Slavery: overrated?

[Tlc2011]

For some reason, i always end up getting cannons in the mid 2000's.

Now, i'm a BtS player so Slavery got nerfed due to the Slave Riot event,
i'd suggest you don't adopt it unless you have no choice and you've got
tanks looking up your capital's ass.

 

[VoiceOfUnreason]

For some reason, i always end up getting cannons in the mid 2000's.

Now, i'm a BtS player so Slavery got nerfed due to the Slave Riot event,
i'd suggest you don't adopt it unless you have no choice and you've got
tanks looking up your capital's ass.

My feeling is that, if the topic is efficiency, then examples from games where cannons appear in the mid 2000s are probably not relevant. You're correct that the risk of the Slave Riot is part of the opportunity cost of using slavery, and should be considered. However, I think most of the old hands here abandoned random events long ago.


A contrived example, intended primarily as a demonstration of how a calculation comparison might go.

Consider please a bad cottage town - lots of green tiles, no food resources.

Happy Cap: 6
Food Bin: 29/32 (size 6).

For the sake of calculating yield, we'll assume we have six cottages (2F), one farm (3F) and two workshops (1F).

First, we look at a stagnant calculation. So 4 cottages + 2 workshops. We net no food in that situation, so can park at this level of food indefinitely. As it happens, we're going to be using a 11 turn cycle, so the yield is 44 cottages + 22 workshops.

Now lets consider the case where we have a two pop whip that will lift the happy cap from 6 to 7. That might be a theater (when we have dyes), a spiritual temple, an industrious forge (plus metal), a garrison unit under HR, or any of a number of things that already have enough hammers invested.

Now let's look at the small whip: we need to avoid the penalty for whipping a fresh item, so @T0 we'll work 6 cottages. That takes the food bin to 31/32. @T1, we whip down to 31/28 (size 4) and work 3 cottages + a farm (surplus:3). We grow - the bin is now at 20/30 (size 5). We work 4 cottages and a farm for 3 turns to bring us to 29/30. in the same configuration, we grow to 17/32 (size 6) - this pop is happy because the higher cap and whip penalty cancel. Now we work 5 cottages + a farm for 4 turns. We're back to 29/32 @T10... but because we did the whip @T1, we have one more turn of unhappy to burn. So to stall out, we can work 4 cottages + 2 workshops.

Total yield: 60 hammers, 6 + 3 + 4 x 3T + 4 + 5 x 4T + 4 = 49 cottages + 2 workshops.

Alternatively, you can trade 2 turns on farms (+3) and 2 turns on workshops (+1) for 4 turns on cottages (+2).

60 hammers + 53 cottages.


So 60 hammers, 9 cottage turns, and the goody 7 turns or so earlier stand against 22 workshop turns.

Edit: nice catch, jmas

 

[jmas]

Total yield: 60 hammers, 6 + 3 + 4x4 + 4 + 5 x 4 + 4 = 49 cottages + 2 workshops.

Thank you for the example. I'm playing my 1st game on Monarch but have been following Slavery threads for probably 2 years or more (spend more time reading than playing ). I just wanted to point out you have an extra 4 in the middle above. Looks like it should be "Total yield: 60 hammers, 6 + 3 + 4x4 + 5 x 4 + 4 = 49 cottages + 2 workshops."

 

[VoiceOfUnreason]

Three further comments on that example.


The reason that I chose 29/32 as my initial condition is that it allows you to work the farm on T0, and grow on the first turn, and then immediately whip away the unhappy citizen 29/32 +3 -> 32/32 (grow) -> 16/34 (whip) 16/30. If you walk through that cycle, you'll find the result is 60 hammers and 52! cottages.

There are a couple ways to think about what happens, but they all essentially come down to P=7 instead of P=6, and therefore the cycle needs two more food, and that costs you one 2F cottage turn.

