Math Riddle: Is it possible to launch infinite rocket on island map?

[pichutarius]

Ans: No. Easiest riddle ever.

But if i use mods to make the following changes:
1. change island size to as big as required, so it can contains arbitrary large amount, but finite resources (excluding fluids);
2. mining productivity now cost the same for all levels, and gives +100% productivity for each level.

Now can i make the island big enough such that it is possible to (mathematically) launch infinitely many rocket as i want?

Bonus point for an explanation/prove.

[DaveMcW]

C = constant mining productivity cost
Total = C/1 + C/2 + C/3 + C/4 + C/5 + C/6 ...
Total = C * (1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 ...)
Total = C * Harmonic series

The harmonic series sums to infinity, therefore a finite island cannot launch infinite rockets.

But the harmonic series diverges very slowly. If we assume some reasonable values, we can calculate the size of the island needed.

• Research 1 level of mining productivity per minute.
• Run at 60 UPS.
• Run the game for 4 years (infinite research was added 4 years ago).

This means we are on term 2,100,000 of the harmonic series, which is equal to 15. So we only need enough resources on the island to research 15 levels of mining productivity with no bonuses!

[pichutarius]

...

bull's eye!

a simple extension is: for any number of bonus productivity, like +10000%, it is still not possible.

this is related to ant on rubber rope puzzle: https://en.wikipedia.org/wiki/Ant_on_a_rubber_rope

[astroshak]

Uh ... its not possible to launch an infinite number of rockets on any vanilla map. The map is finite, 2 million by 2 million. That means the resources available on it are finite as well (except for crude oil and water).

[PunPun]

This riddle was also on reddit. So i will copy some answers from there and add my two cents to them. Kejser Kagespiser: The answer is yes you can get infinite rockets. All you need is to get productivity over 100% and you can get infinite ores by deconstructing the miner before the nonproductivity progress reaches the end and it will restart from zero when you rebuild the miner consuming no resources from the ground while still giving you ores from the productivity progress.

NyaFury: But if deconstructing the miners is not allowed then it becomes impossible just because there is a soft cap for productivity of max one cycle per tick. If you put all the miners in a separate power network with very little power generation then you can make the nonproductivity progress go really really slowly while getting one ore per tick from the productivity progress but that is also capped by the amount of stuff you can fit into the network to slow down the miners.

[pichutarius]

...

nice. Those answer were certainly creative.

[quyxkh]

No, because productivity is always finite and the universe will end, or at least become unable to generate enough energy to run a PC, in finite time.

[NotRexButCaesar]

No, because productivity is always finite and the universe will end, or at least become unable to generate enough energy to run a PC, in finite time.

Your computer will likely run out of memory to store the mining prod level first.

[blazespinnaker]

For real bonus points, what's the theoretical max rockets you can launch assuming I iron, C copper, O oil, and A coal, and S stone. For simplicity, you can create a function just using the most used resource (Copper I believe) if you prefer. Funny hacks for cuteness points, but something a bit more main stream is preferred.

This is likely a maxima problem where you need to determine the point at which to stop researching mining productivity given a finite set of resources. An interesting simplification / first step might be to determine the amount of mining productivity you can generate given a set of resources.

I say theoretical max rockets, as I'd guess that time to launch these rockets will require practically infinite time. Sun is likely to go supernova on most maps before you get it done. The harmonic series diverges, but very slowly. You could change the question to launch in a certain period of time, but then the problem will become a computer that can handle the size of the base(s).

[pichutarius]

...

since harmonic series grow O(ln N), we get:
cost ∝ ln (rocket count)
rocket count ∝ e^cost

so the total rocket count is O(e^resources). Previously Dave show us with cost of 15 rockets, we can actually launch 2,100,000 rockets with mining prod (assume +100% bonus) because it grow exponentially (super fast).

[blazespinnaker]

...

since harmonic series grow O(ln N), we get:
cost ∝ ln (rocket count)
rocket count ∝ e^cost

so the total rocket count is O(e^resources). Previously Dave show us with cost of 15 rockets, we can actually launch 2,100,000 rockets with mining prod (assume +100% bonus) because it grow exponentially (super fast).

With cost of 15 rockets, we can launch 2.1M rockets? Wait what? Before I call crazy, what are you trying to say there?

Assuming 100K copper for rocket launch (simplify), that would mean we need 1.5M copper without bonuses. 1.5M copper can't launch 2.1M rockets, even if we started at 100% prod bonus.

The cost also isn't a simple harmonic series, but rather a recursive function as you need to account for the increasing cost of productivity - a rather critical point if we're going to start looking at limits.

I'd ignore Dave's post and try again. I'll post the solution if no one gets it.

[DaveMcW]

The first post is assuming mods.

In vanilla Factorio, mining productivity and research cost both grow linearly.
cost = 2500 * (L - 3)
productivity = 1 + 0.1*L
resource cost = 2500 * (L - 3) / (1 + 0.1*L)
resource cost = 25000 * (L - 3) / (L + 10)

So in vanilla Factorio, the limit as L goes to infinity is every level of mining productivity costs 25000 science packs of raw resources.

[blazespinnaker]

Yeah, was hoping for something a bit more main stream. Not sure what the answer above is supposed to indicate other than the obvious

[T-A-R]

Not launching would result in 0 launched rockets.

Bonus: Thats more ore to spend on Spidertrons to fill the mean time.