## 0356: "Nerd Sniping"

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### Re: "Nerd Sniping" Discussion

dr_nik wrote:Ok so first off I have to agree that I have been quite successfully sniped...I have to submit a paper to a conference by the end of the day and I have a ton of work to do on it...but instead I decided to solve this problem.

Second, it seems like everyone working on the problem here and historically has been a mathematician, not an EE. Why? Because an EE would make it into a simpler problem. I'm going to go out on a huge limb here and say that everyone who has posted solutions is wrong since the current will not go through *every* resistor in the plane. Path of least resistance yada yada. So heres my solution:

Spoiler:
The equivalent of an infinite series of resistors is an infinite sheet of metal with resistivity of 1 Ohm.m. Then it is a simple solution of finding the distance between the two points (2.236m assuming each segment is 1m long). That makes it 2.236 ohms

Ok, I wanna hear why I am supremely wrong now =)

Oh, and curse you Randal...

Because of the special geometry of this problem, the distance between two points is not Sqrt(x^2+y^2), which is implicit in your method. It's like if you've ever lived in a (grid) city, and you wanted to get somewhere, there are many ways that are equidistant because you can only walk at right angles. The electrons do the same thing. Likewise, it's easier to take a longer route that has less traffic, so the current density will decrease for longer paths, but still be non-zero for every finite path.
DragonMudd

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### Re: "Nerd Sniping" Discussion

Not to derail the thread or anything, but I thought I'd post my favorite true story of this happening. Minus the truck of course.

"Discuss the socio-political ramifications of the Battle of Five Armies."

I was at my local gaming group, and we had just wrapped up a game of Chrononauts (fun game, btw -- check it out), and I got it into my head that it would be absolutely awesome to make a Lord of the Rings version. (Nerd 1 down -- I didn't play any more games for the rest of the night.) About a half hour later, though, I got my revenge with that one simple question.

The best part was, most of the people in the room didn't know what I was working on, yet this completely shut down the group for about an hour as the debates flew around the room.

This was only about a month ago, actually.
queptar

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karthikp wrote:Assume you apply a positive voltage +V to point A. We can then imagine the current that leaves A and propagates in all directions. If you refer to the figure I've attached, you'll see that I/9 of it reaches point B.

Electricity flows in a circuit. Your explanation only holds if the periphery of our infinite grid (what a paradox!) is the reference point, but the problem explicitly uses B as the reference point. The current will not split four ways from point A, because some paths will have a higher resistance than other paths.

Neat idea, though.

LumenPlacidum wrote:Is there anyone who knows of a standard way of calculating equivalent resistance between two points on an infinite continuous sheet of resistive material? I'm willing to bet that such an answer relies fairly heavily on the notion of distance between the two points.

Yes, there is a way to calculate it. Reference here and here

Dr. Howard Johnson wrote:Given a solid, infinite copper plane, the dc resistance in ohms between vias A and B is approximately RDC=(ρ/πt)ln(s/r), where ρ is the dc resistivity of the planes (6.58×10^(-7) Ω-in. for pure electrodeposited copper), t is the thickness of the planes (0.00137 in. for 1-oz copper), s is the separation between vias A and B, and r is the radius of the drilled holes.

According to this, resistance grows as the natural log of the distance between the points, all other factors held constant.

dr_nik wrote:Second, it seems like everyone working on the problem here and historically has been a mathematician, not an EE. Why? Because an EE would make it into a simpler problem. I'm going to go out on a huge limb here and say that everyone who has posted solutions is wrong since the current will not go through *every* resistor in the plane. Path of least resistance yada yada. So heres my solution:

Spoiler:
The equivalent of an infinite series of resistors is an infinite sheet of metal with resistivity of 1 Ohm.m. Then it is a simple solution of finding the distance between the two points (2.236m assuming each segment is 1m long). That makes it 2.236 ohms

Ok, I wanna hear why I am supremely wrong now =)

Your idea is good, except that your understanding of "path of least resistance" is flawed. Technically, electrons take the path of least impedance. At low frequencies like DC, this is indeed the path of least resistance, but at high frequencies it's the path of least inductance. This leads to some neat behavior...the high-frequency return currents on a reference plane near a digital trace follow the shape of the trace; they do not go straight back to the source.

Anyway, there are also other paths that the electricity will take, because the path of absolute least resistance will be crowded. You might be surprised how far it spreads out.

