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charlesfahringer
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Questions

Postby charlesfahringer » Sun Jun 03, 2007 1:00 pm UTC

q
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Re: Questions

Postby Alisto » Sun Jun 03, 2007 3:51 pm UTC

charlesfahringer wrote:What would be the most efficient way to travel for one year keeping the sun above the Horizon the whole year? You can start wherever, whenever you want (including accounting for long term cycles like the axis tilt) and travel up to Mach 3 (actually, you can choose this parameter, but I like this choice). Efficiency can be defined in any reasonable way, but I like shortest total distance travelled on Earth's surface, or least gas used in a specific vehicle. Also, you have to stay within 35000 feet of the surface, so that you can't go out into space and have the sun always above the horizon unless you get eclipsed.


Without doing the math, I would like to say to sit at the north pole for half the year, then as the sun begins to set, haul ass in a spiral path down to the south pole. Rinse and repeat.
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Re: Questions

Postby gmalivuk » Sun Jun 03, 2007 3:56 pm UTC

Are there algebraically closed "fields" where neither the addition nor the multiplication are commutative?


As I'm sure you know, having put "fields" in quotes yourself, a structure like this of course wouldn't be a field. So before anyone can answer this question, you have to tell us what axioms your "fields" do follow.
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Postby charlesfahringer » Sun Jun 03, 2007 5:38 pm UTC

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Postby gmalivuk » Sun Jun 03, 2007 7:56 pm UTC

charlesfahringer wrote:I'm not completely sure that the sun stays over the horizon for half the year at the exact end of the axis.


To bad. It does.

Anyway, you only stay up there for almost half the year. To get to the other pole when the sun starts dipping below the horizon, you just head straight towards it. This will of course involve a spiralling path as soon as you get away from the pole. When you get to the equator (at which point the sun should be directly above you), just do the reverse path to the south pole and you should get there in time for the sun to be above the horizon for the rest of the year.

(At the North Pole, the sun was completely above the horizon starting March 20 this year. It will continue being completely above until September 23. On September 24, it will start being completely above the horizon at the South Pole.)
--

For your math question, we then must suppose you're talking about a set with 0 ≠ 1, which is a nonabelian group under addition and for which all elements other than 0 are a nonabelian group under multiplication. So we have the following for all a in the set:

a+0=0+a=a
a*0=0*a=0
a*1=1*a=a
∃m with a*m=m*a=1
∃s with a+s=s+a=0

And for at least some a in the set:
∃b with a*b≠b*a
∃c with a+b≠c+a

Is this what you're thinking?

(This just gives your "field". More is needed to make it algebraically closed.)
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Postby Cosmologicon » Sun Jun 03, 2007 8:15 pm UTC

gmalivuk wrote:
charlesfahringer wrote:I'm not completely sure that the sun stays over the horizon for half the year at the exact end of the axis.


To bad. It does.

Anyway, you only stay up there for almost half the year. To get to the other pole when the sun starts dipping below the horizon, you just head straight towards it. This will of course involve a spiralling path as soon as you get away from the pole.

Well actually, it matters in this regard: if the sun were below the horizon at both poles at the equinox, then you would have to move away from the pole at the beginning and end of your one-year voyage. This is certainly doable, but you'd need to take it into account as well.

But you're right. The world isn't perfect, but the largest effect due to this imperfection - atmospheric refraction - causes the sun to be above the horizon at both poles on the equinox. The atmsopheric refraction at the horizon is 34 arcminutes, greater than the angular size of the sun. So there's plenty of margin of error here.

Also, it's not necessary to sprial, if we're allowed to go at Mach 3. To trek straight from the North Pole to the South Pole in 12 hours, you'd only need to be able to go Mach 1.4. If you're minimizing the distance travelled, just go straight!

But then charlesfahringer asks what's the path for keeping the sun as high as possible. This, of course, is a completely different question from the one in the OP. For this, you should just start, end, and remain directly under the sun, easily doable at Mach 3.

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Postby Token » Sun Jun 03, 2007 8:17 pm UTC

gmalivuk wrote:And for at least some a in the set:
∃b with a*b≠b*a
∃c with a+b≠c+a

Is it necessary to have both of those holding for some a, or could you have:

∃a,b : a*b≠b*a
∃c,d : c+d≠d+c

(That's an actual question, not a politely worded correction, by the way.)