This is why I'm always suspicious of the "grow to unhappy" approach - in the simple cases, it always takes more food to complete the loop that way, and that food needs to be paid for somewhere.


Although the contrived example shows a reasonable cycle calculation, if those augmented workshops are 1/3 then the entire analysis is broken, because the endpoint is wrong. Running the workshops, you would presumably finish the building to get the extra happy, grow (on the same turn if you can manage it), and the drop the pop back onto the workshops, with the new guy scoring extra cottage turns.

Even with 1/2 workshops, if you are using a theater or spiritual temple, you still have to worry about growing after finishing the construction.


The above example also has that odd little stutter step in it to eliminate the turn 0 whip penalty. Now, in a pure cycle analysis you'll normally assume that the one hammer you need was put in at the end of the previous cycle. However, I decided not to do that here because those two "extra" workshop turns represent something real - that the food used to get the bin to the right starting point is food not used for something else.

After all, the stagnant approach doesn't really care what the food level is - not during the production phase, at least. But whips, where you are worried about how many tiles turns you are losing, are sensitive to the initial food level. It's a fixed cost - you can amortize it across many cycles - but you should be sure to account for it somewhere.

 

[vicawoo]

Good play is whipping inefficiently when you have to. Great play is knowing when you can get away with not being inefficient.

It's like in *gasp* starcraft, bad play is dying because they make too many drones, "good" play is making sure you have enough defense, great play is knowing when you can invest just barely enough in your units to defend/attack while being able to invest the rest.

The value of the minerals of one zergling in zvz battle can be infinite (where infinite is the difference between dying and not dying). The value of the zergling after that is normal. Discontinuous functions are different from continuous functions.

and Voice, ever tried subtracting instead of adding everything up? To me it's less tedious, since you only have to know how many turns it takes to grow, and what you're not working while growing.

So it would be, we lose 2x10 workshop turns (easy to calculate either way), lose 1 cottage turn at size 4, and gain 4 cottage turns, and we have the weird turn 0
4-1+2=5. Edit: nothing revolutionary, it's just a lot less imposing to deal with.

I think the turn 0 is not a good control, since that output has nothing to do with whether you decide to whip or not, and it doesn't change the food.

 

[TheMeInTeam]

You're correct that the risk of the Slave Riot is part of the opportunity cost of using slavery, and should be considered. However, I think most of the old hands here abandoned random events long ago.

The problem is that this particular opportunity cost isn't strong enough to change the relative value of the civic in the majority of cases (IE risk-adjusted ROI is still overwhelmingly favoring slavery for a long time), but the random penalty can still screw you up. Avoiding slavery due to the event chance is EXCESSIVE hedging, but if you don't there's always a small chance of triple slave revolt in the BCs...

They could have made the event leagues better if they tied it to actual whipping in some what, but oh well.

It's like in *gasp* starcraft, bad play is dying because they make too many drones, "good" play is making sure you have enough defense, great play is knowing when you can invest just barely enough in your units to defend/attack while being able to invest the rest.

It's a good analogy for those that know starcraft .

and Voice, ever tried subtracting instead of adding everything up? To me it's less tedious, since you only have to know how many turns it takes to grow, and what you're not working while growing.

I really like this suggestion; it should make doing things in one's head viable more frequently.

 

[VoiceOfUnreason]

and Voice, ever tried subtracting instead of adding everything up? To me it's less tedious, since you only have to know how many turns it takes to grow, and what you're not working while growing.

My feeling is that, when trying to explain an idea to people who may be seeing it for the first time, the clearest approach is to make the calculation match as closely as possible what actually happens, rather than start immediately introducing shortcuts that obscure.

Put another way, I'm deliberately trying to make the calculation as clear as I can because the question under discussion is really "is that the right calculation?"

Or, if you prefer - "how can I write this up so that Iranon and I can figure out if we are talking about the same thing?"

