yujh wrote: ↑April 23rd, 2021, 7:44 pm
I’m here to prove infinitely large scores could be easily constructed (and hashlife friendly???)
((2/5)n)^2-(16+2n)^2=not enough
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x = 85, y = 7, rule = B1e2cen3-aij4ijqrt5cjry6-ei/S1e2ain3cny4nwyz5ijn6ak8
82b2o$82bobo$2o2bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo$obo79b
2o$2o2bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo4bo$82bobo$82b2o!
Seriously, though. Is a positive score even possible? The period of the breeder that creates the guns should be a multiple of the period of the guns, so that the guns are all in phase and hit the minimum population at the same time. We want to maximize the ships created per generation and minimize the size of the guns.
The size of the breeder does not matter because it only adds a linear amount to the negative part of the score, while the guns add a quadratic amount. The period of the breeder does not matter because we can put the gun-deleting spaceship further back to increase n.
Let's pretend that spaceships are allowed to touch and still be considered created. They're not, but ignoring that only serves to show how ridiculous this is. Note that I'm also pretending that we're allowed to shoot on both sides (we're not). If the size of a gun is q cells and the period of a gun is p, then we have a score of approximately ((2/p)n)^2-(qn)^2.
The minimum possible value for p is 2.
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x = 23, y = 2, rule = B2ae3e5y/S1e5i
2o18b2o$19bo2bo!
Oh wait, q is 2, so we have n^2-(2n)^2. Score is negative.
Let's make q equal to 1, and pretend that the ships fired the "wrong way" don't exist, to make things easier for ourselves. In reality, they would become part of q, inflating it to a ridiculous number.
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x = 13, y = 3, rule = B1e2e4cr5y6cn/S2n3nq5y
o10bo$10bobo$10b3o!
p now has to be at least 3, which means that we have (2n/3)^2-n^2. Still negative.
All I can say is good luck. You'll need it.