# Dominion Strategy Forum

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#### SirPeebles

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« Reply #25 on: May 08, 2014, 09:28:49 pm »
+1

why?
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#### sudgy

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« Reply #26 on: May 08, 2014, 09:31:18 pm »
+1

why?

Well, all points have other points that are a rational distance away.  So it can't be one, since one click will not turn the points a rational distance away black.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### heron

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« Reply #27 on: May 08, 2014, 09:33:25 pm »
+2

Well, you cannot color the plane black with just two clicks.
WLOG you click (0,0) and (k,0).
If all of the points are colored black, then there does not exist a point (a,b) such that sqrt(a^2 + b^2) is rational and sqrt((a - k)^2 + b^2) is rational. However, if a = k/2, and b = sqrt(q^2 - a^2), where q is some rational number, then sqrt(a^2 + b^2) is rational and sqrt((a - k)^2 + b^2) = sqrt(a^2 + b^2). So it is impossible to color the entire plane black with just two clicks.

I believe you can color the plane with three clicks.
Consider the clicks (0,0); (1,0); and (sqrt(2),0). Assume that there exists a point (a,b) that is colored white. Then sqrt(a^2 + b^2); sqrt((a - 1)^2 + b^2); and sqrt((a - sqrt(2))^2 + b^2) are all rational.
Note that since the product of two rational numbers is always rational, the square root of an irrational number is never rational. Since we have sqrt((a - 1)^2 + b^2) = sqrt(a^2 + b^2 - 2a + 1) is a rational number, and we know that a^2 + b^2 is rational, we see that a must be rational. (This is because the sum of 2 rational numbers is always rational, and the sum of a rational number and an irrational number is always irrational).

We also have that sqrt((a - sqrt(2))^2 + b^2) = sqrt(a^2 + b^2 - 2a*sqrt(2) + 2) is a rational number. We know that a^2 + b^2 +2 is rational, so 2a*sqrt(2) must be rational as well. However, the product of a rational number and an irrational number is always irrational, so a must be irrational, which is a contradiction.

Therefore the plane can be colored black in a minimum of 3 clicks.

Note: All of my statements about sums and products of rational and irrational numbers are easily proven using the definition of a rational number, or all already given by the closure properties.

Hopefully that is all correct, that was a nice problem.
« Last Edit: May 08, 2014, 09:36:30 pm by heron »
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#### sudgy

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« Reply #28 on: May 08, 2014, 09:36:50 pm »
+1

why?

Well, all points have other points that are a rational distance away.  So it can't be one, since one click will not turn the points a rational distance away black.

Oh, I see what you are saying.  I meant, "I know the answer is at least 2."  Not, "The answer is 2."  Oops.
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Quote from: sudgy on June 31, 2011, 11:47:46 pm

#### SirPeebles

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« Reply #29 on: May 08, 2014, 10:08:38 pm »
+5

heron's argument looks good.

Here's my favorite solution.  Click any point, but consider the portion of the plane which is not turned black.  It will be a collection of concentric circles, one for each rational number.  Call these the red circles.  Now click any other point, and call the rational circles about that point the blue circles.  After these two clicks, the only points which are still white are where a blue and a red circle intersect.  There is a countable number of red circles and a countable number of blue circles, and each pair intersects at most twice, so after two clicks only a countable number of white points remain.  Let's call these points the stubborn points.

Now where should we make the third click?  Let's call a point "good" if clicking there will finish the job, and call it "bad" if it leaves a white point leftover.  Then the "bad" points are precisely the rational-radius circles centered on the stubborn points.  Thus the bad points lie on a countable number of circles.  But the plane is not a countable union of circles*, so there must exist a "good" point.

*Why can't the plane be a countable union of circles?  There are two quick arguments.

1) [Measure theory] A circle has measure zero (Roughly this means that a circle has zero area.  Here I am referring to the circle, not the solid disk.).  A countable union of measure zero sets still has measure zero, and therefore cannot be the full plane.  Indeed, we see that almost all points are "good".

