## 0356: "Nerd Sniping"

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### Re: "Nerd Sniping" Discussion

Showed this problem to my physicist dad while we were at a restaurant. Started working it out on a napkin...lol. Didn't finish it tho, food arrived.

Tachyon

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### Re: "Nerd Sniping" Discussion

wirehead wrote:this caused us great amusement at work.

I seem to be at least partially immune to sniping. At least for this particular problem.

Somewhat OT (actually very OT). Librarians the "Cutter Code" to handle the initial problem mentioned in this link. I've never check out the details of its derivation, but I assume it to be empirical. This post can be considered a response to the early complaints about lack of material for Literature Nerds.

BeetlesBane

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### Re: "Nerd Sniping" Discussion

bjswift wrote:I had this on my 2nd semester physics class midterm. There's a pretty simple, elegant solution to it but I don't remember what it is. Prolly somewhere on the interwebs.

Nice post, Fermat.

JCCyC

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### Re: "Nerd Sniping" Discussion

What else can we use to snipe nerds?
Defenestrator

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### Re: "Nerd Sniping" Discussion

PvNP should do the trick.
"Absolute precision buys the freedom to dream meaningfully." - Donal O' Shea: The Poincaré Conjecture.
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Frimble

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### Re: "Nerd Sniping" Discussion

So i felt it was worth reviving this thread to tell this story. I was visiting this girl who i was considering dating (it didn't end up working out but thats a different story) She wasn't exactly the nerdy type as she was a music major. However her dad was definitly the cliche` nerdy engineer. Enter the Issue. I was hanging out with her watching some movie and her dad comes over (just after the movie ended) and we started talking about various nerd stuff just being general losers. While i was enjoying such it wasn't exactly why i had come there so i tried to get away and go back to Lauren. But he just kept nerd sniping me. In my head i was thinking this is hilarious, he keeps nerd sniping me and probably has no idea that he is doing so. Anyways after a while as in just about any nerd discussion these days i brought up xkcd and said that i think he would really like it. He tells me to hold on for one second and comes back less than 5 seconds latter with a print out of the nerd sniping comic. It was at this point that i realized he knew exactly what he was doing. All i could do was give him a "well played sir!"
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pyroman

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### Re: "Nerd Sniping" Discussion

pyroman wrote:wasn't exactly the nerdy type ... music major.

You must be joking...

Also, this and this.
pegasos989

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### Re: "Nerd Sniping" Discussion

pegasos989 wrote:
pyroman wrote:wasn't exactly the nerdy type ... music major.

You must be joking...

Also, this and this.

The first one is one of my favorite videos on youtube, the musician in me loves it. And to the second... Well played.

nyeguy

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### Re: "Nerd Sniping" Discussion

alright let me rephrase that... she wasn't a TECH nerd

P.S. you suck... dam rick rolls! = P
They who can give up essential liberty to obtain a little temporary safety, deserve neither liberty nor safety. - Benjamin Franklin

pyroman

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### Re: "Nerd Sniping" Discussion

I actually have posters I use to design similar problems to this one. depending on the type of nerd, I can snipe programmer nerds, computer science nerds, electricity & engineering nerds, and my favourite, xkcd nerds.

no I won't give you examples, I might need to use them on you someday.

also, if you have the Firefox add-on that let's you preview a page to see id it is a rickroll, and you see the first bit of it, does that count as being rickrolled?
If all the worlds my stage let's go to intermission
jakerman999

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### Re: "Nerd Sniping" Discussion

russianspy1234

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### Re: "Nerd Sniping" Discussion

I have read the previous posts....
This question has a true answer right?

I saw the one on page one, one a couple couple more pages in... you get the picture.
This has be stuck in my head.
The thing I keep coming back to though is the path of least resistance
in which case the answer for this is 3.
three minimum points to make contact.

I have two more that I have worked out.

#2. It takes 3 runs through all resistors to make it too the second point.
in which case there is an infinate number of resistors in a series and an infinite number in parallel.
RON = Resistance of Network
RON = infinite resistance from series + 1/infinite from parallel
RON = (infinite + 1)/(1 + infinite)
RON = 1
3 uses means 3*RON
Resistance once again = 3 ohms

#3. Calculate for one square of resistors
Resistor Square = 4 in series + 2 sets in parallel.
RS = (1 + 1 + 1 + 1) + ((1/2)*(1/2))
RS = 4 + 1/4
RS = 1
Whole Resistance = infinite * 1 + infinite/1
WR = 1
3 jumps from point to point = 3*WR
Resistance = 3 Ohm

Perhaps all of these are horribly flawed. I have many more. But I just found suspicious(for lack of a better word) it that in the original form that I had worked them out in I could not find an error.
I do understand the other answers, but those solve for the entire grid, not the connected points.

