randomising for a quiz

[WindAndWaves]

Dear All

I am making a quiz with a four-option-multiple-choice scenario.

When you enter the quiz, you provide one answer and three wrong answers.

Then the contestant opens the quiz and I trying to present them the four
answers on the page in a randomized way (i.e. you have no idea which one
comes first).

I know the rnd command, but how do you do rnd without replacement?

TIA

- Nicolaas

[Mike Mc]

This is math not a computer group.
"WindAndWav es" <ac****@ngaru.c om> wrote in message
news:TH******** ************@ne ws.xtra.co.nz.. .

Dear All

I am making a quiz with a four-option-multiple-choice scenario.

When you enter the quiz, you provide one answer and three wrong answers.

Then the contestant opens the quiz and I trying to present them the four
answers on the page in a randomized way (i.e. you have no idea which one
comes first).

I know the rnd command, but how do you do rnd without replacement?

TIA

- Nicolaas

[Michael Gray]

On Fri, 27 Aug 2004 15:14:52 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

Dear All

I am making a quiz with a four-option-multiple-choice scenario.

When you enter the quiz, you provide one answer and three wrong answers.

Then the contestant opens the quiz and I trying to present them the four
answers on the page in a randomized way (i.e. you have no idea which one
comes first).

I know the rnd command, but how do you do rnd without replacement?

TIA

- Nicolaas

Set up an array of 4 elements.
Randomly choose one.
Remove it from the list.
Randomly choose one of the 3.
And so on.

[WindAndWaves]

Dear Michael

I had a go at this and this is what I came up with (slightly different):

Private Sub Form_Current()
'randomises the anwsers
Const ProEro = 12: 'on error GoTo ERR
'---
Dim Ctl As Control
Dim C(3) As Byte
Dim I As Integer
Dim J As Integer
'---assign numbers
For I = 0 To 3
TryAgain:
C(I) = Int(Rnd * 5)
If C(I) = 0 Or C(I) = 5 Then GoTo TryAgain
If I = 0 Then GoTo looper
'--- check if it exists already and try again if this is so
For J = (I - 1) To 0 Step -1
If C(I) = C(J) Then GoTo TryAgain
Next J
looper:
Next I
'---place data
For I = 0 To 3
Set Ctl = Me.Controls("O" & C(I))
Ctl.ControlSour ce = "=[QA" & I & "]"
Ctl.Tag = I
'---reset tickbox
Me.Controls("C" & I + 1) = False
Next I
xit:
Exit Sub
ERR:
Call FerrorLog(ERR.N umber, 0, ProEro + Modero): Resume Next
End Sub
"Michael Gray" <fl****@newsguy .spam.com> wrote in message
news:dh******** *************** *********@4ax.c om...

On Fri, 27 Aug 2004 15:14:52 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

Dear All

I am making a quiz with a four-option-multiple-choice scenario.

When you enter the quiz, you provide one answer and three wrong answers.

Then the contestant opens the quiz and I trying to present them the four
answers on the page in a randomized way (i.e. you have no idea which one
comes first).

I know the rnd command, but how do you do rnd without replacement?

TIA

- Nicolaas

Set up an array of 4 elements.
Randomly choose one.
Remove it from the list.
Randomly choose one of the 3.
And so on.

[Michael Gray]

On Sat, 28 Aug 2004 00:21:41 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

Dear Michael

I had a go at this and this is what I came up with (slightly different):

:

If all you are trying to achieve is to get it to work, by far and away
the most efficient method (but no the most elegant perhaps), is to
pre-populate an array with the 24 possible permutations (of the four
questions), and choose one randomly from this list.
This is much more efficient than any other way.

So you would do something like this the first time the program runs:

Dim Permutation(24) As String
Permutation(0) = "1234"
Permutation(1) = "1243"
Permutation(2) = "1324"
Permutation(3) = "1342"
Permutation(4) = "1432"
Permutation(5) = "1423"
Permutation(6) = "2134"
Permutation(7) = "2143"
Permutation(8) = "2314"
Permutation(9) = "2341"
Permutation(10) = "2431"
Permutation(11) = "2413"
Permutation(12) = "3124"
Permutation(13) = "3142"
Permutation(14) = "3214"
Permutation(15) = "3241"
Permutation(16) = "3412"
Permutation(17) = "3421"
Permutation(18) = "4123"
Permutation(19) = "4132"
Permutation(20) = "4213"
Permutation(21) = "4231"
Permutation(22) = "4312"
Permutation(23) = "4321"

(I have used strings here for ease of demonstration only, and assume
an array base of zero [Option Base 0])

Then your subroutine would select one of these 24 elements at random,
and apply the order chosen.
I will leave it to you work out the mechanics...

[WindAndWaves]

That is a nice solution Michael, thank you for that idea.
"Michael Gray" <fl****@newsguy .spam.com> wrote in message
news:d4******** *************** *********@4ax.c om...

