The op­timal pub­lish mul­ti­plier for T5 is between 2-3 and 6-10, please check with The Sim for ac­cur­acy. The­ory 5 is based on lo­gistic func­tion. This the­ory is slow early, but be­comes very power­ful later on in the game. It is re­com­men­ded to keep push­ing this the­ory as high as pos­sible be­fore reach­ing ee14k ft. Make sure to care­fully read the be­ha­vior of c1 and c2 vari­ables in this the­ory, as the be­ha­vi­ors are quite unique.

$\dot{\rho }=q_1^{1.15}{q}_{2}q$

$\dot{q}=(c_1/c_2)q(c_3^{1.1}-q/c_2)$

The first line states that the rate of change of $\rho$ is the product of some $q_1,q_2,q$. Note that $q_1$ and $q_2$ are vari­ables that you can buy dir­ectly, while $q$ is a sep­ar­ate vari­able that is based off something else.

The second line defines this the­ory. It de­scribes the be­ha­vior of a typ­ical lo­gistic func­tion. A lo­gistic func­tion typ­ic­ally has slow growth at the be­gin­ning, then fast growth in the middle, then it flat­tens out at the end. Here we have $c_1$, which speeds up the rate at which we reach the max­imum value of $q$. Note it DOES NOT in­crease the ac­tual max­imum value of $q$ it­self. We also have $c_2$. This in­creases the max­imum value of $q$. However, it HALVES the speed at which this max­imum value is reached. There­fore we must be care­ful to not buy too many at once.

$c_3$ is sim­ilar to $c_2$ in which buy­ing it in­creases the max­imum value of $q$. However it does not have the draw­back that $c_2$ has. So we can buy this un­con­di­tion­ally.

The max­imum value of $q$ is $c_2{c}_3^{1.1}$. Once $q$ reaches this max­imum value, $q$ dot will be zero and $q$ will not grow any­more un­til we buy either more $c_2,c_3$.

Ap­prox­im­ate vari­able strengths on $\dot{\rho }$ with all mile­stones are as fol­lows:

Think of buy­ing $c_1$ as throt­tling on the bi­cycle faster. Buy­ing $c_2$ is sim­ilar to shift­ing the bi­cycle gear up by 1 gear.

If all one does is buy $c_1$ and never $c_2$, then they are stuck in gear 1 forever and make no pro­gress. However, if all one does is buy $c_2$ and never $c_1$, then this is sim­ilar to try­ing to ride from highest gear from 0 speed, which takes a long time and a lot of ef­fort.

There­fore us­ing the bi­cycle ana­logy, buy $c_2$ only when $\dot{q}$, speed, is enough to sup­port it; not too early and not too late. Fur­ther­more, later in the pub­lic­a­tion, buy only 1 level of $c_2$ at a time. Buy $c_1$ only right after buy­ing $c_2$ (shift­ing up gear).

When de­cid­ing when to buy $c_1,c_2$, think of $c_1$ as throt­tling a bi­cycle, and $c_2$ as shift­ing up gear by 1 level.

The strengths of each vari­able are as fol­lows:

$c_3$ ~= $q_2$ ≈ $c_2 $ > $q_1 $ > $c_1 $

Note that $c_1$ and $c_2$ have spe­cial in­ter­ac­tions.

Manual buy­ing c2 - READ THIS BE­FORE DO­ING THE STRATEGIES #

For step 2 of the semi-idle and act­ive strategies be­low, you should be manu­ally buy­ing $c_2$. If you’re over $e150$ $\rho$, start with buy­ing 10x vari­ables at a time. Oth­er­wise buy 1 level of $c_2$ at a time.

You want to buy $c_2$ un­til the value of $q$ in­creases slowly. The pat­tern should be:

1. Buy $c_2$. $q$ should in­crease.
2. Once $q$ in­creases slowly (or stops in­creas­ing), buy more $c_2$.
   

If you buy a $c_2$ and the value of $q$ is in­creas­ing even slower than be­fore, this means you bought too many levels of $c_2$. Wait for $q$ to sta­bil­ize and then con­tinue buy­ing $c_2$

Once you’ve reached within $e5$ of the pre­vi­ous pub­lic­a­tion point, you can safely auto­buy $c_2$ for the rest of the pub­lic­a­tion. Re­mem­ber to un­tick $c_2$ auto­buy at the end of the pub­lic­a­tion when you click pub­lish!

Idle #

It is not re­com­men­ded to idle the­ory 5 (see ex­plan­a­tion on $c_1$ and $c_2$ above). Do try the semi-idle strategy be­low.

For each pub­lic­a­tion: auto­buy $c_3$ and $q_2$.

1. For the first 10 seconds, auto­buy everything ex­cept $c_2$.
2. Af­ter­wards, simply auto­buy all un­til pub­lish.

Semi-Idle #

Semi-idle is sim­ilar to idle, but once the the­ory is re­covered, manual buy $c_2$ and dis­able $c_1$.

For each pub­lic­a­tion: auto­buy $c_3$ and $q_2$.

