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Author Topic: Maths thread.  (Read 307108 times)

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silverspawn

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Re: Maths thread.
« Reply #600 on: February 07, 2016, 05:46:18 pm »
+1

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.
What about moat?

Yes, that is the solution, although moat moat moat moat moat, moat Moat moat moat note moat. Mote mote mote, mote Mote meat moat moat:

1) moat

2) moat

3) moat moat

Moat, moat moat, moatmoat = moat - moatmoatmoat. Moat moat, moat meat moat moat.


makes me wonder

For which moat ∈ ℝ does the equation moatmoat = moat - moatmoatmoat yield true?

I don't think I'm good enough at maths to calculate that. I guess it can only work for a moat ∈ (0, 1) or maybe also a moat ∈ (-1, 0) ?

Oort Cloud


It doesn't look like it, even in the complex numbers.  Here's a link.

Maybe moat needs a buff?

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Re: Maths thread.
« Reply #601 on: February 07, 2016, 08:39:50 pm »
0

For Easy Mode, I assumed that _ 1 _ would satisfy the requirements.
What about moat?

Yes, that is the solution, although moat moat moat moat moat, moat Moat moat moat note moat. Mote mote mote, mote Mote meat moat moat:

1) moat

2) moat

3) moat moat

Moat, moat moat, moatmoat = moat - moatmoatmoat. Moat moat, moat meat moat moat.


makes me wonder

For which moat ∈ ℝ does the equation moatmoat = moat - moatmoatmoat yield true?

I don't think I'm good enough at maths to calculate that. I guess it can only work for a moat ∈ (0, 1) or maybe also a moat ∈ (-1, 0) ?

Oort Cloud

Moat.
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sudgy

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Re: Maths thread.
« Reply #602 on: February 21, 2016, 03:18:41 pm »
0

So, this is more of a physics problem, but I just want a second opinion.  I'm trying to get an equation that gives the angle to shoot a projectile with the least velocity, given the x displacement and the z displacement.  Through a whole lot of complicated stuff1, it seems like this is the equation I'm looking for.  There's a couple things I'm wondering about though:

1. Wolframalpha originally said that the pi at the beginning of the equation is 2*pi*n, where n is an integer.  However, a lot of answers were only correct with that being pi at the beginning.  Is there any reason this is true (or maybe not true, and what should the equation be)?

2. No matter how small the x is (like 0.0001 type small) and no matter how small the z is (like -9999999 type small), theta is always positive (it just approaches zero).  Is this how it should be?  I would think that when throwing down, throwing it at least some down is a good idea.

1: Starting with equations for x and z velocity, integrating, solving for v as a function of x, z, and theta, then optimizing the function with respect to theta
« Last Edit: February 21, 2016, 04:22:42 pm by sudgy »
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

liopoil

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Re: Maths thread.
« Reply #603 on: February 21, 2016, 06:06:57 pm »
0

So, this is more of a physics problem, but I just want a second opinion.  I'm trying to get an equation that gives the angle to shoot a projectile with the least velocity, given the x displacement and the z displacement.  Through a whole lot of complicated stuff1, it seems like this is the equation I'm looking for.  There's a couple things I'm wondering about though:

1. Wolframalpha originally said that the pi at the beginning of the equation is 2*pi*n, where n is an integer.  However, a lot of answers were only correct with that being pi at the beginning.  Is there any reason this is true (or maybe not true, and what should the equation be)?

2. No matter how small the x is (like 0.0001 type small) and no matter how small the z is (like -9999999 type small), theta is always positive (it just approaches zero).  Is this how it should be?  I would think that when throwing down, throwing it at least some down is a good idea.

1: Starting with equations for x and z velocity, integrating, solving for v as a function of x, z, and theta, then optimizing the function with respect to theta
Theta should always be positive, which is good. The projectile will hit the right z after some amount of time; we would rather this time be long so that we have more time for the projectile to move to the left.

EDIT: The following is silly, don't bother reading

However, your answer should certainly be in terms of g, so something is off here.