 

[VoiceOfUnreason]

If the happy cap doesn't lift during a whip cycle, you have to be a bit more careful - growing back to your original size too soon means feeding an unhappy citizen, a waste of 2 food per turn.

For maximizing the number of tile turns worked in a two pop whip, the rhythm is:
Whip from P to P-2
Grow to size P-1 right away (1 turn)
Slow steady growth (8 turns)
Grow to size P on the final turn (1 turn)

Problem: how much surplus food do we need to be able to generate to make this work?

Initial Condition: 9 + P + X / ( 20 + 2P )

After the whip: 9 + P + X / (16 + 2P )

16 + 2P - (9 + P + X) = 7 + P - X

So to grow on the first turn, we need to be able to generate at least 7 + P - X food on the first turn (using P-2 pop to do it). If we have exactly that much food, our next turn is at (8+P) / (18 + 2P)

Now consider the other end - suppose we are at the last turn, food bin at (17+2P) / (18+2P). We need one food to grow - the granary will take us to (9+P) / (20+2P) - and another X food to bring us to the initial condition. So we need to be able to harvest 1 + X surplus food at the end (with P-1 pop available to work tiles).

At size P-1, how much additional surplus food can we generate over size P-2? We can work all the same tiles, plus one more. In the most common case, that one more is going to be a farm (surplus 1). So we can think of that final turn of work as one new farm plus the tiles we used to grow at size P-2.

So the surplus food from P-2 tiles needs to be more than X, and needs to be more than 7 + P - X. Therefore

Edit: this calculation is wrong, see below

X = 7 + P - X
2X = 7 + P
X = ( 7 + P ) / 2

That tells us both (a) the starting condition that allows us to cycle on a minimum food surplus and (b) exactly what that surplus needs to be.

Well, almost - there's another integer math problem. If P is odd, that calculation is correct. But if P is even, then X is a fraction, and the game doesn't permit a fractional food in the bin. So your minimum surplus will be (8+P)/2 , and your best starting condition will be your choice of ( 6+P ) / 2 and (8+P) / 2.

As the question we're asking is about working tiles at pop P-2, it's sensible to write the equation using that value instead. So

(8+P) / 2 => (10 + (P-2) ) / 2 = 5 + (P-2)/2

In other words, if you can generate 5 surplus with half your pop ( dry corn plus cottages, for example ), and the other half works farms on the turns that need a burst of growth.

As another example, this says that you should be able to maintain an 8->6 whip cycle with pre-biology farms as your only source of surplus food.

If you have a larger food surplus available, you have more liberty to choose the initial condition that you want; the larger surplus gives you options.



Of course, when you have a lot of surplus food, you also need to figure out how to use it effectively without growing too quickly during the 8 turns in the middle.

 

[VoiceOfUnreason]

Some simplifying math...

This isn't quite right, see comments below

9 + P + X
9 + P + (8 + P)/2
9 + P + 4 + P/2
13 + 3P/2
15 - 2 + 3 * 2P / 4
3/4 * 20 - 2 + 3/4 * 2P
3/4 * (20 + 2P) - 2

In other words, the optimum initial condition calculated above is when the granary is 2 food short of three quarters full.

 

[pi-r8]

^ But 5 turns regrow time is way too long. In that case you need a granary or the city is just not fit for whipping. Also you assume it's mines that are whipped away.

Even with mines it's very often good to whip even the capital. Take an early game cap with a big food surplus and a plain tile mine and a grassland mine. Grow to size 6 (or at least until foodbox is full at size 5). Now whip. You gain 60 H, you lose 30 H from not working the grassland mine for 10 turns (you keep on working the plain tile mine). Also you have your worker/settler/building earlier.

Whipping is not marginal by any means, the situation i describe above is not even ideal for whipping.

maybe not ideal, but I'd say that a size 6 city with a large food surplus is a pretty good candidate for whipping. Why not grow the city larger? If you've got a huge food surplus then you've got enough food to quickly grow onto more mines or workshops. Also, getting that large food surplus might require giving up other mines that you could be working.

