for...to...finding factors - almost got it but....

[christophe]

Hi,

My code below generates a random number from -20 to 20. When command2 is clicked, the whole number factors of that number generated are listed in Picture2.

My problem is that I don't want factors to be repeated.

For example, the random number generated in picture1, might be 6.

When command2 is clicked the factors listed will be:

-6 and -1
-3 and -2
-2 and -3
-1 and -6
1 and 6
2 and 3
3 and 2
6 and 1

You'll notice that factors are repeated, just in different order. Any ideas how I could have the above looking like:

-6 and -1
-3 and -2
1 and 6
2 and 3

Thanks for any help! I include my code below.

Dim number As Integer
Dim x As Integer

Private Sub Command1_Click()
Randomize
number = Int(Rnd * 40) - 20
Picture1.Cls
Picture1.Print number
End Sub

Private Sub Command2_Click()
Picture2.Cls
'Prevent backward For - To's like 7 to -7
'if the number generated is less than 0, for example: -7
If number < 0 Then
'd will equal -7
d = number
'and e will equal --7, i.e: +7, so we go from -7 to 7 rather than
'7 to -7, which For- To doesn't deal with
e = -number
Else:
d = -number
e = number
End If
For x = d To e
'avoid division by 0 - gives an error
If x = 0 Then
x = 1
End If
'make sure we get a whole number factor
If number Mod x = 0 Then
'print all factors
Picture2.Print x & " and " & number / x
End If
Next
End Sub

[AKA]

Go from -round(number/2) to round(number/2). Maybe it will be a problem with the rounding if number is negative. If so use the Abs() function to get the absolut value.

[kedaman]

Actually go from -int(sqr(number/2)) to int(sqr(number/2)) and you'll avoid all double factors

[WP]

when i understand your quetion, this will be an answer:

when you ask a random number, its not really a random number, because all the 'random' numbers are listed on your pc, and will always come in the same order.
What I do when i need a random number, i also pick a random number with the rnd function, but i multiply it with the system timer, and i / it trough the date, so i have every day, every time another number

example

number = Int(Rnd * 40) - 20 
number = number * Timer
tmp = Date(Now)
number = number / tmp

WP

[AKA]

kedaman

You cant take a square root from a negative number.

You should not divide by two.

It should maybe look like this.

from -int(sqr(abs(number))) to int(sqr(abs(number)))

[kedaman]

Hehe, actually i can, but VB can't handle imaginary numbers I thought number was a positive number, the 20 he was talking about ;=)

[AKA]

You are so right I had forgoten imaginary numbers ! So long since scool, twelve years and not a single imaginary number.

[ianpaisley]

Everybody thanks,

AKA, what is Abs? Absolute number? (And what is an absolute number again?)

My code is working now, but I'd still like to know what's going on.

[AKA]

Abs function gives you the absoulte number IOW the positive number. Ex Abs(-6) => 6 and Abs(6) => 6.

[kedaman]

Hehe about 1½ year for me, and imaginary numbers showed up last time i talked to sam. You live in sweden? Jag har inte varit där på ett tag, men det är inte så svårt att ta sig dit om man bor på Åland

[christophe]

Hi there,

Thanks for your help. I got my problem sorted with the code below but I hope you can help me with the next part- pick the correct factors from the group of factors.

This is what my program does:

It generates quadratic equations at random and the objective for the student is to find the factors of that equation.

I generate the equations by a mathematical rule that I won't go into here - too complex. If you look at my code you'll see what my program does.

Since you're talking so much about old days in school, you might remember a rule for solving a simple quadratic equation is: find two numbers that multiply to give c, and add to give b.

For example, all equations can be written in the form: ax^2 + bx + c. a,b and c are generated at random, on the basis that the factos will be whole numbers.

If the equation generated is:
x^2+2x-8
Factors are: (x-4)(x+2) (-4 and 2 multiply to give -8, add to give 2)

x^2+ 3x-4
Factors are: (x+4)(x-1) (4 and -1 multiply to give -4 and add to give 3)

So, as you have helped me do, I can list the possible factors of the equation. Now I have to choose the correct factor, and the best way of doing this is by picking the two numbers that add to give b, thus elimating all the errors. But my program crashes every time I try it. If you could take the time to look at my code, I'd greatly appreciate it. In any case, thanks for your help!