I'll try to solve it here myself, maybe the same way you did?

Let's solve for the time it takes to have the right z-coordinate. If z > 0, then we want the time when it is on its way down. We have:

z = vt sin Q - 0.5g t2, where Q is the angle theta and g is the acceleration due to gravity. Then:

t = (1/g){v sin Q + sqrt[(v sin Q)2 - 2gz]}by the quadratic formula. Then:

x/(v cos Q) = t, so x = (v cos Q/g){v sin Q + sqrt[(v sin Q)2 - 2gz]}. Let v be the minimum v such that there exists a Q such that it gets to the point (x,z). Necessarily this Q will be the one which maximizes the right hand side of the equation while keeping it real. At the maximum, dx/dQ = 0, so:

d/dQ[(v^2/g)(sin Q cos Q + sqrt{sin2 Q - 2gv-2z}] = 0, with v, g, z, being constants:
cos2 Q - sin2 Q + d/dQ[sqrt{sin2 Q - c}] = 0, where c = 2gz/v2
1 - 2sin2 Q + sin Q/(sqrt{sin2 Q - c}) = 0
sin Q = (2sin2 Q - 1)(sqrt{sin2 Q - c})
sin2 Q = (4sin4 Q - 4sin2 Q + 1)(sin2 Q - c). Letting u = sin2 Q:
u = (4u2 - 4u + 1)(u - c)
u = 4u3 - 4u2(1 + c) + u(1 + 4c) - c
0 = u3 - (1 + c)u2 + cu - c/4

And, um, not sure what to do now. v is such that c is not more than u; does this mean that c = u? I'll try that I guess, see what happens:

0 = u3 - u3 - u2 + u2 - u/4, uh, nevermind. Wolphram gives the three roots in terms of c; take the real one, square root it and then arcsin it; that's your angle. But there's no way to determine c without v, so I guess I had to isolate v way back up there, which is probably what you did.
« Last Edit: February 21, 2016, 07:23:39 pm by liopoil »
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qmech

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Re: Maths thread.
« Reply #604 on: February 21, 2016, 06:10:58 pm »
+2


I seem to remember there is a nice trick for this question.  I can't remember the details, but I don't feel too bad as you'll get more out of it by working them out yourself.  The idea is that you can do a transformation of the coordinates to make z = 0, so the target is on your level.  You know (or show) that the best angle here is 45 degrees, and pull that back to your original coordinate system to get your answer.  This provides a possible explanation for why you're always aiming up, as halfway between vertical and "really far down" is still above the horizontal.
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sudgy

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Re: Maths thread.
« Reply #605 on: February 21, 2016, 06:35:52 pm »
0

Liopoil: the velocity that you shoot at does depend on g, it's just that the angle doesn't.  I don't see any reason why the angle should depend on g, the best path should be the same regardless of it.  I derived my equations differently from yours, when I have time I'll post them here.

qmech: I didn't quite get how to do what you were saying, and started looking it up, and found this.  That has a really simple answer that I'll probably just use...
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

liopoil

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Re: Maths thread.
« Reply #606 on: February 21, 2016, 09:18:47 pm »
+1

ok, so my previous approach was badly misguided. Here's a nice short solution based on qmech's post and sudgy's link:

Suppose we live in a world where the direction of constant acceleration due to gravity makes an angle 0 < phi < pi with the horizontal. What angle should we fire to get as far as possible on flat ground?

Decompose the velocity vector into components in the direction of gravity and the horizontal (these are not perpendicular, but that is okay). We want to maximize the product of the average horizontal velocity and the time. The average horizontal velocity is given by the velocity at the projectile's peak, and the time is given by twice the time to the peak. Since the only vertical velocity is in the gravity component, at the peak the velocity is just the horizontal component, and the time is proportional to the gravitational component. Thus the distance is directly proportional to the product of the components, so we set the components equal to each other*. Then the direction of initial velocity must be the angle bisector of the direction of gravity and the horizontal, that is, the initial velocity makes an angle of phi/2 with the horizontal.

