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Puzzle n°08 - Display the Numerical Pyramid using straight SQL query (Not PL/SQL) ** [message #294381] Thu, 17 January 2008 07:06 Go to next message
rajavu1
Messages: 1574
Registered: May 2005
Location: Bangalore , India
Senior Member

Hi All,

The aim of this Puzzle is to display a pyramid of Numbers using SQL query . It might be easy with PL/SQL routines. So it should be using the straight SQL. The output will be as follows

SQL> DEFINE LV=10
SQL> /
   A    B    C    D    E    F    G    H    I    J
---- ---- ---- ---- ---- ---- ---- ---- ---- ----
   1    2    3    4    5    6    7    8    9   10
       11   12   13   14   15   16   17   18
            19   20   21   22   23   24
                 25   26   27   28
                      29   30

SQL> DEFINE LV=7
SQL> /
   A    B    C    D    E    F    G    H    I    J
---- ---- ---- ---- ---- ---- ---- ---- ---- ----
   1    2    3    4    5    6    7
        8    9   10   11   12
            13   14   15
                 16

SQL> DEFINE LV=5
SQL> /

   A    B    C    D    E    F    G    H    I    J
---- ---- ---- ---- ---- ---- ---- ---- ---- ----
   1    2    3    4    5
        6    7    8
             9

Thumbs Up
Rajuvan.

[Updated on: Sun, 15 June 2014 00:20] by Moderator

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Re: Puzzle n°08 - Display the Numerical Pyramid using straight SQL query (Not PL/SQL) ** [message #294615 is a reply to message #294381] Fri, 18 January 2008 03:38 Go to previous messageGo to next message
Michel Cadot
Messages: 68597
Registered: March 2007
Location: Nanterre, France, http://...
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Account Moderator
It is more an arithmetic problem than a SQL one.
The question is how to put number in (line,col) coordinates:
SQL> set numwidth 3
SQL> def lv=10
SQL> with 
  2    lines as (
  3      select level line,                      -- line number
  4             &lv-level+1 last_col,            -- last column with a value
  5             (level-1)*(&lv-level+2) nb_prev  -- number of numbers in previous lines
  6      from dual 
  7      connect by level <= trunc((&lv+1)/2)
  8    )
  9  select -- col 1
 10         case when 1 < line then null 
 11              when 1 > last_col then null
 12              else nb_prev+1-(line-1)
 13         end a,
 14         -- col 2
 15         case when 2 < line then null 
 16              when 2 > last_col then null
 17              else nb_prev+2-(line-1)
 18         end b,
 19         -- col 3
 20         case when 3 < line then null 
 21              when 3 > last_col then null
 22              else nb_prev+3-(line-1)
 23         end c,
 24         -- col 4
 25         case when 4 < line then null 
 26              when 4 > last_col then null
 27              else nb_prev+4-(line-1)
 28         end d,
 29         -- col 5
 30         case when 5 < line then null 
 31              when 5 > last_col then null
 32              else nb_prev+5-(line-1)
 33         end e,
 34         -- col 6
 35         case when 6 < line then null 
 36              when 6 > last_col then null
 37              else nb_prev+6-(line-1)
 38         end f,
 39         -- col 7
 40         case when 7 < line then null 
 41              when 7 > last_col then null
 42              else nb_prev+7-(line-1)
 43         end g,
 44         -- col 8
 45         case when 8 < line then null 
 46              when 8 > last_col then null
 47              else nb_prev+8-(line-1)
 48         end h,
 49         -- col 9
 50         case when 9 < line then null 
 51              when 9 > last_col then null
 52              else nb_prev+9-(line-1)
 53         end i,
 54         -- col 10
 55         case when 10 < line then null 
 56              when 10 > last_col then null
 57              else nb_prev+10-(line-1)
 58         end j
 59  from lines
 60  order by line
 61  /
  A   B   C   D   E   F   G   H   I   J
--- --- --- --- --- --- --- --- --- ---
  1   2   3   4   5   6   7   8   9  10
     11  12  13  14  15  16  17  18
         19  20  21  22  23  24
             25  26  27  28
                 29  30

5 rows selected.