Doozo yoroshiku....

Thank you.... I do really need an answer.

Sources:
A.P. Physics.... Not the best, I know. Maybe more in college.
F34R1355

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### Re: "Nerd Sniping" Discussion

F34R1355 wrote:RON = (infinite + 1)/(1 + infinite)
RON = 1

F34R1355 wrote:Whole Resistance = infinite * 1 + infinite/1
WR = 1

I don't think the word "infinite" means what you think that it means. \frac \infty \infty on its own is meaningless... in context, it could have any value, depending on where the infinities come from. For instance, \lim_{n \to \infty} \frac xx = 1 but \lim_{n \to \infty} \frac {2x}x = 2, even though both of these could be seen as \frac \infty \infty. Look up "indeterminate forms" sometime.

Another thing to consider: the "path of least resistance" isn't enough information... two paths in parallel have a lower resistance than either of those paths have on their own. The path of least resistance is only an upper bound on the resistance of the circuit.
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phlip
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### Re: "Nerd Sniping" Discussion

I understand that infinite has no set parameters. But if we are going to bring in the issue of infinite in its meaning then the shape of the grid is unknown as well. I do understand that infinite on its own over itself has no meaning, but at the same time we are dealing with two equal infinite numbers(contradiction, no need two point it out.). If the number of resistors in the grid = N then N=infinite. If you are looking from this at a programing point of view, we have set N. I am kind of in a rush at the moment, I hope this makes a little sense...

phlip wrote:Look up "indeterminate forms" sometime.

F34R1355

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### Re: "Nerd Sniping" Discussion

phlip wrote:Another thing to consider: the "path of least resistance" isn't enough information... two paths in parallel have a lower resistance than either of those paths have on their own. The path of least resistance is only an upper bound on the resistance of the circuit.

I had taken this in to consideration, that is seen where I solve for one square of resistors.
F34R1355 wrote:RS = (1 + 1 + 1 + 1) + ((1/2)*(1/2))

phlip wrote:indeterminate forms

Thank you for this suggestion. I understand what you said about the two infinite numbers not being divisible. But even though they have no set value, in this case they are the same value (unless there is some evil man out there that is placing resistors in while the power is running through them in series, and then removing them before it can run through them in parallel... Impossible, I know.).... But either way I have come up with ways that destroy all three methods I used.

Simetrical wrote:The correct answer was already posted on the first page, twice, and multiple times since then. 4/π − 0.5.

I have seen this a few times, and I have been trying to find on my own the origin of pi. The 4 is for all the resistors in a series the .5 comes from the parallel, but the pi comes from I don't know where.
F34R1355

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### Re: "Nerd Sniping" Discussion

The parts I pointed out in that post were pointed out mostly because they were the only parts of your post that I could actually follow enough to say that they were wrong. Everything else you said I couldn't follow at all.

Except this one that I missed:
F34R1355 wrote:Resistor Square = 4 in series + 2 sets in parallel.
RS = (1 + 1 + 1 + 1) + ((1/2)*(1/2))
RS = 4 + 1/4
RS = 1

I don't see how that first line follows from anything, I don't see how that second line follows from the first, and the fourth line definitely doesn't follow from the third.

In short: try making more sense.
While no one overhear you quickly tell me not cow cow.

phlip
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### Re: "Nerd Sniping" Discussion

F34R1355 wrote:Resistor Square = 4 in series + 2 sets in parallel.
RS = (1 + 1 + 1 + 1) + ((1/2)*(1/2))
RS = 4 + 1/4
RS = 1

