| 1 |
In [url=http://zero-k.info/Forum/Thread/7478]this thread[/url] it has been revealed by @Skasi (actually already 2011) and proven by mostly @TheEloIsALie that current overdrive energy distribution is not ideal. Currently OD energy is distributed proportional to squared mex rates as far as the grid allows it. Connecting mexes can reduce total metal rate, which should actually be avoided by the system in any case.
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1 |
In [url=http://zero-k.info/Forum/Thread/7478]this thread[/url] it has been revealed by @Skasi (actually already 2011) and proven by mostly @TheEloIsALie that current overdrive energy distribution is not ideal. Currently OD energy is distributed proportional to squared mex rates as far as the grid allows it. Connecting mexes can reduce total metal rate, which should actually be avoided by the system in any case.
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2 |
\n
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| 3 |
[color=green][b]1. Ideal Solution[/b][/color]
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3 |
[color=green][b]1. Ideal Solution[/b][/color]
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| 4 |
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4 |
\n
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| 5 |
@TheEloIsALie calculated the ideal distribution of OD energies (e_1,...,e_n) in a grid with n mexes with base rates (m_1,...,m_n) and OD energy E. (I use the same variables as @TheEloIsALie and add some more.) Total mex rates are M_i=gamma(e_i)*m_i with the "Lorentz-factor" gamma(e_i)=sqrt(1+e_i/4).
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5 |
@TheEloIsALie calculated the ideal distribution of OD energies (e_1,...,e_n) in a grid with n mexes with base rates (m_1,...,m_n) and OD energy E. (I use the same variables as @TheEloIsALie and add some more.) Total mex rates are M_i=gamma(e_i)*m_i with the "Lorentz-factor" gamma(e_i)=sqrt(1+e_i/4).
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| 6 |
Current
distribution
is
e_i=E*m_i²/p,
where
p=sum_i(
m_i²)
.
If
the
grid
doesn't
provide
as
much
energy
as
it
"wants",
it
gets
its
maximum
and
the
rest
has
to
be
recalculated.
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6 |
Current
distribution
is{
{
{
e_i=E*m_i²/p,
where
p=sum_i(
m_i²)
.
}
}
}
If
the
grid
doesn't
provide
as
much
energy
as
it
"wants",
it
gets
its
maximum
and
the
rest
has
to
be
recalculated.
|
| 7 |
The
easiest
representation
of
@TheEloIsALie's
ideal
solution
is
e_i
=
m_i²*q/p
-
4,
where
q
=
E
+
4n.
I
have
proven
that
it
is
correct,
so
if
you
want
to
[url=http://zero-k.
info/Forum/Thread/7478#114117]see
a
proof.
.
[/url]
The
problem
is
now
that
it
can
produce
negative
e_i.
Those
e_i
have
to
be
set
to
0
and
the
rest
must
be
recalculated.
Thanks
to
@TheEloIsALie
the
recalculation
is
not
needed,
when
"you
go
through
the
mexes
in
ascending
order
of
base
income
and
(
.
.
.
)
(
i)
f
that
e_i
is
negative,
you
subtract
m_i²
from
p
and
4
from
q
and
continue.
"
If
a
mex'
e_i
is
not
negative,
you
can
be
sure
that
all
following
mexes'
e_i
are
not
negative,
too.
The
current
recalculation
when
grid
is
maxed
is
also
needed.
Former
concerns
about
the
algorithmic
efficiency
of
the
ideal
solution
because
of
many
square
roots
or
recalculations
have
been
eliminated.
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7 |
The
easiest
representation
of
@TheEloIsALie's
ideal
solution
is{
{
{
e_i
=
m_i²*q/p
-
4,
where
q
=
E
+
4n.
}
}
}
I
have
proven
that
it
is
correct,
so
if
you
want
to
[url=http://zero-k.
info/Forum/Thread/7478#114117]see
a
proof.
.
[/url]
The
problem
is
now
that
it
can
produce
negative
e_i.
Those
e_i
have
to
be
set
to
0
and
the
rest
must
be
recalculated.
Thanks
to
@TheEloIsALie
the
recalculation
is
not
needed,
when
"you
go
through
the
mexes
in
ascending
order
of
base
income
and
(
.
.
.
)
(
i)
f
that
e_i
is
negative,
you
subtract
m_i²
from
p
and
4
from
q
and
continue.
"
If
a
mex'
e_i
is
not
negative,
you
can
be
sure
that
all
following
mexes'
e_i
are
not
negative,
too.
