Puzzle of the week #2

Dish washing schedule puzzle:

Four roommate students, Anna, Betty, Carla, and Daniela decided to make a “Dish washing schedule”. Every day one of the girls should do all the dishes. The challenge is to make a schedule for the next month that will spread the responsibilities among the girls as evenly as possible. At the same time the schedule should be completely random.

You need to write a single SELECT statement to generate the schedule for the current calendar month.

The output (for Feb-2016) should look like this:

   Anna      Betty      Carla    Daniela
------- ---------- ---------- ----------
      4          2          1          3
      8          6          5          7
     12         10          9         11
     16         14         13         15
     18         20         19         17
     22         24         23         21
     26         28         27         25
                                      29

or this:

 Anna      Betty      Carla    Daniela
----- ---------- ---------- ----------
    3          1          4          2
    8          6          5          7
   11         10          9         12
   16         13         14         15
   18         19         20         17
   21         23         24         22
   26         28         25         27
              29

Remember, the output is random, which means that subsequent execution of the same query should produce different results.

 

To submit your answer (one or more!) please start following this blog and add a comment to this post.

A correct answer (and workarounds!) will be published here in a week.

If you like this post, you may want to join my new Oracle group on Facebook: https://www.facebook.com/groups/sqlpatterns/

For more tricks and cool techniques check my book “Oracle SQL Tricks and Workarounds” for instructions.


		
Advertisements

Puzzle of the Week Challenge – Solutions to the 1st Puzzle

Last week we started a new contest, Puzzle of the Week. Today we publish correct answers for the 1st puzzle:

jigsaw-puzzle-piece Write a single SELECT statement that would output a calendar for the current month in a traditional tabular format (7 columns: Sun-Sat).

 

Solution #1: No Sub-query solution! We consider it the best solution.

To better understand the following query we suggest you to first check if you can understand Solution #3 (see below).

SELECT MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'), 
                  '1', LEVEL)) SUN,
       MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'),
                  '2', LEVEL)) MON,
       MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'), 
                  '3', LEVEL)) TUE,
       MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'), 
                  '4', LEVEL)) WED,
       MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'), 
                  '5', LEVEL)) THU,
       MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'), 
                  '6', LEVEL)) FRI,
       MIN(DECODE (TO_CHAR (TRUNC(SYSDATE,'MON') + LEVEL - 1, 'd'), 
                  '7', LEVEL)) SAT
FROM DUAL
CONNECT BY LEVEL <= TO_CHAR(LAST_DAY(SYSDATE),'DD')
GROUP BY TRUNC(TRUNC(SYSDATE,'MON') + LEVEL-1, 'DAY')
ORDER BY TRUNC(TRUNC(SYSDATE,'MON') + LEVEL-1, 'DAY');

 

Solution #2: Using PIVOT

SELECT "'SUN'" SU,"'MON'" MO,"'TUE'" TU,"'WED'" WE,
       "'THU'" TH,"'FRI'" FR,"'SAT'" SA
FROM
(
  SELECT TRUNC(TRUNC(SYSDATE,'MON')+LEVEL-1,'DAY') WEEK_START,  
         TO_CHAR(TRUNC(SYSDATE,'MON')+LEVEL-1,'DD') DD, 
         TO_CHAR(TRUNC(SYSDATE,'MON')+LEVEL-1,'DY') DY  
  FROM DUAL
  CONNECT BY TO_CHAR(TRUNC(SYSDATE,'MON')+LEVEL-1,'yyyymm')=
             TO_CHAR(SYSDATE,'yyyymm')
)
PIVOT 
(
  MAX(DD)
  FOR DY IN ('SUN','MON','TUE','WED','THU','FRI','SAT')
)
ORDER BY week_start;

 

Solution #3: Use the power of CONNECT BY clause to generate a range of days for the current month

WITH x AS (
SELECT TRUNC(SYSDATE, 'MON')+level-1 d
FROM DUAL
CONNECT BY MONTHS_BETWEEN(TRUNC(SYSDATE, 'MON')+level-1, TRUNC(SYSDATE, 'MON'))<1
)
SELECT MAX(CASE WHEN TO_CHAR(D,'DY')='SUN' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS SUN,
       MAX(CASE WHEN TO_CHAR(D,'DY')='MON' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS MON,
       MAX(CASE WHEN TO_CHAR(D,'DY')='TUE' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS TUE,
       MAX(CASE WHEN TO_CHAR(D,'DY')='WED' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS WED,
       MAX(CASE WHEN TO_CHAR(D,'DY')='THU' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS THU,
       MAX(CASE WHEN TO_CHAR(D,'DY')='FRI' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS FRI,
       MAX(CASE WHEN TO_CHAR(D,'DY')='SAT' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS SAT
FROM X
GROUP BY TRUNC(D, 'DAY')
ORDER BY TRUNC(D, 'DAY')

