0356: "Nerd Sniping"

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Re: "Nerd Sniping" Discussion

sab39 wrote:I don't think it makes a difference whether the infinite plane "wraps around" at infinity or not.
I'm not certain what it would mean for a plane to wrap around "at infinity". Surely wrapping around means that you must only travel a finite distance to reach where you started. But I don't have a mathematical definition handy.
sab39 wrote:(I'm aware that this isn't a rigorous proof. But it's good enough to convince me in an intuitive sense, especially since an "infinite torus" is pretty darn impossible in the first place).
Yes, I don't think anyone spoke of an infinite torus. An infinite cylinder such as I mentioned is perfectly reasonable to consider, however (or at least as reasonable to consider as an infinite plane).
sab39 wrote:Incidentally, isn't there a projection of the complex numbers onto a sphere which treats infinity in all directions as the point at the north pole? So it is possible to map an infinite grid onto a sphere - but it's not very clear how the resistors would join up at the top, other than being packed infinitely dense.
I believe you're thinking of the Riemann sphere. Regardless, of course you can map any set bijectively onto any set of the same cardinality (and of course the plane and the sphere both have the cardinality of the continuum). The point is that gringer's proposal could not reasonably be considered an infinite grid of resistors, because it contains only finitely many resistors. The reasonable definition of an infinite grid in this case relates to the number of resistors, not their embedding in a metric space, since the latter is irrelevant to the situation: the grid behaves identically however big or small or near or distant the resistors are according to any metric.

Anyway, I think gringer would have gotten more points than most of us here on this question of the Google test, just for coming up with the idea. Possibly even more than someone who actually solved it. I should have thought of it, dammit. I'm a math student, I'm supposed to poke holes in physics questions for fun.
chopps wrote:Isn't it simply 20 ohms? There may be infinite resistors, but I only see 4 parallel paths.
It can't possibly have higher resistance than three ohms, since there is at least one three-ohm path.
Simetrical

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Re: "Nerd Sniping" Discussion

So, I would like someone to explain to me why this method I'm about to explain doesn't work.

First attempting to find the resistance between two adjacent nodes (as a simpler problem) insert the voltage at one node to be o and the other to be 1. Now lets examine the square above the two nodes in question. Due to symmetry, the 1 volt difference will be acquired equivalently by the direct route along the bottom line, or by the path around the top, giving a voltage along the top line of 1/3. Now repeating that process for the square above that, and the square above that, we get that each square has a resistance of 1/3 the previous. As each resistor has unit resistance the currents are the same as the voltages. The same can be applied in the opposite direction. Thus, if we add the current differences across the infinite vertical line separating the two nodes, we get a geometric series which totals 2. Since the difference between the voltage of the two nodes in question is one and we see that exactly 2 amps cross the boundary between the two nodes, the resistance is 1/2.

Now, this fact allows us to replace any cut which crosses an infinite length with a single resistor with half the resistance. Doing this twice gives a resistance of 1 between two inline nodes. Finally, using a similar approach, we see that a resistance of 2/3 exists if the cut goes only in one direction, giving a grand total of 8/3.

(note, as I was typing it I figured out why it doesn't work. for a solution to the problem, check spoiler.)

Spoiler:
The means of finding the resistance between adjacent nodes relied on symmetry, which is broken as soon as the first replacement is done.
Dwalin

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Re: "Nerd Sniping" Discussion

Simetrical wrote:I'm not certain what it would mean for a plane to wrap around "at infinity". Surely wrapping around means that you must only travel a finite distance to reach where you started. But I don't have a mathematical definition handy.
Well, does lim(x->inf) x = lim(x->inf) -x?

I'm no mathematician (although I did study maths at a prestigious university, I wasn't very good at it, and switched to Compsci as soon as I could) but my understanding is that the answer to that question is "it depends". You can come up with a consistent set of axioms in which the answer is yes, or a different set of axioms in which it's no.

My example of the Riemann Sphere (yes, that was exactly what I meant, thank you for finding it) was intended to be an example of that. On the Riemann Sphere, clearly there is only one "point at infinity" no matter which direction you go.

Simetrical wrote:Yes, I don't think anyone spoke of an infinite torus. An infinite cylinder such as I mentioned is perfectly reasonable to consider, however (or at least as reasonable to consider as an infinite plane).