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Postby Cosmologicon » Sun Jun 03, 2007 8:35 pm UTC

All right, here's a 33-omino with a Hamiltonian path that I don't see how to make with the snake. It seems inefficient somehow, though, so I bet there's a better solution:

Code: Select all

 XXXXXXX
XX  XX XX
X XXXXX X
XX XX  XX
 XXXXXXX


Edit: A 30-omino that has only one Hamiltonian path. I would believe it if this one were optimal, but I still doubt it:

Code: Select all

     XXX
     X XX
     XX X
 XXXX  XX
XX  XXXX
X XX     
XX X     
 XXX     

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Postby skeptical scientist » Sun Jun 03, 2007 11:18 pm UTC

For the field question, you probably also need multiplication to distribute, no?

There are algebraically closed skew-fields, but I'm not sure about the existence of algebraically closed field-like objects where neither addition nor multiplication commute. It's not even clear what algebraically closed means any more - of course ax^2+bx+c is a polynomial, but what about xax+bx^2+c+dx+e+(x^2)f?
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Postby gmalivuk » Mon Jun 04, 2007 1:20 am UTC

Token wrote:
gmalivuk wrote:And for at least some a in the set:
∃b with a*b≠b*a
∃c with a+b≠c+a

Is it necessary to have both of those holding for some a, or could you have:

∃a,b : a*b≠b*a
∃c,d : c+d≠d+c

(That's an actual question, not a politely worded correction, by the way.)


Yes, you could have that situation. (I knew when I posted that I probably should have made that bit clearer.)
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Postby Thunderbird4! » Mon Jun 04, 2007 1:42 am UTC

Cosmologicon wrote:All right, here's a 33-omino with a Hamiltonian path that I don't see how to make with the snake. It seems inefficient somehow, though, so I bet there's a better solution:

Code: Select all

 XXXXXXX
XX  XX XX
X XXXXX X
XX XX  XX
 XXXXXXX

If I'm up on how the snake runs here then this can be done:

Code: Select all

 opqr456
mn  s3 78
l xwt21 9
kj vu  ba
 ihgfedc

Follow it from1-9 then from a-x
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Postby Cosmologicon » Mon Jun 04, 2007 1:46 am UTC

I think you might not be up on how the snake runs. Three steps ago, the head was at position 4. Where was the tail then?

It's sort of like a chessboard that's in an apparently legal position, but no way to reach it.

(Anyone know why I can't insert this image?)

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Postby 3.14159265... » Mon Jun 04, 2007 1:58 am UTC

How is that chess setting "theoretically" possible?

The black bishop couldn't have gotten out, and all the pawns are there, so its didn't used to be a pawn either.

Edit: Is it a queen? :oops:
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Postby skeptical scientist » Mon Jun 04, 2007 2:49 am UTC

Cosmo, it's not actually an image; it's a script that automatically generates an image, and the board software doesn't like that.
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Postby Cosmologicon » Mon Jun 04, 2007 2:51 am UTC

Didn't somebody recently have an image in their signature that showed a sign that tells you your ISP and IP address? How did they get away with that?

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Postby skeptical scientist » Mon Jun 04, 2007 2:53 am UTC

They did. The url for that was a url which looks like an image, but they had a script server-side which said "whenever you get a request for this image, dynamically create it and then serve it", so the board software treated it like an image and it worked fine. Your script has a url which looks like a script, and so the board software chokes.
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Postby Cosmologicon » Mon Jun 04, 2007 3:00 am UTC

Well.... hmph. My server won't let me put scripts in anything other than a cgi extension. Don't the phpBB people know that the url isn't a reliable way to distinguish between scripts and images? :-P

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Postby skeptical scientist » Mon Jun 04, 2007 6:17 am UTC

The image file wasn't the script - the machine just had a script set up to run every time it got a request for that particular image. There are various how-tos online to set up something like that if you want.
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Postby charlesfahringer » Mon Jun 04, 2007 11:25 am UTC

q
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Postby Yakk » Mon Jun 04, 2007 3:00 pm UTC

Cosmologicon wrote:Well.... hmph. My server won't let me put scripts in anything other than a cgi extension. Don't the phpBB people know that the url isn't a reliable way to distinguish between scripts and images? :-P


No, they don't. :)

The problem isn't the client-side, but rather the server-side. Someone sets up a script that can be activated with get.

Say, log off, or become friends, or any of the other "click on this to do something on the board".

While by the RFC, get actions shouldn't cause important state changes, lazy web programmers use them too often to do this.

Now you go and form an image link in your post pointing at that get script. You can now make anyone who looks at your post cause the get script to run using their credentials.

By blocking image links that don't look like images, the hope is that fewer such server-side screw ups can be exposed.

Note the right way to do state-changing actions on a web site is with post methods (like the preview/submit buttons on the xkcd posting.php page).

HTH.