Solution #4: Use existing table(s) to generate a range of days for the current month

WITH X AS (
SELECT TRUNC(SYSDATE, 'MON')+ROWNUM-1 D
FROM emp,emp 
WHERE TO_CHAR(TRUNC(SYSDATE, 'MON')+ROWNUM-1, 'YYYYMM')=TO_CHAR(SYSDATE, 'YYYYMM')
  AND ROWNUM<=31
)
SELECT MAX(CASE WHEN TO_CHAR(D,'DY')='SUN' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS SUN,
       MAX(CASE WHEN TO_CHAR(D,'DY')='MON' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS MON,
       MAX(CASE WHEN TO_CHAR(D,'DY')='TUE' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS TUE,
       MAX(CASE WHEN TO_CHAR(D,'DY')='WED' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS WED,
       MAX(CASE WHEN TO_CHAR(D,'DY')='THU' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS THU,
       MAX(CASE WHEN TO_CHAR(D,'DY')='FRI' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS FRI,
       MAX(CASE WHEN TO_CHAR(D,'DY')='SAT' THEN TO_CHAR(D,'DD') 
                ELSE '  ' END) AS SAT
FROM X
GROUP BY TRUNC(D, 'DAY')
ORDER BY TRUNC(D, 'DAY')

 

Solution #5: Present each calendar week as a single column value – using LISTAGG function

WITH X AS (
SELECT TRUNC(SYSDATE, 'MON')+level-1 d
FROM DUAL
CONNECT BY MONTHS_BETWEEN(TRUNC(SYSDATE, 'MON')+LEVEL-1, TRUNC(SYSDATE, 'MON'))<1
), y AS (
SELECT LISTAGG(TO_CHAR(d,'DD'), '  ') WITHIN GROUP(ORDER BY d) AS week, TRUNC(D, 'DAY') wday
FROM X
GROUP BY TRUNC(D, 'DAY')
)
SELECT CASE WHEN week LIKE '01%' THEN LPAD(week, 26)
            ELSE week
       END AS "SUN MON TUE WED THU FRI SAT"
FROM y
ORDER BY wday

 

Solution #6: Present each calendar week as a single column value – using SYS_CONNECT_BY_PATH function

WITH X AS (
SELECT TRUNC(SYSDATE, 'MON')+level-1 d
FROM DUAL
CONNECT BY MONTHS_BETWEEN(TRUNC(SYSDATE, 'MON')+LEVEL-1, TRUNC(SYSDATE, 'MON'))<1
)
SELECT CASE WHEN MAX(SYS_CONNECT_BY_PATH(TO_CHAR(d, 'DD'), ' ')) LIKE ' 01%' THEN
                LPAD(MAX(SYS_CONNECT_BY_PATH(TO_CHAR(d, 'DD'), ' ')), 21)
            ELSE MAX(SYS_CONNECT_BY_PATH(TO_CHAR(d, 'DD'), ' '))
       END " SU MO TU WE TH FR SA"
FROM x
CONNECT BY d=PRIOR d+1 AND TRUNC(d,'DAY')=TRUNC(PRIOR d, 'DAY')
START WITH TO_CHAR(d,'DD')='01' OR d=TRUNC(d,'DAY')
GROUP BY TRUNC(d, 'DAY')
ORDER BY 1

 

Solution #7: A variation of Solution #6

SELECT CASE 
       WHEN MAX(SYS_CONNECT_BY_PATH(TO_CHAR(d, 'DD'), ' ')) LIKE ' 01%' THEN
                LPAD(MAX(SYS_CONNECT_BY_PATH(TO_CHAR(d, 'DD'), ' ')), 21)
            ELSE MAX(SYS_CONNECT_BY_PATH(TO_CHAR(d, 'DD'), ' '))
       END " SU MO TU WE TH FR SA"
FROM (SELECT TRUNC(SYSDATE, 'MON')+level-1 d
      FROM DUAL
      CONNECT BY MONTHS_BETWEEN(TRUNC(SYSDATE, 'MON')+LEVEL-1, TRUNC(SYSDATE, 'MON'))<1) x
CONNECT BY d=PRIOR d+1 AND TRUNC(d,'DAY')=TRUNC(PRIOR d, 'DAY')
START WITH TO_CHAR(d,'DD')='01' OR d=TRUNC(d,'DAY')
GROUP BY TRUNC(d, 'DAY')
ORDER BY 1

If you like this post, you may want to join my new Oracle group on Facebook: https://www.facebook.com/groups/sqlpatterns/

For more tricks and cool techniques check my book “Oracle SQL Tricks and Workarounds” for instructions.

