Simple cases:
f(x,0) = 0 (Since all the Polyraptors will die from state based effects)
f(x,1) = 1 (Since there will be at most 1 Polyraptor in play and it will die in order to make a copy)
f(0,x) = 1 (Since the Forerunner of the Empire dies from state based effects before the Polyraptor resolves)
f(1,x) = 1 (Since the Forerunner of the Empire will trigger exactly once before it dies)
Here's a more realistic example of f(3,5) (like on the actual cards) here is what the battlefield and stack looks like (Fn = Forerunner of the Empire with n damage, Pn = Polyraptor with n damage and a ' denotes the card's trigger):
@chaikov
Unfortunately, things get a lot more complicated if Polyraptors die during this process, since those dead raptors don't duplicate anymore. I've been wrecking my head about this for a while, but still am not close to a formula that doesn't inlcude ever larger sums. I thought, I had solved the puzzle just a few minutes ago, but checking some cases made it fall apart.
@chaikov
Unfortunately, things get a lot more complicated if Polyraptors die during this process, since those dead raptors don't duplicate anymore. I've been wrecking my head about this for a while, but still am not close to a formula that doesn't inlcude ever larger sums. I thought, I had solved the puzzle just a few minutes ago, but checking some cases made it fall apart.
If it helps, Sprouting Phytohydra, with 2 toughness, will follow the Fibonnacci Sequence.
Yeah, I had examined a few examples and had found the Fibonacci sequence for m=2 already. For what it's worth, I found the mistake in my thinking, though due to that, I can only give a formula for m<=n<=2m, because after that, the number of dying Polyraptors in each iteration is not to base 2 at its core anymore. The formula for going beyond n=2m should include f(2m,m) as a constant, but finding the exact formula will be tricky.
For m<=n<=2m: f(n,m) = 2^n - 2^(n-m) - (n-m)*2^(n-m-1)
I've checked this for up to m=5 and n=10 and it holds.
edit:
You can also rewrite the formula into
For m<=n<=2m: f(n,m) = 2^n - (n-m+2)*2^(n-m-1)
the maximal number of polyraptors left on the field after all triggers have resolved is defined as follows:
an n-nacci sequence is defined as: "A Fibonacci sequence of order n" which is "an integer sequence in which each sequence element is the sum of the previous n elements (with the exception of the first n elements in the sequence)." the first n elements in the sequence are defined as n-1 0's and a 1
as an example a normal fibonacci sequence would be a 2-nacci sequence: 0, 1, 1, 2, 3, 5, etc.
a tribonacci sequence would be a 3-nacci sequence: 0, 0, 1, 1, 2, 4, 7, 13, etc.
the 0's at the beginning of the sequence allow the formula to always have a defined answer without needing to add domain restrictions which is nice
examples of the formula in practice: for reference the first 15 numbers in a 5-nacci sequence (pentanacci) are: 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464
for reference the first 10 numbers in a 3-nacci sequence (tribonacci) are: 0, 0, 1, 1, 2, 4, 7, 13, 24, 44
please let me know if I need to add more explanation D:
The problem is, that this is a recurrsive formula, and I got the impression, that the OP wanted a direct formula to calculate the number of Polyraptors.
Also, for m=3, 24 is followed by 44, not 41. If your formula gets 41 here, it is flawed.
The problem is, that this is a recurrsive formula, and I got the impression, that the OP wanted a direct formula to calculate the number of Polyraptors.
Also, for m=3, 24 is followed by 44, not 41. If your formula gets 41 here, it is flawed.
you are correct that is a typo on my part, thanks for pointing it out!
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Pretend that Forerunner of the Empire had n toughness and Polyraptor had m toughness.
If Forerunner of the Empire is on the battlefield and you cast Polyraptor what is the most amount of Polyraptors you would have in play after all triggers resolve?
Let's denote this function as f(n,m).