2) [Baire category]  A circle in the plane is nowhere dense.  The "bad" points therefore form a meager set.  But the plane is not meager.

Note:  The only property of the irrational numbers we use is that their complement is countable.  So for instance, a paintbrush which colors only at transcendental distances will also color the entire plane in three clicks.  And you can choose "almost any" three distinct points to click, in the technical measure theory sense.

edit:  Here's a third way to see that the plane is not a countable union of circles, without using scary words.  (It is secretly just argument 1)

3) [No scary words]  A circle can always be "thickened" into narrow round band whose area is as small as you'd like.  List your countable collection of circles C1, C2, C3, ... .  Thicken C1 to a narrow band of area at most 1/2, thicken C2 to a band of area at most 1/2^2, thicken C3 to at most 1/2^3, and so forth.  Then the union of these thickened circles has area at most 1/2 + 1/2^2 + 1/2^3 + ... = 1.  Thus the union does not cover the entire plane.
« Last Edit: May 08, 2014, 10:35:17 pm by SirPeebles »
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#### pacovf

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« Reply #30 on: May 08, 2014, 10:32:24 pm »
+1

So uh tensorial notation from my string theory course (technically physics, but oh well), something bothers me with the Polyakov action. At some point we want to know the variation of sqrt(-det(g) ) when g varies, g being a 2d metric.

I am told that d(sqrt( -det(g) )) = +1/2 * sqrt( -det(g) ) * gab * dgab = -1/2 * sqrt( -det(g) ) * gab * dgab

So gab * dgab = - gab * dgab, and that sign change is bugging the hell out of me. So, can someone explain that to me, or is it an error?
« Last Edit: May 08, 2014, 10:38:58 pm by pacovf »
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#### AndrewisFTTW

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« Reply #31 on: May 08, 2014, 10:49:06 pm »
0

I got a D in geometry. Then I failed algebra 2. Then when I got to college I had to take basic algebra again for no credit. I think I took another math course and I probably got a C or something.
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#### AndrewisFTTW

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« Reply #32 on: May 08, 2014, 10:49:57 pm »
+2

Ozle was my teacher.
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#### SirPeebles

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« Reply #33 on: May 08, 2014, 10:53:02 pm »
+1

So uh tensorial notation from my string theory course (technically physics, but oh well), something bothers me with the Polyakov action. At some point we want to know the variation of sqrt(-det(g) ) when g varies, g being a 2d metric.

I am told that d(sqrt( -det(g) )) = +1/2 * sqrt( -det(g) ) * gab * dgab = -1/2 * sqrt( -det(g) ) * gab * dgab

So gab * dgab = - gab * dgab, and that sign change is bugging the hell out of me. So, can someone explain that to me, or is it an error?

I'm a bit rusty on this stuff.  If I recall correctly, gab * gab = gbb = tr g.  Is there any reason in your context to assume that the trace of your metric is constant?  Because then your identity just follows from the product rule.  That is, differentiating the trace would give gab * dgab + gab * dgab = 0.

edit:  oh, gbb isn't the trace of g, is it?  It's the trace of the Kronecker delta, or whatever is appropriate for your signature.  So yeah, it should be a constant, so what you're looking for comes right out of the product rule.

edit 2:  The signature doesn't come into play for the scalar gbb.  That is just the trace of the Kronecker symbol, which is the number of dimensions.  In this case, 2.
« Last Edit: May 08, 2014, 11:24:03 pm by SirPeebles »
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#### qmech

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« Reply #34 on: May 09, 2014, 03:28:33 am »
+1

Two very nice solutions.  I managed to get the splitting out with the ham sandwich theorem, and was hoping there would be an elementary way to do it.
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#### WalrusMcFishSr

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« Reply #35 on: May 09, 2014, 04:04:21 am »
+3

Can you extend that irrational paintbrush problem to n dimensions?
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#### qmech

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« Reply #36 on: May 09, 2014, 04:10:46 am »
+2

Can you extend that irrational paintbrush problem to n dimensions?

