Overpush Ratios for Official Custom Theories

Guide written by Mathis S.. Contributions from the Amazing Community.

Feel free to use the glossary or extensions as needed.

Introduction #

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 $Π$ , a notation used by Gaunter in the Laplace Transforms CT.

In this entire article, I will use $\log$ for the base 10 logarithm and $\ln$ for the base $e$ logarithm.

Generalities about the OP coefficient and how to calculate it #

As you have read in the Distribution Overpushing guide, the overpushing coefficient measures the effect of the publication multiplier on $\dot{τ}$ , vs the effect of time on $\dot{τ}$ for each theory. More concretely, the OP coefficient is the base 10 logarithm of the number you should multiply the publication multiplier by to multiply the theory’s $τ$ /hour rates by 10. A simpler way to put it : if you multiply the publication multiplier by $a$ , the $τ$ /hour rate will be multiplied by $a^{\frac{1}{O P}}$ .

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 ρ}{d t} = K ρ^{x} t^{y} ρ_{p u b}^{z}

While this model is great, it has one flaw to analyze official CTs, as it assumes $Π = ρ_{p u b}^{z}$ , or in other words that $\frac{d ρ}{d t} \propto Π$ , which is not the case for every CT. Let’s decompose $z = a b$ , where $a$ is such that $Π \propto ρ^{a}$ . In this case, we have $\frac{d ρ}{d t} \propto Π^{b}$ or $Π^{b} = ρ_{p u b}^{z}$ , a more general model.

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 $ρ_{p u b}^{z} = Π^{b}$ , so the optimal publication multiplier, the optimal multiplier for $Π$ , would instead be:

e^{\frac{z (y + 1)}{b (1 - x)}}

Now, for any non-divergent theory, $z + x \leq 1$ . If a theory can be approximated as balanced, i.e. decay is negligible, then we have $z + x \approx 1$ , then $\frac{z}{1 - x} \approx 1$ . In this case, the optimal publication multiplier can be approximated as:

e^{\frac{y + 1}{b}}

It turns out that $\frac{y + 1}{b}$ , the ratio between the time contribution to the publication multiplier contribution, is exactly the OP factor of the theory. Therefore, we can approximate the optimal publication multiplier as $e^{O P}$ .

Note that in practice, due to $z + x < 1$ along with other factors, the optimal multiplier fluctuates and is generally lower than that.

Proof that $\frac{y + 1}{b}$ is the overpush ratio #

Let’s prove that $\frac{y + 1}{b}$ is indeed the overpush ratio. We’ll prove that multiplying the pub mult $Π$ by some factor $m$ increases the theory speed by $m^{\frac{1}{O P}}$ , which is the same as proving that the time to gain the same amount of $ρ$ is divided by $m^{\frac{1}{O P}}$ when $Π$ is multiplied by $m$ .

To prove this, we’ll start from equation (3-3) of pacowoc’s guide:

ρ = (K^{'} (1 - x) ρ_{p u b}^{z} t^{y + 1})^{\frac{1}{1 - x}}

We substitute $ρ_{p u b}^{z}$ by $Π^{b}$ :

ρ = (K^{'} (1 - x) Π^{b} t^{y + 1})^{\frac{1}{1 - x}}

Now we solve for $t$ :

t = {(\frac{ρ^{1 - x}}{K^{'} (1 - x)} Π^{- b})}^{\frac{1}{y + 1}} = {(\frac{ρ^{1 - x}}{K^{'} (1 - x)})}^{\frac{1}{y + 1}} Π^{- \frac{b}{y + 1}}

Now let’s calculate $t^{'}$ where we replace $Π$ by $m Π$ :

\begin{matrix} \begin{aligned} t^{'} & = {(\frac{ρ^{1 - x}}{K^{'} (1 - x)})}^{\frac{1}{y + 1}} m^{- \frac{b}{y + 1}} Π^{- \frac{b}{y + 1}} \\ \frac{t^{'}}{t} & = \frac{1}{m^{\frac{b}{y + 1}}} \end{aligned} \end{matrix}

We can see that, when multiplying the pub mult by $m$ , the time to reach the same $ρ$ is divided by $m^{\frac{b}{y + 1}}$ .

Therefore, $\frac{y + 1}{b}$ is indeed the overpush ratio of the theory.

