<InsertElement>Bow Quantum Balancing

[Camoninja]

When I'm not earning electrum for a deck I want to make, I like to go to the trainer and build semi-rainbow decks for FG farming, possibly (if they work well). Whenever I test these decks I always come up with quantum problems.

The way I balance mono and duo decks is using 5 quanta for every pillar or pendulum pair, and for rainbow decks I keep all the quantum usage roughly equal and put in quantum towers until they are a third of the deck.

If I have a deck that's mostly mono with a little rainbow in it I suppose I should add some novas, but if it's at least half rainbow I have no idea how to balance it except trial and error, which to me seems too time consuming if there can be a formula made.

[Essence]

Random musing on QI theory.

You want 1 pillar per 5CC of casting cost. 

There are 12 elements.  Quantum Pillars produce 3 random quanta.  On average, then, 25% of elements get a quanta per turn -- or, statistically, each element gets .25 of a quanta per turn.  Therefore, with 4 Quantum Pillars, you get the same net effect as though you had 1 pillar of each kind -- statistically.

So, in order to find the right number of pillars in a half-rainbow deck, you need to find the one sub-element that has the highest total CC, divide that by 5, multiply that by 4, and then add that many Quantum Pillars.  Then, with your main element, subtract 1/4 of your Quantum Pillars from the total number of on-element Pillars you put in.



So, for example, if you have a half-Light deck with 46 total CC in Light, and the other half is rainbow with the highest CC being a Toadfish (5), you need

5/5=1x4=4 Quantum Pillars, and
46/5=9-(4/4=1)=8(-1 for a Light Mark)=7 Light Pillars.  Total of 13 pillars, 4 Quantum and 7 Light.


No idea how to take Novas into account; I usually just swap them out 1-for-1 with QPs.

[Camoninja]

My latest deck is a duo/rainbow, so I assume that it works the same for duos as monos.
So there's 26 in time/death, that makes about 10 pendulums. Then the highest other usage is 8, so that would be 8/5 * 4 = 6. Then I take out 2 pendulums, and I get this:

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6qq 6qq 6qq 6qq 6qq 6qq 6rn 6rn 6rn 6rn 6u8 6u8 712 712 712 712 717 717 71b 71b 74b 77i 77i 77i 7ap 7ap 7do 7k6 7k6 7n0 7n0 7n0 7q5 7q5 7q5 7q5 7q8 7q8 7ri 7ri 7ri 7ri 7ri 7ri 7ri 7ri 7ta 7ta 80h 80h
It works a lot better than my usual first balancing. I'll test it a bit more. Thank you very much!