Before reading this page, I would strongly recommend to read the Distribution Overpushing guide first, as well as pacowoc’s article On the Middle-Term Modeling of Endgame Theories and the Optimal Publication Multiplier.
The overpush ratios are a core part of the endgame strategy for pushing main theories. However, overpushing is not relevant for Custom Theories as they aren’t affected by R9.
It is still possible to calculate the overpushing coefficients for official custom theories. In addition of interesting the curious, calculating them can help understanding how the publication multiplier affects official CTs and explaining the publication multiplier to publish at. This is what I’m gonna do in this article.
In this article, I will refer to the publication multiplier as
$\Pi$
In this entire article, I will use
$\log$
$\ln$
$e$
As you have read in the Distribution Overpushing guide, the overpushing coefficient measures the effect of the publication multiplier on
$\dot{\tau}$
$\dot{\tau}$
$\tau$
$a$
$\tau$
$a^{\frac{1}{OP}}$
The OP coefficient also has a close relationship with the optimal publication multiplier of the theory, as seen in this section of pacowoc’s article.
In this representation, a theory can be modeled as :
$\frac{d\rho}{dt}=K\rho^{x}t^{y}\rho^{z}_{pub}$
While this model is great, it has one flaw to analyze official CTs, as it assumes
$\Pi = \rho^{z}{pub}$
, or in other words that $\frac{d\rho}{dt} \propto \Pi$
, which is not the case for every CT. Let’s decompose $z = ab$
, where $a$
is such that $\Pi \propto \rho^{a}$
. In this case, we have $\frac{d\rho}{dt} \propto \Pi^b$
or $\Pi^{b} = \rho^{z}$
In pacowoc’s article, the optimal publication multiplier was found to be:
$e^{\frac{z(y+1)}{1-x}}$
But this is the optimal multiplier for
$\rho^{z}_{pub} = \Pi^{b}$
$\Pi$
$e^{\frac{z(y+1)}{b(1-x)}}$
Now, for any non-divergent theory,
$z + x \le 1$
$z + x \approx 1$
$\frac{z}{1-x} \approx 1$
$e^{\frac{y+1}{b}}$
It turns out that
$\frac{y+1}{b}$
$e^{OP}$
Note that in practice, due to
$z + x \lt 1$
Let’s prove that
$\frac{y+1}{b}$
$\Pi$
$m$
$m^{\frac{1}{OP}}$
$\rho$
$m^{\frac{1}{OP}}$
$\Pi$
$m$
To prove this, we’ll start from equation (3-3) of pacowoc’s guide:
$\rho = (K’(1-x)\rho^{z}_{pub}t^{y+1})^{\frac{1}{1-x}}$
We substitute
$\rho^{z}_{pub}$
$\Pi^{b}$
$\rho = (K’(1-x)\Pi^{b}t^{y+1})^{\frac{1}{1-x}}$
Now we solve for
$t$
$t = \left(\frac{\rho^{1-x}}{K’(1-x)}\Pi^{-b}\right)^{\frac{1}{y+1}}
= \left(\frac{\rho^{1-x}}{K’(1-x)}\right)^{\frac{1}{y+1}}\Pi^{-\frac{b}{y+1}}$
Now let’s calculate
$t’$
$\Pi$
$m\Pi$
$\begin{gather*}
\begin{aligned}
t’&=\left(\frac{\rho^{1-x}}{K’(1-x)}\right)^{\frac{1}{y+1}}m^{-\frac{b}{y+1}}\Pi^{-\frac{b}{y+1}} \
\frac{t’}{t}&=\frac{1}{m^{\frac{b}{y+1}}}
\end{aligned}
\end{gather*}$
We can see that, when multiplying the pub mult by
$m$
$\rho$
$m^{\frac{b}{y+1}}$
Therefore,
$\frac{y+1}{b}$
From pacowoc’s model:
$\frac{d\rho}{dt}=K\rho^{x}t^{y}\Pi^{b}$
We can immediately find the overpush ratio:
$\frac{y+1}{b}$
We can also treat
$\rho^{x}$
$\rho$
$\rho=K\rho^{x}t^{y+1}\Pi^{b}$
Now the OP ratio is directly visible.