BTW, I'm an EE too. I'm working on designing PCBs so I'm somewhat familiar with this sheet-concept.

trip11 wrote:Another, but much simplier problem that I had some fun with. Imagine a circut like this:
Code: Select all
`.   ._R1_._R3_..   |    |    |.---|    R5   |----.   |    |    |.   ._R2_._R4_.`

This looks a lot like a Wheatstone Bridge...

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### Re: "Nerd Sniping" Discussion

I have, quite clearly, been sniped by this. I even signed up to these boards to comment. I've not figured it out for myself yet, and following all the fourier stuff that's been posted about (which I suspect is the root to the correct answer) would require sufficient concentration that I can't simply spoil it for myself too easily.

I do have to say, to those suggesting it's 0, it's easy to see why it isn't. Any current leaving the initial point has to flow through at least one of the first four resistors before it can go anywhere else, so the resistance can never be less that 3/4 ohms, the resistance those four would have in parallel.

I'm sure I'll comment again when I've got it worked out.
314159

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### Re: "Nerd Sniping" Discussion

Would this be the first comic with non-handwritten* text in it?

* - Help a non-native speaker! What's the more normal word for non-handwritten?

Dingbats

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### Re: "Nerd Sniping" Discussion

Typeset? Or even just typed?
limu->i(patience)=0
Robin S

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### Re: "Nerd Sniping" Discussion

vwlou89 wrote:Wouldn't the resistance be 0? I actually came up with a number that would satisfy this in my Calculus class: 1-.99999... It comes out to .000...0001 (infinite zeros). It can be represented as "the limit of y such that y=1/x as x goes to infinity". Since it could never really reach zero it would be as close to zero as was possible. Just a thought.

This is a rather pedantic point, but 1-.9999... is not .0000....001 and does not require limits to solve. It's just 0. I will demonstrate.

.999... = x
9.999... = 10x (multiply both sides by 10)
9.999... - .999... = 10x - .999 (subtract .999 from the left, x from the right)
9 = 10x - .999... (simplify left side)
9 = 10x - x (.999 = x, see first line)
9 = 9x (simplify right side)
1 = x (divide both sides by 9)
1 = .999... (.999=x, see first line)
QED.

And so your problem is really 1 minus 1, which is just 0. No limits needed
Brandon

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### Re: "Nerd Sniping" Discussion

trip11 wrote:Another, but much simplier problem that I had some fun with. Imagine a circut like this:
Code: Select all
`.   ._R1_._R3_..   |    |    |.---|    R5   |----.   |    |    |.   ._R2_._R4_.`

This looks a lot like a Wheatstone Bridge...

Yup, execpt that instead of a voltmeter in the midle it's another resistor. That means current can flow through the center which makes it hard to figure out. I was actually reading that very article on Wikipedia when I thought to try this.
trip11

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### Re: "Nerd Sniping" Discussion

instead of a voltmeter in the midle it's another resistor
limu->i(patience)=0
Robin S

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### Re: "Nerd Sniping" Discussion

Dingbats wrote:Would this be the first comic with non-handwritten* text in it?

* - Help a non-native speaker! What's the more normal word for non-handwritten?

Non-handwritten text would probably be "typed"

Anyway, see Parody Week: Dinosaur Comics, Action Movies and Search History. (also, Landscape (sketch), although it's not intentional in this case)

schrodingersduck

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### Re: "Nerd Sniping" Discussion

Don't forget "Fixed Width" http://xkcd.com/276/

skitch78

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### Re: "Nerd Sniping" Discussion

.999... = x
9.999... = 10x (multiply both sides by 10)
9.999... - .999... = 10x - .999 (subtract .999 from the left, x from the right)
9 = 10x - .999... (simplify left side)
9 = 10x - x (.999 = x, see first line)
9 = 9x (simplify right side)
1 = x (divide both sides by 9)
1 = .999... (.999=x, see first line)
QED.