How to calculate the overpush ratio in practice #

From pacowoc’s model:

\frac{d ρ}{d t} = K ρ^{x} t^{y} Π^{b}

We can immediately find the overpush ratio: $\frac{y + 1}{b}$ .

We can also treat $ρ^{x}$ as a constant, as it is dependant on upgrades and doesn’t correspond to $ρ$ directly. Therefore, if we integrate:

ρ = K ρ^{x} t^{y + 1} Π^{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 ρ = x \log ρ + (y + 1) \log t + b \log Π

Calculating the OP ratio for Official Custom Theories #

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 $K$ to represent any constant holding parameters that are not relevant to our study, namely parameters that don’t depend on $Π$ nor $t$ . I will reuse the same symbol to avoid having to create many of them.

WSP #

\begin{aligned} \dot{q} & = & K \\ \Rightarrow q & = & K t \\ \dot{ρ} = & K & Π q = K Π t \\ \Rightarrow ρ = & K & Π t^{2} \\ O P & = & 2 \end{aligned}

SL #

\begin{aligned} \dot{ρ_{3}} & = K \\ \Rightarrow ρ_{3} & = K t \end{aligned}

\begin{aligned} {\dot{ρ}}_{2} & = K \times {1.96}^{- \ln ρ_{3}} \\ {\dot{ρ}}_{2} & = K {ρ_{3}}^{- \ln 1.96} \\ {\dot{ρ}}_{2} & = K t^{- \ln 1.96} \\ \Rightarrow ρ_{2} & = K t^{1 - \ln 1.96} \\ \frac{1}{e - γ} & = K ρ_{3} = K t \end{aligned}

Now:

\begin{aligned} \dot{ρ_{1}} & = Π ρ_{2}^{0.53} \times \frac{1}{e - γ} \\ \dot{ρ_{1}} & = K Π t^{0.53 \times (1 - \ln 1.96) + 1} \\ ρ_{1} & = K Π t^{2 + 0.53 \times (1 - \ln 1.96)} \end{aligned}

\begin{aligned} O P & = 2 + 0.53 \times (1 - \ln 1.96) \\ O P & \approx 2.173 \end{aligned}

EF #

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{aligned} \dot{q} & = Π q_{1} q_{2} \\ \Rightarrow q & = K Π q_{1} q_{2} t \end{aligned}

Late-game, in EF’s $\dot{ρ}$ equation’s square root, only the $t q^{2}$ term is significant:

\begin{aligned} \dot{ρ} & = Π (a_{1} a_{2} a_{3})^{1.5} t^{0.5} q \\ \Rightarrow \dot{ρ} & = Π^{2} (a_{1} a_{2} a_{3})^{1.5} q_{1} q_{2} t^{1.5} \\ \Rightarrow ρ & = Π^{2} (a_{1} a_{2} a_{3})^{1.5} q_{1} q_{2} t^{2.5} \\ \dot{R} & = K Π (b_{1} b_{2})^{2} \\ \Rightarrow R & = K Π (b_{1} b_{2})^{2} t \\ \dot{I} & = K Π (c_{1} c_{2})^{2} \\ \Rightarrow I & = K Π (c_{1} c_{2})^{2} t \end{aligned}

Now, let’s express the system in logarithmic form, to turn it into a linear system:

\begin{aligned} \log ρ = & K + 1.5 (\log a_{1} + \log a_{2} + \log a_{3}) \\ + \log q_{1} + \log q_{2} + 2 \log Π + 2.5 \log t \\ \log R = & K + 2 (\log b_{1} + \log b_{2}) + \log Π + \log t \\ \log I = & K + 2 (\log c_{1} + \log c_{2}) + \log Π + \log t \end{aligned}

We have:

\begin{aligned} \log a_{1}, \log q_{1}, \log q_{2} & \propto \log ρ \\ \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{aligned}

For the purpose of this study, we can ignore variables bought with $ρ$ .

\begin{aligned} C \log ρ & = K + 1.5 (\log a_{2} + \log a_{3}) + 2 \log Π + 2.5 \log t \\ \log R & = K + 2 (\log b_{1} + \log b_{2}) + \log Π + \log t \\ \log I & = K + 2 (\log c_{1} + \log c_{2}) + \log Π + \log t \end{aligned}

\begin{aligned} C \log ρ - 1.5 \log a_{2} - 1.5 \log a_{3} & = & 2 \log Π & + & 2.5 \log t & + K \\ (1 - 2 \frac{\log b_{1} + \log b_{2}}{\log R}) \log R & = & \log Π & + & \log t & + K \\ (1 - 2 \frac{\log c_{1} + \log c_{2}}{\log I}) \log I & = & \log Π & + & \log t & + K \end{aligned}

To find EF’s OP factor, we need an expression of the form $\log ρ = α \log Π + β \log t$ . Then, the OP factor will be given by $\frac{β}{α}$ .