For some theories, we’ll have to go to the logarithmic space. If we apply log10 to the previous equation:
$\log{\rho}=x\log{\rho}+(y+1)\log{t}+b\log{\Pi}$
Now that we set the baseline on how to calculate the OP ratio for a given theory, let’s put it in practice to calculate the OP ratio of official custom theories.
Throughout this section, I will use the symbol
$K$
$\Pi$
$t$
$\begin{alignat*}{2}
\dot{q}& &=,& K \
\Rightarrow q& &=,& Kt \
\dot{\rho} =& &K& \Pi q =K\Pi t \
\Rightarrow \rho =& &K& \Pi t^2 \
OP& &=,& 2
\end{alignat*}$
$\begin{alignat*}{1}
\dot{\rho _3} &= K\
\Rightarrow \rho _3 &= Kt
\end{alignat*}$
$\begin{alignat*}{1}
\dot{\rho}_2 &= K \times {1.96}^{-\ln\rho _3} \
\dot{\rho}_2 &= K {\rho _3}^{-\ln1.96} \
\dot{\rho}_2 &= K t^{-\ln1.96} \
\Rightarrow \rho_2 &= Kt^{1-\ln1.96} \
\frac{1}{e - \gamma} &= K\rho _3 = Kt
\end{alignat*}$
Now:
$\begin{alignat*}{1}
\dot{\rho _1} &= \Pi\rho _2^{0.53} \times \frac{1}{e - \gamma} \
\dot{\rho _1} &= K\Pi t^{0.53 \times (1-\ln1.96) + 1} \
\rho _1 &= K\Pi t^{2 + 0.53 \times (1-\ln1.96)}
\end{alignat*}$
$\begin{alignat*}{1}
OP &= 2 + 0.53 \times (1-\ln1.96) \
OP &\approx 2.173
\end{alignat*}$
Calculating EF’s OP factor is comparatively harder due to it having multiple currencies. In general, to study theories with multiple currencies, we use linear algebra in logarithmic space.
$\begin{alignat*}{1}
\dot{q} &= \Pi q_1 q_2\
\Rightarrow q &= K\Pi q_1 q_2 t
\end{alignat*}$
Late-game, in EF’s
$\dot{\rho}$
$tq^2$
$\begin{alignat*}{1}
\dot{\rho} &= \Pi (a_1 a_2 a_3)^{1.5} t^{0.5}q\
\Rightarrow \dot{\rho} &= \Pi^2 (a_1 a_2 a_3)^{1.5} q_1 q_2 t^{1.5}\
\Rightarrow \rho &= \Pi^2 (a_1 a_2 a_3)^{1.5} q_1 q_2 t^{2.5}\
\dot{R} &= K \Pi (b_1 b_2)^2\
\Rightarrow R &= K \Pi (b_1 b_2)^2 t\
\dot{I} &= K \Pi (c_1 c_2)^2\
\Rightarrow I &= K \Pi (c_1 c_2)^2 t
\end{alignat*}$
Now, let’s express the system in logarithmic form, to turn it into a linear system:
$\begin{alignat*}{1}
\log{\rho} =,& K + 1.5(\log{a_1} + \log{a_2} + \log{a_3})\
&+ \log{q_1} + \log{q_2} + 2\log{\Pi} + 2.5\log{t}\
\log{R} =,& K + 2(\log{b_1} + \log{b_2}) + \log{\Pi} + \log{t}\
\log{I} =,& K + 2(\log{c_1} + \log{c_2}) + \log{\Pi} + \log{t}\
\end{alignat*}$
We have:
$\begin{alignat*}{1}
\log{a_1}, \log{q_1}, \log{q_2} &\propto \log{\rho}\
\log{a_2}, \log{b_1}, \log{b_2} &\propto \log{R}\
\log{a_3}, \log{c_1}, \log{c_2} &\propto \log{I}\
\end{alignat*}$
For the purpose of this study, we can ignore variables bought with
$\rho$
$\begin{alignat*}{1}
C\log{\rho} &= K + 1.5(\log{a_2} + \log{a_3}) + 2\log{\Pi} + 2.5\log{t}\
\log{R} &= K + 2(\log{b_1} + \log{b_2}) + \log{\Pi} + \log{t}\
\log{I} &= K + 2(\log{c_1} + \log{c_2}) + \log{\Pi} + \log{t}\
\end{alignat*}$
$\begin{alignat*}{5}
C\log{\rho} - 1.5\log{a_2} - 1.5\log{a_3} &,=,& 2\log{\Pi} &,+,& 2.5\log{t} &+ K\
\left(1 - 2\frac{\log{b_1} + \log{b_2}}{\log{R}}\right)\log{R} &,=,& \log{\Pi} &,+,& \log{t} &+ K\
\left(1 - 2\frac{\log{c_1} + \log{c_2}}{\log{I}}\right)\log{I} &,=,& \log{\Pi} &,+,& \log{t} &+ K\
\end{alignat*}$
To find EF’s OP factor, we need an expression of the form
$\log{\rho} = \alpha\log{\Pi} + \beta\log{t}$
$\frac{\beta}{\alpha}$
Let’s now compute how the variables we need scale with their currencies. To do so, we use this model:
- An exponential variable
$u$
of power $p$
and cost scaling $c$
, bought with currency $P$
is approximated as $\log{u} = \frac{\log{p}}{\log{c}} \cdot \log{P}$
.
- A stepwise variable
$v$
of power $p$
, cycle length $n$
and cost scaling $c$
, bought with currency $P$
is approximated as $\log{v} = \frac{\log{p}}{n\log{c}} \cdot \log{P}$
.
Therefore we can calculate the following:
$a_2$
is a stepwise variable of power 40, cycle length 10 and cost scaling $2^{2.2}$
: $\log{a_2} = \frac{\log{40}}{22\log{2}} \cdot \log{R}$
.$a_3$
is an exponential variable of power 2 and cost scaling $2^{2.2}$
: $\log{a_3} = \frac{1}{2.2} \cdot \log{I}$
.$b_1$
is a stepwise variable of power 2, cycle length 10 and cost scaling 200: $\log{b_1} = \frac{\log{2}}{10\log{200}} \cdot \log{R}$
.$b_2$
is an exponential variable of power 1.12 and cost scaling 2: $\log{b_2} = \frac{\log{1.12}}{\log{2}} \cdot \log{R}$
.$c_1$
is a stepwise variable of power 2, cycle length 10 and cost scaling 200: $\log{b_1} = \frac{\log{2}}{10\log{200}} \cdot \log{I}$
.$c_2$
is an exponential variable of power 1.125 and cost scaling 2: $\log{c_2} = \frac{\log{1.125}}{\log{2}} \cdot \log{I}$
.