I think I misunderstood what gringer was getting at. I didn't realize he meant an actual finite torus - I thought he meant something like what I was suggesting. An infinite number of resistors, in a grid, but where lim(x->inf) and lim(y->inf) are equivalent to lim(x->-inf) and lim(y->-inf). That would have the topology of an "infinite torus" to the extent that such a thing is possible, right?
sab39

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Re: "Nerd Sniping" Discussion

sab39 wrote:
Simetrical wrote:I'm not certain what it would mean for a plane to wrap around "at infinity". Surely wrapping around means that you must only travel a finite distance to reach where you started. But I don't have a mathematical definition handy.
Well, does lim(x->inf) x = lim(x->inf) -x?

I'm no mathematician (although I did study maths at a prestigious university, I wasn't very good at it, and switched to Compsci as soon as I could) but my understanding is that the answer to that question is "it depends". You can come up with a consistent set of axioms in which the answer is yes, or a different set of axioms in which it's no.

Well, your statement of the problem doesn't make a lot of sense to me, formally. The definition of limits uses, conventionally, +∞ and −∞ to denote the limits of sequences or functions that exceed any positive (respectively, negative) finite bound if you go sufficiently far along the sequence. You would have to redefine the meaning of the expression lim sn = ±∞; I'm not aware of alternative definitions for real-valued sequences or functions.

I do see the general point you're making. There is such a thing as a system of numbers that has a single quantity ∞ with no sign or direction, as the Riemann sphere illustrates. (There are other examples too, but that's possibly the most mainstream and useful.) You're saying you should try taking an infinite plane, and in some sense identify one direction with another, just as you would identify finite edges of a rectangle with one another to form a torus. But while I can readily see how to mathematically identify one set of points (say, the finite edge of a rectangle, or the infinite positive and negative branches of the complex log on the negative x-axis), I can't see how to identify two directions at infinity. To identify sets of points, you can create an equivalence relation, and express coordinates in terms of equivalence classes. You can declare two directions to be equivalent, but then in what equations or formulas would you use the resulting equivalence classes?
sab39 wrote:I think I misunderstood what gringer was getting at. I didn't realize he meant an actual finite torus - I thought he meant something like what I was suggesting. An infinite number of resistors, in a grid, but where lim(x->inf) and lim(y->inf) are equivalent to lim(x->-inf) and lim(y->-inf). That would have the topology of an "infinite torus" to the extent that such a thing is possible, right?
As I say, I don't know how to make sense of the possibility. In the plane you could take the limit at infinity by taking the limit in one direction along an infinite curve that gets and stays arbitrarily far from the origin. You could identify the limits of all of them easily enough, but I don't see how this would affect the geometry or topology of the plane, since things like coordinates, curvature, smoothness, connectedness, etc., etc. would all seem to be unaffected by what you chose to call the values at infinity. On the other hand if you identify two sets of points, that immediately changes how you need to express any point's coordinates, which will immediately change the geometry.
Simetrical

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Re: "Nerd Sniping" Discussion

chopps wrote:Isn't it simply 20 ohms? There may be infinite resistors, but I only see 4 parallel paths.
Simetrical wrote:It can't possibly have higher resistance than three ohms, since there is at least one three-ohm path.

Yeah I actually was guessing at what equiv resistance was, now that I've looked it up, I'll revise my answer and suggest 21/20 ohms 1/(2/7 + 2/3).
chopps

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Re: "Nerd Sniping" Discussion

Hmmm.
I can see an easy lower and upper bound.
There is a direct chain of 3 resistors, so it cannot be more than 3ohm (3x1ohm resistors in series).
If you consider a single node, it has four resistors connected to it. If you reduced the resistance of all others to zero (which cannot increase resistance between the two nodes) you would have just these four resistors in parallel, so (1+1+1+1)^-1 = 1/4ohms is a lower bound.
By considering the fact that both nodes have this constraint (and if non-adjacent do not share a resistor), and are in series, the lower bound can be improved to 1/2ohms.

I'm not a sucker for doing Fourier Transforms, so I'll just assume the above solutions to be correct.