Cute but decimals are not a unique representation so it isn't valid.
Also, as a nerdy physicist the fourier x-form solution is the correct one. After you take math methods sophomore/junior year of college this problem becomes doable. Apparently (according to my canadian grad student friend) it is a common problem up in Canadia.
tglems

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### Re: "Nerd Sniping" Discussion

[quote="314159"]I do have to say, to those suggesting it's 0, it's easy to see why it isn't. Any current leaving the initial point has to flow through at least one of the first four resistors before it can go anywhere else, so the resistance can never be less that 3/4 ohms, the resistance those four would have in parallel. quote]

My suggestion wasn't technically 0, just infinitely small. Also, who said any current is flowing? :p

Syphon

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### Re: "Nerd Sniping" Discussion

Infinitely small is zero, as has been discussed to death elsewhere.
limu->i(patience)=0
Robin S

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### Re: "Nerd Sniping" Discussion

trip11 wrote:
trip11 wrote:Another, but much simplier problem that I had some fun with. Imagine a circut like this:
Code: Select all
`. ._R1_._R3_.. | | |.---| R5 |----. | | |. ._R2_._R4_.`

This looks a lot like a Wheatstone Bridge...

Yup, execpt that instead of a voltmeter in the midle it's another resistor. That means current can flow through the center which makes it hard to figure out. I was actually reading that very article on Wikipedia when I thought to try this.

Yeah, it is. There's something called delta-wye resistor transformations. They take a triangle resistor formation and turn it into a Y shape. To solve wheatstone bridge shapes using resistor transformations, you grab one of the resistor triangles (half of the diamond shape) and convert it to a Y shape, this gives you a circuit with only series/parallel connections, and you simplify from there.

ricree wrote:
karthikp wrote:The answer is R/9. While I could claim the answer is all mine (solved it on my shower tiles, no less), I can't. Most students in India preparing for the dreaded IIT Joint Entrance Exams will be familiar with the solution because a similar problem shows up in a book of physics problems by Irodov.

Assume you apply a positive voltage +V to point A. We can then imagine the current that leaves A and propagates in all directions. If you refer to the figure I've attached, you'll see that I/9 of it reaches point B.
Attachment:
xkcd.png

Superimpose this problem on top of another problem in which the potential farfield is still zero, but you apply a -V voltage at point B. The currents "leaving" A for B will be I/9, too.

You seemed to start with the assumption that current is going to flow equally away from the node in all directions, and that is flat out a wrong assumption.

Edit: Never mind about the KCL part, as you are obviously assuming that the current comes from an leaves through your source lines. Still, I don't see why it is valid to assume that all currents will flow equally from the node, as the voltage difference across the adjectent resistors is going to be different for each resistor.

We solved this in class for two nodes that are only 1 resistor separated, and the method was very similar to what karthikp showed, using a test voltage at one node and figuring out what the current would be, and then doing the same for the other node.. I'm a little fuzzy on it though. To solve the reference voltage problem, we just assigned a node (not A or B which are the nodes in question) a voltage of 0 and computed from there. I think we said current is evenly divided in these situations because all paths look like they have the same resistance. I don't actually know where my notes are though, and that point about different voltages seems justified. Though really, if you are JUST applying the voltage and the first electrons are applied to the circuit, there isn't really any current going through the resistors, so they all have the same voltage, and theres no reason why the resistance at time zero would be different from the resistance at a later time. I know the answer for 1 node separation definitely didn't contain any irrationals though. Maybe I can find my notes...Huk, sniped!
Tuuli

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### Re: "Nerd Sniping" Discussion

Syphon wrote:
314159 wrote:I do have to say, to those suggesting it's 0, it's easy to see why it isn't. Any current leaving the initial point has to flow through at least one of the first four resistors before it can go anywhere else, so the resistance can never be less that 3/4 ohms, the resistance those four would have in parallel.

My suggestion wasn't technically 0, just infinitely small. Also, who said any current is flowing? :p

The whole analysis of this problem depends on the idea that there is current flowing. "Equivalent Resistance" is a relationship between the potential between the two nodes and the current going into and out of the circuit at those points - assumed (quite correctly in this case, of course) to be linear. The quantity doesn't have any meaning outside of the context of that kind of measurement.

If you had an infinitely small resistance, you'd better believe there'd be some current flowing.
Last edited by tetsujin on Wed Dec 12, 2007 8:12 pm UTC, edited 1 time in total.
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tetsujin

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### Re: "Nerd Sniping" Discussion

I like a lot of people have been stumped by this discussion.

At first I tried to use current superposition to try and solve it (ie. inject current in to node A, this splits four-ways and then for each neighbouring node calculate the current that flows in to it and repeat), but this technique is flawed since it doesn't take in to account voltage potential, and so the current radiates outward to infinity.