What I said was in SOME cases it's marginal. Of course there are situations where it's great, and other situations where it's terrible. But you can't just mindlessly whip every city every 10 turns for magical extra production.

Just that this situation rarely happens in a time where slavery is still > Caste.
You grew this city to it's food cap and it's working at least 3 mines, sounds like this took you some time + workers.
How can you compare the "loss" of hammers to the loss of time, that's the thing.
Was it worth it to wait x turns for being able to work all mines?
How many hammers is time worth? How much is a hammer worth?
You can't tell without seeing a map, and it's hardly compareable.

I agree, these are all good questions, so you have to look at the specific instance to decide if slavery is worth it. Don't just robotically follow a rule of "always whip" or "never whip".

That's for a two pop whip at the happy cap. What I said was a 2 pop whip at -1 happiness. This way you only lose 1 mine, since the other pop you whipped was doing nothing.
This way you give up 30P in 10 turns for 60P right now... how bad of a trade is that?

Ok so you start 1 over the happy cap, and after losing 2 pop, now you're at exactly the happy cap. As soon as you grow back the first pop, you're losing food because of the unhappy citizen, so you'll need to switch off a mine to compensate for that. Also, the food basket that you need to regrow will be larger, and you had to spend the extra turns growing it into unhappiness.

 

[JammerUno]

Whipping isn't good because it's more efficient than workshops. And any talk about efficiency will have to take into account the gain of the earlier hammers. That's extremely hard. What's the value of getting a settler 5 turns faster? A library? Oxford? A worker? An attack force?

The maximum number of hammers every turn averaged over the entire game is utterly useless.

Also, whipping as a long term strategy uses natural food surplus, it doesn't depend on improvements that much. You 2 pop-whip when you're about to go into uphappiness, then grow to the new cap the next turn, then grow to the old one in 9 turns. Just improve more tiles than you have population so you can switch between fast growth and fast hammers.

Besides, smaller sizes favour slavery even in efficiency:
A size 6 city with a 5F tile and 5 grassland workshops under caste produces 110H in ten turns.

If you whip 2 pop every ten turns, you have 49 citizen turns vs. 60 from caste, and you only need to produce 47F (for growth), 98F for sustenance, and 50H in those turns. Your city tile takes care of 20F10H, your 5F tile takes care of 50F in 10 citizen turns. That leaves 75F40H for 39 citizen turns. So something close to 2F1H. That's very doable.

And you get the extra benefit of getting your 60H earlier, which is hard to quantify, but certainly better than getting them later.

 

[cabert]

slavery is a clear winner in some situations :
1) you may have no other labor civic available (!!!)
2) you have some food, but not loads of, in several cities (usual for archipelago maps, for example)
3) you just don't have any other production option

 

[Iranon]

@ VoiceOfUnreason:

I'm going to lose some geek cred, but after the effort you put in you deserve an answer

Unfortunately I can't get Civ4 to run at the moment, and I don't want to claim things I can't check. I'm used to some simplifications like 'current pop+10 food to grow by 1' and eyeballing over 100% accurate optimisation for rigidly defined situations.

Whip from P to P-2
Grow to size P-1 right away (1 turn)
Slow steady growth (8 turns)
Grow to size P on the final turn (1 turn)

If I understand it correctly, P is your stable cap and this cycle needs a considerable food surplus on line 2 or 4. You also need a way to work efficiently without a food surplus during the remaining time. I generally settle my cities tightly and often like to keep some forests around so that's a steep requirement for me.

Trying to use the same format, the cycle I had in mind would be:

Whip from P+1 to P-1
Slow steady growth (9 turns)
Grow to size P as anger wears off (1 turn)
Grow to size P+1 at whatever rate you like (x turns)

The unhappy citizen won't survive until end of turn and so won't have any time to gobble up any of our precious food. This version works more citizens on average, but has more expensive regrowths and requires a longer cycle. I usually use it with a relatively long x (we were originally comparing it to 1-pop whips, suggesting we care about efficiency rather than volume). If we want a short cycle and have the spare food tiles to enable it, yours may be better.

