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#### Kirian

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« Reply #150 on: June 27, 2014, 08:40:38 am »
+1

OK, strange question: in UK and non-US post-colonies, you shorten "mathematics" to "maths."  Do you also shorten "economics" to "econs"?
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#### Kirian

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« Reply #151 on: June 27, 2014, 08:44:48 am »
0

Consider the complete graph on 7 vertices (that is, 7 points, all of which have an edge connecting them to every other point). It has 21 edges. It's possible to pick 7 triangles of 3 edges such that the edges form a triangle, and each edge is part of only one triangle (an example is shown below).

How many different sets of triangles can be picked like this?

My answer: I made the answer 30, which I think is correct, but I think it's an interesting enough question to have a go at.

Do rotations and reflections count as unique sets?  I'm assuming not, but maybe you counted them.
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#### navical

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« Reply #152 on: June 27, 2014, 08:46:34 am »
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OK, strange question: in UK and non-US post-colonies, you shorten "mathematics" to "maths."  Do you also shorten "economics" to "econs"?
If it gets shortened at all, it would be "econ", but I've not really heard that from people other than economics students (or at least, students who study some economics)
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#### Tables

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« Reply #153 on: June 27, 2014, 09:03:33 am »
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Consider the complete graph on 7 vertices (that is, 7 points, all of which have an edge connecting them to every other point). It has 21 edges. It's possible to pick 7 triangles of 3 edges such that the edges form a triangle, and each edge is part of only one triangle (an example is shown below).

How many different sets of triangles can be picked like this?

My answer: I made the answer 30, which I think is correct, but I think it's an interesting enough question to have a go at.

Do rotations and reflections count as unique sets?  I'm assuming not, but maybe you counted them.

Yes, provided it's actually unique and not just a recolouring (e.g. if you'd be rotating/reflecting in a way that made the 7 same triangles I did but with different colours, that doesn't count).

OK, strange question: in UK and non-US post-colonies, you shorten "mathematics" to "maths."  Do you also shorten "economics" to "econs"?

I don't think I've ever heard economics shortened, but econ would seem the natural way to shorten it if it were.
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#### Kirian

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« Reply #154 on: June 27, 2014, 10:56:02 am »
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Consider the complete graph on 7 vertices (that is, 7 points, all of which have an edge connecting them to every other point). It has 21 edges. It's possible to pick 7 triangles of 3 edges such that the edges form a triangle, and each edge is part of only one triangle (an example is shown below).

How many different sets of triangles can be picked like this?

My answer: I made the answer 30, which I think is correct, but I think it's an interesting enough question to have a go at.

Do rotations and reflections count as unique sets?  I'm assuming not, but maybe you counted them.

Yes, provided it's actually unique and not just a recolouring (e.g. if you'd be rotating/reflecting in a way that made the 7 same triangles I did but with different colours, that doesn't count).

Let me rephrase: Your solution contains triangles 123, 145, 167, 246, 257, 347, and 357.  A tau/7 rotation clockwise is the set of triangles 234, 256, 127, 357, 136, 145, and 146.

If these are unique solutions, then you've provided seven solutions with your diagram, because it doesn't have sevenfold symmetry.  Is it correct that your diagram is really seven unique solutions?

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#### sudgy

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« Reply #155 on: June 27, 2014, 03:13:40 pm »
0

Consider the complete graph on 7 vertices (that is, 7 points, all of which have an edge connecting them to every other point). It has 21 edges. It's possible to pick 7 triangles of 3 edges such that the edges form a triangle, and each edge is part of only one triangle (an example is shown below).

How many different sets of triangles can be picked like this?

My answer: I made the answer 30, which I think is correct, but I think it's an interesting enough question to have a go at.

I remember my mathbook last year had a formula for this, but I can't remember it...
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#### heron

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« Reply #156 on: June 27, 2014, 03:17:59 pm »
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#### soulnet

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« Reply #157 on: June 28, 2014, 02:44:49 am »
+1

I agree with Kirian's request for clarification. Are the nodes labeled or indistinguishable? Rotations and symmetry in graphs are strange, because it is not just rotating and symmetry, since the topology is not geometric.
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#### florrat

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« Reply #158 on: June 28, 2014, 08:42:47 am »
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Assuming the two colorings Kirian gives count as different solutions, I get -135- colorings (I can write an explanation later)
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#### silverspawn

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« Reply #159 on: June 28, 2014, 10:19:12 am »
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i have this task I'm struggling with. I'm sure I can selfishly abuse this thread to get some help... right?

