[extropy-chat] glat test
Adrian Tymes
wingcat at pacbell.net
Sat Oct 30 19:37:58 UTC 2004
--- Simon Dawson <extropians at perception.co.nz> wrote:
> you know, the thought occurs to me. Often these
> kinds of tests are the "aha" kind.
>
> ie, there's a trick.
True. Those tests in practice check more for
foreknowledge of the trick than anything else, despite
what is claimed about them. (In fact, it'd usually be
a fairer test to make sure everyone knows, or has had
ample opportunity and notice to learn, about the trick
then apply it. This version often finds good use in
school, although some students fail to understand why
the particular information being communicated to them
is of use - or, sometimes, even that information of
potential value is being communicated to them in the
first place. But I digress...)
> 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.
Electricity doesn't quite work like that. If there
are two paths along a circuit, electricity will flow
along both paths, not exclusively the one with the
least resistance (barring superconductors with
exactly zero resistance). If all of it went along one
path, it'd encounter a higher total resistance than if
a little bit of it went along a nominally higher
resistance path.
It may help to think of it like cars moving from point
A to point B along a freeway system. Put in a wide
freeway directly linking those two, designed to handle
lots of traffic, and many people will use it...and
cause traffic jams. A few commuters will take side
streets that, while narrower and not as direct, still
get them to their destination faster than if they'd
taken the freeway (and added to that mess, slowing
everyone on the freeway down).
For example, in the infinite grid, connecting two
nodes diagonally adjacent: the most direct paths (one
up then one across, or one across then one up) both
have a resistance of two. But the total effective
resistance, after calculating in all paths, is
(according to another post) the square root of two.
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