Yes, it works straightforwardly.

(Post 1000 is in the maths thread!  I mean, whatever.)
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#### Joseph2302

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« Reply #37 on: May 09, 2014, 05:06:17 am »
+5

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#### pacovf

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« Reply #38 on: May 09, 2014, 06:13:38 am »
+1

So uh tensorial notation from my string theory course (technically physics, but oh well), something bothers me with the Polyakov action. At some point we want to know the variation of sqrt(-det(g) ) when g varies, g being a 2d metric.

I am told that d(sqrt( -det(g) )) = +1/2 * sqrt( -det(g) ) * gab * dgab = -1/2 * sqrt( -det(g) ) * gab * dgab

So gab * dgab = - gab * dgab, and that sign change is bugging the hell out of me. So, can someone explain that to me, or is it an error?

I'm a bit rusty on this stuff.  If I recall correctly, gab * gab = gbb = tr g.  Is there any reason in your context to assume that the trace of your metric is constant?  Because then your identity just follows from the product rule.  That is, differentiating the trace would give gab * dgab + gab * dgab = 0.

edit:  oh, gbb isn't the trace of g, is it?  It's the trace of the Kronecker delta, or whatever is appropriate for your signature.  So yeah, it should be a constant, so what you're looking for comes right out of the product rule.

edit 2:  The signature doesn't come into play for the scalar gbb.  That is just the trace of the Kronecker symbol, which is the number of dimensions.  In this case, 2.

Ugh, you are obviously right.

[ By definition in the tensor notation, gab= (gab)-1, where the -1 means the inverse of the metric, not the inverse of gab, tensor notation can be very unclear. So yeah, gab * gab = gab * gba = Iaa = n (where n dimension), and derivation indeed gives the result. ]

What bothered me in the equation is that, "usually", Aa*Ba = Aa*Ba because of the symmetry of the metric (sometimes you don't use a metric to lower and raise indices though). Somehow I can't make sense of why this is different when using dg. Note that in this case, g is a variable that happens to be a metric, so there's no reason why you should use g to raise and lower indices, in fact we had defined another metric beforehand. If we call hab the object that we use to lower indices (probably the predefined metric, but when reading the notes again, I noticed it wasn't specified), we get:

hac*hbd*gab*dgcd = - hca*hdb*gab*dgcd

And somehow that doesn't feel right, 95% of the time h is symmetric, and in the cases where it isn't, I've only encountered antisymmetric h. I guess there must be a reason that prevents me from raising and lowering indices for dg, but I don't see why.
« Last Edit: May 09, 2014, 06:42:34 am by pacovf »
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#### SirPeebles

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« Reply #39 on: May 09, 2014, 10:30:30 am »
+2

Ugh, you are obviously right.

[ By definition in the tensor notation, gab= (gab)-1, where the -1 means the inverse of the metric, not the inverse of gab, tensor notation can be very unclear. So yeah, gab * gab = gab * gba = Iaa = n (where n dimension), and derivation indeed gives the result. ]

What bothered me in the equation is that, "usually", Aa*Ba = Aa*Ba because of the symmetry of the metric (sometimes you don't use a metric to lower and raise indices though). Somehow I can't make sense of why this is different when using dg. Note that in this case, g is a variable that happens to be a metric, so there's no reason why you should use g to raise and lower indices, in fact we had defined another metric beforehand. If we call hab the object that we use to lower indices (probably the predefined metric, but when reading the notes again, I noticed it wasn't specified), we get:

hac*hbd*gab*dgcd = - hca*hdb*gab*dgcd

And somehow that doesn't feel right, 95% of the time h is symmetric, and in the cases where it isn't, I've only encountered antisymmetric h. I guess there must be a reason that prevents me from raising and lowering indices for dg, but I don't see why.

Suppose you have a matrix A, and you change it by a small amount dA.  How much does the inverse of A change by?  Well, A-1A = I, after the change we will have (A-1 + d[A-1])(A + dA) = I to first order, so A-1 dA + d[A-1] A = 0. That is essentially what we just proved if you put in A = g and contract indices.