Now, on to the N dimensional variant proposed by Alsadius, I have a proof that that it is impossible to get zero resistance with finite dimension.
It is trivial to say that an N-dimensional system has 2N resistors attached to each node. By applying the same logic as above, the lower bound for resistance is (2N)^-1, even with adjacent nodes. This is >0 for finite N.
Q.E.D.

Now, what about non-square layouts, or non-ideal resistors...

FOOM
Ouch.
jack b

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Re: "Nerd Sniping" Discussion

The mathematician's solution: Treat this problem as a relatively trivial case of the more interesting problem of finding the resistance between ANY two points on the resistor plane. Assign to grad student.

The applied mathematician's solution: Find reasonably good upper and lower bounds. Have a beer.

The physicist's solution: Solve by symmetry.

The engineer's solution: Simulate a finite 1000 X 1000 grid in a program and then spit out a pretty good answer.

The biologist's solution: "Get... a... freaking... multimeter!"

Signed,
A biologist
jwwells

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Re: "Nerd Sniping" Discussion

Well, while i have no idea how to do a rigorous proof, this will give a 'best estimate' to a high enough degree of accuracy.

First, start off with a grid of 4 nodes. Using basic circuit theory it is possible to calculate the equivalent resistance (or you can use a circuit simulator such as pspice). Then, add an extra row onto each side of the initial grid. Solve for this. Then add another row of resistors to each side and get a third point.

As you can see, a circuit simulator would be best for solving this as it gets pretty big pretty fast. After you have a few points, just extrapolate from those to find the limit as you approach infinity.

Alternatively, take the engineering approach:
1) There are 4 resistors connected to each node, therefore the resistance seen while trying to enter the node is (at the very least) 0.25 Ohm. (combine resistors in parallel)
2) Using (1), you can see the very minimum impedance seen leaving one node and entering another is 0.5 Ohm. (Assume that there are just two nodes and the four exit ports in node A are connected to the corresponding input ports in node B).
3) An upper bound of 1 Ohm can be easily shown to exist. 1 Ohm is a nice round number.
4) Lower bound is 0.5, upper bound is 1Ohm - Guesstimate 0.75Ohm as the true answer. Worst case scenario, you're only 50% off - a good approximation in engineering terms

EDIT: Heh heh heh, looks like i'm both an engineer and an applied mathemagician
Mutant_Fruit

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Re: "Nerd Sniping" Discussion

Simetrical wrote:As I say, I don't know how to make sense of the possibility. In the plane you could take the limit at infinity by taking the limit in one direction along an infinite curve that gets and stays arbitrarily far from the origin. You could identify the limits of all of them easily enough, but I don't see how this would affect the geometry or topology of the plane, since things like coordinates, curvature, smoothness, connectedness, etc., etc. would all seem to be unaffected by what you chose to call the values at infinity. On the other hand if you identify two sets of points, that immediately changes how you need to express any point's coordinates, which will immediately change the geometry.
Ok, here's a simple way of phrasing the question which in fact does make sense, I think.

The most obvious way of thinking of an infinite grid of resistors is to start with a finite grid and grow it, and take the limit of the answer to the question as the grid size grows to infinity. So we start with an n by n grid, for say n=3 so we can fit the two points inside it, and then take the limit of that grid as n->infinity.

But suppose we instead take a n by n torus, as it were, by joining the left and right sides of the grid to each other, and the top and bottom sides to each other. And then take the limit of THAT as n->infinity.

The question is, do both of those approaches converge on the same solution to the problem as n gets sufficiently large. My intuition says that they do, because the conditions at very large distances from the points at which voltage is being applied are effectively the same anyway, so it doesn't matter whether they're directly joined up or not. But that formulation does actually lead to a question that people who could figure out the fourier transforms and stuff (ie not me) could give a definite answer to.

Have I successfully sniped anyone yet?
sab39

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Re: "Nerd Sniping" Discussion

jwwells wrote:[...]
The biologist's solution: "Get... a... freaking... multimeter!"

Signed,
A biologist

Obviously, no beer for you then :]
PAStheLoD

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Re: "Nerd Sniping" Discussion

sab39 wrote:Ok, here's a simple way of phrasing the question which in fact does make sense, I think.