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{aligned} C \log ρ - \frac{1.5 \log 40}{22 \log 2} \log R - \frac{1.5}{2.2} \log I & = & 2 \log Π & + & 2.5 \log t & + K \\ [1 - 2 (\frac{\log 2}{10 \log 200} + \frac{\log 1.12}{\log 2})] \log R & = & \log Π & + & \log t & + K \\ [1 - 2 (\frac{\log 2}{10 \log 200} + \frac{\log 1.125}{\log 2})] \log I & = & \log Π & + & \log t & + K \end{aligned}

These expressions will be heavy to manipulate so let’s set it to:

\begin{aligned} C \log ρ & - & d_{12} \log R & - & d_{13} \log I & = & 2 \log Π & + & 2.5 \log t & + K \\ d_{22} \log R & = & \log Π & + & \log t & + K \\ d_{33} \log I & = & \log Π & + & \log t & + K \end{aligned}

Now if we add $\frac{d_{12}}{d_{22}}$ times equation (2) and $\frac{d_{13}}{d_{33}}$ times equation (3) to equation (1) we get:

\begin{aligned} C \log ρ & = & (2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}) \log Π \\ + & (2.5 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}) \log t + K \end{aligned}

We can finally express EF’s OP factor:

\begin{aligned} O P & = \frac{2.5 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}}{2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}} \\ O P & = 1 + \frac{0.5}{2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{33}}} \\ O P & = 1 + \frac{1}{2 (2 + \frac{d_{12}}{d_{22}} + \frac{d_{13}}{d_{23}})} \\ O P & = 1 + \frac{1}{4 + 2 \frac{d_{12}}{d_{22}} + 2 \frac{d_{13}}{d_{23}}} \end{aligned}

For those who want to see what it looks like if we replace those $d_{i}$ :

O P = 1 + {[\begin{array}{l} 4 + \frac{3 \log 40}{22 \log 2} \frac{1}{1 - 2 (\frac{\log 2}{10 \log 200} + \frac{\log 1.12}{\log 2})} \\ + \frac{3}{2.2} \frac{1}{1 - 2 (\frac{\log 2}{10 \log 200} + \frac{\log 1.125}{\log 2})} \end{array}]}^{- 1}

Finally,

O P \approx 1.137

CSR2 #

\begin{aligned} \dot{q} & = & K Π \\ \Rightarrow q & = & K Π t \\ \dot{ρ} = & K & Π q = K Π^{2} t \\ \Rightarrow ρ = & K & Π^{2} t^{2} \\ O P & = & 1 \end{aligned}

FI #

\begin{aligned} \dot{q} & = K \\ \Rightarrow q & = K t \\ \dot{r} & = K \\ \Rightarrow r & = K t \end{aligned}

\int_{0}^{q} g (x) d x = K q^{6} = K t^{6}

\sqrt[π]{\int_{0}^{q} g (x) d x} = K t^{\frac{6}{π}}

\begin{aligned} \dot{ρ} & = K Π r t \sqrt[π]{\int_{0}^{q} g (x) d x} \\ \dot{ρ} & = K Π t^{2 + \frac{6}{π}} \\ \Rightarrow ρ & = K Π t^{3 + \frac{6}{π}} \end{aligned}

\begin{aligned} O P & = 3 + \frac{6}{π} \\ O P & \approx 4.91 \end{aligned}

FP #

\begin{aligned} \dot{q} & = K \\ \Rightarrow q & = K t \\ \dot{r} & = K \\ \Rightarrow r & = K t \end{aligned}

\begin{aligned} \dot{ρ} & = K Π q r t \\ \dot{ρ} & = K Π t^{3} \\ \Rightarrow ρ & = K Π t^{4} \end{aligned}

