## How to generate a list of first N binary numbers in Oracle SQL?

In my recent post I showed how to convert a decimal number (i.e. an integer) into a binary string. We can build upon that technique to answer the question:

```WITH x AS (
SELECT LEVEL n
FROM dual
CONNECT BY LEVEL<=50
)
SELECT x.N, y.bin
FROM x, LATERAL (SELECT LISTAGG(SIGN(BITAND(x.N, POWER(2,LEVEL-1))),'')
WITHIN GROUP(ORDER BY LEVEL DESC) bin
FROM dual
CONNECT BY POWER(2, LEVEL-1)<=x.N) y
```

Note the LATERAL keyword (Oracle 12c new feature) that enables us to reference “x” in the inline view “y”. In pre-12c world, we would have to use TABLE/CAST/MULTISET function composition to achieve the same result:

```WITH x AS (
SELECT LEVEL n
FROM dual
CONNECT BY LEVEL<=50
)
SELECT x.N, y.column_value bin
FROM x, TABLE(CAST(MULTISET(
SELECT LISTAGG(SIGN(BITAND(x.N, POWER(2,LEVEL-1))),'')
WITHIN GROUP(ORDER BY LEVEL DESC) bin
FROM dual
CONNECT BY POWER(2, LEVEL-1)<=x.N) AS sys.odcivarchar2list)) y
```

The idea used in the following query is based on a totally different approach. It builds a string of “0”s and “1”s in a loop until its length reaches a desired value:

```WITH x(v, n) AS (
SELECT column_value, 1
FROM TABLE(sys.odcivarchar2list('0','1'))
UNION ALL
SELECT x.v || t.column_value, x.n+1
FROM TABLE(sys.odcivarchar2list('0','1')) t JOIN x on LENGTH(x.v)=n
WHERE n<=CEIL(LOG(2,50))
), y AS (
SELECT NVL(LTRIM(x.v,'0'),'0') bin, ROWNUM-1 dec
FROM x
WHERE n=(SELECT MAX(n) FROM x)
)
SELECT *
FROM y
WHERE dec<=50
```

To better understand the above query, try the following one:

```SELECT *
FROM TABLE(sys.odcivarchar2list('0','1')),
TABLE(sys.odcivarchar2list('0','1')),
TABLE(sys.odcivarchar2list('0','1')),
TABLE(sys.odcivarchar2list('0','1'))
```

If we put enough tables in the Cartesian product and concatenate all column_value columns in a single character string expression, we will achieve our goal. The challenge with this approach is to dynamically change the number of the tables in the FROM clause. This can be simulated in the recursive WITH clause by repeatedly adding more and more collections of bits (0 and 1).

### My Oracle Group on Facebook:

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

### Would you like to read about many more tricks and puzzles?

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

## How to Simulate SIGN Function

### Puzzle of the day:

How to simulate the SIGN function in Oracle SQL by only using CEIL, FLOOR, and ABS Oracle SQL functions along with arithmetic operators? No PL/SQL.

### Solution:

```SIGN(x)=CEIL(x/(1+ABS(x)))+FLOOR(x/(1+ABS(x)))
```

In SQL, we can demonstrate it as follows:

```WITH r AS (
SELECT dbms_random.VALUE(-999,999) rnd
FROM dual
CONNECT BY LEVEL<=10
UNION ALL
SELECT 0
FROM dual
)
SELECT rnd, SIGN(rnd), CEIL(rnd/(1+ABS(rnd)))+FLOOR(rnd/(1+ABS(rnd))) "MySign"
FROM r
```

Result:

```       RND  SIGN(RND)     MySign
---------- ---------- ----------
-519.606         -1         -1
-657.62692         -1         -1
414.625079          1          1
736.175183          1          1
268.689074          1          1
-647.12649         -1         -1
338.192233          1          1
784.780876          1          1
-529.69184         -1         -1
-596.56803         -1         -1
0          0          0
```

As you can see, “MySign” column perfectly matches SIGN column.