1. For the first 10 seconds, auto­buy everything ex­cept $c_2$.
2. Then we want to manu­ally buy $c_2$. See Manual Buy­ing c2. Do this un­til it slows down and you’re within ~$e5\rho$ un­der last pub­lic­a­tion mark.
3. Then we auto­buy all un­til $\rho$ has reached its pre­vi­ous pub­lic­a­tion value (fin­ished re­cov­ery).
4. Af­ter­wards, de­ac­tiv­ate $c_1$ and auto­buy the rest un­til pub­lish.

Act­ive #

Here’s a simple yet ef­fect­ive act­ive strategy that can be used right un­til en­dgame. To find more op­tim­ized strategies, please see the List of the­ory strategies.

For each pub­lic­a­tion: auto­buy $c_3$ and $q_2$. Buy $c_1$ ONLY right after buy­ing $c_2$ from steps 2 on­wards. Prac­tic­ally, everytime level of $c_2$ is bought from steps 2 on­wards, ~5-6 levels of $c_1$ would be bought.

Note that for faster speed, dur­ing the first part of step 2, 10 levels can be bought at a time.

1. For the first 10 seconds, auto­buy everything ex­cept $c_2$.
2. Then we want to manu­ally buy $c_2$. See Manual Buy­ing c2.
3. Then we auto­buy $c_3,q_2,c_2$. Out of these 3 vari­ables, find the one with the cheapest cost. Then buy $q_1$ un­til its cost ex­ceeds 15% of the cheapest vari­able found above. Buy $c_1$ ONLY right after buy­ing a level of $c_2$.
4. Once the the­ory has re­covered to its pre­vi­ous pub­lic­a­tion mark, slowly put less em­phasis on $c_1$. When in doubt, have $c_1$’s cost be sim­ilar to $q_1$’s cost. Con­tinue do­ing step 2 un­til pub­lish.

All mile­stones into the 2nd mile­stone to un­lock $c_3$. Then into 1st mile­stone be­cause $q_1$ vari­able is higher value than $c_3$ vari­able, fi­nally put the rest into the last mile­stones. This the­ory does not have a known ef­fect­ive mile­stone swap­ping strategy.

Com­ment­ary

(By Snaeky)

No com­ment­ary

(By Snaeky)

(By Playspout)

T5 will al­ways give its best res­ults from act­ive play. However, after step 3, you can still get good res­ults while auto­buy­ing $q_1$ and manu­ally pur­chas­ing $c_1$ every 10-15min. This makes the the­ory slightly less act­ive and easier to deal with.

Warn­ing: Do not overnight this the­ory. It has ter­rible de­cay after passing a good pub­lic­a­tion mark and will not give good res­ults. T5i is only vi­able very late/​en­dgame.

Ad­di­tional in­form­a­tion #

Pur­chase $c_2$ when $1.5q>c_2\ast c_3^{m_3}$. $m_3$ is the num­ber of mile­stone 3.

$q$ be­gins to slow down upon reach­ing $2q>c_2\ast c_3^{m_3}$.

Strategy con­struc­ted by: Snaeky, Marks, Baldy, and Nerdy

T6 has the low­est de­cay of all the the­or­ies. It will be second place to T5 un­til ~$e750$ and is the only the­ory that can get to $>e1300\tau$. You should overnight this and T4 after you get your T2 to $e350+$.

At first, T6 only finds the area un­der the curve of the graph $f(q)$ from $0$ to $q$, which is es­sen­tially a 2d plane. This is done us­ing a def­in­ite in­teg­ral, an in­teg­ral that is bound between 2 val­ues giv­ing a single out­put. This does the op­pos­ite of what a de­riv­at­ive does, but within a spe­cific bound. With later mile­stones, this will in­clude the vari­able $r$, this new equa­tion is find­ing the volume of the graph within the bounds of planes $f(q)$ and $f(r)$ from $0$ to $q$ and $0$ to $r$ re­spect­ively.

$C$ is equal to the sum of all in­stant­an­eous changes in $\rho$, or equi­val­ently in­stant­an­eous in­creases to the in­teg­ral, caused by buy­ing $c_i$ up­grades*. When a $c_i$ up­grade is pur­chased, the value of the in­teg­ral in­creases by a re­spect­ive amount.

For ex­ample, let the in­teg­ral equal 10, and you buy an up­grade that costs $5\rho$. $\rho$ should de­crease to $\rho =10-5=5$. If the in­teg­ral in­creases to 20, then to for $\rho =5$, $C=$in­teg­ral$-\rho =20-5=15$.

$C$ acts as the mech­an­ism which de­creases $\rho$ the cor­rect amount after a pur­chase. If $C$ did not counter-bal­ance, then $\rho$ would di­verge as every up­grade would in­crease $\rho$ in­stead of de­crease (also the reason it is $-C$ and not $+C$).

* This is not the same as the sum of the cost of all $c_i$ up­grades as not every up­grade in­creases the in­teg­ral by the same amount, i.e. an up­grade cost­ing $5\rho$ will not in­crease the in­teg­ral by $5\rho$, but more or less.

