(Updating)
I think I can prove that this puzzle cannot be solved
Assume (0,0) is on
Code: Select all
x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$A19D!
Think about 1c. if it is saved by the 1c rule, then:
Code: Select all
x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$DA$A19D!
That will create a born on (1,0) and (0,1). If (2,0) , (2,1) , (0,2) and (1,2) is on:
Code: Select all
x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$2A$D2A$ADA17D!
(2,1), (1,2) and (1,1) will die.
If (1,0) and (0,1) is on:
Code: Select all
x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$2A$2A18D!
、
A block appears.
So (1,1) and (0,0) can't be on at the same time.
If it is saved by the S2 rule, there are two possibilities, but (1,1) must be off, so (0,1) and (1,0) must be on.
Code: Select all
x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$AB$2A18D!
(0,1) and (1,0) already had two neighbour, so the blue cells must be off:
Code: Select all
x = 20, y = 20, rule = LifeHistory
D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$D$2B$A2B$2AB17D!
This will cause a B3a on (1,1), so (2,2) must be on
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