The most obvious way of thinking of an infinite grid of resistors is to start with a finite grid and grow it, and take the limit of the answer to the question as the grid size grows to infinity. So we start with an n by n grid, for say n=3 so we can fit the two points inside it, and then take the limit of that grid as n->infinity.

I wouldn't consider that the obvious way at all. I would consider the obvious way to simply start with an infinite set with the stated geometry. Of course some calculations would be identical either way, such as adding things up over the set, but there are many properties of infinite sets (and graphs, etc.) that don't arise from taking limits. How can you show that the number of nodes on your plane remains the same if you delete every second node (and reconnect the remaining ones somehow), for instance, using limits? In the limit, the result is half as large, always.

I don't know if any of these properties affect the problem at all, however. Probably not, since it only involves infinite arithmetic. I still wouldn't say that conceiving of the infinite plane as the limit of finite squares is the most obvious way, at least to me, and it's certainly not correct formally by conventional definitions.
sab39 wrote:But suppose we instead take a n by n torus, as it were, by joining the left and right sides of the grid to each other, and the top and bottom sides to each other. And then take the limit of THAT as n->infinity.

Well, okay, how do you define a limit of a sequence of sets? Or, say, graphs, which is of more immediate relevance? Consider a sequence of graphs (G1, G2, . . .), with Gi = (Vi, Ei). Let a graph L to be a limit of the sequence if and only if
1. for any two nodes n, mV(L) so that {n, m} ∈ E(L), there exists some N so that if n > N, n, mVn and {n, m} ∈ En, and
2. for any two nodes n, mV(L) so that {n, m} ∉ E(L), there exists some N so that if n > N, n, mVn and {n, m} ∉ En.
with appropriate provision for the one-node graph (which is irrelevant here). This definition is perhaps wrongheaded, but it appears relatively reasonable to me, if not necessarily stated as well as it could be. Under it, an infinite torus is precisely identical to a plane, I believe. Intuitively, this makes sense: as the torus grows arbitrarily large, the fact that it wraps becomes academic, because the distance until you get back to where you started is impractically large.
Simetrical

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Re: "Nerd Sniping" Discussion

Another sniping victim here.
Simetrical wrote:Yes, I don't think anyone spoke of an infinite torus. An infinite cylinder such as I mentioned is perfectly reasonable to consider, however (or at least as reasonable to consider as an infinite plane).

Some quick numerical approximations using a cylinder n resistors around (and 5000 total resistors), with the long axis of the figure (the two-moves direction) parallel to or perpendicular to the axis of the cylinder:

Spoiler:
n = 1: Req = 2 parallel, 1 perpendicular (degenerate, a line of 1-Ohm resistors with self-connections)
n = 2: Req = 1.182 parallel, 0.646 perpendicular (semi-degenerate, a ladder with 1/2-Ohm rungs and 1-Ohm rails)
n = 3: Req = 0.963 parallel, 0.655 perpendicular
n = 5: Req = 0.840 parallel, 0.715 perpendicular
n = 10: Req = 0.789 parallel, 0.758 perpendicular
n = 20: Req = 0.777 parallel, 0.769 perpendicular
n = 50: Req = 0.774 parallel, 0.773 perpendicular
Last edited by smq on Thu Dec 13, 2007 10:53 pm UTC, edited 1 time in total.

smq

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Re: "Nerd Sniping" Discussion

sab39 wrote:But suppose we instead take a n by n torus, as it were, by joining the left and right sides of the grid to each other, and the top and bottom sides to each other. And then take the limit of THAT as n->infinity.

The question is, do both of those approaches converge on the same solution to the problem as n gets sufficiently large.

Numerically, yes:
Spoiler:
n = 3: Req = 0.556
n = 5: Req = 0.680
n = 10: Req = 0.749
n = 20: Req = 0.767
n = 50: Req = 0.772
n = 100: Req = 0.773

Have I successfully sniped anyone yet?

smq

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Re: "Nerd Sniping" Discussion

Yesterday I posted a link to a bunch of references that solve the problem. Unfortunately due to a bug in someone else's code, none of you actually saw the references. So to either make myself look more stupid or less stupid, here they are again

dirtysouthafrican

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Re: "Nerd Sniping" Discussion

Simetrical wrote:How can you show that the number of nodes on your plane remains the same if you delete every second node (and reconnect the remaining ones somehow), for instance, using limits? In the limit, the result is half as large, always.