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Postby German Sausage » Mon Jun 04, 2007 4:45 pm UTC

gmalivuk wrote:
(At the North Pole, the sun was completely above the horizon starting March 20 this year. It will continue being completely above until September 23. On September 24, it will start being completely above the horizon at the South Pole.)

surely there will be a stage where it is half-visible at each pole for a few days? your solution doesn't seem right.
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Postby Cosmologicon » Mon Jun 04, 2007 4:47 pm UTC

That's an interesting theory, Yakk, but I bet it was just more of a vague sense of prohibiting scripting -> good security on their part than any actual calculated hazard. Either way, it doesn't change the fact that if such a vulnerability due to improper method choice existed, anyone with a web server of their own could easily circumvent the "protection" they have implemented. *I* even knew how to do it, and I don't even know what I'm talking about!

3.14159265... wrote:How is that chess setting "theoretically" possible?

The black bishop couldn't have gotten out, and all the pawns are there, so its didn't used to be a pawn either.

Yeah, that was the whole point. I never said it was possible, theoretically or otherwise. And what do you mean by "theoretically" possible? Do you mean theoretically possible, or the opposite, or something else entirely? I often feel like I'm missing the point when people use those scare quotes.

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Postby gmalivuk » Tue Jun 05, 2007 2:10 am UTC

German Sausage wrote:surely there will be a stage where it is half-visible at each pole for a few days? your solution doesn't seem right.


atmospheric refraction
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Postby charlesfahringer » Tue Jun 05, 2007 3:45 pm UTC

q
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Postby Buttons » Tue Jun 05, 2007 4:12 pm UTC

Back to the "noncommutative field" question, I assume you want distributivity? On the left and right? That is, (a+b)*c = a*c+b*c, and a*(b+c) = a*b + a*c. If you don't have this, it's going to be awfully tricky to say much more about how addition and multiplication interact.

And, well, if you do have left- and right-distributivity, then you can't have such a set. The issue is that the ring axioms say nothing about the order of how you compute things, nor can they without becoming very silly (a*(b+c) = a*b + a*c if and only if a isn't already a sum of two things?). So expanding in either order, we've got

(a+b)*(c+d) = a*(c+d) + b*(c+d) = a*c + a*d + b*c + b*d

OR

(a+b)*(c+d) = (a+b)*c + (a+b)*d = a*c + b*c + a*d + b*d

This yields a*d + b*c = b*c + a*d for all a, b, c, d in your "ring". Choosing c and d to be 1 gives a+b=b+a for all a and b, which isn't what you want.

If you don't require addition to be noncommutative, you're talking about what's called a division ring, i.e. a ring with unity such that every nonzero element is a unit. While there are noncommutative division rings, they are all infinite: Wedderburn's theorem says that every finite division ring is a field.

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Postby joeframbach » Tue Jun 05, 2007 5:12 pm UTC

Cosmologicon wrote:I think you might not be up on how the snake runs. Three steps ago, the head was at position 4. Where was the tail then?

It's sort of like a chessboard that's in an apparently legal position, but no way to reach it.

(Anyone know why I can't insert this image?)


What if white started at the top and worked its way down, and black started at the bottom and went up?

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Postby Cosmologicon » Tue Jun 05, 2007 5:18 pm UTC

Hm, good call. If I'd had the letters on the side that would have cleared it up, but until I get those, this should cover it. The reason I didn't add more pieces to begin with was so that it was clear what I was getting at, is all.

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Postby charlesfahringer » Wed Jun 06, 2007 7:11 pm UTC

q
Last edited by charlesfahringer on Thu Apr 20, 2017 3:17 am UTC, edited 1 time in total.

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Postby Woxor » Wed Jun 06, 2007 8:23 pm UTC

charlesfahringer wrote:So distributivity implies additive commutativity?

Distributivity and the existence of a multiplicative identity are both required for the given proof. It might be possible to have a "ring" without unity that had a non-commutative additive structure, but I don't know for sure.

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Awwww

Postby PainGod » Wed Jun 06, 2007 9:06 pm UTC

Distributivity and the existence of a multiplicative identity are both required for the given proof. It might be possible to have a "ring" without unity that had a non-commutative additive structure, but I don't know for sure.


Awwww, I was totally going to make that point, and with that proof no less. Anyway, the point remains valid:

distributivity + ring with a 1 + abelean addition => commutative multiplication

Otherwise, you can make such a "field". However, not we aren't necessarily talking about rings, since rings have abelean addition. Chose one to throw out.

But I guess we've probably lost the question at this point.

Oh, and you do definitely need some way for multiplication and addition to interact, otherwise you do get two unrelateable structures being confounded in the same set. This, my friends, is why groups have all four of their properties: to be able to do algebra. Otherwise, we don't get anything interesting.