---------------------------------------

'Five command buttons:command1,command2,command3,command4
'command5
'two picture boxes, picture1 and picture2

Option Explicit
Dim a As Integer
Dim b As Integer
Dim c As Integer
Dim x As Integer

Dim bplus As String
Dim cplus As String



Private Sub Command1_Click()
Dim factorsfound As Boolean
'this is a mathematical rule. Roots of a quadratic
'are always:

'x= -b + or - Sqr (b^2-4ac)/2a

Do

a = 1
b = randomfactors
c = randomfactors

bplus = IIf(b > 0, "+" & b & " ", b & " ")
cplus = IIf(c > 0, "+" & c & " ", c & " ")

If b ^ 2 - 4 * a * c >= 0 Then
If b ^ 2 - 4 * a * c >= 0 Then
If (b ^ 2 - 4 * a * c) = 1 Or (b ^ 2 - 4 * a * c) = 4 _
Or (b ^ 2 - 4 * a * c) = 9 Or (b ^ 2 - 4 * a * c) = 16 _
Or (b ^ 2 - 4 * a * c) = 25 Or (b ^ 2 - 4 * a * c) = 36 _
Or (b ^ 2 - 4 * a * c) = 49 Or (b ^ 2 - 4 * a * c) = 64 _
Or (b ^ 2 - 4 * a * c) = 81 Or (b ^ 2 - 4 * a * c) = 100 Then
If (-b + Sqr(b ^ 2 - 4 * a * c)) Mod 2 * a = 0 Then
If (-b - Sqr(b ^ 2 - 4 * a * c)) Mod 2 * a = 0 Then


factorsfound = True

End If
End If
End If
End If
End If

Loop Until factorsfound

Picture1.Cls
Picture1.Print " Factorise: x² " & bplus & cplus

Picture2.Cls
Picture2.Print a & " " & b & " " & c

End Sub


Private Function randomfactors() As Integer
Randomize
randomfactors = Int(Rnd * 40) - 20
If randomfactors = 0 Then
randomfactors = 1 + Int(Rnd * 10)
End If
End Function


Private Sub Command2_Click()
Picture2.Cls
Picture2.Print "Our factors will always be of the form (x+?)(x+?)."
Picture2.Print
Picture2.Print "The numbers that end up in the brackets will always"
Picture2.Print "multiply to give " & c & " and add to give " & b & "."
Picture2.Print "What numbers can you think of?"
End Sub


Private Sub Command3_Click()
Picture2.Cls
Picture2.Print "These pairs multiply to give " & c & ":"
Picture2.Print
For x = -Int(Sqr(Abs(c))) To Int(Sqr(Abs(c)))
'avoid division by 0 - gives an error
If x = 0 Then
x = 1
End If
'make sure we get a whole number factor
If c Mod x = 0 Then
'print all factors
Picture2.Print x & " and " & c / x
End If
Next
Picture2.Print
Picture2.Print "Which pair adds to give " & b & "?"
End Sub
'Program crashes here
Private Sub Command4_Click()
Dim addfactors As Boolean

Picture2.Cls
Do

If x = 0 Then
x = 1
If x + (c / x) = b Then

addfactors = True
End If
End If

Loop Until addfactors
Picture2.Print x & " " & c / x
End Sub

[Sam Finch]

Christophe,

You're going about the whole thing the wrong way mate, with all the questions start of with the answer and then work out the question.


pick an answer in the form

(ax - b)(cx - d) = 0


then you can work out the equation

(ac)x^2 - (ad + bc)x + (bd) = 0

so have something like this

Dim a As Integer
Dim b As Integer
Dim c As Integer
Dim d As Integer


Public Sub getquestion()

a = Fix(Rnd * 4) + 1
b = Fix(Rnd * 9) + 1
c = Fix(Rnd * 4) + 1
d = Fix(Rnd * 9) + 1

Form1.Pictureblackbrd.Cls
Form1.Pictureblackbrd.Print 
Form1.Pictureblackbrd.Print "We are given the equation"
Form1.Pictureblackbrd.Print 
Form1.Pictureblackbrd.Print CStr(a*c) & "x² - " & CStr((a * d) + (b * c)) & " x + " & CStr(b * d)
Form1.Pictureblackbrd.Print 
Form1.Pictureblackbrd.Print " Find the value of x."

End Sub

Then the rest is easy.

[AKA]

This thread show a for me as a system developer at an "inhouse" IT department a common problem. The users always comes with the question can you do this change to this list/app. You normaly can do it but you always will do a better solution if you ask the question "Wath do you want to achive ?".

Christophe's last mail that the one he realu wanted help with. But he of course learnt more from all answers this way .

[kedaman]

yep, for an equation
ax^2+bx+c=0

where the discriminand, the part under the root is:
D=b^2-4ac

So you have three formulas depending on D

if D>0 then
  x1=(-b-sqr(D)/2a
  x2=(-b+sqr(D)/2a
elseif D=0
  x1i=sqr(-D)/2a 'imaginary 1
  x2i=-sqr(-D)/2a 'imaginary 2
  xr=sqr(-D)/2a 'real
else
  x=-b/2a
end if

If you have a negative discriminand, you get two roots, wich should be their conjugate, (hmm is this the correct term?) imaginary values.