O P = 4

RZ #

Since RZ also has multiple currencies, we’ll follow the same steps as with EF. For the purpose of this study, $ζ$ and $\dot{ζ}$ values can be approximated as constants. The black hole has no effect on the overpush factor as the optimal $t$ to activate the black hole is roughly proportional to the publication time.

\begin{aligned} \dot{ρ} & = K Π c_{1}^{1.25} c_{2} w_{1} t \\ \Rightarrow ρ & = K Π c_{1}^{1.25} c_{2} w_{1} t^{2} \\ \dot{δ} & = K Π w_{1} w_{2} w_{3} \\ \Rightarrow δ & = K Π w_{1} w_{2} w_{3} t \end{aligned}

Let’s express this system on its logarithmic form:

\begin{aligned} \log ρ & = & 1.25 \log c_{1} + \log c_{2} + \log w_{1} \\ + & \log Π + 2 \log t + K \\ \log δ & = & \log w_{1} + \log w_{2} + \log w_{3} \\ + & \log Π + \log t + K \end{aligned}

We have:

\begin{aligned} \log c_{i} & \propto \log ρ \\ \log w_{i} & \propto \log δ \end{aligned}

We can also ignore variables bought with $ρ$ .

\begin{aligned} C \log ρ - \log w_{1} & = & \log Π & + & 2 \log t & + K \\ (1 - \frac{\log w_{1} + \log w_{2} + \log w_{3}}{\log δ}) \log δ & = & \log Π & + & \log t & + K \end{aligned}

Let’s now compute how the variables we need scale with their currencies.

$w_{1}$ is a steowise 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: $\log w_{1} = \frac{\log 2}{8} \cdot \frac{6}{\frac{1}{3} \log 100} \cdot δ = \frac{9}{8} \log 2 \cdot \log δ$ .
$w_{2}$ is an exponential variable of power 2 and cost scaling 10. Its power is given by $\log w_{2} = \log 2 \cdot \log δ$ .
$w_{3}$ is an exponential variable of power 2 and cost scaling $1 e 30$ . Its power is given by $\log w_{3} = \frac{\log 2}{30} \cdot \log δ$ .

Now we can substitute in the system:

\begin{aligned} - \frac{9}{8} \log 2 \log δ & + & C \log ρ & = & \log Π & + & 2 \log t & + K \\ [1 - (1 + \frac{9}{8} + \frac{1}{30}) \log 2] \log δ & = & \log Π & + & \log t & + K \\ (1) & - \frac{9}{8} \log 2 \log δ & + & C \log ρ & = & \log Π & + & 2 \log t & + K \\ (2) & (1 - \frac{259}{120} \log 2) \log δ & = & \log Π & + & \log t & + K \end{aligned}

Now if we add $\frac{\frac{9}{8} \log 2}{1 - \frac{259}{120} \log 2}$ times equation (2) to equation (1) we get:

\begin{aligned} C \log ρ & = & (1 + \frac{\frac{9}{8} \log 2}{1 - \frac{259}{120} \log 2}) \log Π \\ + & (2 + \frac{\frac{9}{8} \log 2}{1 - \frac{259}{120} \log 2}) \log t + K \end{aligned}

We can now express RZ’s OP factor:

\begin{aligned} O P & = \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}} \\ O P & = 1 + {[1 + \frac{\frac{9}{8} \log 2}{1 - \frac{259}{120} \log 2}]}^{- 1} \end{aligned}

Finally,

O P \approx 1.508

MF #

To determine the OP factor of MF, we first need to determine if, long-term, $I$ is capped or not.

\begin{aligned} \dot{I} & = \frac{a_{1}^{1.01}}{400} (10^{- 15} - \frac{I}{a_{2}}) \\ \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_{cap} - I) \end{aligned}

To determine if $I$ is capped or not, we need to study the behavior of $\frac{a_{1}^{1.01}}{a_{2}}$ at large rho.

If it goes to $+ \infty$ , then $I$ will almost instantly cap at large rho.
If it converges to $0$ , then $I ≪ I_{cap}$ at large rho.

Let’s now calculate how $a_{1}$ and $a_{2}$ scale with rho.

