Eyes on Final Fantasy Forums > Reference > General Archive > EoFF Archived Awards > The Mid-Year Ciddies 2007 - Nominate now!

[Evastio]

I hope we have different people winning this time and not the same people winning Ciddies repeatedly.

[Tavrobel]

And the mome rath Evastio outgrabe:

I hope we have different people winning this time and not the same people winning Ciddies repeatedly.

Impossible! By using the derivation of the "Evastio" formula and the Elitism Postulate as theorized by Maxx Power, owner of all things humble and holy, the Ciddies is one giant popularity contest, based mostly upon people who have been here before 2003, and those in power or recognition. Popular contenders include Rye, Psychotic, BoB, and so on, even if they have done nothing recently. And I do mean nothing. I would not be surprised if Samuraid gets a "Techies" votes, simply because he can plant 600 damage Remote Mines for the low cost of 300 MP, lasting approximately 4 minutes.

Now, since I have told you, we shall explore how we came to this conclusion.

By assuming that all polynomials have a value, and that a² + b² = c², one can reduce all sides by a parenthetical value, eventually winding up with the Determinant value, (b²-4ac)/2a. If b is negative, then you have complex, conjugate roots. By substituting the following values for b, determine the function in terms of p and q. The function will look similar to this: 5p(4a²-3q) + radical(21).

Since p and q have definite values, we may now apply the Elitism Postulate, which states that all Elitists will vote only for other Elitists, sending every real value to a deadlock. However, one in a while, when one vote goes astray, a new Pirate King is elected, and crowned by the anti-Elitist, Maxx Power. a is to be given the imaginary value of -i. (Pi^e - p) * q should give you an astronomical vlue. However, it can be reduced by using Limit formulae.

If p^p-q = a, then all values for b will be correct, as long as b < 2.00E3 +4. When one brings the equation is graphed, it will look like a negative cubic, starting from the top, with a burp curve. However, hypothetically, if a lovehurts reference is thrown into the mix, it is possible that the answer will come out impossibly different from what is normally gathered, resulting in an Out-of-bounds error, or a Domain of log(x)^y. But I'm talking applied math, not theoretical math.

Therefore, by re-applying the Elitism Postulate with the now reconcilable Christmas formula, and adding the answer to the Evastio formula, you will have the answer of: p/q, with upper bounds of 4, and a lower bound of -5. By re-applying these numbers to the original equation, synthetic divison with a remainder of zero, the answer will come out to be "less than 50% of all Ciddies winners deserve what they have been recognized for." Therefore, it is also possible to assume from this proof and derivation that Lekana will still nominate and vote for me.

Be proud Maxx! Your efforts and research have finally come to a head!

Knowledge is power.

[Jessweeee♪]

o.o

[Fonz]

And the mome rath Tavrobel outgrabe:

And the mome rath Evastio outgrabe:

I hope we have different people winning this time and not the same people winning Ciddies repeatedly.

Impossible! By using the derivation of the "Evastio" formula and the Elitism Postulate as theorized by Maxx Power, owner of all things humble and holy, the Ciddies is one giant popularity contest, based mostly upon people who have been here before 2003, and those in power or recognition. Popular contenders include Rye, Psychotic, BoB, and so on, even if they have done nothing recently. And I do mean nothing. I would not be surprised if Samuraid gets a "Techies" votes, simply because he can plant 600 damage Remote Mines for the low cost of 300 MP, lasting approximately 4 minutes.

Now, since I have told you, we shall explore how we came to this conclusion.

By assuming that all polynomials have a value, and that a² + b² = c², one can reduce all sides by a parenthetical value, eventually winding up with the Determinant value, (b²-4ac)/2a. If b is negative, then you have complex, conjugate roots. By substituting the following values for b, determine the function in terms of p and q. The function will look similar to this: 5p(4a²-3q) + radical(21).

Since p and q have definite values, we may now apply the Elitism Postulate, which states that all Elitists will vote only for other Elitists, sending every real value to a deadlock. However, one in a while, when one vote goes astray, a new Pirate King is elected, and crowned by the anti-Elitist, Maxx Power. a is to be given the imaginary value of -i. (Pi^e - p) * q should give you an astronomical vlue. However, it can be reduced by using Limit formulae.

If p^p-q = a, then all values for b will be correct, as long as b < 2.00E3 +4. When one brings the equation is graphed, it will look like a negative cubic, starting from the top, with a burp curve. However, hypothetically, if a lovehurts reference is thrown into the mix, it is possible that the answer will come out impossibly different from what is normally gathered, resulting in an Out-of-bounds error, or a Domain of log(x)^y. But I'm talking applied math, not theoretical math.

Therefore, by re-applying the Elitism Postulate with the now reconcilable Christmas formula, and adding the answer to the Evastio formula, you will have the answer of: p/q, with upper bounds of 4, and a lower bound of -5. By re-applying these numbers to the original equation, synthetic divison with a remainder of zero, the answer will come out to be "less than 50% of all Ciddies winners deserve what they have been recognized for." Therefore, it is also possible to assume from this proof and derivation that Lekana will still nominate and vote for me.

Be proud Maxx! Your efforts and research have finally come to a head!

Knowledge is power.

Don't fill there heads with fairy tales.

[Moon Rabbits]

Awesome, something to do on the computer again!