On Sat, 28 Aug 2004 00:21:41 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

Dear Michael

I had a go at this and this is what I came up with (slightly different):

:

If all you are trying to achieve is to get it to work, by far and away
the most efficient method (but no the most elegant perhaps), is to
pre-populate an array with the 24 possible permutations (of the four
questions), and choose one randomly from this list.
This is much more efficient than any other way.

So you would do something like this the first time the program runs:

Dim Permutation(24) As String
Permutation(0) = "1234"
Permutation(1) = "1243"
Permutation(2) = "1324"
Permutation(3) = "1342"
Permutation(4) = "1432"
Permutation(5) = "1423"
Permutation(6) = "2134"
Permutation(7) = "2143"
Permutation(8) = "2314"
Permutation(9) = "2341"
Permutation(10) = "2431"
Permutation(11) = "2413"
Permutation(12) = "3124"
Permutation(13) = "3142"
Permutation(14) = "3214"
Permutation(15) = "3241"
Permutation(16) = "3412"
Permutation(17) = "3421"
Permutation(18) = "4123"
Permutation(19) = "4132"
Permutation(20) = "4213"
Permutation(21) = "4231"
Permutation(22) = "4312"
Permutation(23) = "4321"

(I have used strings here for ease of demonstration only, and assume
an array base of zero [Option Base 0])

Then your subroutine would select one of these 24 elements at random,
and apply the order chosen.
I will leave it to you work out the mechanics...

[Michael Gray]

On Sun, 29 Aug 2004 08:46:02 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

That is a nice solution Michael, thank you for that idea.

Actually it's not my idea.
It dates from ancient times.
Logarithm tables are a good example, as are "Ready Reckoners" which
were essential in the days of Pounds, Shillings & Pence.
Often pre-populated tables are the most effective way to go with a lot
of problems, although it's surprising how many programmers shy away
from them, or don't even think of them.
They are particularly handy in programs that perform intensive
trigonometry. (Even if a F.P. processor is available)
A table of sin(x) covering an 8th of a circle to the required
precision is all that one needs. The rest of the circle can be
synthesised without having to call transcendental functions
repeatedly.
It's used a lot by people who program microprocessors in machine code,
who *do* tend to think this way.

Michael G.

[David Bandel]

Michael Gray <fl****@newsguy .spam.com> wrote in message news:<d4******* *************** **********@4ax. com>...

On Sat, 28 Aug 2004 00:21:41 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

Dear Michael

I had a go at this and this is what I came up with (slightly different):

:

If all you are trying to achieve is to get it to work, by far and away
the most efficient method (but no the most elegant perhaps), is to
pre-populate an array with the 24 possible permutations (of the four
questions), and choose one randomly from this list.
This is much more efficient than any other way.

So you would do something like this the first time the program runs:

Dim Permutation(24) As String
Permutation(0) = "1234"
Permutation(1) = "1243"
Permutation(2) = "1324"
Permutation(3) = "1342"
Permutation(4) = "1432"
Permutation(5) = "1423"
Permutation(6) = "2134"
Permutation(7) = "2143"
Permutation(8) = "2314"
Permutation(9) = "2341"
Permutation(10) = "2431"
Permutation(11) = "2413"
Permutation(12) = "3124"
Permutation(13) = "3142"
Permutation(14) = "3214"
Permutation(15) = "3241"
Permutation(16) = "3412"
Permutation(17) = "3421"
Permutation(18) = "4123"
Permutation(19) = "4132"
Permutation(20) = "4213"
Permutation(21) = "4231"
Permutation(22) = "4312"
Permutation(23) = "4321"

(I have used strings here for ease of demonstration only, and assume
an array base of zero [Option Base 0])

Then your subroutine would select one of these 24 elements at random,
and apply the order chosen.
I will leave it to you work out the mechanics...

or.. more simply and elegantly.. pick a random number from 1 to 24..

then u can sieve through the # to find out which permutation it is..

if it's in the first 6.. 1 is first
second 6 2 is first
etc.

within that group of six..
if it's in the first 2.. etc.

in the 2.. the order of the two will determine the order of the last two...
it's very simple..

[Michael Gray]

On 15 Sep 2004 19:21:56 -0700, dw*****@yahoo.c om (David Bandel) wrote:

Michael Gray <fl****@newsguy .spam.com> wrote in message news:<d4******* *************** **********@4ax. com>...

On Sat, 28 Aug 2004 00:21:41 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

>Dear Michael
>
>I had a go at this and this is what I came up with (slightly different):
>

:

If all you are trying to achieve is to get it to work, by far and away
the most efficient method (but no the most elegant perhaps), is to
pre-populate an array with the 24 possible permutations (of the four
questions), and choose one randomly from this list.
This is much more efficient than any other way.

