Talk, Nosh and Gridlock

I paid a flying visit yesterday to the beautiful city of Edinburgh in order to give a seminar at the Institute for Astronomy, which is situated with the historic Royal Observatory. I was there not long ago, in fact, to do a PhD examination but on this occasion all I had to was stand up in a lecture room and rabbit on for an hour or so. That part of it seemed to go reasonably well, in that no more than half the audience fell asleep while I wittered away.

The morning flight from Cardiff to Edinburgh was uneventful and got me there in time to chat with various people and have lunch before the talk. I elected not to rush straight from the seminar to the airport in order to return the same day, but stayed overnight giving some of  the locals the dubious pleasure of paying for my dinner and enduring my company during it, which they did with great patience. I’d like to thank Alan, John, Alina, Stefano and Brendan for rounding off such a nice day with such a pleasant evening.

In the restaurant we ended up setting each other little geometry problems drawn on napkins, to the palpable disdain of our waiter who clearly wanted us to leave.  However, since I had to get up at 5am the following morning (i.e. this morning) to get the flight back to Cardiff, we didn’t stay out too late. I got back to the B&B where I was billeted in good time to check last night’s football results  before retiring to grab some shut-eye. Newcastle United 4 Coventry City 1 was the result, so it was good news to end the day…

I had to get up at the ungodly hour of 5am in order to catch the flight at Edinburgh airport, but the return flight was right on time. This was fortunate because, not long after the plane landed, a blizzard descended on Cardiff. Snow has fallen intermittently all day. Although I’m a bit tired after getting up so early – hence the brevity of this post -  I’m relieved I managed to get back to work without any major travel hitches.

Anyway, my contribution to the little problem-setting session that took place between the plates and wine glasses was this one, which I was asked during the interview I had to undergo to get a place to study at Cambridge:

Consider an infinite square grid made as shown above from 1Ω resistors. What is the resistance between any two adjacent nodes of this network?

If you’re really interested, a general solution for the resistance between any two (not necessarily adjacent) nodes is given here but you should be able to get the answer for adjacent nodes by a much simpler line of reasoning!

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16 Responses to “Talk, Nosh and Gridlock”

  1. Anton Garrett Says:

    Surely you were asked only the adjacent-node special case at interview, because it’s the only case you can do in your head? Was that put to you by John Field?

    Having cracked that a few years ago I tried to find the general solution and came to the conclusion that it woud take more than a bit of doodling. Nice to have it in closed form.

    I remember the first time a computer algebra routine cracked an ordinary differential equation that I had failed to. At that point I got some insight into how Kasparov felt after defeat by Deep Blue in 1997. It was

    dx/dt + x^4 = A sin^2 bt

    which specifies the time-variation of the temperature x of a light bulb filament; the x^4 term is radation, the sin62 term is heating due to electrical resistance (b is the angular frequency of the ac current) and dx/dt is a conduction term. Maple got a closed-form, albeit impilcit, solution of this, which I hadn’t (and which, strangely, an update of Maple failed to find). It involved the function inverse to x \exp (-x).

    You actually only want periodic solutions, not transients, but the Fourier coefficients of such a solution are coupled by the nonlinearity.

    Anyway, I was impressed.

    Anton

  2. telescoper Says:

    Yes, it was John Field that interviewed me and it was the special case of adjacent nodes only. I started trying to do it by summing series, but he gave me a big hint about the trick needed to get the solution easily. Maybe everyone knows it, because nobody seems to have offered an answer!

    Much later I tried to solve the general solution, but I never managed to get it in the neat form given in the link (which I only found yesterday). The case for diagonal nodes has an interesting answer: 2/pi. I wonder if there’s a trick for that one?

  3. Anton Garrett Says:

    Is this problem a piece de resistance?

  4. telescoper Says:

    Is there such a thing as a piece de capacitance?

  5. Anton Garrett Says:

    I’l tackle that integral for R(M,N) in 2D when I’m back off holiday. The generalisation that interests me is not to higher dimensions, but to regular triangular and hexagonal networks (rather than a square network as here).
    Anton

  6. What was the big hint you got in the interview?

    I’ve seen the answer but not the trick. I therefore know the effective resistance of all the other paths, but I don’t see how that is worked out…

  7. John Peacock Says:

    Hint: superposition.

  8. Here is an alternative solution:

    (skip to 3 minutes… the guy is amusing, but has trouble getting to the point)

  9. 0.5 R

    (the wonders of google)

  10. Anton Garrett Says:

    A literature search reveals that the solution has also been found for the resistance between arbitrary nodes of a 3D cubic lattice of identical resistors, and also for the 2D regular triangular lattice and 2D hexagonal lattice (these two are trivially related by the Y-Delta transformation). The general solution has not yet been found for a 3D regular tetrahedral lattice of identical resistors, ie diamond structure.

    • telescoper Says:

      It seems the kind of thing that would be straightforward to find numerically, and is surely interesting for many reasons, but it doesn’t surprise me that there isn’t a nice analytic form.

  11. [...] A few days ago I posted a little puzzle about the resistance measured between two adjacent nodes in an infinite square grid made of of 1Ω [...]

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