SQL> def lv=7
SQL> /
  A   B   C   D   E   F   G   H   I   J
--- --- --- --- --- --- --- --- --- ---
  1   2   3   4   5   6   7
      8   9  10  11  12
         13  14  15
             16

4 rows selected.

SQL> def lv=5
SQL> /
  A   B   C   D   E   F   G   H   I   J
--- --- --- --- --- --- --- --- --- ---
  1   2   3   4   5
      6   7   8
          9

3 rows selected.

To add a little bit more of SQL chalenge and trying to get the exact number of columns displayed.
Build a query that give the correct number for each line and column:
SQL> with 
  2    lines as ( -- row generator for each output line
  3      select level line,                      -- line number
  4             &lv-level+1 last_col,            -- last column with a value
  5             (level-1)*(&lv-level+2) nb_prev  -- number of numbers in previous lines
  6      from dual 
  7      connect by level <= trunc((&lv+1)/2)
  8    ),
  9    cols as ( -- row generator for each output column
 10      select level col from dual connect by level <= &lv
 11    )
 12      select line, col,
 13             case when col < line then '   ' 
 14                  when col > last_col then '   '
 15                  else to_char(nb_prev+col-(line-1),'99')
 16             end data
 17      from lines, cols
 18  order by line, col
 19  /
LINE COL DAT
---- --- ---
   1   1   1
   1   2   2
   1   3   3
   1   4   4
   1   5   5
   2   1
   2   2   6
   2   3   7
   2   4   8
   2   5
   3   1
   3   2
   3   3   9
   3   4
   3   5

15 rows selected.

Now use any pivot method to put them line by line, for instance (adding the header):
SQL> set head off
SQL> col nop noprint
SQL> col line format a50
SQL> def lv=10
SQL> with 
  2    lines as ( -- row generator for each output line
  3      select level line,                      -- line number
  4             &lv-level+1 last_col,            -- last column with a value
  5             (level-1)*(&lv-level+2) nb_prev  -- number of numbers in previous lines
  6      from dual 
  7      connect by level <= trunc((&lv+1)/2)
  8    ),
  9    cols as ( -- row generator for each output column
 10      select level col from dual connect by level <= &lv
 11    ),
 12    results as ( -- result values distributed into lines and columns
 13      select line, col,
 14             case when col < line then '   ' 
 15                  when col > last_col then '   '
 16                  else to_char(nb_prev+col-(line-1),'99')
 17             end data
 18      from lines, cols
 19    )
 20  -- Display header
 21  select 1 nop, 
 22        replace(substr(sys_connect_by_path(lpad(chr(ascii('A')+col-1),3),'/'),2),'/',' ')
 23  from cols
 24  where col = &lv
 25  connect by prior col = col-1
 26  start with col = 1
 27  union all
 28  -- Display "-" line
 29  select 2, replace(substr(sys_connect_by_path('---','/'),2),'/',' ')
 30  from cols
 31  where col = &lv
 32  connect by prior col = col-1
 33  start with col = 1
 34  union all
 35  -- Display result lines
 36  select 2+line,
 37         replace(substr(sys_connect_by_path(data,'/'),2),'/',' ') line
 38  from results
 39  where col = &lv
 40  connect by prior line = line and prior col = col-1
 41  start with col = 1
 42  order by 1
 43  /
  A   B   C   D   E   F   G   H   I   J
--- --- --- --- --- --- --- --- --- ---
  1   2   3   4   5   6   7   8   9  10
     11  12  13  14  15  16  17  18
         19  20  21  22  23  24
             25  26  27  28
                 29  30

7 rows selected.

SQL> def lv=7
SQL> /
  A   B   C   D   E   F   G
--- --- --- --- --- --- ---
  1   2   3   4   5   6   7
      8   9  10  11  12
         13  14  15
             16

6 rows selected.