Horrible math, ain't it.
I do realize that it makes no sense. Partly due to me copying from what I had originally written it on, partly due to not having near enough sleep.
When I used it above I was making two points. Point one was that the math did in fact have mistakes. The second point was that I did realize how a parallel circuit would have less resistance. It was included as the ((1/2)(1/2)) line. Each parallel in the square of resistors would contain two one ohm resistors. Making each of the two parallels have 1/2 ohm resistance. (1/(r1+r2)) (1/(r3/r3). Every equation I posted contained fraction addition (among other) errors, I do realize this and apologize for lowering your IQ a few notches.
To tell you the truth, I have no idea as to what I was trying to say there either. This question bugs me. Perhaps when I find my notebook again I will re-post the unsimplified/mistake free ones that I have.
I do however think that I understand what I was trying to get at with the 4 resistors in a square(which you would solve for once.)
Resistor Square = RS
RS = r1+r2+r3+r4 + (1/(r1+r3)) + (1/(r2+r4))
RS = 1ohm + 1ohm +1ohm +1ohm + (1/(1ohm +1ohm)) + (1/(1ohm + 1ohm))
RS = 4ohm + (1/2ohm) + (1/2ohm)
RS = 5ohm
That may have been what I was trying to get at with that. However that cannot be used to solve for the grid. It can solver for the center grid of the square. From there you would have a three squares in which r1=5ohm r2,r3,r4=1ohm (each)... and on and on it goes.

Is this clearer?
F34R1355

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### Re: "Nerd Sniping" Discussion

Perhaps you are misunderstanding the definition of "series" and "parallel"?

Two resistors are in series iff one end of one is connected to one end of the other, and nothing else is connected to the same point. No two resistors in the infinite grid are in series, as any connection between two resistors has two other resistors also connected to the same node.
Two resistors are in parallel iff both ends of one resistor are directly connected to the respective ends of the other resistor. No two resistors in the infinite grid are parallel, as any two resistors that are directly connected at one end, are not directly connected at the other end.

The only way I can make your equation even almost make sense, is if you're looking at a square of resistors, and seeing it as 4 resistors in series, and two pairs of resistors in parallel... which is wrong in several ways. Not least of which is that even if it was, it wouldn't be useful for looking at the infinite grid, since it has extra resistors hanging of every node of the square - if you want to try to look at a subcircuit, and simplify it with series/parallel/etc stuff, then it has to be isolated except on the nodes you're finding the effective resistance across... it can't have extra resistors hanging off everywhere...
While no one overhear you quickly tell me not cow cow.

phlip
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### Re: "Nerd Sniping" Discussion

I think that you jumped to conclusions before reading the post.

F34R1355 wrote:I do however think that I understand what I was trying to get at with the 4 resistors in a square(which you would solve for once.)
...That may have been what I was trying to get at with that. However that cannot be used to solve for the grid...

I did say that this is unusable... This is what the way that I was thinking of it. That could have been clearer. My apologies.

F34R1355 wrote: Every equation I posted contained fraction addition (among other) errors...

I pointed out that the most visible error was the fraction addition. The among others meant that there where a few logic error, and wrong equations. Again could have been clearer, my apologies.

Definition of Series Resistance: Set of resistors in which the same level of current flow through.
Definition of Parallel Resistance: Set of resistor in which the current of the system is divided.
Are those wrong?

I have also acknowledged 4/π − 0.5 as being the right answer. I do not however understand where the pi came from.
I did not come here to argue in favor of my first post, I came here rather to publicly acknowledge that my first post contained many mistakes, and should never be used as a reference for anyone attempting to work with circuits. If I have some how offended you, I apologize for it.
F34R1355

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### Re: "Nerd Sniping" Discussion

The resistance would be 0. Resistance is 1/conductance and an infinate grid of resistors would have infinate conductance in any direction. 1/infinity = 0
Agamemnom

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### Re: "Nerd Sniping" Discussion

# I love this strip: it's in (or on? or at? Damn it!) my top five. \o\

# I prefer snipe not nerd phisicians and mathematicians, but literary critics. We just have to use the sign "Say your opinion about *put a name of a book with difficult language and philosophic questions in the plot*".

# Literary critics score 2, Arts professors score 3. \o/
(...)
(...)

StrixVanAllen

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### Re: "Nerd Sniping" Discussion

I have no idea how, I am just very gifted at physics, but I saw it and instantly knew it was zero (an infinite amount of parallel paths give the 1/total R to be infinity, so total R is zero).

Try a better one if you ever want to get me, Randall!
'; DROP DATABASE;-- wrote:The problem with imagination is it does a lousy job of interacting with the physical world. And you look crazy when you talk to it.
Whyareall

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### Re: "Nerd Sniping" Discussion

... except that the total resistance isn't zero, and indeed it's simple to prove that it must be >1/2.

Note that while there's an infinite number of paths between the two points, most of them are really long, and thus have really high resistance. The total tesistance is roughly (not exactly, as the paths overlap) equal to the (harmonic) average resistance of the paths, divided by the number of paths... that's \infty \over \infty, which doesn't necessarily equal 0.
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phlip
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### Re: "Nerd Sniping" Discussion

phlip wrote:... except that the total resistance isn't zero, and indeed it's simple to prove that it must be >1/2.