The reason that the indices don't behave the way you are expecting is because d[A-1] is NOT just the matrix inverse of dA.  If you raise the indices on dgab then you get the equivalent of (dg)-1, which is not the same thing as d(g-1).
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#### Kirian

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« Reply #40 on: May 09, 2014, 10:33:10 am »
+2

I took two years of college physics and two years of calculus... and I have exactly no idea what y'all are talking about.
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#### SirPeebles

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« Reply #41 on: May 09, 2014, 10:39:54 am »
+2

I took two years of college physics and two years of calculus... and I have exactly no idea what y'all are talking about.

Newton's laws of motion only apply to inertial frames.  For example, if you consider motion from the noninertial frame where the Earth is at rest, then F = ma gets cluttered up with extra terms which are often interpreted as fictitious forces, namely centrifugal force and Coriolis force.  The tensor notation was developed by Albert Einstein during his work on general relativity as a way of writing down laws of physics which hold in any frame, inertial or not.  Nowadays it is used throughout physics, not only in general relativity.
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#### pacovf

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« Reply #42 on: May 09, 2014, 10:48:50 am »
+2

The reason that the indices don't behave the way you are expecting is because d[A-1] is NOT just the matrix inverse of dA.  If you raise the indices on dgab then you get the equivalent of (dg)-1, which is not the same thing as d(g-1).

Oh, ok, so basically dgab and dgab as "defined" in that result are not the same tensor modulo index raising. Man, that could definitely use a footnote. Thanks for the explanation, though! Much appreciated.

I took two years of college physics and two years of calculus... and I have exactly no idea what y'all are talking about.

Me neither.
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#### SirPeebles

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« Reply #43 on: May 09, 2014, 10:51:00 am »
+1

I don't think they are usually even tensors.

Edit:  I take that back.  If g and g+dg are both tensors, then so is dg.
« Last Edit: May 09, 2014, 10:54:27 am by SirPeebles »
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#### DStu

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« Reply #44 on: May 09, 2014, 10:56:08 am »
+3

Quote
I took two years of college physics and two years of calculus... and I have exactly no idea what y'all are talking about.

Me neither.

I'm about to defend my math phd thesis, and I quitted this subthread at "tensor"...
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#### Kirian

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« Reply #45 on: May 09, 2014, 11:04:41 am »
+5

I'm trying to mostly ignore it, it'll just make me tensor.
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#### pacovf

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« Reply #46 on: May 09, 2014, 11:26:07 am »
+1

Quote
I took two years of college physics and two years of calculus... and I have exactly no idea what y'all are talking about.

Me neither.

I'm about to defend my math phd thesis, and I quitted this subthread at "tensor"...

I nearly quitted this thread at "trascendental". Somehow I feel that if tensors want to make you quit, you know how to use that. Or you've had to work with non-commutative algebras.

Is there any other big pure maths sector? Differential Geometry, Number Theory, Algebra and its numerous spin-offs... Topology, Set Theory...
« Last Edit: May 09, 2014, 11:35:22 am by pacovf »
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#### DStu

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« Reply #47 on: May 09, 2014, 11:29:22 am »
+1

Quote
I took two years of college physics and two years of calculus... and I have exactly no idea what y'all are talking about.

Me neither.

I'm about to defend my math phd thesis, and I quitted this subthread at "tensor"...

I nearly quitted this thread at "trascendental". Somehow I feel that if tensors want to make you quit, you know how to use that. Or you've had to work with non-commutative algebras.

Is there any other big pure maths sector?
Actually, applied math.  But actually from all of these, I probably can handle tensors if I must, but the point is that here I don't need to, so...
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#### pacovf

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« Reply #48 on: May 09, 2014, 11:31:23 am »
+1

Want kind of applied?

I'm doing mostly theoretical physics (but I did some engineering before, so I've forgotten some advanced probability and statistics), so that's differential geometry, group theory and gauge theory (not sure if that last one is actually a math discipline).
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#### DStu

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