SQL> def lv=5
SQL> /
  A   B   C   D   E
--- --- --- --- ---
  1   2   3   4   5
      6   7   8
          9

5 rows selected.

SQL> def lv=12
SQL> /
  A   B   C   D   E   F   G   H   I   J   K   L
--- --- --- --- --- --- --- --- --- --- --- ---
  1   2   3   4   5   6   7   8   9  10  11  12
     13  14  15  16  17  18  19  20  21  22
         23  24  25  26  27  28  29  30
             31  32  33  34  35  36
                 37  38  39  40
                     41  42

8 rows selected.

Then we are no more limited in the number of columns.

Regards
Michel
Re: Puzzle n°08 - Display the Numerical Pyramid using straight SQL query (Not PL/SQL) ** [message #294938 is a reply to message #294381] Sun, 20 January 2008 23:24 Go to previous messageGo to next message
rajavu1
Messages: 1574
Registered: May 2005
Location: Bangalore , India
Senior Member

Hi Michel ,


It is simply Awesome .... You are rocking .

Thumbs Up
Rajuvan
Re: Puzzle n°08 - Display the Numerical Pyramid using straight SQL query (Not PL/SQL) ** [message #294944 is a reply to message #294381] Sun, 20 January 2008 23:44 Go to previous messageGo to next message
rajavu1
Messages: 1574
Registered: May 2005
Location: Bangalore , India
Senior Member

I have just tried code for the pyramid before i posted the actual Post . Let me share my code which is slightly complex than Michel's one .

SQL> WITH ASET AS
  2                (
  3                     SELECT du.*,
  4                               LEVEL L,
  5                                NVL(LAG((x-1) - (level -1)*2) OVER (PARTITION BY null ORDER BY LeVEL),0) n
  6                                FROM (
  7                 SELECT  max(DECODE(LEVEL,1,1)) A,
  8                                max(DECODE(LEVEL,2,2)) B,
  9                                max(DECODE(LEVEL,3,3)) C,
 10                                max(DECODE(LEVEL,4,4)) D,
 11                                max(DECODE(LEVEL,5,5)) E,
 12                                max(DECODE(LEVEL,6,6)) F,
 13                                max(DECODE(LEVEL,7,7)) G,
 14                                max(DECODE(LEVEL,8,8)) H,
 15                                max(DECODE(LEVEL,9,9)) I,
 16                                max(DECODE(LEVEL,10,10)) J,
 17                                max(LV) x
 18                      FROM( SELECT &LV LV
 19                                                     FROM DUAL )
 20                      CONNECt by LEVEL <= LV) DU
 21                      CONNECT by LEVEL < (X/2)+1 ),
 22               bset as ( select a,b,c,d,e,f,g,h,i,j,x,L ,n , sum(n) over (partition by null order by  l) p
 23                from aset )
 24                 select   DECODE(SIGN(a- (L -1) ),1 , DECODE(SIGN((a+l-1)-x), 0, a+P ,-1 ,a+P) ) A ,
 25                          DECODE(SIGN(b- (L -1) ),1 ,DECODE(SIGN((b+l-1)-x), 0, b+P,-1, b+P) ) B ,
 26                          DECODE(SIGN(c- (L -1) ),1 , DECODE(SIGN((c+l-1)-x), 0,c+P,-1,c+P) ) C  ,
 27                          DECODE(SIGN(d- (L -1) ),1 , DECODE(SIGN((d+l-1)-x), 0,d+P,-1,d+P) ) D  ,
 28                          DECODE(SIGN(e- (L -1) ),1 , DECODE(SIGN((e+l-1)-x), 0,e+P,-1,e+P) ) E  ,
 29                          DECODE(SIGN(f- (L -1) ),1 , DECODE(SIGN((f+l-1)-x), 0,f+P,-1,F+P) ) F  ,
 30                          DECODE(SIGN(g- (L -1) ),1 , DECODE(SIGN((g+l-1)-x), 0,g+P,-1,g+P) ) G  ,
 31                          DECODE(SIGN(h- (L -1) ),1 , DECODE(SIGN((h+l-1)-x), 0,h+P,-1,h+P) ) H  ,
 32                          DECODE(SIGN(i- (L -1) ),1 , DECODE(SIGN((i+l-1)-x), 0,i+P,-1,i+P) ) I  ,
 33                          DECODE(SIGN(j- (L -1) ),1 , DECODE(SIGN((j+l-1)-x), 0,j+P,-1,j+P) ) J
 34              from  Bset;