_____________________________________________________________

Note: Being Abelean is the same thing as being commutate, but I usually use the term "Abelean" to refer to groups and addition, and the term "commutative" to refer to rings and to its multiplication.

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Re: Awwww

Postby Woxor » Wed Jun 06, 2007 9:24 pm UTC

PainGod wrote:distributivity + ring with a 1 + abelean addition => commutative multiplication

Wait, that's not true. There are noncommutative rings with unity (e.g., matrices).

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Postby Buttons » Wed Jun 06, 2007 11:06 pm UTC

Woxor wrote:It might be possible to have a "ring" without unity that had a non-commutative additive structure, but I don't know for sure.

I think I might have found one (albeit with commutative multiplication), but the structure is the sort of thing I'm not used to dealing with, and stuff keeps surprising me. Do look for holes in this.

Let x be such that x^2 = x, 4x = 4, 2x ? 2. The underlying set of our ring R is 2(Z[x]), that is to say the smallest ring containing the integers and x with even coefficients. Addition (denoted "+") is not our usual addition (denoted "+"). Instead we define a + b = a + bx. Multiplication works the same, so (a+bx)*(c+dx) = ac+adx+bcx+bdx = ac+(ad+bc+bd)x. Associativity and distributivity seem to hold fine. Also note that for elements which can be written as products (e.g. 4 and 8), commutativity of addition holds: 4 + 8 = 4 + 8x = 4 + 2*4x = 4 + 2*4 = 12. 8 + 4 yields the same. This matches up with what we saw above. A general element of the "ring" looks like a+bx with a in 2Z and b either 0 or 2.

Note that addition isn't always abelian: 2 + 4 = 2 + 4x = 2 + 4 = 6, but 4 + 2 = 4 + 2x != 6.

I could probably be more rigorous. Please point anything that isn't well-defined. And don't feel bad if it ruins my whole argument. Oh, and if you want multiplication to be noncommutative, I think considering the ring of 2x2 matrices with these elements as entries and addition as defined above should work.

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Postby Woxor » Wed Jun 06, 2007 11:29 pm UTC

If you have left and right additive inverses, then ax = a for all a.

-a+a = 0
-a+ax = 0
a-a+ax = a+0
0+ax = a
ax = a

Thus your addition ends up being commutative after all! Maybe if we used the multiplicative structure of invertible matrices as our "addition" and defined A*B=exp(log(A)log(B)), where exp and log are defined by their series representations. I'll have to think on that.

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Postby Buttons » Thu Jun 07, 2007 1:22 am UTC

Woxor wrote:If you have left and right additive inverses, then ax = a for all a.

Well, no, that would be assuming that the inverse of a under + is the same as it is under +, which is quite an assumption. But you're right that something goes wrong with additive inverses: about half of all elements don't have one! 2 + anything cannot be 0, since the xs only "cancel out" in fours. Darn.

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Postby Woxor » Thu Jun 07, 2007 2:06 am UTC

Buttons wrote:Well, no, that would be assuming that the inverse of a under + is the same as it is under +, which is quite an assumption. But you're right that something goes wrong with additive inverses: about half of all elements don't have one! 2 + anything cannot be 0, since the xs only "cancel out" in fours. Darn.

My browser font can't tell the difference between a bold + and a regular one, so I'll call the bold one #.

If a#b = b#a = 0 (which is usually the case with inverses even in nonabelian groups), then the proof I gave becomes

b#a=0
a#(b#a)=a#0
(a#b)#a=a
0#a=a
0+ax=a
ax=a

I'm essentially proving that -a=b, rather than assuming it.

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Re: Awwww

Postby PainGod » Thu Jun 07, 2007 2:40 am UTC

PainGod wrote:
Distributivity and the existence of a multiplicative identity are both required for the given proof. It might be possible to have a "ring" without unity that had a non-commutative additive structure, but I don't know for sure.


Awwww, I was totally going to make that point, and with that proof no less. Anyway, the point remains valid:

(Edited)
distributivity + ring with a 1 => abelean addition

Otherwise, you can make such a "field". However, not we aren't necessarily talking about rings, since rings have abelean addition. Chose one to throw out.

But I guess we've probably lost the question at this point.

Oh, and you do definitely need some way for multiplication and addition to interact, otherwise you do get two unrelateable structures being confounded in the same set. This, my friends, is why groups have all four of their properties: to be able to do algebra. Otherwise, we don't get anything interesting.

_____________________________________________________________

Note: Being Abelean is the same thing as being commutate, but I usually use the term "Abelean" to refer to groups and addition, and the term "commutative" to refer to rings and to its multiplication.


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