so the growth rate of a plant is proportional to its height and anti proportional to the third power (do you say it like that?) of the time passed. the height is y and y(1) = 5.

so I made the equation y'(t) = y * 1/t³ * K

and I have this formula that says
Quote
y' = g(t)*h(y) => INTEGRAL(1/h(y)) = INTEGRAL(g(t))

so i define g(t) := K/t³ and h(y) := y, and use the formula and then I get

INTEGRAL(1/y) = INTEGRAL(1/t³*K)
and so
ln(y) = K * (-1/2 (t^-2)) | e^
y = e^(-1/2 K*(t^-2))

so y(1) = e^(-1/2K) = 5 | ln
-1/2 K = ln(5) | *(-2)
K = -2ln(5)

and so

y(t) = e^[-1/2(-2ln(5)) * t^-2)]
y(t) = [e^(ln(5))]^(t^-2)
y(t) = 5^(t^-2)
y(t) = 5^(1/t²)

so the plant has infinite height right after it spawns and then shrinks really fast? that doesn't sound right
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#### Kirian

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« Reply #160 on: June 28, 2014, 10:34:13 am »
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Assuming the two colorings Kirian gives count as different solutions, I get -135- colorings (I can write an explanation later)

If all seven rotations are unique, the final answer has to be divisible by 7, doesn't it?
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#### pacovf

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« Reply #161 on: June 28, 2014, 11:32:30 am »
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i have this task I'm struggling with. I'm sure I can selfishly abuse this thread to get some help... right?

so the growth rate of a plant is proportional to its height and anti proportional to the third power (do you say it like that?) of the time passed. the height is y and y(1) = 5.

so I made the equation y'(t) = y * 1/t³ * K

and I have this formula that says
Quote
y' = g(t)*h(y) => INTEGRAL(1/h(y)) = INTEGRAL(g(t))

so i define g(t) := K/t³ and h(y) := y, and use the formula and then I get

INTEGRAL(1/y) = INTEGRAL(1/t³*K)
and so
ln(y) = K * (-1/2 (t^-2)) | e^
y = e^(-1/2 K*(t^-2))

so y(1) = e^(-1/2K) = 5 | ln
-1/2 K = ln(5) | *(-2)
K = -2ln(5)

and so

y(t) = e^[-1/2(-2ln(5)) * t^-2)]
y(t) = [e^(ln(5))]^(t^-2)
y(t) = 5^(t^-2)
y(t) = 5^(1/t²)

so the plant has infinite height right after it spawns and then shrinks really fast? that doesn't sound right

Setting t = 0 doesn't make any sense because of your differential equation. It spawns at t = 1 (or t0 =/= 0, whatever) in the model you are using, so it has a limited height when it "spawns". And for t very large, the plant is actually decreasing in size very slowly...

On the other hand, if you choose y(t0) <  1, K becomes positive. Your plant then increases in size, but never getting larger than 1. I would guess that makes more intuitive sense (since we are working in arbitrary units, you can choose 1 to be anything you like with the relevant rescaling).
« Last Edit: June 28, 2014, 11:41:32 am by pacovf »
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#### Kirian

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« Reply #162 on: June 28, 2014, 11:34:09 am »
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Silverspawn, the problem as given needs a second initial condition; you've forgotten to introduce a constant of integration, but once you have that there's no second initial condition.

Are we supposes to assume that y(0)=0?
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#### pacovf

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« Reply #163 on: June 28, 2014, 11:38:12 am »
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Silverspawn, the problem as given needs a second initial condition; you've forgotten to introduce a constant of integration, but once you have that there's no second initial condition.

Are we supposes to assume that y(0)=0?

AFAIK, he's not missing any condition. It's a first order differential equation, one condition is enough. His integration is somewhat dirty (lacks integration constant indeed), but the result is correct.
inb4 I am showed the error in my ways.

y(0) = 0 is not a very interesting case though
« Last Edit: June 28, 2014, 11:45:49 am by pacovf »
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#### Kirian

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« Reply #164 on: June 28, 2014, 11:45:29 am »
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Silverspawn, the problem as given needs a second initial condition; you've forgotten to introduce a constant of integration, but once you have that there's no second initial condition.

Are we supposes to assume that y(0)=0?

