) and prior x = x - 1
  3   group by x
  4  having x - reverse(to_char(x)) = sum(substr(x,level,1)) + round(exp(sum(ln(decode(substr(x,level,1),0,1,substr(x,level,1))))))
  5   order by x
  6  / 
 
         X
----------
        63
       726
      8937
 
Abgelaufen: 00:00:02.29