I need help with algorithms if possible

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exoticTaytay606
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Joined: 19 Jun 2018, 21:02
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I need help with algorithms if possible

19 Jun 2018, 21:09

So I have been working on a script but I need this:
A, b = 0, 1
While b > 5 A = 0 [5 is 3x5, 3 is whatever you want.]
Else A = 1 [1 is B-3, B is 37/50(74%) of the total of 5, B = A + 1]
[Divide the value of 5 by 26 to get B]
[A is a 3 stage fraction, of the final values]
3=4 so 5 3's = 20
At that point b would be 16
So b-3
Is 16-4
So 12
turned into a script that will output the different numbers created by it. Could someone please help once I'm done the script I intend on sharing it I just got stuck on this part.
Thanks in advance!
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DyaTactic
Posts: 183
Joined: 04 Apr 2017, 05:52

Re: I need help with algorithms if possible

20 Jun 2018, 15:26

I do understand mathemethics at a certain level but I belive this is not really ment to be a 'mathematical' explanation, is it? In other words, it's hard to understand what you want. Can you tell more about the goal of this algarithm? Maybe that clarifies one and other.
I'll write down in AHK what I do get, maybe that makes it easier to explain, just giving some handles:

Code: [Select all] [Download] GeSHi © Codebox Plus

A = 0
B = 1 ; Set the startvalues of A and B to 0 and 1.

If (B > 5) { ; In the case B is more then 5.
A = 1 ; A gets the value 1.
} Else If (A=1) { ; In the other case, in which A is equal to 1.
B := 5 / 26 ; Divide 5 by 26 and save the value in B.
}
; And the rest is even harder...

I'll be happy to hear from you.
exoticTaytay606
Posts: 2
Joined: 19 Jun 2018, 21:02
Facebook: https://www.facebook.com/profile.php?id=1000196606
Google: https://plus.google.com/u/0/102851039320256239379
GitHub: exoticTaytay606

Re: I need help with algorithms if possible

21 Jun 2018, 08:46

The goal of this algorithm is to output the number it produces through the 3 stage fraction using the other numbers produced by a,b, and I don't really care what b equals; and then to have it press enter. To be honest even I struggle with this math it's my good friends equation that he figured out all I'm trying to do is to trick the thing the equation comes from to just about always get the number it needs to advance.
3 = 4 so times 5 which is 3 by 5 so you get 20 (the total of B) then b -3 so 20 -4 is 16 then 16 - 3(4) so you get 12
I suck at explaining things XD

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