The
current
recalculation
when
grid
is
maxed
is
also
needed.
Former
concerns
about
the
algorithmic
efficiency
of
the
ideal
solution
because
of
many
square
roots
or
recalculations
have
been
eliminated.
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\n
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8 |
\n
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| 9 |
I
had
a
look
at
[url=https://github.
com/ZeroK-RTS/Zero-K/blob/master/LuaRules/Gadgets/unit_mex_overdrive.
lua#L689]the
source
code[/url]
and
it
seems
as
if
overdrive
does
obey
emp
and
disarm,
but
not
slow
dmg.
Something
like
"orgMetal*=spGetUnitRulesParam(
unitID,
"slowRate")
"
or
else
should
be
written
there
in
line
725
and
755
and
maybe
others.
How
can
allyTeamMexes
be
sorted
by
orgMetal
after
emp,
disarm
and
slow
have
been
applied
and
then
used
in
another
for-structure?
Then
you
only
need
to
replace
current
"mexE
=
allyE*(
orgMetal
*
orgMetal)
/
allyMetalSquared"
by
the
ideal
solution
and
check
if
you
have
to
change
p
and
q.
Maybe
you
can
save
a
little
if
you
first
check
if
orgMetal
of
current
mex
=
orgMetal
of
previous
mex,
because
this
will
often
be
true.
Then
you
can
just
use
e_i
of
previous
mex
and
if
it's
0,
change
p
and
q.
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9 |
I
had
a
look
at
[url=https://github.
com/ZeroK-RTS/Zero-K/blob/master/LuaRules/Gadgets/unit_mex_overdrive.
lua#L689]the
source
code[/url]
and
it
seems
as
if
overdrive
does
obey
emp
and
disarm,
but
not
slow
dmg.
Something
like
"orgMetal*=spGetUnitRulesParam(
unitID,
"slowRate")
"
or
else
should
be
written
there
in
line
725
and
755
and
maybe
others.
How
can
allyTeamMexes
be
sorted
by
orgMetal
after
emp,
disarm
and
slow
have
been
applied
and
then
used
in
another
for-structure?
Then
you
only
need
to
replace
current{
{
{
"mexE
=
allyE*(
orgMetal
*
orgMetal)
/
allyMetalSquared"}
}
}
by
the
ideal
solution
and
check
if
you
have
to
change
p
and
q.
Maybe
you
can
save
a
little
if
you
first
check
if
orgMetal
of
current
mex
=
orgMetal
of
previous
mex,
because
this
will
often
be
true.
Then
you
can
just
use
e_i
of
previous
mex
and
if
it's
0,
change
p
and
q.
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(Btw why is there "1+mexE/5" instead of "energyToExtraM(mexE)" in lines 761 and 795?)
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10 |
(Btw why is there "1+mexE/5" instead of "energyToExtraM(mexE)" in lines 761 and 795?)
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11 |
\n
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[color=green][b]2. Underdrive[/b][/color]
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12 |
[color=green][b]2. Underdrive[/b][/color]
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13 |
\n
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When
I
thought
about
how
to
reduce
recalculations
and
sortings
due
to
negative
OD
energies
e_i
in
@TheEloIsALie's
solution,
a
really
crazy
idea
came
to
my
mind:
Why
not
leave
negative
e_i
as
it
is?
Call
it
underdrive.
This
would
only
require
replacing
|
14 |
When
I
thought
about
how
to
reduce
recalculations
and
sortings
due
to
negative
OD
energies
e_i
in
@TheEloIsALie's
solution,
a
really
crazy
idea
came
to
my
mind:
Why
not
leave
negative
e_i
as
it
is?
Call
it
underdrive.
This
would
only
require
replacing{
{
{
"mexE
=
allyE*(
orgMetal
*
orgMetal)
/
allyMetalSquared"
by
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| 15 |
"mexE
=
allyE*(
orgMetal
*
orgMetal)
/
allyMetalSquared"
by
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15 |
"mexE
=
(
allyE+4*allyTeamMexes.
number(
)
)
*(
orgMetal
*
orgMetal)
/
allyMetalSquared
-
4"
(
not
sure
about
.
number(
)
)
}
}
}
In
most
cases
it
would
make
no
difference.
Only
when
small
and
big
mexes
are
connected
in
a
grid
with
low
energy,
small
mexes
would
be
underdriven
to
provide
a
little
OD
energy
for
the
big
mexes.
