[extropy-chat] glat test
Simon Dawson
extropians at perception.co.nz
Fri Oct 29 21:21:14 UTC 2004
you know, the thought occurs to me. Often these kinds of tests are the "aha" kind.
ie, there's a trick.
in this problem, resistance across a knight's move, perhaps the clue is really "the grid of resistors"
a grid implies square pattern, right?
which means you'd have to go UP two resistors, and ACROSS one, to get the knight's move.
this would mean the sum of 3 resistances.
Just a thought.
Si
At 07:08 30/10/2004, you wrote:
>Well, I don't care to try solving this thing, but had some musings to share:
>
>1. Solving the general problem of R(x0,y0;x1,y1) might be simpler or same difficulty as solving the knights move instance.
>
>2. Such a solution requires enumerating all of the current carrying paths and constraints thereto, looks nasty, like a infinite series of summations of infinite series, and how to count these in a 1 for 1 way is tricky.
>
>3. What is the test actually looking for. If there is a nice closed mathematical expression that's one thing, but will "good enough" do? If "good enough" does, then it is simple enough, though tedious, to write a simulator, say a grid of 100 x 100 resistors and run till it converges. If I did that I would email the solution, how I found it and why I did it this way.
>
>4. Of course there's the story about Von Neumann when asked to sum the numbers 1 - 1000 and he did't see the obvious pairings of 1,999; 2,998 and so on, and when asked how he did the problem he said he just added them all up. Wonder if something similar lurks here.
>
>5. Knights move is a co-ordinate transformation. Think of it as a diagonal on a matrix of resistors whose diagonal is the knights move. Probably a PITA, just a thought.
>
>> At 08:22 PM 10/28/2004 -0700, Spike wrote:
>>> >
>>> >
>>> >If you set up a subset of the grid with only 7 resistors
>>> >and calculate the hard way, the knight-move nodes Req is 7/5.
>
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