  A   B   C   D   E   F   G   H   I   J
--- --- --- --- --- --- --- --- --- ---
  1   2   3   4   5   6   7   8   9  10
     11  12  13  14  15  16  17  18
         19  20  21  22  23  24
             25  26  27  28
                 29  30

SQL>


Michel's solution is more elegant
Still expecting much simpler method from experts Smile

Thumbs Up
Rajuvan

[Updated on: Sun, 20 January 2008 23:45]

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Re: Puzzle n°08 - Display the Numerical Pyramid using straight SQL query (Not PL/SQL) ** [message #295829 is a reply to message #294381] Wed, 23 January 2008 12:30 Go to previous messageGo to next message
varu123
Messages: 754
Registered: October 2007
Senior Member
What is this language called?
I thought SQl is simple Shocked
Re: Puzzle n°08 - Display the Numerical Pyramid using straight SQL query (Not PL/SQL) ** [message #687421 is a reply to message #295829] Tue, 07 March 2023 13:37 Go to previous message
mathguy
Messages: 105
Registered: January 2023
Senior Member
I will take the view that the result must be in ten columns (rather than just a single trompe l'oeil column that, when rendered graphically, looks like multiple columns). That is, view it as a "data processing" problem rather than a "displaying" or "reporting" problem.

In that case the number of columns must be given in advance, it can't depend on the input LVL (unless we use dynamic SQL, which - let's leave that alone).

The computation can be all done in closed form, which is tedious but perhaps seventh- or eighth- grade level. (Or college level in some countries, depending on the education system.) But since this is a forum for SQL, not for arithmetic, perhaps it is better to let SQL do most of the work.

In the solution below I generate column numbers from 1 to 10 (hard-coded), then row numbers from 1 to CEIL(LVL/2) (this will depend on LVL). This models the two-dimensional array required in the output, but modeled by row number and column number; in the last step, we will pivot to get the desired format. Then the problem is simply to determine which "cells" will be non-NULL, which is a trivial task, and then assign consecutive integers to the cells - that is trivial with the ROW_NUMBER() analytic function.

In the solution below I model the input LVL as a bind variable, I use the WITH clause with column aliases in each declaration (available only since Oracle 11.2, but this is not essential - everything can be written with old-style subqueries), and I use the PIVOT operator available only since Oracle 11.1, but that too can be done the old way, with conditional aggregation. Even the analytic function is not critical - we can get the same result with an ORDER BY clause and an outer query where we select ROWNUM. The solution can be written in very, very old versions of Oracle.

As an aside, why is the result called a pyramid? A pyramid is a three-dimensional thing; what we have here is a triangle. (By analogy, in combinatorics we have Pascal's triangle, nobody calls it Pascal's "pyramid".)

with
  c (cn) as (select level from dual connect by level <= 10),
  r (rn) as (select cn from c where cn <= ceil(:lvl/2)),
  prep (rn, cn, val) as (
    select rn, cn, row_number() over (order by rn, cn)
    from   c join r on cn between rn and :lvl + 1- rn
  )
select a, b, c, d, e, f, g, h, i, j
from   prep
pivot  (min(val) for cn in (1 a, 2 b, 3 c, 4 d, 5 e, 6 f, 7 g, 8 h, 9 i, 10 j))
order  by rn
;
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