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Need algorithm
Message
From
07/07/2011 09:33:09
 
General information
Forum:
Games
Category:
Trivia
Title:
Miscellaneous
Thread ID:
01517506
Message ID:
01517572
Views:
40
>>>>>>I need to create algorithm (based on an integer number) that by applying some formula or equation to this number I can identify if the number belongs to group 1, 2, or 3 or any combination of them (e.g. 1 and 2, 1 and 3, 1 and 2 and 3, 2 and 3)
>>>>>>For example, number 8 would indicate belonging to all three groups (since I can divide this number by the group number without a remainder.
>>>>>>Any suggestions?
>>>>>
>>>>>It just occured to me that the answer is very simple.
>>>>
>>>>
>>>>Is this about your options ? (from yesterday)
>>>>
>>>>Then the groups are just bits. Test witth BitTest()
>>>>
>>>>And belonging to all three groups = 7
>>>
>>>Yes, this is a follow up to the options thread from yesterday. But I can't seem to grasp how to apply BitTest() to this issue. I can't understand how the value of 7 by using BitTest() indicates that 7 belongs to all three groups? Could you please explain?
>>
>>
>>option1 is 2^0 , ie 0 or 1
>>option2 is 2^1 , ie 0 or 2
>>option3 is 2^2 , ie 0 or 4
>>
>>
>>So, if all options are set, you get 4+2+1 = 7
>>
>>
>>n = 7
>>
>>hasOption1 = bittest(n, 0)  && is bit 0 set  ?
>>
>>hasOption2 = bittest(n, 1) && is bit 1 set ?
>>
>>hasOption3 = bittest(n, 2) && is bit 2 set ?
>>
>>
>>
>>If n is 5 ( 1+4) then it's option1 and option3
>
>Thank you. I think between the explanation you kindly provided and the articles Naomi provided I will finally get it.


It's just the sum of some powers of 2. Each bit represents a power of 2

On a 32 bit system:

The sum of 2^n, for n = 0 to 31, is 2^32 -1

In your case

3 options, going from 0 to 2

The sum of 2^n, for n = 0 to 2, is 2^3 - 1 (8-1 = 7) . In total 2^3 possibilities
Gregory
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