[Dynast-Kid]

And the mome rath Tavrobel outgrabe:

And the mome rath Evastio outgrabe:

I hope we have different people winning this time and not the same people winning Ciddies repeatedly.

Impossible! By using the derivation of the "Evastio" formula and the Elitism Postulate as theorized by Maxx Power, owner of all things humble and holy, the Ciddies is one giant popularity contest, based mostly upon people who have been here before 2003, and those in power or recognition. Popular contenders include Rye, Psychotic, BoB, and so on, even if they have done nothing recently. And I do mean nothing. I would not be surprised if Samuraid gets a "Techies" votes, simply because he can plant 600 damage Remote Mines for the low cost of 300 MP, lasting approximately 4 minutes.

Now, since I have told you, we shall explore how we came to this conclusion.

By assuming that all polynomials have a value, and that a² + b² = c², one can reduce all sides by a parenthetical value, eventually winding up with the Determinant value, (b²-4ac)/2a. If b is negative, then you have complex, conjugate roots. By substituting the following values for b, determine the function in terms of p and q. The function will look similar to this: 5p(4a²-3q) + radical(21).

Since p and q have definite values, we may now apply the Elitism Postulate, which states that all Elitists will vote only for other Elitists, sending every real value to a deadlock. However, one in a while, when one vote goes astray, a new Pirate King is elected, and crowned by the anti-Elitist, Maxx Power. a is to be given the imaginary value of -i. (Pi^e - p) * q should give you an astronomical vlue. However, it can be reduced by using Limit formulae.

If p^p-q = a, then all values for b will be correct, as long as b < 2.00E3 +4. When one brings the equation is graphed, it will look like a negative cubic, starting from the top, with a burp curve. However, hypothetically, if a lovehurts reference is thrown into the mix, it is possible that the answer will come out impossibly different from what is normally gathered, resulting in an Out-of-bounds error, or a Domain of log(x)^y. But I'm talking applied math, not theoretical math.

Therefore, by re-applying the Elitism Postulate with the now reconcilable Christmas formula, and adding the answer to the Evastio formula, you will have the answer of: p/q, with upper bounds of 4, and a lower bound of -5. By re-applying these numbers to the original equation, synthetic divison with a remainder of zero, the answer will come out to be "less than 50% of all Ciddies winners deserve what they have been recognized for." Therefore, it is also possible to assume from this proof and derivation that Lekana will still nominate and vote for me.

Be proud Maxx! Your efforts and research have finally come to a head!

Knowledge is power.

So only oldbies can win?

[Evastio]

And the mome rath Tavrobel outgrabe:

And the mome rath Evastio outgrabe:

I hope we have different people winning this time and not the same people winning Ciddies repeatedly.

Impossible! By using the derivation of the "Evastio" formula and the Elitism Postulate as theorized by Maxx Power, owner of all things humble and holy, the Ciddies is one giant popularity contest, based mostly upon people who have been here before 2003, and those in power or recognition. Popular contenders include Rye, Psychotic, BoB, and so on, even if they have done nothing recently. And I do mean nothing. I would not be surprised if Samuraid gets a "Techies" votes, simply because he can plant 600 damage Remote Mines for the low cost of 300 MP, lasting approximately 4 minutes.

Now, since I have told you, we shall explore how we came to this conclusion.

By assuming that all polynomials have a value, and that a² + b² = c², one can reduce all sides by a parenthetical value, eventually winding up with the Determinant value, (b²-4ac)/2a. If b is negative, then you have complex, conjugate roots. By substituting the following values for b, determine the function in terms of p and q. The function will look similar to this: 5p(4a²-3q) + radical(21).

Since p and q have definite values, we may now apply the Elitism Postulate, which states that all Elitists will vote only for other Elitists, sending every real value to a deadlock. However, one in a while, when one vote goes astray, a new Pirate King is elected, and crowned by the anti-Elitist, Maxx Power. a is to be given the imaginary value of -i. (Pi^e - p) * q should give you an astronomical vlue. However, it can be reduced by using Limit formulae.

If p^p-q = a, then all values for b will be correct, as long as b < 2.00E3 +4. When one brings the equation is graphed, it will look like a negative cubic, starting from the top, with a burp curve. However, hypothetically, if a lovehurts reference is thrown into the mix, it is possible that the answer will come out impossibly different from what is normally gathered, resulting in an Out-of-bounds error, or a Domain of log(x)^y. But I'm talking applied math, not theoretical math.

Therefore, by re-applying the Elitism Postulate with the now reconcilable Christmas formula, and adding the answer to the Evastio formula, you will have the answer of: p/q, with upper bounds of 4, and a lower bound of -5. By re-applying these numbers to the original equation, synthetic divison with a remainder of zero, the answer will come out to be "less than 50% of all Ciddies winners deserve what they have been recognized for." Therefore, it is also possible to assume from this proof and derivation that Lekana will still nominate and vote for me.

Be proud Maxx! Your efforts and research have finally come to a head!

Knowledge is power.

It came to my head all right. And it made it explode.

Basically as long as 25% or more people that win Ciddies are people that never got Ciddies before I'm happy.

[Shiny]

And the mome rath Dynast-Kid outgrabe:

So only oldbies can win?

No. That would defeat the purpose of the best newbie Ciddie. Unless you consider 1999-2003 new.