A guide on breeding Unown's

Pokemon can get eggs, in order to do so you have to put them into the daycare - keep in mind: One Pokemon has to be male and the other has to be female. There are a few exceptions to that like breeding with Ditto but in most cases you want a male and a female pokemon. Also besides having the right gender they need to have the same egg group in order to actually get some eggs (from what I've seen the DayCare checks that for you ... after putting the first Pokemon into the DayCare you can only choose fitting Pokemon). Once those two are in the DayCare you'll get a message on how likely those Pokemon are to get an egg. The higher the chance the better. Now that you've placed those Pokemon into the DayCare you'll need to interact with other Pokemon (basic interactions, hold the egg, give a berry ... each interaction earns you one interaction point). After you've hit a certain amount of interactions (I think it's 64 or 128 ... not to sure so ... just interact a lot and you'll be fine) your pokemon will try to breed an egg. Now it's luck based on how likely your pokemon actually get an egg ... but when you interact a lot you'll get your egg eventually. Then adopt the egg and tada, you succesfully breeded a Pokemon, have fun hatching the egg.

Remember what I said about genders? Screw that! Unowns are special! They can breed with each other, just put two Unown's into the Daycare and have fun breeding them. But that's not all there is to it, here comes the important part: A certain pair of Unowns will always have the same child Unown! For example: If you put "A" and "B" into the DayCare you'll ALWAYS get "C". Now you might ask: Is there an easy system to that childreen system? Yes there is! There is an easy to memorize patern that you can use to get all Unowns once you own two (unless you got really unlucky) here is the pattern: A+B=C A+C=E A+D=G and so on. If someone's able to word it better you are welcome to do so ... I'll quote it and add it to the original post. Also yeah, this also goes for other pairs, like this: B+C=D X+Y=Z and so on. Have fun breeding Unowns and make sure to do it now, cause I have the feeling there'll be a lot of Unown's in the shelter to start this soon!

I'm not that great with explanations, but... I hope that helped. If anyone cares to explain more concisely, please do. Note: This is purely based on the pattern given by one test performed (for an overshot scenario) and the information given in the first post (after verification using two different breeding pairs)

So are you saying that it goes by the number that corresponds to the letter of the alphabet? Like A would be 1 and B would be 2 and C would be 3. A+B=C and 1+2=3, so is it like that? If I'm correct If I breeded, say, M(13) and P(16) I would always get C(13+16=29-26=3)?

It looks more to be the difference in number of letters, A+B=C and 1+2=3, but A+C=E and 1+3=/=5 or A+D=G and 1+4=/=7 or X+Y=Z and 24+25=/=26. It seems that the "larger" of the parents (number wise) is the median/average between the child and the smaller parent (eg D is the midpoint between A and G, C is the midpoint between A and E, and P would be the midpoint between M and S.) If the information is correct and I'm thinking right, M and P should breed S, not C. As far as a pattern for parents far enough apart to overshoot Z, I don't know how they are handled. Edit: I'm testing E and Z to check for an overshoot, the child should be U (if I counted right) if the pattern above holds and overshooting wraps the alphabet. ReEdit: I forgot about ? and !, looking at S now instead of U.

*shrugs* Maybe there is more experimenting to be done. I should work on this at some point.

Looks like it is based off the amount of separation between letters and it wraps on overshoot, E+Z=S (after taking into account ? and !)

That's interesting. I can't find any mathematical sense in that, but that may only be because I haven't learned enough math. If you know the mathematical sense could you please explain?

If numbers are assigned to each unown (1-28 for a-z, ?, and !), for a given breeding pair, the difference between the number values is also the difference between the higher value parent and the child. For example parents = A & J or 1 & 10, 10-1=9 and 10+9=19, the 19th letter is S, therefore a breeding pair of A & J would produce S offspring. In the case of my test pair of E & Z or 5 & 26, the difference is 21 and 26+21=47, which is larger than the maximum of 28 (it overshoots the maximum by 19), if the alphabet restarts/wraps (as appears to be the case) any amount of excess (19 here) starts counting at the beginning, A (since the overshoot in this example is 19, the child is S, the 19th letter).

Ah, so I was right about it wrapping around. So by these calculations, B(2) and Q(17) would breed to make D(4). (17-2=15, 15+17=32, 28-32=4) Hmm... I'm tempted to collect a bunch of each Unown and test all this out.

No, B(2) and Q(17) would breed to make T(19) I believe, since B+Q in this case equals 19. I think I've followed the post well enough to make that deduction.

Oh, well I apologize if I'm wrong. I might've done my math wrong. X3 I'm going to go over it again real quick... B = 2 Q = 17 17 - 2 =15 15 + 17 = 32 32 - 28 = 4 4 = D O-O I got the same thing. Could you try working your equation out so I could see what I might've done wrong..?