Now, if the point (x,z) you want to get to makes an angle phi = arctan(z/x) with the horizontal, then rotating so that the vector (x,z) points horizontally, now gravity makes an angle of pi/2 - phi with the horizontal, which for positive x will always be between 0 and pi. Therefore we should shoot at an angle of pi/4 - arctan(z/x)/2.

*I'm having trouble articulating exactly why setting the components equal to each other is best, especially because their sum could be increased by having them different.


I can't remember the details, but I don't feel too bad as you'll get more out of it by working them out yourself.
Thanks for wasting my evening.
« Last Edit: February 21, 2016, 09:23:02 pm by liopoil »
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sudgy

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Re: Maths thread.
« Reply #607 on: February 24, 2016, 12:40:01 am »
0

Another random problem I just had (that I gave up for not being important whatsoever, not knowing the required material, and knowing that I would be nerd sniped if I did it):

In minecraft, sugarcane grows at a somewhat weird rate.  Every tick, a block of sugarcane has a chance to try to grow.  This chance is determined by taking all of the blocks in a 16x16x16 area and choosing three of those blocks.  If a sugarcane happens to be chosen, it increases its age by one.  When the age hits 16, it grows one block.

If you don't know anything about minecraft, when you get a new block of sugarcane, you can harvest it without affecting the old one, and can either save it or plant it again.  So, assume that you harvest the new sugarcane instantly, and at first, you replant it, but later, you save it.  Also assume that you won't use sugarcane you planted already.  Say you want to save 138 sugarcane1.  What is the optimal point to switch from replanting sugarcane to saving it?

I know this question might seem complicated just because of the way I worded it, but in the end, it should make sense.  I just realized that the solution gets really complicated.

1: Although it's not important for this problem, this is the amount needed to make an enchantment table and 15 bookshelves
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

liopoil

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Re: Maths thread.
« Reply #608 on: February 24, 2016, 12:57:15 am »
+1

It takes on average 164/3 ticks for any given block of sugar cane to grow, so you should plant as long you expect it to be that many ticks until you get to 138, which is easily calculated. In fact I think that you will always just end up planting 138 total sugarcane and wait until the last one finishes growing, which will take floor(log2(138))*164/3 = 174763 ticks on average. Unless I am misunderstanding the question.
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sudgy

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Re: Maths thread.
« Reply #609 on: February 24, 2016, 01:45:35 am »
0

They don't grow when that 164/3 chance happens, they only age by one.  When the age reaches 16, they grow.  One thing I forgot to mention that's important, is that then the age goes back to zero.  So, each one doesn't have a chance to grow on each tick, each is a part of the way through.

What do you mean by "wait until the last one finished growing"?  I feel like something's not right there.
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   Quote from: sudgy on June 31, 2011, 11:47:46 pm

DStu

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Re: Maths thread.
« Reply #610 on: February 24, 2016, 05:57:37 am »
0

They don't grow when that 164/3 chance happens, they only age by one.  When the age reaches 16, they grow.  One thing I forgot to mention that's important, is that then the age goes back to zero.  So, each one doesn't have a chance to grow on each tick, each is a part of the way through.

What do you mean by "wait until the last one finished growing"?  I feel like something's not right there.
They age with 16^3/3, and they need 16 ages to grow, that makes it 16^4/3, or?
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liopoil

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Re: Maths thread.
« Reply #611 on: February 24, 2016, 07:29:34 am »
0

They don't grow when that 164/3 chance happens, they only age by one.  When the age reaches 16, they grow.  One thing I forgot to mention that's important, is that then the age goes back to zero.  So, each one doesn't have a chance to grow on each tick, each is a part of the way through.

What do you mean by "wait until the last one finished growing"?  I feel like something's not right there.
Right, but as DStu said we need a 3/16^3 chance to happen 16 times, which takes 16^4/3 ticks on average. Once you have 138 planted and no sugarcane saved, wait for all 138 to finish growing.
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DStu

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Re: Maths thread.
« Reply #612 on: February 24, 2016, 12:30:10 pm »
0

Do you know the age of the plant? That would change things a bit...