Now we can substitute in the system:
$\begin{alignat*}{5}
C\log{\rho} - \frac{1.5\log{40}}{22\log{2}}\log{R} - \frac{1.5}{2.2}\log{I} &,=,& 2\log{\Pi} &,+,& 2.5\log{t} &+ K\
\left[1 - 2\left(\frac{\log{2}}{10\log{200}} + \frac{\log{1.12}}{\log{2}}\right)\right]\log{R} &,=,& \log{\Pi} &,+,& \log{t} &+ K\
\left[1 - 2\left(\frac{\log{2}}{10\log{200}} + \frac{\log{1.125}}{\log{2}}\right)\right]\log{I} &,=,& \log{\Pi} &,+,& \log{t} &+ K\
\end{alignat*}$
These expressions will be heavy to manipulate so let’s set it to:
$\begin{alignat}{9}
C\log{\rho} &,-,& d_{12}\log{R} &,-,& d_{13}\log{I} &,=,& 2\log{\Pi} &,+,& 2.5\log{t} &+ K\
&&d_{22}\log{R} && &,=,& \log{\Pi} &,+,& \log{t} &+ K\
&&&&d_{33}\log{I} &,=,& \log{\Pi} &,+,& \log{t} &+ K\
\end{alignat}$
Now if we add
$\frac{d_{12}}{d_{22}}$
$\frac{d_{13}}{d_{33}}$
$\begin{alignat*}{2}
C\log{\rho} &=& &\left(2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}\right)\log{\Pi} \
&&+ &\left(2.5 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}\right)\log{t} + K
\end{alignat*}$
We can finally express EF’s OP factor:
$\begin{alignat*}{1}
OP &= \frac{2.5 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}}{2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}} \
OP &= 1 + \frac{0.5}{2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}} \
OP &= 1 + \frac{1}{2\left(2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{23}}\right)} \
OP &= 1 + \frac{1}{4 + 2\frac{d_{12}}{d_{22}} + 2\frac{d_{13}}{d_{23}}} \
\end{alignat*}$
For those who want to see what it looks like if we replace those
$d_i$
$
OP = 1 +
\begin{bmatrix*}[l]
4 + \frac{3\log{40}}{22\log{2}}\frac{1}{1 - 2\left(\frac{\log{2}}{10\log{200}} + \frac{\log{1.12}}{\log{2}}\right)} \
+ \frac{3}{2.2}\frac{1}{1 - 2\left(\frac{\log{2}}{10\log{200}} + \frac{\log{1.125}}{\log{2}}\right)}
\end{bmatrix*}^{-1}
$
Finally,
$OP \approx 1.137$
$\begin{alignat*}{2}
\dot{q}& &=,& K\Pi \
\Rightarrow q& &=,& K\Pi t \
\dot{\rho} =& &K& \Pi q =K\Pi^2 t \
\Rightarrow \rho =& &K& \Pi^2 t^2 \
OP& &=,& 1
\end{alignat*}$
$\begin{alignat*}{1}
\dot{q} &= K\
\Rightarrow q &= Kt\
\dot{r} &= K\
\Rightarrow r &= Kt
\end{alignat*}$
$\int_{0}^{q}{g(x)dx} = Kq^6 = Kt^6$
$\sqrt[\pi]{\int_{0}^{ q }g(x)dx} = Kt^{\frac{6}{\pi}}$
$\begin{alignat*}{1}
\dot{\rho} &= K\Pi rt\sqrt[\pi]{\int_{0}^{ q }g(x)dx}\
\dot{\rho} &= K\Pi t^{2+\frac{6}{\pi}}\
\Rightarrow \rho &= K\Pi t^{3+\frac{6}{\pi}}
\end{alignat*}$
$\begin{alignat*}{1}
OP &= 3 + \frac{6}{\pi} \
OP &\approx 4.91
\end{alignat*}$
$\begin{alignat*}{1}
\dot{q} &= K\
\Rightarrow q &= Kt\
\dot{r} &= K\
\Rightarrow r &= Kt
\end{alignat*}$
$\begin{alignat*}{1}
\dot{\rho} &= K\Pi qrt\
\dot{\rho} &= K\Pi t^3\
\Rightarrow \rho &= K\Pi t^4
\end{alignat*}$
$OP = 4$
Since RZ also has multiple currencies, we’ll follow the same steps as with EF. For the purpose of this study,
$\zeta$
$\dot{\zeta}$
$t$
$\begin{alignat*}{1}
\dot{\rho} &= K\Pi c_1^{1.25}c_2w_1t\
\Rightarrow \rho &= K \Pi c_1^{1.25}c_2w_1t^2\
\dot{\delta} &= K\Pi w_1w_2w_3\
\Rightarrow \delta &= K\Pi w_1w_2w_3t
\end{alignat*}$
Let’s express this system on its logarithmic form:
$\begin{alignat*}{2}
\log{\rho} &=& &1.25\log{c_1} + \log{c_2} + \log{w_1}\
&&+ &\log{\Pi} + 2\log{t} + K\
\log{\delta} &=& &\log{w_1} + \log{w_2} + \log{w_3}\
&&+ &\log{\Pi} + \log{t} + K
\end{alignat*}$
We have:
$\begin{alignat*}{1}
\log{c_i} &\propto \log{\rho}\
\log{w_i} &\propto \log{\delta}\
\end{alignat*}$
We can also ignore variables bought with
$\rho$
$\begin{alignat*}{5}
C\log{\rho} - \log{w_1} &=,& \log{\Pi} &,+,& 2\log{t} &+ K\
\left(1- \frac{\log{w_1} + \log{w_2} + \log{w_3}}{\log{\delta}}\right)\log{\delta} &=,& \log{\Pi} &,+& \log{t} &+ K
\end{alignat*}$
Let’s now compute how the variables we need scale with their currencies.