AFAIK, he's not missing any condition. It's a first order differential equation, one condition is enough.
inb4 I am showing the error in my ways.

y(0) = 0 is not a very interesting case though

Then it's not a completely defined equation... because we don't know the initial rate.  We have y' = ky/t^3 and not y' = y/t^3
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#### WalrusMcFishSr

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« Reply #165 on: June 28, 2014, 11:46:06 am »
+1

i have this task I'm struggling with. I'm sure I can selfishly abuse this thread to get some help... right?

so the growth rate of a plant is proportional to its height and anti proportional to the third power (do you say it like that?) of the time passed. the height is y and y(1) = 5.

so I made the equation y'(t) = y * 1/t³ * K

and I have this formula that says
Quote
y' = g(t)*h(y) => INTEGRAL(1/h(y)) = INTEGRAL(g(t))

so i define g(t) := K/t³ and h(y) := y, and use the formula and then I get

INTEGRAL(1/y) = INTEGRAL(1/t³*K)
and so
ln(y) = K * (-1/2 (t^-2)) | e^
y = e^(-1/2 K*(t^-2))

so y(1) = e^(-1/2K) = 5 | ln
-1/2 K = ln(5) | *(-2)
K = -2ln(5)

and so

y(t) = e^[-1/2(-2ln(5)) * t^-2)]
y(t) = [e^(ln(5))]^(t^-2)
y(t) = 5^(t^-2)
y(t) = 5^(1/t²)

so the plant has infinite height right after it spawns and then shrinks really fast? that doesn't sound right

It's been a while, but I'll try...

I think part of the problem is after

INTEGRAL(1/y) = INTEGRAL(1/t³*K)

we need an integration constant C. So

ln(y) = K * (-1/2 (t^-2)) + C

or going to the more usual way

y = C * e^(-1/2 K*(t^-2))

The problem is I'm not sure you can solve C and K at the same time without some extra information. Even with y(1)=5 you get different answers for different values of K.

Solving in terms of K you end up with a solution like

y = 5e ^ ((K * (t^2 - 1)) / 2t^2)

which for positive K gives a behavior more like what you'd expect for a plant. You might want to wait for one of the real mathematicians to chime in though.

PPE: I agree with Kirian
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#### Tables

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« Reply #166 on: June 28, 2014, 11:48:43 am »
+1

I agree with Kirian's request for clarification. Are the nodes labeled or indistinguishable? Rotations and symmetry in graphs are strange, because it is not just rotating and symmetry, since the topology is not geometric.

Yeah, sorry, I knew it wasn't entirely clear, I hoped numbering the vertices made it clearer but apparently not. The two examples Kirian gave would be distinct.

Assuming the two colorings Kirian gives count as different solutions, I get -135- colorings (I can write an explanation later)

If all seven rotations are unique, the final answer has to be divisible by 7, doesn't it?

No, not necessarily. Some rotations could give solutions that had already been counted, I believe.

(Also I'm just going to throw this here, how I came up with the question. I'm currently playing a game called Xenoblade Chronicles, for about the 10th time, and it has 7 playable characters. You always have 3 active characters at a time, which are the ones who participate in battles and stuff. I noticed you could form 7 parties of characters such that everyone was in a party with everyone else, exactly once. I then started wondering how many possible ways you could arrange the people into 7 battle parties, as above. And so the question was born. Hopefully if you get what I'm saying here you'll also get what things I'm looking for that count and what doesn't).
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#### sudgy

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« Reply #167 on: June 28, 2014, 11:49:19 am »
+4

Here's an integral symbol for you all to be able to copy and paste: ∫

(I'm getting tired of seeing INTEGRAL everywhere)
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### Tables

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« Reply #168 on: June 28, 2014, 11:54:22 am »
+6

Here's an integral symbol for you all to be able to copy and paste: ∫

(I'm getting tired of seeing INTEGRAL everywhere)

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#### SirPeebles

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« Reply #169 on: June 28, 2014, 11:55:29 am »
+4

Here's an integral symbol for you all to be able to copy and paste: ∫

(I'm getting tired of seeing INTEGRAL everywhere)

I knew sum one would make that joke.
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#### pacovf

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« Reply #170 on: June 28, 2014, 11:55:48 am »
0

y' = dy/dt = K*y/t3

=> dy/y = K * dt/t3  (yes yes physicists are eeevil)

=> ∫y(1)y(t) dy/y = ∫1t K*dt/t3

=> ln(y(t)/y(1)) = K/2 * (1-1/t2)

Somehow when I did this in my mind knowing y(1) actually helped me find K. So once again I bow to the wisdom of Kirian.
I stand by my previous result that you need K positive to get a sensible progression for the plant, though.