Video of T6 at En­dgame

The op­timal mul­ti­plier var­ies between 6-12, but spikes de­pend­ing on what vari­able is dom­in­ant at the time and how close you are to a mile­stone. For an ac­cur­ate mul­ti­plier, check with the sim.

T7 can be sum­mar­ized as a max­im­iz­a­tion prob­lem : given a sur­face in 3-di­men­sional space, you want to find its highest alti­tude by mov­ing along the sur­face, al­ways in the dir­ec­tion of steep­est as­cent (that’s ba­sic­ally a gradi­ent as­cent). The func­tion $g(x,y)$ can be seen as a sur­face in ${\mathbb{R}}^3$ (con­sid­er­ing the set of points $(x,y,g(x,y))$, see at­tached im­age). $({\rho }_1,{\rho }_2,g({\rho }_1,{\rho }_2))$ is a point on this sur­face. Our goal is to max­im­ize $g({\rho }_1,{\rho }_2)$, i.e. to find $({\rho }_1,{\rho }_2)$ that max­im­ize $g({\rho }_1,{\rho }_2)$. No­tice that the func­tion $g$ is un­boun­ded, i.e. you can’t find a proper max­imum (we say that the max­im­iz­a­tion prob­lem is ill-con­di­tioned); so one way to max­im­ize $g({\rho }_1,{\rho }_2)$ is to move $({\rho }_1,{\rho }_2)$ to­wards the dir­ec­tion of steep­est as­cent. This is what is pre­cisely done by set­ting $\dot{\rho }$ (which is the dir­ec­tion the point $\rho =({\rho }_1,{\rho }_2)$ will move to­ward) to $\mathrm{\nabla }g({\rho }_1,{\rho }_2)$ (i.e. the gradi­ent of $g$ eval­u­ated at $({\rho }_1,{\rho }_2)$, which gives the dir­ec­tion of steep­est as­cent of $g$ at the point $({\rho }_1,{\rho }_2)$.

This is the graph of the func­tion $g$, taken after the first four mile­stones have been un­locked (Note: here, coef­fi­cients like $c_1,c_2\dots$ are ig­nored. The ef­fect of those coef­fi­cients is simply mak­ing the graph steeper in $x$ or $y$ dir­ec­tion, de­pend­ing on the value of each coef).

The op­timal pub­lic­a­tion mul­ti­plier is $4$-$6$. You will swap from 0/​1/​1 → 0/​0/​2 near $e67$. The strategy for manual buy be­fore 4 mile­stones is to only manual buy $q_1$ and $c_1$ cheap ($e1$ less $\rho$) and the rest full auto. After mile­stone 5, turn it on full auto­buy for idle. For act­ive, fol­low the strat de­scribed in the the­ory sim guide or watch the video be­low.

Mem­or­ize your stu­dent dis­tri­bu­tions with and without 10/​20/​30 R9 stu­dents. Use the stu­dent cal­cu­lator if needed. You will com­monly see people refer to this as R9 seap­ing as a long held name of the strategy.

1. Wait till $f(t)$ stops grow­ing with stu­dents in R9 push­ing $\tau$.
2. Start ac­cel (prefer­ably keep it between prestiges).
3. Po­ten­tially sit here to stack t for big­ger ${\phi }_2$ when you have stu­dents in ${\phi }_2$. Only do this when you are near a gradu­ation mark. This is not use­ful if you will not swap into ${\phi }_2$.
4. Re­spec all 10/​20/​30 stu­dents from R9.
5. Wait for the auto­prestige to prestige and swap back stu­dents to R9.
6. Re­peat.

This method al­lows you to push $f(t)$ with al­most no loss of R9 up­time or push­ing power. This is harder with fewer levels of R9 but still helps if you get used to it.

You can find the auto­prestige used for R9 Swap­ping here: Equa­tion. If you don’t have this ex­pres­sion, then you will have to manu­ally prestige each swap.

Ref­er­ence R9 Swap­ping Auto­prestige Ex­plan­a­tion

The op­timal pub­lic­a­tion mul­ti­plier is 2.5-5 de­pend­ing on how close you are to the next mile­stone. This the­ory is ex­tremely slow at the start which is why we skip un­til we ob­tain R9. It is also the only one with a $1e20$ mile­stone step. It will speed up once you hit $1e60$ and even faster at $1e80$ and $1e100$ etc. un­til ~$e250$-$e300$. The worst part is the $1e50$-$1e60$ grind. The grind to $1e60$ will take a good bit of time but is faster with R9.

At the start, manual buy pri­or­it­izes $c_2$ then $c_1$ then rest. Once you get to 0/​0/​0/​2, pri­or­it­ize $c_2$ and $c_5$ then $c_1$ then the rest. Once at 2/​0/​3/​0, you will pri­or­it­ize $c_2$ and $c_4$ then $c_1$ then the rest after. This con­tin­ues to max at 2/​3/​3/​3.