Note that while there's an infinite number of paths between the two points, most of them are really long, and thus have really high resistance. The total tesistance is roughly (not exactly, as the paths overlap) equal to the (harmonic) average resistance of the paths, divided by the number of paths... that's \infty \over \infty, which doesn't necessarily equal 0.

Well, just to annoy everyone here with an invalid proof, I will now prove that infinity over infinity IS zero. But I don't know how to use the maths formatting, so I'll just put it in words.

Infinity squared is still infinity, therefore infinity squared take infinity is zero. So infinity outside of infinity take one is zero. By the null factor law, infinity equals zero and infinity take one equals zero. Therefore infinity is one or zero. By substitution into infinity over infinity gives 0 over 1, which anyone will tell you DOES equal zero.
'; DROP DATABASE;-- wrote:The problem with imagination is it does a lousy job of interacting with the physical world. And you look crazy when you talk to it.
Whyareall

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### Re: "Nerd Sniping" Discussion

The fault in that proof is that you're assuming that "an infinite number" = "another infinite number" (or, more specifically, "an infinite number" - "a different infinite number" = 0), which it doesn't, in general.

In general, you can't treat infinity as a number. There are some special cases where it does what you want, but this isn't one of them.

Take, say, \lim_{x \to \infty} \frac{2x}{x}
Both the top and bottom of the limit go to infinity. However, the limit is clearly 2, not 0 (intuitively: the numerator goes to infinity twice as fast as the denominator, so even at infinity, their ratio is 2. Note however that intuition is often mistaken when it comes to infinities, and it's always best to back it up with rigorous maths whenever possible).

Perhaps more convincingly... say that \infty - \infty = 0 was true, as you claim that it is. We also have \infty + 1 = \infty, pretty much by definition of \infty. Now we have \infty + 1 - \infty = 0, \infty - \infty + 1 = 0, 0 + 1 = 0, which is clearly false.
So one of our assumptions must be false... here, our assumptions are: \infty - \infty = 0, \infty + 1 = \infty, (a + b) + c = a + (b + c), a + b = b + a, and 0 ≠ 1. The only one of those we can really declare to be false is the first one... the others are too fundamental.
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phlip
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### Re: "Nerd Sniping" Discussion

phlip wrote:The fault in that proof is that you're assuming that "an infinite number" = "another infinite number" (or, more specifically, "an infinite number" - "a different infinite number" = 0), which it doesn't, in general.

In general, you can't treat infinity as a number. There are some special cases where it does what you want, but this isn't one of them.

Take, say, \lim_{x \to \infty} \frac{2x}{x}
Both the top and bottom of the limit go to infinity. However, the limit is clearly 2, not 0 (intuitively: the numerator goes to infinity twice as fast as the denominator, so even at infinity, their ratio is 2. Note however that intuition is often mistaken when it comes to infinities, and it's always best to back it up with rigorous maths whenever possible).

Perhaps more convincingly... say that \infty - \infty = 0 was true, as you claim that it is. We also have \infty + 1 = \infty, pretty much by definition of \infty. Now we have \infty + 1 - \infty = 0, \infty - \infty + 1 = 0, 0 + 1 = 0, which is clearly false.
So one of our assumptions must be false... here, our assumptions are: \infty - \infty = 0, \infty + 1 = \infty, (a + b) + c = a + (b + c), a + b = b + a, and 0 ≠ 1. The only one of those we can really declare to be false is the first one... the others are too fundamental.

Hence the INVALID proof.
'; DROP DATABASE;-- wrote:The problem with imagination is it does a lousy job of interacting with the physical world. And you look crazy when you talk to it.
Whyareall

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### Re: "Nerd Sniping" Discussion

Whyareall wrote:
phlip wrote:The fault in that proof is that you're assuming that "an infinite number" = "another infinite number" (or, more specifically, "an infinite number" - "a different infinite number" = 0), which it doesn't, in general.

In general, you can't treat infinity as a number. There are some special cases where it does what you want, but this isn't one of them.

Take, say, \lim_{x \to \infty} \frac{2x}{x}
Both the top and bottom of the limit go to infinity. However, the limit is clearly 2, not 0 (intuitively: the numerator goes to infinity twice as fast as the denominator, so even at infinity, their ratio is 2. Note however that intuition is often mistaken when it comes to infinities, and it's always best to back it up with rigorous maths whenever possible).