If limn->inf f(n) exists, then limn->inf f(n/2) exists, and has the same value. So while it's not trivial show the cardinalities match if you remove half the nodes, you can show that removing half the nodes doesn't affect any of the limits that you're using.
While no one overhear you quickly tell me not cow cow.

phlip
Restorer of Worlds

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Re: "Nerd Sniping" Discussion

Okay, I first came to the engineer's answer: Calc for a 2x3 grid, then a 4x5 grid, so on and so forth. I was actually just about to start coding it (TiBasic, oh baby yes) when I then thought:

As the infinite grid gets larger, and larger, it becomes more as a solid, a 2d plane. No definition from resistor to resistor. Thus, could not we measure the resistance across a surface?

Experement #1:

Gather a large sample of copper-clad PCB. Cut a strip of it as wide as the clad is thick. Measure a distance N so that the resistance is 1 Ohm. Take note of this distance.

Now, with the remaining PCB, approximate the center, and displace both multimeter probes by root(5)/2 from it in opposite directions. The measured resistance would be the best approximation for the answer, would it not?

Edit: I just realized, I got sniped!! I came home, saw this problem, and... I've been sitting here thinknig for about 30 minutes straight.
Wizzard1

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Re: "Nerd Sniping" Discussion

[First try with attachments didn't work, trying with jpeg] Folks, I setup a simple circuit to test my understanding of how this should work, and why the infinite resistors don't really matter. I've probably missed something, but I think only fully divergent paths between the nodes count as 'parallel paths' for which the equiv resistance should be calculated from.
circuit.jpg (71.21 KiB) Viewed 6534 times
circuit.jpg (71.21 KiB) Viewed 6534 times
results.jpg (28.29 KiB) Viewed 6516 times
chopps

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Re: "Nerd Sniping" Discussion

the resistance is 4/pi

happy?

look it up
the problem has already been worked
http://www.geocities.com/frooha/grid/node2.html
PhoenixGeimer

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Re: "Nerd Sniping" Discussion

jwwells wrote:The biologist's solution: "Get... a... freaking... multimeter!"

Signed,
A biologist

Silly biologist. Where, exactly, are you going to find an infinite grid of resistors to measure WITH your freaking multimeter?

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Re: "Nerd Sniping" Discussion

So why did the Physicist cross the road?
jasticE

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Re: "Nerd Sniping" Discussion

Simetrical wrote:Regardless, of course you can map any set bijectively onto any set of the same cardinality (and of course the plane and the sphere both have the cardinality of the continuum).

Actually, you can't bijectively map a sphere into plane. That's the topology for you. You can map a sphere-minus-two-points or a torus, though.

I don't know the correct terms in English, so I'll try to translate from Russian: the atlas of the sphere contains at least two maps. It's intuitively easy to understand if you try to bijectively map a circle to a segment.
Cyberax

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Re: "Nerd Sniping" Discussion

jwwells wrote:The biologist's solution: "Get... a... freaking... multimeter!"

Silly biologist. Where, exactly, are you going to find an infinite grid of resistors to measure WITH your freaking multimeter?

In the equipment cabinet, next to where they keep the infinitely long test leads for the multimeter, of course.

-----

PhoenixGeimer wrote:http://www.geocities.com/frooha/grid/node2.html

Besides, quite a few people have posted links to various things that claim to be the right answer, but that doesn't mean they *are* the right answer. Just because it's on a web page somewhere doesn't mean it's true.
Ben'); DROP TABLE Users;--

GENERATION 42: The first time you see this, copy it into yοur sig on any forum and stick a fork in yοur еyе. Social experiment.

DragonHawk

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Re: "Nerd Sniping" Discussion

Here is a solution to a similar problem I had posted. I will work out the same problem as on the sign though and see what's different.

http://interblag.blogspot.com/2007/01/challenge-problem.html

http://interblag.blogspot.com/2007/01/challenge-problem-solution.html
itln

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Re: "Nerd Sniping" Discussion

I'm still struggling with the length of the circuit.
Assuming an infinite circuit path (and an infinite power source), wouldn't the net resistance of the grid approach zero over time?
The speed of light is a constant which would force the solution to be only an instantaneous measurement when power was first applied...
The "right" answer can never be found as long as e limits the time it takes for an infinite circuit to energize.

snurfle

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Re: "Nerd Sniping" Discussion

It's like me and a rubik's cube. Someone scrambles my rubik's cube and I stop everything I'm doing to set it right. The bad thing is I'm not really good at them so it's usually a 2 minute waste of my time, 10 minutes if it's 4x4x4.
Guys guys guys! I found Russel's teapot! . . . nevermind, it was just Jesus flying to Mars again.