Eventually I collapsed in to a ball of hate and confusion, but then I came across this paper which explains it all very nicely:

http://atkinson.fmns.rug.nl/public_html/resist.pdf

The author solves it using a combination of voltage and current superposition, and fourier transforms which are going beyond my ability.

Spoiler:
( (-1/2) + (4/pi) ) Ohms or 0.7732395 Ohms

I feel bad for posting the answer without fully understanding the proof, but anyway, there it is!
Last edited by Spudstercool on Wed Dec 12, 2007 8:19 pm UTC, edited 1 time in total.
Spudstercool

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### Re: "Nerd Sniping" Discussion

Tuuli wrote:Yeah, it is. There's something called delta-wye resistor transformations. They take a triangle resistor formation and turn it into a Y shape. To solve wheatstone bridge shapes using resistor transformations, you grab one of the resistor triangles (half of the diamond shape) and convert it to a Y shape, this gives you a circuit with only series/parallel connections, and you simplify from there.

Wow, I totally forgot about that transformation. That's digging way back into Circuits 1 like five years ago...kudos for reminding me!

Tuuli wrote:Though really, if you are JUST applying the voltage and the first electrons are applied to the circuit, there isn't really any current going through the resistors, so they all have the same voltage, and theres no reason why the resistance at time zero would be different from the resistance at a later time.

While you might be technically correct that the resistance doesn't change, that is only because this grid is composed of ideal resistors. Once you add parasitic inductance and capacitance, you now have reactive elements. As a result, the impedance will certainly change with respect to time (EDIT: actually, the impedance would change with respect to frequency...)

tglems wrote:
.999... = x
9.999... = 10x (multiply both sides by 10)
9.999... - .999... = 10x - .999 (subtract .999 from the left, x from the right)
9 = 10x - .999... (simplify left side)
9 = 10x - x (.999 = x, see first line)
9 = 9x (simplify right side)
1 = x (divide both sides by 9)
1 = .999... (.999=x, see first line)
QED.

Cute but decimals are not a unique representation so it isn't valid.

The problem is similar in other representations. 1.111.. in binary would be indistinguishable from 10

Spudstercool wrote:I feel bad for posting the answer without fully understanding the proof, but anyway, there it is!

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### Re: "Nerd Sniping" Discussion

Spudstercool wrote:I feel bad for posting the answer without fully understanding the proof, but anyway, there it is!

Meh, knowing the numeric answer is meaningless. Understanding how you get that answer is everything...
---GEC
I want to create a truly new command-line shell for Unix.
Anybody want to place bets on whether I ever get any code written?

tetsujin

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### Re: "Nerd Sniping" Discussion

That Nerd Sniping describes me on a lesser scale...
Or something to that effect. Hell, I don't know.

i like pi

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### Its only a flesh wound...

After some sleepless consideration from my immediate answer of 0, I have reconsidered. I agree with the fourier series that sets the answer at 4/pi-.5, because shared resistors would not contribute to paths totals. It would appear then that I would have avoided being hit by the truck smug in the confidence that I had figured it out before it came, only to die as I kept turning it over and over in my head reconsidering it on a later date (likly crossing another street) a delayed reaction sniping it would seem...
*ME*

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### Re: "Nerd Sniping" Discussion

DragonHawk wrote:
I have discovered a truly marvelous answer to this, which this post is too small to contain...

How much time passed between when Fermat's Last Theorem was proposed and his death? Does DragonHawk similarly only has that much time left to live? DragonHawk, you will be missed...
Lathe

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### Re: "Nerd Sniping" Discussion

Well, I found out something amazing while trying to prove .99999...=1

I found out it doesn't equal one.

Too bad, I was hoping to shove my little calculations in people's faces, now I hide in shame.
Felstaff wrote:I actually see what religion is to social, economical and perhaps political progress in a similar way to what war is to technological progress.