Perhaps more convincingly... say that \infty - \infty = 0 was true, as you claim that it is. We also have \infty + 1 = \infty, pretty much by definition of \infty. Now we have \infty + 1 - \infty = 0, \infty - \infty + 1 = 0, 0 + 1 = 0, which is clearly false.
So one of our assumptions must be false... here, our assumptions are: \infty - \infty = 0, \infty + 1 = \infty, (a + b) + c = a + (b + c), a + b = b + a, and 0 ≠ 1. The only one of those we can really declare to be false is the first one... the others are too fundamental.

Hence the INVALID proof.

Yeah, but given your aim was to "annoy", I think it failed. I think that most mathematically minded people look at those kind of things and have a little chuckle at the person who's misusing infinity. Still, it's not a bad thing to look at, since infinity / infinity is one of the indeterminate forms that can sometimes be tackled by l'Hopital's Theorem, which is basically a way of finding limits of functions that have two terms that *might* cancel each other out - e.g. infinity/infinity, infinity * 0, and infinity - infinity. However, those are just notations to identify the type of indeterminate form (which can all be transformed into each other anyway), and the actual formulation involves f(x)/g(x) where both f and g grow without bound in the appropriate limit.
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### Re: "Nerd Sniping" Discussion

Whyareall wrote:I have no idea how, I am just very gifted at physics, but I saw it and instantly knew it was zero (an infinite amount of parallel paths give the 1/total R to be infinity, so total R is zero).

Try a better one if you ever want to get me, Randall!

Apparently not so gifted. Anyone who has at least an undergraduate degree in physics would realize that an infinite grid of resistors is periodic, and can be simplified using Fourier Transforms, which leads quite easily to the 4/pi-.5 answer. (I'm just a freshman and I could get this answer, so it can't be too hard).
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qinwamascot

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### Re: "Nerd Sniping" Discussion

qinwamascot wrote:
Whyareall wrote:I have no idea how, I am just very gifted at physics, but I saw it and instantly knew it was zero (an infinite amount of parallel paths give the 1/total R to be infinity, so total R is zero).

Try a better one if you ever want to get me, Randall!

Apparently not so gifted. Anyone who has at least an undergraduate degree in physics would realize that an infinite grid of resistors is periodic, and can be simplified using Fourier Transforms, which leads quite easily to the 4/pi-.5 answer. (I'm just a freshman and I could get this answer, so it can't be too hard).

I have no such undergraduate degree in physics. I'm in year 11 (dont know how your American school system works, but I'm in my 2nd last year of high school), and have already taught my physics and chemistry teachers multiple things (such as light going from medium A to B to C, the refracted angle in C would be the same as if the light just went from A to C at the same incident angle as A to B from before). We have barely done any work in electricity; and I have NEVER studied for a physics test and I'm coming 2nd in the year at it.

If that isn't gifted, I don't know what is.
'; DROP DATABASE;-- wrote:The problem with imagination is it does a lousy job of interacting with the physical world. And you look crazy when you talk to it.
Whyareall

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### Re: "Nerd Sniping" Discussion

I had someone do this to me. They gave me a 42-bit binary integer and told me to convert it to decimal.

When I realised I'd made a mistake at bit 34, I gave up
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pcmattman

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### Re: "Nerd Sniping" Discussion

Whyareall wrote:
qinwamascot wrote:
Whyareall wrote:I have no idea how, I am just very gifted at physics, but I saw it and instantly knew it was zero (an infinite amount of parallel paths give the 1/total R to be infinity, so total R is zero).

Try a better one if you ever want to get me, Randall!

Apparently not so gifted. Anyone who has at least an undergraduate degree in physics would realize that an infinite grid of resistors is periodic, and can be simplified using Fourier Transforms, which leads quite easily to the 4/pi-.5 answer. (I'm just a freshman and I could get this answer, so it can't be too hard).

I have no such undergraduate degree in physics. I'm in year 11 (dont know how your American school system works, but I'm in my 2nd last year of high school), and have already taught my physics and chemistry teachers multiple things (such as light going from medium A to B to C, the refracted angle in C would be the same as if the light just went from A to C at the same incident angle as A to B from before). We have barely done any work in electricity; and I have NEVER studied for a physics test and I'm coming 2nd in the year at it.

If that isn't gifted, I don't know what is.