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Re: "Nerd Sniping" Discussion

I half expected to get hit by a truck after solving the 5x5x5.

Asad: If you don't expect the biologist to ever find an infinite grid to test this on, then why does the problem even matter?
mikebolt

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Re: "Nerd Sniping" Discussion

mikebolt wrote:If you don't expect the biologist to ever find an infinite grid to test this on, then why does the problem even matter?

You sound like http://xkcd.com/230/
sab39

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Re: "Nerd Sniping" Discussion

jwwells wrote:The biologist's solution: "Get... a... freaking... multimeter!"

Signed,
A biologist

Multimeter is easy. The infinite grid of resistors might cost a bit more.

Edit: Dagnabbit, about 47 people posted this before me. Howbout this: An infinite grid of ideal resistors costs even more than an infinite grid of your normal everyday 5% tolerance resistors.
Some people look at what is, and ask "why?"
I think about it for a while, then say "Oh, that's why not."
kwan3217

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Re: "Nerd Sniping" Discussion

kwan3217 wrote:Howbout this: An infinite grid of ideal resistors costs even more than an infinite grid of your normal everyday 5% tolerance resistors.

Due to the nature of infinity, they bost cost the same:P
mikebolt

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Re: "Nerd Sniping" Discussion

As a pure mathematician, I am interested in solving this rigorously.

As a pure mathematician senior in high school with not much experience in fourier series, I am interested in solving this rigorously in a different manner.

This makes me a masochist (I think) and also throughly sniped, but...

So far I think the most promising stab at a non-fourier-series solution was the method involving keeping a 1 and -1 voltage anchored in the plane and repeatedly changing each point into the average of its neighbors.

This can be solved using combinatorics. I offer no proof here, but consider a function:

f(n) is the number of ways you can get to the point (1,0) starting at (0,0) in exactly n moves (where a move constitutes going from one vertex to an adjacent vertex), such that the path never returns to (0,0) or goes through (2,1).
Similarly, q(n) is the number of ways you can get to the point (1,0) starting at (2,1) in exactly n moves without returning to (2,1) or going through (0,0).

Then the voltage at (1,0) is...

sum(n=0, infinity) (f(n)-q(n))/4^n

Then we use a similar method to find voltages at (0,1), (-1,0), (0,-1) and then subtract 1 from their sum for victory!

Note: This does not seem like a very good method of solution, unless it simplifies very nicely or helpful approximations can be made.
Note #2: The computation is much easier if you have 0 voltage at (2,1), then you don't need to have q(n) but you still need to avoid (2,1) when counting paths for f(n).
Hey baby, I'm proving love at nth sight by induction and you're my base case.

imatrendytotebag

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Re: "Nerd Sniping" Discussion

He should have had the man in the Beret driving the truck... That would have made this comic the funniest one ever.
~AWOL~ Elendil

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Re: "Nerd Sniping" Discussion

Aradae wrote:It's like me and a rubik's cube. Someone scrambles my rubik's cube and I stop everything I'm doing to set it right. The bad thing is I'm not really good at them so it's usually a 2 minute waste of my time, 10 minutes if it's 4x4x4.

Yes! I had to hide my cubes in a drawer in my dorm room. People would walk in, mess one of them up, then I wouldn't be able to do anything else until it was right again.
doomofdoom

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Re: "Nerd Sniping" Discussion

imatrendytotebag wrote:As a pure mathematician, I am interested in solving this rigorously.

As a pure mathematician senior in high school with not much experience in fourier series, I am interested in solving this rigorously in a different manner.

This makes me a masochist (I think) and also throughly sniped, but...

So far I think the most promising stab at a non-fourier-series solution was the method involving keeping a 1 and -1 voltage anchored in the plane and repeatedly changing each point into the average of its neighbors.