Basically you would want to stop earlier, because your expected time to grow 138 cones given the ages is smaller than without knowing the ages.*


* This might be wrong, but in this case liopoils approach is also wrong. There is a small ( I think) inaccuracy in it, but if * is wrong it is too larger to ignore.
« Last Edit: February 24, 2016, 12:32:26 pm by DStu »
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Re: Maths thread.
« Reply #613 on: February 24, 2016, 12:33:54 pm »
0

Do you know the age of the plant? That would change things a bit...

Basically you would want to stop earlier, because your expected time to grow 138 cones given the ages is smaller than without knowing the ages.

Not unless you're cheating. Of course, you could estimate them, since you know when you planted it.

Of course this is very theoretical, because in real Minecraft life you just plant all of them since you can retrieve them after you've planted them.
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DStu

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Re: Maths thread.
« Reply #614 on: February 24, 2016, 01:05:43 pm »
0

Do you know the age of the plant? That would change things a bit...

Basically you would want to stop earlier, because your expected time to grow 138 cones given the ages is smaller than without knowing the ages.

Not unless you're cheating. Of course, you could estimate them, since you know when you planted it.

Of course this is very theoretical, because in real Minecraft life you just plant all of them since you can retrieve them after you've planted them.

I think it's even true if you don't know the age.  Liopoli assumes that it takes on average 16^4/3 ticks to spawn a cone for every plant. That's only true if the time to spawn is geometrically distributed, here it's the sum of 16 geometric distributions.  Basically, for a random plant far enough in the future, you would guess that on average it has age 8, so it does only need 8*16^3/3 ticks to spawn a new cone.
It now get's a bit more complicated, because at the time you want to start harvesting probably most of the plants a quiet young (because you just planted them, but anyway even if they have age 2 they produce a new cone a but faster compared to the geometric situation), and only the older ones can be assumed to have a random age.  But basically I would guess because of this you want to start harvesting a bit earlier, as the first new cones spawn a bit faster, so you have less time for a new plant to earn back its investment (which now really takes 16^4/3 ticks on average).

:e also, if my simulations are correct, it doesn't really matter. Stop somewhere between 100 and 150, the variance is much larger than the difference in expectation values...
« Last Edit: February 24, 2016, 01:22:23 pm by DStu »
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liopoil

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Re: Maths thread.
« Reply #615 on: February 24, 2016, 05:08:06 pm »
0

Yes, there is a slight innaccuracy, but it is negligible. However I see no point in ever going past 138.
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Re: Maths thread.
« Reply #616 on: February 25, 2016, 04:57:08 am »
0

However I see no point in ever going past 138.
Obviously.

However, seems like the the optimum is somewhere around 100.

(100 samples each, coupled until the first one starts harvesting to reduce variance and improve speed)

Code: [Select]
import numpy as np
import numpy.random as rnd
import pandas as pd
import matplotlib.pyplot as plt


def iterate(field, planted, harvested, stop):
    def grow(planted, i):
        if planted > i:
            field[i] = field[i] + 1
            if field[i]==16:
                field[i]=0
                return 1
        return 0
       
    i = rnd.randint(N)
    j=-1
    k=-1
    while (j<0) or (i==j):
        j = rnd.randint(N)
    while (k<0) or (i==k) or (j==k):
        k = rnd.randint(N)
   
    new_plants = 0
    new_plants += grow(planted, i)
    new_plants += grow(planted, j)
    new_plants += grow(planted, k)
    harvested += max(0, planted + new_plants - stop)
    planted = min(planted + new_plants, stop)
   
    return field, planted, harvested


rep = 100
res = 2
goal = 138

start = 90
end = 150


result = pd.Series(index=range(start, end, res))
samples = pd.DataFrame(columns = range(rep), index=result.index)
stops = np.asarray(result.index)
count = np.zeros(len(result.index))

