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Last edited by charlesfahringer on Thu Apr 20, 2017 3:37 am UTC, edited 1 time in total.
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Re: Questions
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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 gmalivuk
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Re: Questions
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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 gmalivuk
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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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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 oneyear 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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All right, here's a 33omino 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:
Edit: A 30omino that has only one Hamiltonian path. I would believe it if this one were optimal, but I still doubt it:
Code: Select all
XXXXXXX
XX XX XX
X XXXXX X
XX XX XX
XXXXXXX
Edit: A 30omino 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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For the field question, you probably also need multiplication to distribute, no?
There are algebraically closed skewfields, but I'm not sure about the existence of algebraically closed fieldlike 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?
There are algebraically closed skewfields, but I'm not sure about the existence of algebraically closed fieldlike 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?
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
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 gmalivuk
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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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Cosmologicon wrote:All right, here's a 33omino 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 from19 then from ax
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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?)
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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Cosmo, it's not actually an image; it's a script that automatically generates an image, and the board software doesn't like that.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
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They did. The url for that was a url which looks like an image, but they had a script serverside 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.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
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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 howtos online to set up something like that if you want.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.
"With math, all things are possible." —Rebecca Watson
"With math, all things are possible." —Rebecca Watson

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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 clientside, but rather the serverside. 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 serverside screw ups can be exposed.
Note the right way to do statechanging 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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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 halfvisible at each pole for a few days? your solution doesn't seem right.
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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!
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.
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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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 rightdistributivity, 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.
And, well, if you do have left and rightdistributivity, 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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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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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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Awwww
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 noncommutative 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.
Woxor wrote:It might be possible to have a "ring" without unity that had a noncommutative 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 welldefined. 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.
If you have left and right additive inverses, then ax = a for all a.
a+a = 0
a+ax = 0
aa+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.
a+a = 0
a+ax = 0
aa+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.
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.
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.
Re: Awwww
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 noncommutative 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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