This can be solved using combinatorics. I offer no proof here, but consider a function:

f(n) is the number of ways you can get to the point (1,0) starting at (0,0) in exactly n moves (where a move constitutes going from one vertex to an adjacent vertex), such that the path never returns to (0,0) or goes through (2,1).
Similarly, q(n) is the number of ways you can get to the point (1,0) starting at (2,1) in exactly n moves without returning to (2,1) or going through (0,0).

Then the voltage at (1,0) is...

sum(n=0, infinity) (f(n)-q(n))/4^n

Then we use a similar method to find voltages at (0,1), (-1,0), (0,-1) and then subtract 1 from their sum for victory!

Note: This does not seem like a very good method of solution, unless it simplifies very nicely or helpful approximations can be made.
Note #2: The computation is much easier if you have 0 voltage at (2,1), then you don't need to have q(n) but you still need to avoid (2,1) when counting paths for f(n).

I tried thinking about this in terms of paths, but it turned out to be nontrivial for a simple problem.

Consider comparing a series-parallel circuit (one wire splits into two, each has a resistor, then they combine together and pass through one more resistor) to a parallel circuit (one wire splits into two, each has two resistors before coming back together). By a purely path summing formula, both circuits have the same number of paths, and the same number of resistors on each path. But the equivalent resistance is different.

In order to get the path sum approach working, you'd have to take into account how many paths share the same resistor as well. I'm not sure how it would work in general.
Chalnoth

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Re: "Nerd Sniping" Discussion

The classhole is canonically listed as "The Hat Guy", and is based on Aram from Men in Hats, as shown by the alt text of http://xkcd.com/29/

pyrosim

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Re: "Nerd Sniping" Discussion

This almost happened to a friend of mine once. We were walking out of our Automata Theory midterm exam, talking about a particular problem (I don't recall what the problem was), and I was drawing a diagram in my notebook as we were walking. We were about to cross the street, and there was a car coming; the friend was paying too much attention to the notebook, and didn't stop at the, not noticing that there was a car about to hit her. I quickly called out "watch out, car!", to which she backed up and onto the curb. The car was within three feet of her while she was still in the street.

We then crossed the street and continued talking about the problem.
Cynical

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Re: "Nerd Sniping" Discussion

GRRR all these people who don't read before posting....

Here's the link to the ANALYTICAL SOLUTION of the problem worked out:

http://www.geocities.com/frooha/grid/node2.html

If you want to skip the derivation and jump to solving the problem scroll down to equation 37. M=1, N=2. Use the simplification immediately prior to equation 39 and integrate to get the answer:

Spoiler:
The answer is 4/pi-.5 Ohms == .773 Ohms.

It is not infinity nor zero. Neither of those makes sense since there's always a path that is a finite resistance and never a path with no resistance. As a physicist the spread of misinformation saddens me.
floyd4one

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Re: "Nerd Sniping" Discussion

floyd4one wrote:GRRR all these people who don't read before posting....

The fact that the actual solution has already been posted in this thread several times, including that exact link once, makes this comment all the funnier.
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:
floyd4one wrote:GRRR all these people who don't read before posting....

The fact that the actual solution has already been posted in this thread several times, including that exact link once, makes this comment all the funnier.

if you would like to go back to page one, you will see that indeed both were posted there, the solution by me, the link by someone else. i was just reiterating the fact, since despite the answer being on page 1 of the posts, there are 5 pages following of incorrect reasoning, answers and references.

hence, there's no irony in me saying "people who don't read before they post." thanks for playing though.
floyd4one

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Re: "Nerd Sniping" Discussion

Maybe, just maybe, people wanted to figure it out on their own? Because they're nerds?
Wait. With a SPOON?!

Geekthras
3) What if it's delicious?

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Re: "Nerd Sniping" Discussion

itln wrote:Here is a solution to a similar problem I had posted. I will work out the same problem as on the sign though and see what's different.

http://interblag.blogspot.com/2007/01/challenge-problem.html

http://interblag.blogspot.com/2007/01/challenge-problem-solution.html

So I posed a solution I think is correct, let me know if anyone finds mistakes.
http://interblag.blogspot.com/
itln

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