Coloured tones?

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Chaos soda
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Coloured tones?

Post by Chaos soda »

I used to think the system worked like that. Like you could breed a white chao with an orange chao and end up with an orange with white tips. I think this would be cool and would bring a whole new level to the variety bringing the known chao count up from 82 to 1148 personality I was sad when I realized that the orange white toned chao that I had been excited to get for so long would never happen, but I want to see what everyone else thinks about this
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Mamkute
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Re: Coloured tones?

Post by Mamkute »

I agree, having color-tone mixing would have been an awesome feature. I can see how it would be very hard to implement, since so many evolutions have certain color features already that are present (like how two-tone dark chao are almost universally black base with highlights of their actual color.)

Maybe it would have been too many different types of Chao.
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Jeffery Mewtamer
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Re: Coloured tones?

Post by Jeffery Mewtamer »

^I don't know, I can think of a way it could be done fairly simply.

Monotone:
normal/normal are the monotone normal we're familiar with.
colorA/colorA or colorA/normal are the monotone color chao we're familiar with.
colorA/colorB are Monotones whose color is the mixture of the two color alleles(e.g. red/white makes pink, though not necessarily the same pink as the pink/pink or pink/normal monotone).

Two-Tone:
normal/normal are the normal two-tones we're familiar with.
ColorA/colorA look monotone
ColorA/normal either have a base color that changes like a monotone normal and markings that are always colorA or colorA as base color and markings that change color(I never bred a colored two-tone, so I'm not sure but I think the latter might be the two-tones we're familiar with).
colorA/colorB One color is used as base color, the other as marking color.

If I'm doing my math right, this would increase the number of monotones to
14 -13 + 14*13/2 = 925
and the number of two-tones to:
14*14 = 196

The logic being:
14 homozygous monotones
13 half-normal monotones indistinguishable from homozygous monotones that are otherwise counted twice in the next step.
14*13 ways of getting heterozygous alleles.
/2 because the order of heterozygous alleles doesn't matter.
Note: this assumes all the colors formed from blending are distinguishable from the ones produce by homozygous and half-normal monotones.

For two-tones, the math is simpler, you just take the square since being half-normal actually does something and the order of alleles actually matters.

If we also allowed Jewel coats to blend, that would add 12*11/2 = 66 blended jewels.

If we let all for appearance genes interact... then we might really have an unmanageable combinatorial explosion on our hands.

Also, if I made any mistakes above, please correct me.
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