So you would do something like this the first time the program runs:

Dim Permutation(24) As String
Permutation(0) = "1234"
Permutation(1) = "1243"
Permutation(2) = "1324"
Permutation(3) = "1342"
Permutation(4) = "1432"
Permutation(5) = "1423"
Permutation(6) = "2134"
Permutation(7) = "2143"
Permutation(8) = "2314"
Permutation(9) = "2341"
Permutation(10) = "2431"
Permutation(11) = "2413"
Permutation(12) = "3124"
Permutation(13) = "3142"
Permutation(14) = "3214"
Permutation(15) = "3241"
Permutation(16) = "3412"
Permutation(17) = "3421"
Permutation(18) = "4123"
Permutation(19) = "4132"
Permutation(20) = "4213"
Permutation(21) = "4231"
Permutation(22) = "4312"
Permutation(23) = "4321"

(I have used strings here for ease of demonstration only, and assume
an array base of zero [Option Base 0])

Then your subroutine would select one of these 24 elements at random,
and apply the order chosen.
I will leave it to you work out the mechanics...

or.. more simply and elegantly.. pick a random number from 1 to 24..

then u can sieve through the # to find out which permutation it is..

if it's in the first 6.. 1 is first
second 6 2 is first
etc.

within that group of six..
if it's in the first 2.. etc.

in the 2.. the order of the two will determine the order of the last two...
it's very simple..

I agree.

But for time critical routines, there's no substitute for a
pre-populated table.
I don't implement it as above, of course, but as an array of arrays.
One random look-up gives you the order without having to go through
seiving ranges.
They are "Pre-seived" as it were.

But you are right: your way is more elegant,
but I dispute that it is "simpler" to actually implement.

[G.J. v.d. Kamp]

Here's what i came up with:

I assume the table with the questions has the following format:

Id
Question
RightAnswer
WrongAnswer1
WrongAnswer2
WrongAnswer3

Now the following query will give you the answers to question 1 in
random order:

SELECT Answer, Correct
FROM
(SELECT RND() AS Position, RightAnswer AS Answer, True AS Correct FROM
Questions WHERE Id = 1
UNION ALL
SELECT RND() AS Position, WrongAnswer1, False AS Correct FROM
Questions WHERE Id = 1
UNION ALL
SELECT RND() AS Position, WrongAnswer2, False AS Correct FROM
Questions WHERE Id = 1
UNION ALL SELECT RND() AS Position, WrongAnswer3, False AS Correct
FROM Questions WHERE Id = 1)
ORDER BY Position;

I hardcoded question 1 in there to keep it simple, when you open this
in VBA you replace that with the Id of the question you want to ask.
It almost as lengthy as the other alternative, it just looks more
'pure' to me.

Regards,

GJ
Michael Gray <fl****@newsguy .spam.com> wrote in message news:<k8******* *************** **********@4ax. com>...

On 15 Sep 2004 19:21:56 -0700, dw*****@yahoo.c om (David Bandel) wrote:

Michael Gray <fl****@newsguy .spam.com> wrote in message news:<d4******* *************** **********@4ax. com>...

On Sat, 28 Aug 2004 00:21:41 +1200, "WindAndWav es" <ac****@ngaru.c om>
wrote:

>Dear Michael
>
>I had a go at this and this is what I came up with (slightly different):
>
:

If all you are trying to achieve is to get it to work, by far and away
the most efficient method (but no the most elegant perhaps), is to
pre-populate an array with the 24 possible permutations (of the four
questions), and choose one randomly from this list.
This is much more efficient than any other way.

So you would do something like this the first time the program runs:

Dim Permutation(24) As String
Permutation(0) = "1234"
Permutation(1) = "1243"
Permutation(2) = "1324"
Permutation(3) = "1342"
Permutation(4) = "1432"
Permutation(5) = "1423"
Permutation(6) = "2134"
Permutation(7) = "2143"
Permutation(8) = "2314"
Permutation(9) = "2341"
Permutation(10) = "2431"
Permutation(11) = "2413"
Permutation(12) = "3124"
Permutation(13) = "3142"
Permutation(14) = "3214"
Permutation(15) = "3241"
Permutation(16) = "3412"
Permutation(17) = "3421"
Permutation(18) = "4123"
Permutation(19) = "4132"
Permutation(20) = "4213"
Permutation(21) = "4231"
Permutation(22) = "4312"
Permutation(23) = "4321"

(I have used strings here for ease of demonstration only, and assume
an array base of zero [Option Base 0])

Then your subroutine would select one of these 24 elements at random,
and apply the order chosen.
I will leave it to you work out the mechanics...

or.. more simply and elegantly.. pick a random number from 1 to 24..

then u can sieve through the # to find out which permutation it is..

if it's in the first 6.. 1 is first
second 6 2 is first
etc.

within that group of six..
if it's in the first 2.. etc.

in the 2.. the order of the two will determine the order of the last two...
it's very simple..

I agree.

But for time critical routines, there's no substitute for a
pre-populated table.
I don't implement it as above, of course, but as an array of arrays.
One random look-up gives you the order without having to go through
seiving ranges.
They are "Pre-seived" as it were.

But you are right: your way is more elegant,
but I dispute that it is "simpler" to actually implement.