Very difficult question, help would be appreciated... Good luck

u can do a straight forward simulation (assuming constant flow)
few calculations (discrete problem), but u can abuse symmetry ...

>Some buckets will have two spouts filling it up while other buckets will only have one filling it up.
So what? It's still linear.

Actually do it the other way around. Take the top bucket. Its input is X1. Its output is (X1-2)/2 on each side. So the buckets on the second row, say buckets 2 and 3, have as inputs X2 = (X1-2)/2 and X3 = (X1-2)/2. Their output on each side is for bucket 2 (X2-2)/2, etc. So you can calculate the inputs of the buckets of the next row in function of X1 only. Do that for all the rows until you reach an expression linking Xb to X1 and solve X1 for Xb = 2

This makes the most sense, attempting this now.

>says it's 47.4119
brainlet here

how the fuck do you get that answer. the only way i can get as low as 48 is if i count up all the buckets, add two to that, and multiply by 2

but if my brainlet mind actually tries to think about it i keep getting like 58 or some shit

I also got 56, but I'm sure there's some shit we're not seeing.

solve this:
[math]f(t) \quad\dots \quad\text{flow} \\
f_\text{in}^{B_{i,j}}(t)=\frac{1}{2}\left ( f_\text{out}^{B_{i-1,j-1}}(t)+f_\text{out}^{B_{i-1,j}}(t) \right )\\
f_\text{in}^{B_{0,0}}(t)=1\\
f_\text{out}^{B_{i,j}}(t)=f_\text{in}^{B_{0,0}}(t-a^{B_{i,j}})\\
2L=\int_{-\infty}^{a^{B_{i,j}}}f_\text{in}^{B_{i,j}}(t)dt[\math]
(i - row, j - column)
then you can get time when B is filled => how much water is poured ...

three of us got 56, i can't see any possible way the answer wouldn't be an integer

and 47 makes absolutely no sense

solve this:
$$ f(t) \quad\dots \quad\text{flow} \\
f_\text{in}^{B_{i,j}}(t)=\frac{1}{2}\left ( f_\text{out}^{B_{i-1,j-1}}(t)+f_\text{out}^{B_{i-1,j}}(t) \right )\\
f_\text{in}^{B_{0,0}}(t)=1\\
f_\text{out}^{B_{i,j}}(t)=f_\text{in}^{B_{0,0}}(t-a^{B_{i,j}})\\
2L=\int_{-\infty}^{a^{B_{i,j}}}f_\text{in}^{B_{i,j}}(t)dt $$
(i - row, j - column)
then you can get time when B is filled => how much water is poured ...

solve this:
[math] f(t) \quad\dots \quad\text{flow} \\
f_\text{in}^{B_{i,j}}(t)=\frac{1}{2}\left ( f_\text{out}^{B_{i-1,j-1}}(t)+f_\text{out}^{B_{i-1,j}}(t) \right )\\
f_\text{in}^{B_{0,0}}(t)=1\\
f_\text{out}^{B_{i,j}}(t)=f_\text{in}^{B_{0,0}}(t-a^{B_{i,j}})\\
2L=\int_{-\infty}^{a^{B_{i,j}}}f_\text{in}^{B_{i,j}}(t)dt [/math]
(i - row, j - column)
then you can get time when B is filled => how much water is poured ...

Yeah I have no idea what basically any of this means...what is a? And d? And t? I'm lost

40

Guy you're responding to: thing is the middle ones would fill quicker due to 2 buckets funneling in, and the subsequent buckets underneath those would fill faster as well, so you need to take into account that theyd be all filling up at different intervals/speeds, so it's very likely that a bucket is 1/4 or 1/8 full or so. I think I could figure it out with enough time and patience and some assumptions but it's work.

i have 500000 IQ

bow to me you fucking plebians

...

error:
[math] f_\text{in}^{B_{0,0}}(t)=1 [/math] for t >= 0

here
I fucked up a few things in this.
Obviously there are not 72 buckets first off, but there would be 36 buckets and 72 liters if it were a pyramid and all water was conserved. It's not.
Second, good point made by that I forgot to consider. The buckets receiving input from 2 instead of just one, fucks even the first point up I was trying to make when it comes to determining how much water has been poured into the first bucket by the time B is full, the flow rates for the buckets above it, if time were a factor.

I don't know, maybe my picture helps though.
If nothing else you can talk about buckets as letters and groups of them in terms of the colors I've used? I'll think about this some more.

that is actually hilarious
that has got to be the biggest bastardization of notation i've ever seen

47.84375 L

How did you get this, I got 47.875

Please explain your process, I thought I had it too but I didn't take into account there's progressively less water going down

No water will reach bucket B because there's a blockage in the pipe flowing to it.

Actually I goofed on the math. The answer is 47 + 7/17 = 47.411764705882352941176470588235. Will post how in a moment.

I used a computer program to simulate the waterfall of buckets, which increments 1L at a time. In actuality, I fill up 128 subunits of a litre at a time. Since the flowrate into each lower bucket is the sum of half of the flowrates into the buckets above we won't get fractions with denominator more than 128, and they will all be powers of 2.

Going step by step, this program shows that after 47 L the B bucket is one more pouring of 1L away from being full. Specifically it is 121/64 L full.

At this stage none of the buckets that feed into B will be filled, before then if they aren't already. So the flowrate into B remains constant. Running one more step of the program overflows B to 69/32 L, so we can now find the flow rate.

1L into the top means 17/64 L into B. We only need 7/64 to go into B to fill it up. So we need only pour an additional 7/17 into the top to get B to fill. 47 + 7/17 = 47.411764705882352941176470588235

isn't this just counting the number of buckets. to get to the next level all the buckets have to be filled so you just count all them then double it to get 44 liters

...

But only if the previous buckets are full, and they won't all be full at the same time.

>17/64 = 2

Double check this, going at a rate of 128/4352 L per step, it really does fill up after 1612 steps, which is 47 7/17 L. Problem solved.

gotta think about rates of flow into each bucket and how it varies based on their position relative to the buckets above the rates that water flow into them when the ones above it are full, different buckets will fill at different times and those flow rates don't begin to contribute to the buckets below until that bucket is filled.
see

No this is the last liter he adds, 1 in at the top is equal to 17/64 at the bottom

How did you double check?

But it's not. That would mean that you have to pour 64/17 liters on top, i.e. around 3 liters, to get 1 liter at the bottom...

it's flow rate from the buckets above, not capacity.
but, as stated, they need to have filled before they contribute.
I like 's example of 1L intervals into the top bucket, but that is imperfect.
There should also be some stages introduced when the flow rate out of a bucket >0.

>128/4352 L per step
That's where I was going.
Well done.

What program did you use?

Java.

>trying to solve this without differential equations

i lose more respect for this board every time I visit