RE: SQL RAND function question -- MIDRANGE-L

That doesn't work. Using just (INT(RAND() * 12) gives 0..12. But only a handful of 12's. Which I
don't want anyway :) (INT(RAND() * 11) gives 0..11 but just a handful of 11's.

It appears that IBM's RAND() function returns [0,1] inclusive.

Adam's rewrite of my working solution, floor( rand() * (n+0.99999) ) seems to be the best bet for
getting an even distribution between 0 and 11.

Look back at my results and Adams posts.

Charles Wilt
Software Engineer
CINTAS Corporation - IT 92B
513.701.1307

wiltc@xxxxxxxxxx

-----Original Message-----
From: midrange-l-bounces@xxxxxxxxxxxx [mailto:midrange-l-
bounces@xxxxxxxxxxxx] On Behalf Of Michael_Schutte@xxxxxxxxxxxx
Sent: Tuesday, February 12, 2008 3:42 PM
To: Midrange Systems Technical Discussion
Subject: RE: SQL RAND function question

So don't add 1.

Michael Schutte
Admin Professional
Bob Evans Farms, Inc.
"The Secret's the Sauce! Enjoy our new Bob-B-Q Pulled Pork Knife & Fork
Sandwich!"

midrange-l-bounces@xxxxxxxxxxxx wrote on 02/12/2008 03:40:34 PM:

Michael,

In this case I want 0..11 not 1..12.

Charles Wilt
Software Engineer
CINTAS Corporation - IT 92B
513.701.1307

wiltc@xxxxxxxxxx

-----Original Message-----
From: midrange-l-bounces@xxxxxxxxxxxx [mailto:midrange-l-
bounces@xxxxxxxxxxxx] On Behalf Of Michael_Schutte@xxxxxxxxxxxx
Sent: Tuesday, February 12, 2008 2:51 PM
To: Midrange Systems Technical Discussion
Subject: RE: SQL RAND function question

If I remember correctly.... in TI-83 calculator (I know this is

going
to

make me sound real geeky!!!!), if I wanted to get a random number

between

1 and 100. I had to take the random function and multiply by 100 then

add

1. Because int(rand() * 100) resulted in 0 through

99.9999999999999999

etc... I believe that's where I was doing that calculation.

So to answer your question... shouldn't it be...

(INT(RAND() * 12) + 1) as RandonNumber

Michael Schutte
Admin Professional
Bob Evans Farms, Inc.
"The Secret's the Sauce! Enjoy our new Bob-B-Q Pulled Pork Knife &

Fork

Sandwich!"

midrange-l-bounces@xxxxxxxxxxxx wrote on 02/12/2008 02:44:30 PM:

Adam,

Thanks for the explanation, that's what I was thinking, but you
spelled it out nicely.

Like you, I'm a little confused about getting half the 1's I
expected. Getting half the 0's would
have made sense.

Still I'm confused, IIRC every example I've seen about using _any_
built in random number function has
said basically, they return [0,1) or [0,1] and to get a number
between a given range multiply by the
range. (IBM's SQL manual included!)

Obviously, that doesn't hold true for the reasons you've explained.

So what is the "right" answer to using such a built in random number
function?

Charles Wilt
Software Engineer
CINTAS Corporation - IT 92B
513.701.1307
wiltc@xxxxxxxxxx

-----Original Message-----
From: midrange-l-bounces@xxxxxxxxxxxx [mailto:midrange-l-
bounces@xxxxxxxxxxxx] On Behalf Of Adam Glauser
Sent: Tuesday, February 12, 2008 1:59 PM
To: midrange-l@xxxxxxxxxxxx
Subject: Re: SQL RAND function question

Wilt, Charles wrote:

I'm thinking perhaps the way I'm going from the RAND() results

to
a

integer from 0-11 is the problem.

Anybody understand what I'm doing wrong?

myDate = date('2008-03-01') + cast(round((rand(12) * 11),0) as

int)

months

I don't think round() is what you want, although I'm confused a

bit
by

the distribution of results that you stated. Anyway, here is why

I

think that you got about half as many 11s as you expected:

rand(12) * 11 gives a double float number r, in the range [0,11].
round(r) gives an integer in the set Z' (0,1,2,...,11).

What values result in each number in the set Z'?
if r is in [0,0.5), r' = 0
if r is in [0.5,1.5), r' = 1
if r is in [1.5,2.5), r' = 2
...
if r is in [9.5 ,10.5), r' = 10
if r is in [10.5,11] , r' = 11

So, there are only half as many values that will lead to a result

of

either 0 or 11 as there are that lead to any of the other numbers.

Of course, the problem with this conclusion is that it doesn't

explain

why you got half as many 1s as you were expecting.
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