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Cryopyre

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### Re: "Nerd Sniping" Discussion

mathematician down, 3pts. (registered just to be tallied, though i had been meaning to for a while)
PluTonyum

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### Re: "Nerd Sniping" Discussion

skitch78
Don't forget "Fixed Width" http://xkcd.com/276/

You know, 'definitely for real' would also have kept the alignment...just a thought from a nerd.
tglems

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### Re: "Nerd Sniping" Discussion

i have no clue what the answer to the problem is... seeing as im just starting to study physics...
BUT. i was wondering if it actually matters that there is an infinite grid surrounding it. could you solve the problem by just isolating that small section of the grid?
_______
|___|___|

it probably does matter though. because then the solution of 1.4666666 would be a little too simple to stop a physicist in the middle of the road.
unless it was just an aspiring physicist like me. then, yes, alter-me would have been roadkill.

somegirlyouknew

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### CalTech (Feinstein) solution to hexagonal problem

This reminded me of a chapter in Carver Mead's book on biomorphic engineering in analog VLSI that discussed arificial retinas (large expanses of sensors and resistors). It was based on work by CalTech's David Feinstein.

Here is a link to the paper at CalTech. "The Hexagonal Resistive Network and the Circular Approximation" (1988).

sw

Sam Weir

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### Re: "Nerd Sniping" Discussion

I was all ready to start on this problem, until...

Am I the only one thrown off by the differing values given for Physicists and Mathematicians?! Does that mean we Mathematicians don't think involuntarily think hard on problems or something?! *shakes fist*
Delanda est Carthago.

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fephisto

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### Re: "Nerd Sniping" Discussion

fephisto wrote:I was all ready to start on this problem, until...

Am I the only one thrown off by the differing values given for Physicists and Mathematicians?! Does that mean we Mathematicians don't think involuntarily think hard on problems or something?! *shakes fist*

Not at all! My guess is that you get more points for mathematicians because you first have to get them off whatever they were already absorbed in.
[This space intentionally left blank.]

Owehn

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### Re: "Nerd Sniping" Discussion

I suck at analytical solutions. Fortunately, I'm typing this at a computer rather than a blackboard. Pure mathematicians, stop reading now. I'm about to get computational on your ass.

Attach the network to a constant voltage source connected at the two marked test nodes 1 and 2. Set the voltage at these points equal to +1 volt and -1 volt. (You can pick any voltage, but these are convenient.)

Then the equivalent resistance of the circuit is

Req = 2 / Itotal

Where Itotal is the total current flowing out of node 1.

At any node *except* the test nodes, the sum of currents flowing in and out equals zero:
Inorth + Isouth + Ieast + Iwest = 0
The current Inorth is equal to the voltage drop between "here" and the node just to the north:
(Vnorth - Vhere) = Inorth R = Inorth
...and so on. This gives us:
(Vnorth + Vsouth + Veast + Vwest - 4*Vhere = 0

So we've got an (infinite) series of coupled linear equations which give each voltage in terms of its neighbors. Now, once we get "very far away", all the voltages are near zero and little current flows: one would hope that a large *finite* grid would approximate the answer.

We can rewrite the formula as:
Vhere = (Vnorth + Vsouth + Veast + Vwest)/4

That is, the "correct solution" has the voltage at each point equal to the average of its neighbors. One might guess that one could solve the problem by successive approximation: just set each point (except nodes 1 and 2) equal to the average of its neighbors, keep doing that over and over, and hopefully you'll converge on a consistent solution. (This isn't an idle hope: the equation is equivalent to a finite-difference approximation to Laplace's equation, which can be solved this way.)

So, here's what we're gonna do:
0) Make a large but finite grid of nodes. Set all but nodes 1 and 2 equal to zero voltage to start.
1) Set each node's voltage equal to the average of its neighbors.
2) Exception: set nodes 1 and 2 to the known externally-applied voltages.
3) Repeat 1-2 a whole lot.
4) Itotal = (Vnorthofnode1 + Vsouthofnode1 + ...) - 4*Vnode1
5) Requivalent = 2/Itotal

DONE! But just to make the poor wincing mathematicians happy:

6) Confirm that the solution converges as you increase the number of averaging steps
7) Confirm that the solution converges as you increase the size of the finite grid.

Solution (for a 400x400 grid, 800 averaging steps):

Requivalent = 0.7722

Which is the right answer. Okay, the right answer is a cool algebraic expression with pis in it. I don't care.

400x400x800 is total overkill: for 100x100x200, I get 0.7694, which is pretty close.