$w_1$
is a stepwise variable of power 2, cycle length 8, following a stepwise cost of power $100^{1/3}$
and cycle length 6. Its power is given by: $\log{w_1} = \frac{\log{2}}{8}\cdot\frac{6}{\frac{1}{3}\log{100}}\cdot\log{\delta} = \frac{9}{8}\log{2}\cdot\log{\delta}$
.$w_2$
is an exponential variable of power 2 and cost scaling 10. Its power is given by $\log{w_2}=\log{2}\cdot\log{\delta}$
.$w_3$
is an exponential variable of power 2 and cost scaling $1e30$
. Its power is given by $\log{w_3}=\frac{\log{2}}{30}\cdot\log{\delta}$
.
Now we can substitute in the system:
$\begin{alignat}{7}
-\frac{9}{8}\log{2}\log{\delta} &,+,& C\log{\rho} &,=,& \log{\Pi} &,+,& 2\log{t} &+ K \nonumber\
\left[1-(1+\frac{9}{8}+\frac{1}{30})\log{2}\right]\log{\delta} &&&,=,& \log{\Pi} &,+,& \log{t} &+ K \nonumber\
-\frac{9}{8}\log{2}\log{\delta} &,+,& C\log{\rho}&,=,& \log{\Pi} &,+,& 2\log{t} &+ K \tag{1}\
\left(1-\frac{259}{120}\log{2}\right)\log{\delta} &&&,=,& \log{\Pi} &,+,& \log{t} &+ K \tag{2}
\end{alignat}$
Now if we add
$\frac{\frac{9}{8}\log{2}}{1-\frac{259}{120}\log{2}}$
$\begin{alignat*}{2}
C\log{\rho} &=& &\left(1 + \frac{\frac{9}{8}\log{2}}{1-\frac{259}{120}\log{2}}\right)\log{\Pi} \
&&+ &\left(2 + \frac{\frac{9}{8}\log{2}}{1-\frac{259}{120}\log{2}}\right)\log{t} + K
\end{alignat*}$
We can now express RZ’s OP factor:
$\begin{alignat*}{1}
OP &= \frac{2 + \frac{\frac{9}{8}\log{2}}{1-\frac{259}{120}\log{2}}}{1 + \frac{\frac{9}{8}\log{2}}{1-\frac{259}{120}\log{2}}} \
OP &= 1 + \left[1 + \frac{\frac{9}{8}\log{2}}{1-\frac{259}{120}\log{2}}\right]^{-1}
\end{alignat*}$
Finally,
$OP \approx 1.508$
To determine the OP factor of MF, we first need to determine if, long-term,
$I$
$\begin{alignat*}{1}
\dot{I} &= \frac{a_1^{1.01}}{400}\left(10^{-15}-\frac{I}{a_2}\right) \
\dot{I} &= \frac{a_1^{1.01}}{400 a_2}(10^{-15}a_2-I) \
\dot{I} &= K\frac{a_1^{1.01}}{a_2}(I_\text{cap}-I)
\end{alignat*}$
To determine if
$I$
$\frac{a_1^{1.01}}{a_2}$
- If it goes to
$+\infty$
, then $I$
will almost instantly cap at large rho.