Puzzle of the Week Challenge

Today we are starting a new challenge – every week we will publish a puzzle (various levels).

There will be some awards and signs of recognition to the winners.

Puzzle of the week:
Level: Advanced

jigsaw-puzzle-piece Write a single SELECT statement that would output a calendar for the current month in a traditional tabular format (7 columns: Sun-Sat).

Expected result may look like this (in SQL*Plus):

p01-2

or this:

p01-2

To submit your answer (one or more!) please start following this blog and add a comment to this post.

A correct answer (and workarounds!) will be published here in a week.

 

Common rule to create or validate the GROUP BY clause

Level: Intermediate

Do you know that the content of the GROUP BY clause can be derived from the rest of the clauses in SELECT statement?

Well, it is true to a great extent. What can be derived is a combination of columns and expressions that must be included in GROUP BY.

We will start a demonstration with a simple puzzle:

Complete GROUP BY Clause in the following query:

SELECT 'Salesmen Report' AS "Report Title", 
       d.deptno, COUNT(e.empno) cnt
FROM dept d, emp e 
WHERE d.deptno=e.deptno
GROUP BY _______________________________
HAVING e.job=’SALESMAN’
   AND SUM(e.sal)>2000
ORDER BY TO_CHAR(e.hiredate,’YYYY’), SUM(e.sal) DESC;

If you are not sure about the answer, learn the following rule.

Common rule to create or validate the GROUP BY clause:

It should at least include all columns and scalar expressions (constants, aggregate functions, and analytical functions are excluded) referenced in SELECT, HAVING, and ORDER BY clauses. Other columns can be specified as well, but they don’t have to.

Here is a solution to the above puzzle:

SELECT 'Salesmen Report' AS "Report Title", 
       d.deptno, COUNT(e.empno) cnt
FROM dept d, emp e 
WHERE d.deptno=e.deptno
GROUP BY d.deptno, e.job, TO_CHAR(e.hiredate,'YYYY')
HAVING e.job=’SALESMAN’
   AND SUM(e.sal)>2000
ORDER BY TO_CHAR(e.hiredate,’YYYY’), SUM(e.sal) DESC;

Note that logically e.job=’SALESMAN’ condition does not belong to HAVING clause. If we push it to WHERE clause (where it belongs to), the group by clause may change:

SELECT 'Salesmen Report' AS "Report Title", 
       d.deptno, COUNT(e.empno) cnt
FROM dept d, emp e 
WHERE d.deptno=e.deptno AND e.job=’SALESMAN’
GROUP BY d.deptno, TO_CHAR(e.hiredate,'YYYY')
HAVING SUM(e.sal)>2000
ORDER BY TO_CHAR(e.hiredate,’YYYY’), SUM(e.sal) DESC;

Since e.job is no longer in the HAVING clause, we don’t have to put it in GROUP BY, but we CAN. The Rule on GROUP BY clause says that at very least, the columns referenced in SELECT, HAVING, and ORDER BY clauses must be referenced in GROUP BY as well, but there might be other columns.

If you like this post, you may want to join my new Oracle group on Facebook: https://www.facebook.com/groups/sqlpatterns/

For more tricks and cool techniques check my book “Oracle SQL Tricks and Workarounds” for instructions.

Interview Question: Can a Select statement have HAVING clause and no GROUP BY?

Question: Can a Select statement have HAVING clause and no GROUP BY?
Level: Intermediate/Advanced.

I was first asked this question about 20 years ago and I have to admit I did not provide the right answer which seemed to be very counter intuitive.

The correct answer: Yes, it can.

Example:

SELECT SUM(sal)
FROM emp
HAVING COUNT(*)>10

Essentially, when we don’t use GROUP BY clause we treat the entire table as a single group. As with any group, we can reference various aggregate functions in SELECT, HAVING, and ORDER BY clauses.

We can even use some analytical functions as follows:

SELECT COUNT(*), SUM(COUNT(*)) OVER() sm
FROM emp
HAVING COUNT(*)<100

If you like this post, you may want to join my new Oracle group on Facebook: https://www.facebook.com/groups/sqlpatterns/

For more tricks and cool techniques check my book “Oracle SQL Tricks and Workarounds” for instructions.