### Comment:

WITH clause is needed to generate 10 random values in the range of -999 .. +999. “0” value is added to demonstrate a special case as it is unlikely that zero will be randomly generated.

### My Oracle Group on Facebook:

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

### Would you like to read about many more tricks and puzzles?

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

## Puzzle of the Week #9:

All employees are sorted by employee number and need to be split in 3 groups equal in size (as close as possible).  Employees with the smallest numbers will get into the 1st group, the 2nd group will have employees with the next (bigger) range of numbers, etc.

Write a single SELECT statement (against emp table) that would show group number, the range of employee numbers, and the size of each group.

Expected Result:

```Group RANGE           Count
----- ---------- ----------
1 7369-7654           5
2 7698-7844           5
3 7876-7934           4
```

Solutions:

#1: Using NTile Analytic function

```WITH x AS (
SELECT empno, NTILE(3)OVER(ORDER BY empno) nt
FROM emp
)
SELECT nt "Group", MIN(empno)||'-'||MAX(empno) "Range", COUNT(*) "Count"
FROM x
GROUP BY nt
ORDER BY 1;

Group Range           Count
------ ---------- ----------
1 7369-7654           5
2 7698-7844           5
3 7876-7934           4```

#2: Simulating NTile function

```WITH x AS (
SELECT empno, CEIL(ROW_NUMBER()OVER(ORDER BY empno)/CEIL(COUNT(*)OVER()/3)) nt
FROM emp
)
SELECT nt "Group", MIN(empno)||'-'||MAX(empno) "Range", COUNT(*) "Count"
FROM x
GROUP BY nt
ORDER BY 1;

Group Range           Count
------ ---------- ----------
1 7369-7654           5
2 7698-7844           5
3 7876-7934           4```

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.

## Round 1 Playoff Schedule

N teams (N between 1 and 32) just finished the season and are all qualified for the playoff. If the number of teams were 2, 4, 8, 16, or 32 (powers of 2), the playoff schedule would be trivial: 1st team plays vs last team, 2nd – vs 2nd from the last, etc. However, there is no guarantee that the number of teams would be a power of 2. The challenge is to write a single SELECT statement that accepts the number of teams as a parameter and generates the round 1 pairings.

There should be 1, 2, 4, 6, or 16 teams (power of 2) in the 2nd round.

## Solution

```WITH x AS (
SELECT &teams p
FROM dual
), y AS (
SELECT ROWNUM home, POWER(2,CEIL(LOG(2,p)))-ROWNUM+1 away,
POWER(2,CEIL(LOG(2,p))) maxp
FROM dual, x
CONNECT BY LEVEL<=POWER(2,CEIL(LOG(2,p))-1)
)
SELECT CASE WHEN away<=p AND p>1 THEN ROWNUM+p-maxp END AS "Game #",
CASE WHEN away>p AND p>1 THEN 'Team-' || home || ' advances to Round 2'
WHEN p=1 THEN 'Team-1 is a Champion!'
ELSE 'Team-' || home || ' vs Team-' || away
END AS "Playoff Round 1 Pairings"
FROM y, x
ORDER BY 1 NULLS LAST, home
```

## Explanation

The trick here was to figure out which teams should be playing and which simply advance to the Round 2. Suppose that we have a power of 2 number of teams. Then top 1st team plays against the bottom 1st, top 2ng vs bottom 2nd, etc. If there are N teams, then we will have N/2 games. This is the simplest case. What if we have 6 teams? We should add 2 fake teams to “round up” the number of teams to the nearest power of 2 that is greater or equal to N. For 6 teams, we round up to 8. Those 2 fake teams should be paired against top 2 teams, and this gives us an answer which teams should advance to the Round 2 without playing. General rule, if we have to add K “faked teams” to “round up” to the nearest power of 2 number, this means that top K teams advance to the next round without playing.

We used CEIL, LOG, and POWER functions to get the next power of 2 for any whole N:

```POWER(2,CEIL(LOG(2,p)))
```

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.