Talk:Hidden Power (move)/Calculation: Difference between revisions

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--[[User:Null Set|Null Set]] 02:42, 11 November 2009 (UTC)
--[[User:Null Set|Null Set]] 02:42, 11 November 2009 (UTC)
== Possible HP Combos ==
The Section about the number of possible Hidden Power combinations is not entirely correct. The section says that since there are 32 possible values for each IV, the number of possible combinations is 32^6. If you get down to purely what the outcome is, then there are only 4^6 combinations. If you graph the outcomes for for the possible numbers this is what you get.
{|
|
|0
|1
|2
|3
|4
|5
|6
|7
|8
|9
|10
|-
|Type
|0
|1
|0
|1
|0
|1
|0
|1
|0
|1
|0
|-
|Power
|0
|0
|1
|1
|0
|0
|1
|1
|0
|0
|1
|}
There are four possible out comes for each IV Value; 00, 01, 10 and, 11. For any one who doesn't know binary these values equal the numbers 0, 1, 2 and, 3. Basically, you take the binary version of the IV and look at the last two bits to get the the outcome. This means that even though each IV is a number 0-31, the outcome boils down to a number 0-3. So there are 4 significant values for each IV, and there are 6 IV's so there are 4^6 (2^12) combinations for Hidden Power. That comes out to be 4096 possible outcomes. Far less than this article's stated 32^6(1073741824) outcomes. For any one who is confused about all of this I can explain it in further detail. Tell me watcha think! [[User:Professor|The Professor]] 20:21, 25 June 2010 (UTC)
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