MATLAB code: (Ugh, MATLAB!)
Code: Select all
`Nx = 400; Ny = 400;                  % 400 equals infinity!V = zeros(Nx,Ny);x1 = Nx/2-1; y1 = Ny/2;            % If you make Nx or Ny odd, you suck.x2 = Nx/2+1; y2 = Ny/2+1;V(x1,y1) = 1;                           % Set voltages at nodes 1 and 2V(x2,y2) = -1;nsteps = 800;                           % 800 equals infinity too!for i = 1:nsteps    % averaging step:    V(2:end-1,2:end-1) = (V(1:end-2,2:end-1) + V(3:end,2:end-1) + V(2:end-1,1:end-2)+V(2:end-1,3:end))/4;    % Reset voltages at the test nodes    V(x1,y1) = 1;    V(x2,y2) = -1;    Iatnode1(i) = 4*V(x1,y1) - (V(x1+1,y1) + V(x1,y1+1) + V(x1-1,y1) + V(x1,y1-1));endRequiv = 2/Iatnode1(end)plot(Iatnode1)   % to appease paranoid mathematicians`
Last edited by argonaut on Wed Dec 12, 2007 10:45 pm UTC, edited 2 times in total.
argonaut

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### Re: "Nerd Sniping" Discussion

Heh... I hadn't read today's comic yet, but I stumbled across another "sniped" nerd asking about it at the Wikipedia science reference desk. The discussion is currently here:
http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Science#Today.27s_xkcd

But will be archived in a few days here:
http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Science/2007_December_12#Today.27s_xkcd

Discussion includes spoilers, so don't read it if you don't want the answer.
The difference between intelligence and stupidity is that intelligence has its limits.

HiEv

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### Re: "Nerd Sniping" Discussion

argonaut wrote:400 equals infinity!

If I was the kind of person who put random quotes in sigs, I'd quote this in my sig.

I was thinking of trying to solve it that way... except without the approximations, finding an actual expression for the resistance in an nxn grid, and taking the limit... but it was a bit too much...

While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

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### Re: "Nerd Sniping" Discussion

phlip wrote:I was thinking of trying to solve it that way... except without the approximations, finding an actual expression for the resistance in an nxn grid, and taking the limit... but it was a bit too much...

You can construct a large vector in which each element is the voltage at a node in your grid, and then write it out as a large linear algebra problem. Then you "just" invert the matrix to find the answer. It's a giant pain in the butt to do, you're going to need a computer to do the inversion anyway. The exact matrix inversion for a finite grid takes *forever*: for a 400x400 grid, you've got a 16,000x16,000 matrix to invert. So you start thinking about approximate, iterative matrix inversion techniques, which puts you right back at my solution.

Me, I solved the problem and *then* read the forums.
argonaut

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### Re: "Nerd Sniping" Discussion

argonaut wrote:You can construct a large vector in which each element is the voltage at a node in your grid, and then write it out as a large linear algebra problem.

Yeah, I know... I was trying to find a way to generalise it to get the growing matrix out of there... so that I could take a real limn -> inf, of a formula that doesn't get absurdly complicated when n is huge... (I'm one of those pure mathematicians you mentioned... approximations are fine, and all, but they're no substitute for the real thing.)
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phlip
Restorer of Worlds

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### Re: "Nerd Sniping" Discussion

tetsujin wrote:You know what this comic reminds me of? Captain Kirk.

Destroying automata with circular logic and contradictions was practically a specialty of his. For instance Nomad ("You must destroy anything flawed: you are flawed, you haven't destroyed yourself, therefore you are flawed."), Norman from Mudd's Planet (Kirk: Anything he says is a lie. Mudd: I am lying.) - I don't remember if there were other cases I've forgotten... (He pretty much just walked the M-5 through a morality problem - not really in the same vein I guess... and Rayna just died of a broken heart...)

Isn't there also that episode with the computer Landreu (sp?)? I don't remember verbatim, but it was something to the effect of, "You exist to protect these people. Your existence is putting them in danger. Therefore you must destroy yourself to fulfill your reason for existing."

I had to share this comic, especially after seeing mathematicians and physicists are worth differing amounts of points. I lean more towards the mathematician side, but I have enough physics background that I could swing that direction if I wanted to. Does that make me worth about 2.5 points, about 5 points, or about 1.2 points? That is, are the points for a given individual averaged, added in series, or added in parallel?

Also, it was amusing to see the post count increase as I read each page of the thread. It was about 140 when I started. This is number 155. What? I had distractions (e.g. wiki)
LOWA

Nimz

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### Re: "Nerd Sniping" Discussion

Cryopyre wrote:Well, I found out something amazing while trying to prove .99999...=1

I found out it doesn't equal one.

