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[Question] VB6 random code handling

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mathewthe2
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Joined: 28 Aug 2008
Posts: 562
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Posted: Fri Jan 30, 2009 3:09 am Post subject: [Question] VB6 random code handling

Ok I made a game in vb6, a fairly simple game, what I want to do is to let the computer pick a random number from variable(integer) x to y, and so I put the following code:

Code:

Dim x As Integer
Dim y As Integer
Dim computerpick As Integer
Dim aroundnumber As Integer
x = Val(maximumnumber.Caption)
y = Val(minimumnumber.Caption)
aroundnumber = x - y
computerpick = (Rnd()) *aroundnumber + y
Show.Caption = computerpick

And it worked fine, the computer picks a number between x and y(for example if x = 1, y = 20, it would choose 3,4,7,19 etc.)
According to theology, if
aroundnumber = x - y and computer pick = aroundnumber + y,
computer pick will equal to x - y + y(which is x), so I changed the code to

Code:

Dim x As Integer
Dim y As Integer
Dim computerpick As Integer
Dim aroundnumber As Integer
x = Val(maximumnumber.Caption)
y = Val(minimumnumber.Caption)
computerpick = (Rnd()) *x
Show.Caption = computerpick

which doesn't work and picked a number smaller than minimum, I wonder how does the random code(the number after * is the maximum the number could go)work.
Can anyone tell me I know you are reading this!
Also, I included the game in attachment, read instructions!

The Extension 'rar' was deactivated by an board admin, therefore this Attachment is not displayed.

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Snootae
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Posts: 969
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Posted: Fri Jan 30, 2009 5:48 am Post subject:

this works because

Code:

computerpick = (Rnd()) *aroundnumber + y

you have clearly found the difference, hence how small or big the random number can be, then added the minimum

this fails because

Code:

computerpick = (Rnd()) *x

you completely ignored the minimum, this will give you a number between 0 and x, hence not helpful

Code:

computerpick = ((Rnd()) *x) + y

that will work

if you want it even shorter

Code:

Show.Caption = ( (Rnd()) *(Val(maximumnumber.Caption) - Val(minimumnumber.Caption)) + Val(minimumnumber.Caption) )

more confusing but you dont need to make variables for things you only use once really

edit: found the hole

Quote:

aroundnumber = x - y and computer pick = aroundnumber + y

computer pick = ((random value) * aroundnumber) + y, hence why you cant simplify it
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mathewthe2
Grandmaster Cheater

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Joined: 28 Aug 2008
Posts: 562
Location: behind you... BOO!

Posted: Fri Jan 30, 2009 9:12 am Post subject:

Snootae wrote:

this works because

Code:

computerpick = (Rnd()) *aroundnumber + y

you have clearly found the difference, hence how small or big the random number can be, then added the minimum

this fails because

Code:

computerpick = (Rnd()) *x

you completely ignored the minimum, this will give you a number between 0 and x, hence not helpful

Code:

computerpick = ((Rnd()) *x) + y

that will work

if you want it even shorter

Code:

Show.Caption = ( (Rnd()) *(Val(maximumnumber.Caption) - Val(minimumnumber.Caption)) + Val(minimumnumber.Caption) )

more confusing but you dont need to make variables for things you only use once really

edit: found the hole

Quote:

aroundnumber = x - y and computer pick = aroundnumber + y

computer pick = ((random value) * aroundnumber) + y, hence why you cant simplify it

yes i know that code number2 has ignored minium, thank you for replying, now i know that there is brackets in between even if it is the same number in mathematics.
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