- If it converges to
$0$
, then $I \ll I_\text{cap}$
at large rho.
Let’s now calculate how
$a_1$
$a_2$
$a_1$
$\log{a_1} = \frac{\log{2}}{2\log{25}} \times \log{\rho}$
$a_2$
$\begin{alignat*}{1}
\log{a_2} &= \frac{\log{1.25}}{\log{100}} \times \log{\rho} \
&= \frac{\log{1.25}}{2} \times \log{\rho}
\end{alignat*}$
Now:
$\begin{alignat*}{1}
\log{\left(\frac{a_1^{1.01}}{a_2}\right)} &= 1.01\log{a_1} - \log{a_2} \
&= \log{\rho} \times \left(\frac{1.01\log{2}}{2\log{25}} - \frac{\log{1.25}}{2}\right)
\end{alignat*}$
If we compute this, we find that
$\log{\left(\frac{a_1^{1.01}}{a_2}\right)} \approx -0.005 \times \log{\rho}$
Therefore, as
$\rho \to +\infty$
$\log{\left(\frac{a_1^{1.01}}{a_2}\right)} \to -\infty$
$\frac{a_1^{1.01}}{a_2} \to 0$
We can then conclude that
$I \ll I_\text{cap}$
$\dot{I}$
$\dot{I} = K\frac{a_1^{1.01}}{a_2}I_\text{cap} = K$
Therefore
$I = Kt$
With that out of the way, let’s calculate MF’s OP factor.
$\begin{alignat*}{1}
\dot{x} &= K\
\Rightarrow x &= Kt\
\end{alignat*}$
$\omega = KI = Kt$
$\begin{alignat*}{1}
\dot{\rho} &= K\Pi \omega^{4.4}x^{3.4}\
\dot{\rho} &= K\Pi t^{7.8}\
\Rightarrow \rho &= K\Pi t^{8.8}
\end{alignat*}$
$OP = 8.8$
To compute BaP’s overpush factor, we first need to determine the limit of
$a$
$\rho \to \infty$
After the last milestone, we have:
$a=2\cdot\frac{6}{\pi^2}-\left(\sum_{i=1}^n\frac{1}{i^2}\right)^{-1}$
As
$n \to \infty$
$\left(\sum_{i=1}^n\frac{1}{i^2}\right) \to \frac{\pi^2}{6}$
$\left(\sum_{i=1}^n\frac{1}{i^2}\right)^{-1} \to \frac{6}{\pi^2}$
Therefore, as
$\rho \to \infty$
$a \to \frac{6}{\pi^2}$
Now let’s calculate BaP’s OP factor.
$\begin{alignat*}{1}
\dot{q_9} &= K\
\Rightarrow q_9 &= Kt\
\dot{q_8} &= Kq_9 = Kt\
\Rightarrow q_8 &= Kt^2\
&\dots\
q_1 &= Kt^9\
\dot{r} &= K\
\Rightarrow r &= Kt
\end{alignat*}$
$\begin{alignat*}{1}
\dot{\rho} &= \Pi t(q_1 r)^a \
&= K\Pi t\cdot (t^{10})^{\frac{6}{\pi^2}}\
&= K\Pi t^{1+\frac{60}{\pi^2}}\
\Rightarrow \rho &= K\Pi t^{2+\frac{60}{\pi^2}}
\end{alignat*}$
$\begin{alignat*}{1}
OP &= 2+\frac{60}{\pi^2}\
OP &\approx 8.079
\end{alignat*}$