Previous: Theory 9 to Endgame
Disclaimer: This is a simplified version of the guide. The guide will skip over things, and is not completely optimal. Click here for a more polished, in-depth, and optimal guide.
When discussion strategy, this section may use more jargon then the rest of the guide. If you are confused, please check out the discord server and ask.
Since it was started before the official BaP guide was written, this guide is not yet simplified. Should still be understandable though, it might just need a bit more effort to read.
Thanks to Mathis S for help with this guide
Basel Problem, abbreviated BaP to differentiate from unofficial CT Bin Packing (BiP), is based around the Basel series: $$\sum_{n=1}^{\infty}\frac{1}{n^2}$$ The subject of the famous mathematical Basel Problem which was solver by Euler, who showed the series converged to \(\frac{\pi^2}{6}\). This series converges linearly, with a 2x increase in the number of terms summed corresponding to a 2x decrease in the difference between the sum and \(\frac{\pi^2}{6}\). For example, if the first 10 terms sum are 0.001 away from being perfectly \(\frac{\pi^2}{6}\), then the first 20 terms will be about 0.0005 way from perfectly \(\frac{\pi^2}{6}\)
BaP also makes use of many nested cumulative variables, similar to T2. This makes the theory similar to T2 is many regards, with long(ish) publications, MS, and stuff generally taking a long time to happen
Before first 2 milestones
$$\dot{\rho} = (tq_1r)^a$$
$$\dot{q_1} = c_2$$
$$\dot{r} = \sum_{i=1}^{c_1}\frac{1}{i^2}$$
After first 2 milestones
$$\dot{\rho} = t(q_1r)^a$$
$$\dot{q_i} = c_{i+1}q_{i+1}, 1 \leq i \leq C$$
$$\dot{q}_{C+1} = c_{C+2}$$
$$\dot{r} = (\sum_{i=c_1}^{\infty}\frac{1}{i^2})^{-1}$$
Where \(C\) is a number based on your level of milestone 4
Pre-Milestone 1 & 2 the equations should be straight forward. Note that \(c_1\) here is practically pointless, as after only a few upgrade of \(c_1\) you will have reached very near \(\frac{pi^2}{6}\), and no more upgrades will help. From now on, only the post 2 milestones will be discussed
These equations can be quite abstracted away from what is actually happening (which is quite simple), as they combine many lines into few for brevity. This first equation determines \(\rho\) with \(a\) and \(t\) being self explanatory.
The equations governing r are also quite simple. The form of the basil problem presented is actually a condensed form of the difference between the first n terms and \(\frac{pi^2}{6}\) that was discussed earlier. This difference cen be modelled with the equation $$(\frac{\pi^2}{6} - \sum_{i=1}^{n}\frac{1}{i^2})^{-1}$$ for the first n terms. Here, we take the reciprocal of the difference, turning a 2x decrease into a 2x increase, effectively measuring how close this difference is to zero. Re-substituting the Basel Series Identity $$(\sum_{i=1}^{\infty}\frac{1}{i^2} - \sum_{i=1}^{n}\frac{1}{i^2})^{-1}$$ Here, notice the first n terms of the series actually cancel. This is more visible if you write the subtraction out in full $$(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2} + \frac{1}{(n+1)^2} + ...)$$ $$-(\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + ... + \frac{1}{n^2})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ $$ Thus, our equations can be written as $$(\frac{1}{(n+1)^2} + \frac{1}{(n+2)^2} + ...)^{-1}$$ $$(\sum_{i=n+1}^{\infty}\frac{1}{i^2})^{-1}$$ Which is the written equation, with \(n + 1 = c_1\)
Thus, we can see that the \(\dot{r}\) equation is actually the SAME as the equation for the difference between the first \(c_1\) terms of the basil series and what it converges too, that is \(\frac{\pi^2}{6}\), so one can conclude that increasing \(c_1\) will increase \(\dot{r}\) linearly
As an interesting consequence, \(\dot{r} = c_1\) is not a bad approximation
$$\dot{q_i} = c_{i+1}q_{i+1}, 1 \leq i \leq C$$
$$\dot{q}_{C+1} = c_{C+2}$$
For these equations, we will expand the notation that has been used to condense the many similar q equations. This first equation can be thought of as repeating this \(\dot{q_i} = c_{i+1}q_{i+1}\) for \(i = 1, 2, 3 ... C\) where C is based on your milestone levels. If we expand it we see $$\dot{q_1} = c_{2}q_{2}$$ $$\dot{q_2} = c_{3}q_{3}$$ $$...$$ $$\dot{q}_C = c_{C+1}q_{C+1}$$ $$\dot{q}_{C+1} = c_{C+2}$$ I hope this is looking more understandable. We see that each q level increases by the q level BELOW multiplied with the c upgrade with the corresponding subscript to that level, with \(q_1\) being increased by \(q_2\) and \(c_2\) and so on, with your final unlocked level of q only being increased by the purchasable upgrade
This is actually the same as the T2 equations, where each q/r level is influenced by the one below it and a purchasable upgrade, only in T2's case the upgrades were baked into the equations.
| \(\dot{t}\) | Increases \(\dot{t}\) by 1, max 5 |
| \(c_1\) | Increases \(\dot{r}\) by 7% (imagine \(\dot{r} = c_1\) directly). Powerful boost of x1024 every 64 level (due to stepwise variables) |
| \(c_2\) | Increases \(\dot{q_1}\) by 2x |
| \(c_3\) | Increases \(\dot{q_2}\) by 3x |
| \(c_4\) | Increases \(\dot{q_3}\) by 4x |
| \(c_5\) | Increases \(\dot{q_4}\) by 5x |
| \(c_6\) | Increases \(\dot{q_5}\) by 6x |
| \(c_7\) | Increases \(\dot{q_6}\) by 7x |
| \(c_8\) | Increases \(\dot{q_7}\) by 8x |
| \(c_9\) | Increases \(\dot{q_8}\) by 9x |
| \(c_{10}\) | Increases \(\dot{q_9}\) by 10x |
Being very T2 like, the best strategy is BaP is often autobuy all with some MS. The a milestones increase \(\dot{\rho}\) and the q milestones give you a new layer of q. For milestone swapping, you want to swap into the \(q_i\) milestone, and wait for \(q_i\) and \(q_{i-1}\) to increase for a while, before swapping back into \(a\) milestones. As there are large gaps between \(c_i\) upgrades for later i (3/4+), you will generally only need to do this swap few or one time per pub. Furthermore, the waiting times are much longer. Whilst T2 has cycles of 50s, BaP will have cycles measured in hours. The exact best timing is unknown
Furthermore, BaP has Doubling Chasing and chasing \(c_1\) boosts. As \(c_1\) get's a huge x1024 boost every 64 levels, if you are close to a boost (divide by 64 and find the remainder to determine how close you are. If you get something like 56 you are close) then you should autobuy \(c_1\) to reach it. When you are not chasing a boost, you should buy \(c_1\) at \(c_1 \% 64 / 2\) less then other variables
Go from Autobuy All to Never Buy Anything (full coast) at around 25x lower then you desired pub multi, which is 1e26 \(\rho\)