 

[vicawoo]

If the happy cap doesn't lift during a whip cycle, you have to be a bit more careful - growing back to your original size too soon means feeding an unhappy citizen, a waste of 2 food per turn.

For maximizing the number of tile turns worked in a two pop whip, the rhythm is:
Whip from P to P-2
Grow to size P-1 right away (1 turn)
Slow steady growth (8 turns)
Grow to size P on the final turn (1 turn)

Problem: how much surplus food do we need to be able to generate to make this work?

Initial Condition: 9 + P + X / ( 20 + 2P )

After the whip: 9 + P + X / (16 + 2P )

16 + 2P - (9 + P + X) = 7 + P - X

So to grow on the first turn, we need to be able to generate at least 7 + P - X food on the first turn (using P-2 pop to do it). If we have exactly that much food, our next turn is at (8+P) / (18 + 2P)

Now consider the other end - suppose we are at the last turn, food bin at (17+2P) / (18+2P). We need one food to grow - the granary will take us to (9+P) / (20+2P) - and another X food to bring us to the initial condition. So we need to be able to harvest 1 + X surplus food at the end (with P-1 pop available to work tiles).

At size P-1, how much additional surplus food can we generate over size P-2? We can work all the same tiles, plus one more. In the most common case, that one more is going to be a farm (surplus 1). So we can think of that final turn of work as one new farm plus the tiles we used to grow at size P-2.

So the surplus food from P-2 tiles needs to be more than X, and needs to be more than 7 + P - X. Therefore

X = 7 + P - X
2X = 7 + P
X = ( 7 + P ) / 2

That tells us both (a) the starting condition that allows us to cycle on a minimum food surplus and (b) exactly what that surplus needs to be.

Well, almost - there's another integer math problem. If P is odd, that calculation is correct. But if P is even, then X is a fraction, and the game doesn't permit a fractional food in the bin. So your minimum surplus will be (8+P)/2 , and your best starting condition will be your choice of ( 6+P ) / 2 and (8+P) / 2.

As the question we're asking is about working tiles at pop P-2, it's sensible to write the equation using that value instead. So

(8+P) / 2 => (10 + (P-2) ) / 2 = 5 + (P-2)/2

In other words, if you can generate 5 surplus with half your pop ( dry corn plus cottages, for example ), and the other half works farms on the turns that need a burst of growth.

As another example, this says that you should be able to maintain an 8->6 whip cycle with pre-biology farms as your only source of surplus food.

If you have a larger food surplus available, you have more liberty to choose the initial condition that you want; the larger surplus gives you options.



Of course, when you have a lot of surplus food, you also need to figure out how to use it effectively without growing too quickly during the 8 turns in the middle.

The way I'd treat this:
Total food needed = ( food to grow at size P-2 ) + (food to grow at size P-1).
Total tile turns lost = 10 (or whatever your cycle) + 1 (or however many you spend growing from size P - 2 to P - 1).
Divide food surplus by food needed.
Any remainder, you convert other tile turns to farms or vice versa.

If you compare it to not whipping, you convert all your food surplus away and multiply by tile turns.
So you would compare converted surplus x 10 to 60 hammers + 11 turns lost + extra food surplus remainder converted.

So, let's compare 1 turn regrowth to growing to size P+1.
If we grow to size P+1 instead of P, we need 2 extra food. We also take extra turns to grow from P-1 to P (as opposed to P-2 to P-1). We receive 10 extra tile turns (some of which is lost to growth).
So overall we trade 10 turns - extra turns to grow and 2 extra food.
2 extra food has different conversions; if you have grassland mines, 2 extra turns of farms means 2 less turns of grassland mines. Or it is 2/3 of a plains hill turn. If you had too much food and had to stall say an unirrigated rice for a cottage, you get that rice back.