EDIT: groan...
« Last Edit: June 28, 2014, 12:05:59 pm by pacovf »
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#### SirPeebles

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« Reply #171 on: June 28, 2014, 11:58:05 am »
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I think we've given plenty of hints towards how to do his homework problem already, we don't need to present a detailed write up.
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#### silverspawn

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« Reply #172 on: June 28, 2014, 01:10:14 pm »
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I think we've given plenty of hints towards how to do his homework problem already, we don't need to present a detailed write up.
it's not a homework. i've had maths 2 semesters ago, but the lecture was at 8am, and i wasn't capable of standing up at 6am regularly, so I stopped visiting it at some point, and then I didn't want to visit it again because I had already skipped too many lectures (and the lecture was kind of bad too); now I want to write the exam at the end of this semester. unfortunately, I now have just a script to learn from.

which means I don't have get something done or anything, I just need to be able to do it properly.

I think what you're supposed to do is not calculate K but do it relative to K instead, like so

Quote
Solving in terms of K you end up with a solution like

y = 5e ^ ((K * (t^2 - 1)) / 2t^2)

which for positive K gives a behavior more like what you'd expect for a plant. You might want to wait for one of the real mathematicians to chime in though.

based on how it's formulated

thanks *)
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#### florrat

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« Reply #173 on: June 28, 2014, 06:06:22 pm »
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[note: spoilers for triangle puzzle below]

@silverspawn: Not sure if the integral problem was already sorted out completely, but Kirian is right, you need an constant of integration. So the equation you get is (ln is for barbarians, I use log with base e):

log(y) = K * (-1/2 (t^-2)) + C

(note that you only need a constant on one side of the equation, since if you have constants on both sides, you could just define a new constant as the difference between the two constants you had, and use that instead)

Now you want to use the information y(1) = 5 to eliminate the constant C, and then you'll be able to get a formula

y(t) = *something which depends on t and K*

If all seven rotations are unique, the final answer has to be divisible by 7, doesn't it?
As Tables said, some solutions will be rotated to themselves. However, this is only possible if every rotation of that solution will be identical to it.

[sidenote]You can see this as follows. Suppose that if by rotating 3 times (3/7tau) you get to the same solution, then if you rotate 3*5 times, you'll get to the same solution as well (because you did 3 rotations 5 times, which doesn't change anything), but rotating 15 times is the same as rotating once. And if rotating once doesn't change anything, any rotation doesn't change anything. On a more technical note: the reason is that since 7 is prime, (Z/7Z,+) is a group.[/sidenote]

Indeed, there are two solutions which are rotated to themselves, namely:
124 and its rotations (124,235,346,457,561,672,713)
134 and its rotations (134,245,356,467,571,612,723)
And the remaining number of solutions - 133 - is indeed divisible by 7.

---

Back to the original puzzle, to see that the answer is 135, consider the following. First note that the problem is equivalent to finding the number of ways to pick 7 triangles such that no two triangles have 2 nodes in common (since a triangle has an edge in common if and only if it has 2 nodes in common).

For convenience we don't have names for the 7 nodes yet. Pick two arbitrary nodes, and call them node 1 and node 2. First let's see how many ways there are to put node 1 in three triangles with the other six nodes. There has to be a triangle with both nodes 1 and 2, and there are 5 possibilities for that. Call the third node of this triangle node 3, and call the other four nodes 4-7 in any order. For the triangle with both nodes 1 and nodes 4, there are 3 possibilities (by picking any of nodes 5-7), and for the remaining triangle with 1 as node there's just 1 possibility (picking the remaning two nodes).

Now every node other than node 1 still needs two more triangles. Let's focus on node 2, which still needs triangles with nodes 4,5,6,7. There are 3 ways to split these four nodes in two groups of two nodes, which is the number of ways to pick the remaining two triangles containing node 2. Also node 3 needs two more triangles with nodes 4,5,6,7, so there are also 3 possibilities for that.

In total there are 5*3*3*3=135 possible colorings.
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#### heron

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« Reply #174 on: June 28, 2014, 10:12:04 pm »
+2

Now every node other than node 1 still needs two more triangles. Let's focus on node 2, which still needs triangles with nodes 4,5,6,7. There are 3 ways to split these four nodes in two groups of two nodes, which is the number of ways to pick the remaining two triangles containing node 2.

Except, one of the ways you split those four nodes will result in a triangle with two nodes that were used in one of the triangles with node one.
(e.g. if 145 is a triangle, then then the other triangles with node 2 must be 246 and 257 or 247 and 256, but not 245 and 267.)
So, in fact there are just two ways to pick the remaining two triangles for node 2.

Similarly, there is just 1 way to pick all of the remaining triangles, so I stand by my answer of thirty.
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