On Tue, Nov 9, 2010 at 7:25 PM, Greg <[email protected]> wrote:
>> I remember that I simply gave up and assumed that there were no such
>> sequence. I admire your systematic approach to this!
>
> Thanks! And it turns out that I was a bit mistaken!
>
> Simon messaged me off-list to let me know that in fact that code is actually
> enterable at that location, but I was just not use the machinery properly...
> (won't go into details as I don't want to spoil it for anyone who happens to
> be playing the game).
No spoiler here, just a bit of reasoning: as it happens, *any*
operation on the numbers other than the two already available is
likely to make (2 2 1) reachable. (And any that doesn't is redundant
with the two operations you mentioned previously -- this follows from
the vector space representation of the problem, as the operation
becomes addition of a vector that, since (1 1 0) and (0 1 1) span a
hyperplane, must either lie in that span or extend {(1 1 0), (0 1 1)}
to a basis. The latter makes (2 2 1) reachable and the former makes
the new operation equivalent to performing some sequence composed of
the original two moves.)
In particular, if a third operation exists that bumps any single
number in isolation OR bumps all three OR bumps the first and the
third, (2 2 1) becomes reachable. (All three operations cycle (a - b +
c) modulo 3, so move out of that hyperplane -- changing only one of a,
b, and c by 1 changes (a - b + c) either up or down by 1, whereas
increasing all three by 1 increases a + c twice as much as it
decreases -b, and so increases (a - b + c) by one.
Back to the vector space: with the (1 1 0) and (0 1 1) moves, any
state y is reachable from x if and only if y dot (1 -1 1) = x dot (1
-1 1) (same hyperplane parallel to the one through the origin those
vectors span); (a - b + c) is just (a b c) dot (1 -1 1). In fact you
can compute this for other (linearly independent) pairs of moves; if
the moves (represented as vectors) are p and q then y is reachable
from x with them if and only if y dot (p x q) = x dot (p x q).
So if you encounter any other puzzles in Myst that seem similar, it
may be that the same math can be used to analyze them.
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