Simple cases:
f(x,0) = 0 (Since all the Polyraptors will die from state based effects)
f(x,1) = 1 (Since there will be at most 1 Polyraptor in play and it will die in order to make a copy)
f(0,x) = 1 (Since the Forerunner of the Empire dies from state based effects before the Polyraptor resolves)
f(1,x) = 1 (Since the Forerunner of the Empire will trigger exactly once before it dies)
Here's a more realistic example of f(3,5) (like on the actual cards) here is what the battlefield and stack looks like (Fn = Forerunner of the Empire with n damage, Pn = Polyraptor with n damage and a ' denotes the card's trigger):
Thus we can get a maximum of 8 Polyraptors
Therefore f(3,5) = 8
I believe that f(n,m) = 2^n when n < m since no Polyraptors die and they can double on each Forerunner of the Empire trigger
So what is f(n,m) in general?
http://www.mtgsalvation.com/forums/magic-fundamentals/magic-rulings/788441-forerunner-of-the-empire-and-polyraptor-and
RULES OF MAGIC :
http://magic.wizards.com/en/game-info/gameplay/rules-and-formats/rules
Unfortunately, things get a lot more complicated if Polyraptors die during this process, since those dead raptors don't duplicate anymore. I've been wrecking my head about this for a while, but still am not close to a formula that doesn't inlcude ever larger sums. I thought, I had solved the puzzle just a few minutes ago, but checking some cases made it fall apart.
Former Rules Advisor
"Everything's better with pirates." - Lodge
(The Gamers: Dorkness Rising)
"Any sufficiently analyzed magic is indistinguishable from science."
(Girl Genius - Fairy Tale Theater Break - Cinderella, end of volume 8)
Two Score, Minus Two or: A Stargate Tail
(Image by totallynotabrony)
For m<=n<=2m: f(n,m) = 2^n - 2^(n-m) - (n-m)*2^(n-m-1)
I've checked this for up to m=5 and n=10 and it holds.
edit:
You can also rewrite the formula into
For m<=n<=2m: f(n,m) = 2^n - (n-m+2)*2^(n-m-1)
Former Rules Advisor
"Everything's better with pirates." - Lodge
(The Gamers: Dorkness Rising)
"Any sufficiently analyzed magic is indistinguishable from science."
(Girl Genius - Fairy Tale Theater Break - Cinderella, end of volume 8)
an n-nacci sequence is defined as: "A Fibonacci sequence of order n" which is "an integer sequence in which each sequence element is the sum of the previous n elements (with the exception of the first n elements in the sequence)." the first n elements in the sequence are defined as n-1 0's and a 1
as an example a normal fibonacci sequence would be a 2-nacci sequence: 0, 1, 1, 2, 3, 5, etc.
a tribonacci sequence would be a 3-nacci sequence: 0, 0, 1, 1, 2, 4, 7, 13, etc.
the 0's at the beginning of the sequence allow the formula to always have a defined answer without needing to add domain restrictions which is nice
examples of the formula in practice:
for reference the first 15 numbers in a 5-nacci sequence (pentanacci) are: 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 31, 61, 120, 236, 464
for reference the first 10 numbers in a 3-nacci sequence (tribonacci) are: 0, 0, 1, 1, 2, 4, 7, 13, 24, 44
please let me know if I need to add more explanation D:
m-nacci(x) =
{ 0 if x < m
{ F(m)x - (m - 1) otherwise.
Also, for m=3, 24 is followed by 44, not 41. If your formula gets 41 here, it is flawed.
Former Rules Advisor
"Everything's better with pirates." - Lodge
(The Gamers: Dorkness Rising)
"Any sufficiently analyzed magic is indistinguishable from science."
(Girl Genius - Fairy Tale Theater Break - Cinderella, end of volume 8)
not only is it related, that is precisely what it is, just by another name
(I found the name "n-nacci" on OEIS so thats what I know it by)
you are correct that is a typo on my part, thanks for pointing it out!