Did you sum the numbers from 1 to 100 inclusive in a matter of seconds in 1st grade?
MSTK

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### Re: "Nerd Sniping" Discussion

MSTK wrote:
Whyareall wrote:
qinwamascot wrote:
Whyareall wrote:I have no idea how, I am just very gifted at physics, but I saw it and instantly knew it was zero (an infinite amount of parallel paths give the 1/total R to be infinity, so total R is zero).

Try a better one if you ever want to get me, Randall!

Apparently not so gifted. Anyone who has at least an undergraduate degree in physics would realize that an infinite grid of resistors is periodic, and can be simplified using Fourier Transforms, which leads quite easily to the 4/pi-.5 answer. (I'm just a freshman and I could get this answer, so it can't be too hard).

I have no such undergraduate degree in physics. I'm in year 11 (dont know how your American school system works, but I'm in my 2nd last year of high school), and have already taught my physics and chemistry teachers multiple things (such as light going from medium A to B to C, the refracted angle in C would be the same as if the light just went from A to C at the same incident angle as A to B from before). We have barely done any work in electricity; and I have NEVER studied for a physics test and I'm coming 2nd in the year at it.

If that isn't gifted, I don't know what is.

Did you sum the numbers from 1 to 100 inclusive in a matter of seconds in 1st grade?

guys...seriously?
it's so pathetic when people get into an argument over who's the smartest.

P.S. knowing how to do Gaussian summations in first grade doesn't make you the smartest person in the world. Teaching your teachers (who, fyi, are people who make mistakes) something doesn't, either. And if one of you (or both) really is (are) gifted, then do you have to convince the rest of the world? Isn't being gifted independent of anyone else's beliefs? And does one of you being gifted make the other one not gifted? And don't gifted people make mistakes (since they, like teachers, are also people)?!???!??!
luolimao

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### Re: "Nerd Sniping" Discussion

Okay getting back to the infinite grid problem, being slightly more practically minded about things & having a hypothesis I tried my theory and experimentation seems to support it.

The equivalent resistance tends to 0 as grid size tends to infinity
The idea behind it was simple: when you connect resistors in parallel the equivalent resistance is lower than the smallest resistance
In this grid you have an infinite amount of parallel paths to get between each point hence 0 total resistance

Now to test this i took a bread board and using 1k resistors i built grids of progressively larger sizes
each time I increase the size of the grid the total resistance drops, which would bear out my hypothesis
dabraude

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### Re: "Nerd Sniping" Discussion

dabraude wrote:each time I increase the size of the grid the total resistance drops, which would bear out my hypothesis

Sure, it drops, but does it drop all the way to zero? Or does it approach another limit?

The sequence {17,9,5,3,2,1.5,1.25,1.125,...} gets smaller each time, but its limit is definitely not 0.

And as has been said in the thread already, the resistance must be at least half an ohm (or 500 ohms in your version with 1k resistors), by a quite simple argument.

It's not like the right answer is hard to find in this thread... it's been mentioned multiple times on every page...
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phlip
Restorer of Worlds

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### Re: "Nerd Sniping" Discussion

I know im not supposed to post links unitl i get a certain post count, but it is hackaday, and it IS awesome, its also on topic.

youtube video of someone solving the infinite resistor grid by building a 14 x 14 grid of resistors as well as explaining solutions etc.

quietus

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### Re: "Nerd Sniping" Discussion

Khalid

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### Re: "Nerd Sniping" Discussion

Spoiler:
3 Ohms
Nintenman

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### Re: "Nerd Sniping" Discussion

Spoiler:
3 Ohms

my heart just stopped. a conbination of trolling and nerdsniping, like saying pie is exactly 3...
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merseyless

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### Re: 0356: "Nerd Sniping"

I have two dumb questions.

First, would it be possible to simulate this by simulating the first circle of resistors, then the next ring, and the next ring (like the last five photos in the Postby iNap » Wed Dec 19, 2007 11:36 pm UTC:
http://forums.xkcd.com/viewtopic.php?f=7&t=16104&start=240&st=0&sk=t&sd=a

Then calculate the rate of decrease per added ring (on a logarithmic scale or something?) and have software extrapolate that out to infinity?

Second, can someone link to a solution? I'm not interested in the math because I am very unlikely to understand it, I just want to know the end result.
webgrunt

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### Re: 0356: "Nerd Sniping"

The limit of a finite grid is not necessarily going to be the value for an infinite grid. In this case I don't know whether it is or not.