for i in range(rep):
    burn_in = 0
    planted = 1
    harvested = 0
    field = np.zeros([N], dtype=int)
    avg = np.zeros(len(result.index))
    stop = result.index.min()
    while planted<stop:
        burn_in += 1
        field, planted, harvested = iterate(field, planted, harvested, stop)
    #until here the same for everyone
    save_field = field.copy()
    save_planted = planted
    for stop in range(len(stops)):
        s = stops[stop]
        count[stop] += burn_in
        while (harvested<goal):
            field, planted, harvested = iterate(field, planted, harvested, s)
            count[stop] += 1
       
        field = save_field.copy()
        planted = save_planted
        harvested = 0
   
avg = pd.Series(count/rep, index=result.index)
avg

:edit Went a bit further with the coupling, and did 300 samples


Code: [Select]
...
rep = 300
res = 2
goal = 138

start = 60
end = 140


result = pd.Series(index=range(start, end, res))
samples = pd.DataFrame(columns = range(rep), index=result.index)
stops = np.asarray(result.index)
count = np.zeros(len(result.index))

for i in range(rep):
    print(i)
    burn_in = 0
    planted = 1
    harvested = 0
    field = np.zeros([N], dtype=int)
    avg = np.zeros(len(result.index))
    for stop in range(len(stops)):
        s = stops[stop]
        while planted<stop:
            burn_in += 1
            field, planted, harvested = iterate(field, planted, harvested, s)
        #until here the same for everyone following
        np.copyto(save_field, field)
        save_planted = planted
        save_harvested = harvested
        count[stop] += burn_in
        while (harvested<goal):
            field, planted, harvested = iterate(field, planted, harvested, s)
            count[stop] += 1
        np.copyto(field, save_field)
        planted = save_planted + save_harvested
        harvested = 0
   
avg = pd.Series(count/rep, index=result.index)
avg
« Last Edit: February 25, 2016, 07:05:35 am by DStu »
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liopoil

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Re: Maths thread.
« Reply #617 on: February 25, 2016, 11:36:39 am »
0

Why is it all spiky? It should change direction just once right? Just not enough trials?

You want to plant as long as the probability it finishes growing twice before you hit 138 is higher than the probability it doesn't finish growing before you hit 138. Even this will not be entirely accurate, however. This leads me to estimate that around 3/4 of the number of blocks you want is optimal.
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Re: Maths thread.
« Reply #618 on: February 25, 2016, 12:53:25 pm »
0

Why is it all spiky? It should change direction just once right? Just not enough trials?

Yeah, 300 is still a bit low, even if the coupling is already quite strong.

I think I greatly increase the number of trials I can simulate by not drawing the 3 fields every tic (which will miss most of the time), but just jumping over all these by simulating the time it takes until some plant is drawn (which is mostly geometric with mean 16^3/3, slightly off from this because the 3 fields are not independent).

Just don't have time at the moment, but I think one sees what one should do anyway, somewhere around 100 stop planting...
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Re: Maths thread.
« Reply #619 on: February 25, 2016, 03:06:35 pm »
0

Why is it all spiky? It should change direction just once right? Just not enough trials?
the time it takes until some plant is drawn (which is mostly geometric with mean 16^3/3, slightly off from this because the 3 fields are not independent).
Due to linearity of expectation, it is exactly this for any given plant plot, right?
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Re: Maths thread.
« Reply #620 on: February 25, 2016, 04:47:58 pm »
0

Why is it all spiky? It should change direction just once right? Just not enough trials?
the time it takes until some plant is drawn (which is mostly geometric with mean 16^3/3, slightly off from this because the 3 fields are not independent).
Due to linearity of expectation, it is exactly this for any given plant plot, right?
It's not exactly geometric, as the three chosen fields are not independent. You can't chose one field twice.