 

[VoiceOfUnreason]

If I understand it correctly, P is your stable cap and this cycle needs a considerable food surplus on line 2 or 4. You also need a way to work efficiently without a food surplus during the remaining time.

Not without a surplus - you do need to fill the food bin in step 3 - but you've got the idea. I'm trying to better understand what sorts of conditions are necessary to produce the huge surpluses during the feasts while "efficiently" keeping things under control during the famines.



I generally settle my cities tightly and often like to keep some forests around so that's a steep requirement for me.

The unhappy citizen won't survive until end of turn and so won't have any time to gobble up any of our precious food. This version works more citizens on average, but has more expensive regrowths and requires a longer cycle. I usually use it with a relatively long x (we were originally comparing it to 1-pop whips, suggesting we care about efficiency rather than volume).

Same camel, different hump - you are primarily interested in the cases where there isn't enough food around for the feast periods. Roughly speaking, whipping in cities that have no food resources and/or sufficient high yield tiles to chew through it all.

The way I'd treat this...

I don't think you've persuaded me that you've treated anything at all. As far as I can tell, you are looking at "what's the value of an idealized whip", where I'm looking at something more like "can an idealized whip be executed with these tiles?" or "what's the maximum yield given the tiles that are available.


For example, the analysis you quoted teaches me that an 11 turn 6->4 cycle isn't possible if you have no bonus food lying around.

Where does your treatment teach me that?

 

[vicawoo]

I'm simplifying the calculations so that they could be done in a few seconds. The food calculations is just cost to grow/food surplus.

If you grow from size 4 to 6, it costs 29. If your normal food surplus is 3 or more, you can do it in 10 turns.

I'm not sure why if it's not possible to grow in 1 turn and having to take two turns prevents it from being "optimal". If you take two turns to grow instead of one, you lose one additional tile turn, which may give you less production ... or maybe it would give you more production if you whipped from the grow to 1 above your usable tiles example.

 

[pi-r8]

For maximizing the number of tile turns worked in a two pop whip, the rhythm is:
Whip from P to P-2
Grow to size P-1 right away (1 turn)
Slow steady growth (8 turns)
Grow to size P on the final turn (1 turn)

Doesn't this shortchange the food in the city? You'd have to start with a food bin that's nearly full, to grow to P-1 right away. But then it would end up just half hull, from growing to P on the final turn. So if you wanted to repeat this cycle, you'd have to spend some extra turns refilling at size P, and it's no longer a 10 turn cycle. Which is OK, but I'm not sure it's optimal.

 

[pi-r8]

Whipping isn't good because it's more efficient than workshops. And any talk about efficiency will have to take into account the gain of the earlier hammers. That's extremely hard. What's the value of getting a settler 5 turns faster? A library? Oxford? A worker? An attack force?

The maximum number of hammers every turn averaged over the entire game is utterly useless.

You can decide on a case by case basis whether you need it right now, or you can afford to wait. Obviously if you need something right away then you have to whip it, regardless of efficiency. But if you can afford to delay it a few turns, then you do have to consider efficiency, and that's where all this math becomes relevant.

 

[VoiceOfUnreason]

Doesn't this shortchange the food in the city? You'd have to start with a food bin that's nearly full, to grow to P-1 right away. But then it would end up just half hull, from growing to P on the final turn. So if you wanted to repeat this cycle, you'd have to spend some extra turns refilling at size P, and it's no longer a 10 turn cycle. Which is OK, but I'm not sure it's optimal.

At least one of us doesn't understand the question. Try again?

It might be that you are missing the food overflow that occurs on both ends.