The
theoretical
maximum
for
the
underdrive
energy
a
mex
can
provide
is
4,
but
usually
it
will
be
lower.
This
could
give
pylons
a
little
more
use,
but
building
generators
instead
will
still
be
better.
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"mexE = (allyE+4*allyTeamMexes.number())*(orgMetal * orgMetal)/ allyMetalSquared - 4" (not sure about .number())
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In most cases it would make no difference. Only when small and big mexes are connected in a grid with low energy, small mexes would be underdriven to provide a little OD energy for the big mexes. The theoretical maximum for the underdrive energy a mex can provide is 4, but usually it will be lower. This could give pylons a little more use, but building generators instead will still be better.
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\n
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16 |
\n
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| 19 |
[color=green][b]3. Symmetrization[/b][/color]
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17 |
[color=green][b]3. Symmetrization[/b][/color]
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| 20 |
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18 |
\n
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Negative e_i is only possible, because negativity of e_i is not equivalent to negativity of the OD Lorentz-factor's radicand gamma(e_i)=sqrt(1+e_i/4). This can be changed with a symmetrization of the OD function gamma_b(e_i):=1+b*sqrt(e_i), where b is a constant factor that determines the effectivity of OD, for example b=0.3. I found the ideal solution for this and it is exactly the currently implemented solution independently of b! So you would only have to change
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19 |
Negative e_i is only possible, because negativity of e_i is not equivalent to negativity of the OD Lorentz-factor's radicand gamma(e_i)=sqrt(1+e_i/4). This can be changed with a symmetrization of the OD function gamma_b(e_i):=1+b*sqrt(e_i), where b is a constant factor that determines the effectivity of OD, for example b=0.3. I found the ideal solution for this and it is exactly the currently implemented solution independently of b! So you would only have to change
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| 22 |
"return -1+sqrt(1+(energy*0.25))" in line 256 to
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20 |
"return -1+sqrt(1+(energy*0.25))" in line 256 to
|
| 23 |
"return b*sqrt(energy)" with a certain b.
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21 |
"return b*sqrt(energy)" with a certain b.
|
| 24 |
This would also solve the problem of extreme OD being too effective atm as it would make low OD more effective and high OD less effective than currently without introducing more complex functions than sqrt (like log). b could also be changed at any time for balance reasons. The break even point of energy, where current OD becomes equally efficient as symmetric OD for a single mex, is at e_i= 4b²/(b²-1/4)² or b=(sqrt(1+e_i/4)-1)/sqrt(e_i).
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22 |
This would also solve the problem of extreme OD being too effective atm as it would make low OD more effective and high OD less effective than currently without introducing more complex functions than sqrt (like log). b could also be changed at any time for balance reasons. The break even point of energy, where current OD becomes equally efficient as symmetric OD for a single mex, is at e_i= 4b²/(b²-1/4)² or b=(sqrt(1+e_i/4)-1)/sqrt(e_i).
|
| 25 |
\n
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23 |
\n
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| 26 |
[color=green][b]4. Quantum Mechanics[/b][/color]
|
24 |
[color=green][b]4. Quantum Mechanics[/b][/color]
|
| 27 |
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|
25 |
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| 28 |
The whole calculation and especially how adding an assymetry term to the energy (which means the existence of mass in a relativistic sense) creates antimatter (underdrive) reminds me a lot of relativistic quantum mechanics. In the early times of quantum mechanics people also struggled a lot for eliminating the negative energy (underdrive) until they finally had to accept that antimatter exists. The symmetrization is the non-relativistic approximation. It is interesting that the current solution solves the "non-relativistic" approx.
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26 |
The whole calculation and especially how adding an assymetry term to the energy (which means the existence of mass in a relativistic sense) creates antimatter (underdrive) reminds me a lot of relativistic quantum mechanics. In the early times of quantum mechanics people also struggled a lot for eliminating the negative energy (underdrive) until they finally had to accept that antimatter exists. The symmetrization is the non-relativistic approximation. It is interesting that the current solution solves the "non-relativistic" approx.
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27 |
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| 30 |
So which of the 3 systems do you prefer? The lower a system's number, the less elegant and more difficult to implement (if not too much) it is, but closer to the current system.
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28 |
So which of the 3 systems do you prefer? The lower a system's number, the less elegant and more difficult to implement (if not too much) it is, but closer to the current system.
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Overdrive, Underdrive and Quantum Mechanics