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I don't understand why this is an objection to balancing cost. Of course units should be designed to be different and have interesting counter structures. But when you have chosen such units then balancing their cost is exactly what is needed to balance choices. If you choose a unit because it has no counters or is a good counter vs specific other units this would be reflected in those cost formulas if they are correct. It doesn't mean that a game with perfect costs is perfect when unit variety iself isn't good.
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I don't understand why this is an objection to balancing cost. Of course units should be designed to be different and have interesting counter structures. But when you have chosen such units then balancing their cost is exactly what is needed to balance choices. If you choose a unit because it has no counters or is a good counter vs specific other units this would be reflected in those cost formulas if they are correct. It doesn't mean that a game with perfect costs is perfect when unit variety iself isn't good.
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| 3 |
A thing I have just thought of, too, would be specifically/only considering unit groups within the same factory because a choice for a factory is indeed something that has to be considered.
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A thing I have just thought of, too, would be specifically/only considering unit groups within the same factory because a choice for a factory is indeed something that has to be considered.
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| 5 |
Actually those integrals are "sumtegrals" which means sums if Omega is at most countable and integrals otherwise. Your requested Omegas for Y=(hp, dps) in Omega: {{{Omega_0 = {Y}
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Actually those integrals are "sumtegrals" which means sums if Omega is at most countable and integrals otherwise. Your requested Omegas for Y=(hp, dps) in Omega: {{{Omega_0 = {Y}
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Omega_1 = {(100, 30), (100, 10)}
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Omega_1 = {(100, 30), (100, 10)}
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Omega_2 = {(100, 30), (101, 10)} }}} If we sumtegrate those over the 1-function, the results are: {{{sumtegral_Omega_0 1dY = sum_(Y in Omega_0) 1 = 1
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Omega_2 = {(100, 30), (101, 10)} }}} If we sumtegrate those over the 1-function, the results are: {{{sumtegral_Omega_0 1dY = sum_(Y in Omega_0) 1 = 1
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| 8 |
sumtegral_Omega_1 1dY = sum_(Y in Omega_1) 1 = 2
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sumtegral_Omega_1 1dY = sum_(Y in Omega_1) 1 = 2
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| 9 |
= sumtegral_Omega_2 1dY}}} Now this doesn't make much sense, because we only sumtegrated over the 1-function which doesn't even depend in Omega. According to my formula, we have to sumtegrate over s or more general {{{ (g*s*f). }}} Then the results are as follows: {{{sumtegral_Omega_0 s(X,Y)(g*f)(Y)dY = sum_(Y in Omega_0) s(X,Y)*(g*f)(Y) = s(X,Y)(g*f)(Y) = s(X,Y)g(Y)f(Y)
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= sumtegral_Omega_2 1dY}}} Now this doesn't make much sense, because we only sumtegrated over the 1-function which doesn't even depend in Omega. According to my formula, we have to sumtegrate over s or more general {{{ (g*s*f). }}} Then the results are as follows: {{{sumtegral_Omega_0 s(X,Y)(g*f)(Y)dY = sum_(Y in Omega_0) s(X,Y)*(g*f)(Y) = s(X,Y)(g*f)(Y) = s(X,Y)g(Y)f(Y)
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sumtegral_Omega_1 s(X,Y)(g*f)(Y)dY = sum_(Y in Omega_1) s(X,Y)*(g*f)(Y) = s(X,(100, 30))(g*f)(100, 30) + s(X,(100, 10))(g*f)(100, 10)
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sumtegral_Omega_1 s(X,Y)(g*f)(Y)dY = sum_(Y in Omega_1) s(X,Y)*(g*f)(Y) = s(X,(100, 30))(g*f)(100, 30) + s(X,(100, 10))(g*f)(100, 10)
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sumtegral_Omega_2 s(X,Y)(g*f)(Y)dY = sum_(Y in Omega_2) s(X,Y)*(g*f)(Y) = s(X,(100, 30))(g*f)(100, 30) + s(X,(101, 10))(g*f)(100, 10) }}} For finite Omega where all units are cheap enough to be actually used, I recommend g=1. For everywhere uncountable Omega, I recommend g=1/f. In those cases it should be defined and consistent. For countably infinite Omega, 1/f is defined, but I'm not sure about consistency for the 2*2=4 problem @TheEloIsALie mentioned.
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sumtegral_Omega_2 s(X,Y)(g*f)(Y)dY = sum_(Y in Omega_2) s(X,Y)*(g*f)(Y) = s(X,(100, 30))(g*f)(100, 30) + s(X,(101, 10))(g*f)(100, 10) }}} For finite Omega where all units are cheap enough to be actually used, I recommend g=1. For everywhere uncountable Omega, I recommend g=1/f. In those cases it should be defined and consistent. For countably infinite Omega, 1/f is defined, but I'm not sure about consistency for the 2*2=4 problem @TheEloIsALie mentioned.
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| 13 |
PS:
My
formulas
would
also
allow
to
make
some
units
OP
by
design
if
you
think
that
would
improve
"choices".
I
would
only
make
mex
op
and
balance
fight
units,
though.
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13 |
PS:
My
formulas
would
also
allow
to
make
some
units
op
by
design
if
you
think
that
would
improve
"choices".
I
would
only
make
mex
op
and
balance
fight
units,
though.
|
How to balance (calculate strength of) units in any strategy game