Initial condition: 9 + P + X / ( 20 + 2P )
After whip: 9 + P + X / ( 16 + 2P )
Surplus required on first turn of cycle: 7 + P - X
Foodbin after first turn: ( 8 + P ) / (18+2P)
Surplus required during next nine turns: 9 + P
Foodbin after tenth turn: ( 17 + 2P )
Surplus reuqired on last turn of cycle: 1 + X
Final Condition: 9 + P + X / (20 + 2P)

So roughly speaking, a harvest at the beginning and the end, with 9 turns in the middle where you are net positive food per turn, but not very much.

And then the rest of the exercise is just demonstrating that the arrangement that requires the lowest peak surplus is not to start with the granary nearly full (because that requires a lot of food on the last turn), but rather with the granary just short of 34 full.

With numbers, at size six.

If you start with the granary almost full (28/32), you don't need any food surplus on the first turn (28/28), but you need 14 surplus on the last turn (from 29/30 to 28/32).

If you start with the granary at 15/32 (the result of an exact growth), then you only need 1 surplus on the last turn (29/30 to 15/32), but you need 13 surplus on the first turn (15/28 to 14/30).

But if you start with the granary at 21/32, the whip brings you to 21/28, you need 7 surplus to grow on the first turn, and then you need 7 surplus to grow from 29/30 to 21/32.



A Ha! Found an error. In earlier posts, I had talked about needing a different equation for odds and evens, but I misunderstood the result - I got confused by a piece that was an artifact of my model. So the ideal point is 3/4 bin - 3 when P is even and... something else when P is odd, because 3/4 bin is not an integer in that case.

In other words - as long as you hit the right end points for the middle turns (growing exactly to P-1 on the first turn, exactly one food short of growing on the last), any initial condition that doesn't require starving to set it up will need the same food surplus on the first and last turn combined.

You need the minimum (at the 3/4 point) to show you what's possible -- so that you can abandon a plan that can't possible work, and so that you don't put yourself in a position where you are "forced" into working lower yield tiles than is optimal.

Here's an extreme illustration of the problem: suppose you start the cycle at 31/32. you whip (31/28), and arrange a 3 food deficit on the initial turn. Over the next 9 turns you slowly grow to 29/30. Now on the final turn, you need a 17 food surplus.

That's possible - the city is +2, so if each pop is on a 5F tile, you produce 27 food and consume 10. Perfect. But, umm - what tiles were you working when you weren't growing at 17 food / turn? How on earth did you manage a -3 deficit when a single tile and the city would provide 7 of the 8 food you needed? etc.

 

[VoiceOfUnreason]

When calculating the idealized starting point, I made a couple of mistakes that didn't quite cancel out. Redoing that part of the calculation.

When P is even

We need 7 + P - X surplus on the first turn, and 1 + X surplus on the last turn. We'd like to find the value of X that minimizes both of these, and the straight forward (!) approach is to set them equal.

7 + P - X = 1 + X
6 + P = 2X
3 + P/2 = X

When P is even, X is an integer, and we can calculate the target in a straight forward way.

9 + P + X
= 9 + P + 3 + P/2
= 12 + 3P/2
= 15 + (3/4)2P - 3
= (3/4) ( 20 + 2P ) - 3

in other words, a little short of 3/4 full. Or....

= (3/4) ( 20 + 2P ) - 3
= (3/4) ( 20 + 2P - 4 )
= (3/4) ( 20 + 2(P-2) )

alternatively, we can think that we are whipping down to exactly 3/4 full.

When P is odd, all of the basic ideas are the same except that 3/4 of the granary means a fractional food. So we have our choice to round up or down. If we round down, then we need extra food on the first turn and one food less than that at the last turn. But that's backwards - it makes much more sense to try to get the extra food when we have one more pop. In practical turns, we work whatever assortment of tiles we need to grow exactly on the first turn, and then on the last turn we work those same tiles plus a 3F tile.

So when P is odd, we're going to whip the food bin so that it ends up exactly 3/4 full after the whip - rounding UP.