$a_{1}$ is a stepwise variable with a power of 2, a cycle length of 5 and a cost scaling of 25. Its power is given by:

\log a_{1} = \frac{\log 2}{2 \log 25} \times \log ρ

$a_{2}$ is an exponential variable with a power of 1.25 and a cost scaling of 100. Its power is given by:

\begin{aligned} \log a_{2} & = \frac{\log 1.25}{\log 100} \times \log ρ \\ = \frac{\log 1.25}{2} \times \log ρ \end{aligned}

Now:

\begin{aligned} \log (\frac{a_{1}^{1.01}}{a_{2}}) & = 1.01 \log a_{1} - \log a_{2} \\ = \log ρ \times (\frac{1.01 \log 2}{2 \log 25} - \frac{\log 1.25}{2}) \end{aligned}

If we compute this, we find that $\log (\frac{a_{1}^{1.01}}{a_{2}}) \approx - 0.005 \times \log ρ$ .

Therefore, as $ρ \to + \infty$ , $\log (\frac{a_{1}^{1.01}}{a_{2}}) \to - \infty$ so $\frac{a_{1}^{1.01}}{a_{2}} \to 0$ .

We can then conclude that $I ≪ I_{cap}$ at large rho, which means we can simplify $\dot{I}$ :

\dot{I} = K \frac{a_{1}^{1.01}}{a_{2}} I_{cap} = K

Therefore $I = K t$ .

With that out of the way, let’s calculate MF’s OP factor.

\begin{aligned} \dot{x} & = K \\ \Rightarrow x & = K t \end{aligned}

ω = K I = K t

\begin{aligned} \dot{ρ} & = K Π ω^{4.4} x^{3.4} \\ \dot{ρ} & = K Π t^{7.8} \\ \Rightarrow ρ & = K Π t^{8.8} \end{aligned}

O P = 8.8

BaP #

To compute BaP’s overpush factor, we first need to determine the limit of $a$ as $ρ \to \infty$ .

After the last milestone, we have:

a = 2 \cdot \frac{6}{π^{2}} - {(\sum_{i = 1}^{n} \frac{1}{i^{2}})}^{- 1}

As $n \to \infty$ , $(\sum_{i = 1}^{n} \frac{1}{i^{2}}) \to \frac{π^{2}}{6}$ , then ${(\sum_{i = 1}^{n} \frac{1}{i^{2}})}^{- 1} \to \frac{6}{π^{2}}$ .

Therefore, as $ρ \to \infty$ , $a \to \frac{6}{π^{2}}$ .

Now let’s calculate BaP’s OP factor.

\begin{aligned} \dot{q_{9}} & = K \\ \Rightarrow q_{9} & = K t \\ \dot{q_{8}} & = K q_{9} = K t \\ \Rightarrow q_{8} & = K t^{2} \\ \dots \\ q_{1} & = K t^{9} \\ \dot{r} & = K \\ \Rightarrow r & = K t \end{aligned}

\begin{aligned} \dot{ρ} & = Π t (q_{1} r)^{a} \\ = K Π t \cdot (t^{10})^{\frac{6}{π^{2}}} \\ = K Π t^{1 + \frac{60}{π^{2}}} \\ \Rightarrow ρ & = K Π t^{2 + \frac{60}{π^{2}}} \end{aligned}

\begin{aligned} O P & = 2 + \frac{60}{π^{2}} \\ O P & \approx 8.079 \end{aligned}

Overpushing Coefficients #

From previous workings, the approximate overpushing coefficients of official custom theories are:

Class: breakdown;
last_row: false;

			Overpushing Coefficients
WSP	2	[class=“bheader”;]FP	4
SL	2.17	[class=“bheader”;]RZ	1.51
EF	1.14	[class=“bheader”;]MF	8.8
CSR2	1	[class=“bheader”;]BaP	8.08
FI	4.91

Now, let’s calculate $e^{O P}$ for each CT, which gives an estimate on the optimal publication multiplier of these CTs

Class: breakdown;
last_row: false;

			Pub Multi
WSP	7.39	[class=“bheader”;]FP	54.6
SL	8.79	[class=“bheader”;]RZ	4.52
EF	3.12	[class=“bheader”;]MF	6634
CSR2	2.72	[class=“bheader”;]BaP	3227
FI	136

As we can see, the expected publication multipliers from the model are close to the results given by the sim.

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