Too bad, I was hoping to shove my little calculations in people's faces, now I hide in shame.

Oh?

I don't see how you can possibly disprove it...

Syphon

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### Re: "Nerd Sniping" Discussion

a possible reason for extra point for mathematicians being 2 defense mechanisms: "boring, someone else has already done that one" or "a solution exists, what do you mean you want me to find it?" i am not aware of parallel for physicists
PluTonyum

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### Re: "Nerd Sniping" Discussion

I may have a solution, but I probably overlooked some detail or other. Here it is anyway:

Spoiler:
So in the grid, let's go east 2, north one. Resistance of 3.
Then, go south 1, east 2, north 2. Resistance of 5.
Then, go south 2, east 2, north 3. Resistance of 7.
Etc.
Now, go north 1, and east 2. Resistance of 3.
Then, go north 2, east 2, and south 1. Resistance of 5
Then, go north 3, east 2, and south 2. Resistance of 7

ETC. When adding stuff in parallel, we take the reciprocal of the sum of the reciprocals of the resistance of each individual path.

So that's one over: 1/3 + 1/3 + 1/5 + 1/5 + 1/5 + 1/7 +1/7...
Now, we take the second 1/3, the second 1/5, the second 1/5, and the second 1/7 etc... And we know that 1/3>1/4, 1/5>1/6, and 1/7>1/8. Therefore, the summation is greater than this: 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

Which, by the way, dominates the harmonic series, which diverges to infinity. Since the sum of the reciprocals of the resistance of the paths in infinity, the total resistance is ZERO!!!

What the hell? Superconducting with resistors and without low temps. Of course, the harmonic series diverges ridiculously slowly, so while that may seem unreasonable, keep in mind that it's still not anywhere close to being practical.

Of course, I don't know if the paths are allowed to touch each other or if there's some nuance I'm missing, but that's my 2 cents.
Chx

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### Re: "Nerd Sniping" Discussion

ckt wrote:I have been successfully sniped (I notice I'm not the only one who posted for the first time in response to this comic).

For those that are obsessed with the problem and need some help, I did a quick literature search on the problem, and I found this:
"Infinite Resistive Lattices" by D. Atkinson and F.J. van Steenwijk (it's in the American Journal of Physics, but a google search also turns up a citeseer copy).

Spoiler:
The poster (floyd4one) who said the answer is 4/pi - 1/2 is correct, I'm just supplying a reference that shows the full solution.

Those were my teachers; get the paper here:
http://atkinson.fmns.rug.nl/public_html/resist.pdf

Steenwijk gave me courses about electromagnetism... I feel I failed him
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theraven1982

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### Re: "Nerd Sniping" Discussion

Chx wrote:I may have a solution, but I probably overlooked some detail or other. Here it is anyway:

Spoiler:
So in the grid, let's go east 2, north one. Resistance of 3.
Then, go south 1, east 2, north 2. Resistance of 5.
Then, go south 2, east 2, north 3. Resistance of 7.
Etc.
Now, go north 1, and east 2. Resistance of 3.
Then, go north 2, east 2, and south 1. Resistance of 5
Then, go north 3, east 2, and south 2. Resistance of 7

ETC. When adding stuff in parallel, we take the reciprocal of the sum of the reciprocals of the resistance of each individual path.

So that's one over: 1/3 + 1/3 + 1/5 + 1/5 + 1/5 + 1/7 +1/7...
Now, we take the second 1/3, the second 1/5, the second 1/5, and the second 1/7 etc... And we know that 1/3>1/4, 1/5>1/6, and 1/7>1/8. Therefore, the summation is greater than this: 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ...

Which, by the way, dominates the harmonic series, which diverges to infinity. Since the sum of the reciprocals of the resistance of the paths in infinity, the total resistance is ZERO!!!

What the hell? Superconducting with resistors and without low temps. Of course, the harmonic series diverges ridiculously slowly, so while that may seem unreasonable, keep in mind that it's still not anywhere close to being practical.

Of course, I don't know if the paths are allowed to touch each other or if there's some nuance I'm missing, but that's my 2 cents.

That solution would work iff the paths involved didn't share resistors. They do, which complicates things.
Murgatroyd

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