You can see this in the extreme case with only three fields , where you would always chose every field, so the time to chose field 1 is constant 1, and not geometric with mean 1.
But with 4k fields and just 100 targets the deviation should be quite small, and I also think I know how to make it exact.
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Re: Maths thread.
« Reply #621 on: February 25, 2016, 05:06:16 pm »
0

Why is it all spiky? It should change direction just once right? Just not enough trials?
the time it takes until some plant is drawn (which is mostly geometric with mean 16^3/3, slightly off from this because the 3 fields are not independent).
Due to linearity of expectation, it is exactly this for any given plant plot, right?
It's not exactly geometric, as the three chosen fields are not independent. You can't chose one field twice.

You can see this in the extreme case with only three fields , where you would always chose every field, so the time to chose field 1 is constant 1, and not geometric with mean 1.
But with 4k fields and just 100 targets the deviation should be quite small, and I also think I know how to make it exact.
Why does the distribution matter rather than just the mean? Wait, duh. Nevermind.
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Re: Maths thread.
« Reply #622 on: February 26, 2016, 01:23:22 am »
0

Why is it all spiky? It should change direction just once right? Just not enough trials?
the time it takes until some plant is drawn (which is mostly geometric with mean 16^3/3, slightly off from this because the 3 fields are not independent).
Due to linearity of expectation, it is exactly this for any given plant plot, right?
It's not exactly geometric, as the three chosen fields are not independent. You can't chose one field twice.

You can see this in the extreme case with only three fields , where you would always chose every field, so the time to chose field 1 is constant 1, and not geometric with mean 1.
But with 4k fields and just 100 targets the deviation should be quite small, and I also think I know how to make it exact.
Why does the distribution matter rather than just the mean? Wait, duh. Nevermind.

I think you are right they might really be exactly geometric, I had a wrong assumption before.

I'm not sure if just the expectation matters, as the distribution controls the spawning of new plants, and these again spawn plants, you have some exponential behaviour here, which is nonlinear. Not sure if this is relevant in this context.

:e I think basically the question is, is https://en.wikipedia.org/wiki/Wald%27s_equation applicable here. And I don't think it's obvious to see that condition 2) is satisfied.
« Last Edit: February 26, 2016, 01:40:18 am by DStu »
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skip wooznum

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Re: Maths thread.
« Reply #623 on: March 18, 2016, 01:23:23 am »
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Is anyone familiar with the game Set? I was wondering, what is the most cards you could have out on the table where it would still be possible for there to be no sets?
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Titandrake

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Re: Maths thread.
« Reply #624 on: March 18, 2016, 02:02:32 am »
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Is anyone familiar with the game Set? I was wondering, what is the most cards you could have out on the table where it would still be possible for there to be no sets?

If you Google this, the answer is 20. It looks like the actual proof you can't use 21 or more is a bit complicated, unfortunately.

Side note, I learned the following variant from people who play Set a lot: when you get to the end of the deck, leave the last card face down. You can figure out what the last card is from the face-up cards, and winning by taking a set with the facedown card is much more stylish.

Every card has 4 attributes: color, shape, shading, and number. Whenever you have a set, for every attribute
  • All of that attribute match, or
  • All of that attribute are different.

So, considering just the number attribute for now. A set either has three 1s, three 2s, three 3s, or one of each. At the beginning of the game, # cards with 1 thing = # cards with 2 thing = # cards with 3 things. Whenever you take a set, this still holds true...in mod 3 (remainder when dividing by 3 if you're unfamiliar with mod)

The same argument applies for every other attribute. So, at the end of the game, all the cards left must satisfy:
  • # cards with 1 thing = # cards with 2 things = # cards with 3 thing (mod 3)
  • # cards with ovals = # cards with diamonds = # cards with squiggles (mod 3)
  • # cards with red = # cards with blue = # cards with green (mod 3)
  • # cards with empty = # cards with striped = # cards with solid (mod 3)

And with 1 card hidden, you can figure out what it needs to be to make all the above hold true.

PPE: Turns out you can't have a list inside a spoiler.
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