
The Parallel Climbers Puzzle - luisb
http://fermatslibrary.com/s/the-parallel-climbers-puzzle
======
SilasX
Sorry if OT, but this article hits a pet peeve of mine: unclear examples. In
the initial image for the problem, it shows a mountain that is practically
symmetrical and thus doesn't convey any of the intuition for why it would be
"hard" for the climbers to stay at the same height for the whole trip.

They should have used an illustration where the sides of the mountain were
more obviously different.

An example should be chosen to highlight as much variation as possible so the
reader can tell what is and isn't essential to the problem.

~~~
elbigbad
I didn't get the impression that it was symmetrical from the picture, and I
was browsing on a phone. I guess it could have been more clearly marked
though. Plus, as the other reply noted, the "problem" wouldn't have been a
problem if both sides were symmetrical.

~~~
tgb
I don't think the comment is pointing out the difficulty in telling whether
it's meant to be symmetric, it's pointing out that the symmetric case is
trivial and so any example should be really different from the symmetric case
in order to serve well as an example. It's hard to picture how the climbers
might have any difficulty whatsoever when seeing an example that's almost
trivial.

------
Smudge
Intuitively the solution kind of makes sense.

M must always be the highest point within the range. A and Z are at the same
height and are the lowest point within the range.

Imagine that the hikers are somehow connected (quantum entanglement, etc) such
that one cannot physically move up or down if other cannot, and as one moves
up and down the other must follow at the same height.

Perhaps a better metaphor would be if the paths are cut into a wall, and you
have inserted two pegs that are connected by a horizontal backing bar behind
the wall. The bar may move up and down but will always remain horizontal. The
pegs can slide left and right along the bar, and up and down along the paths,
but must always remain horizontally level.

Now, imagine that whenever a peg reaches a local max or min (peak or valley),
a change in vertical direction may also cause a traversal along the opposite
size of the peak/valley, thus allowing for forward progression.

While one peg hits a vertical stop and makes horizontal progression, the other
peg will simply move up and down along the same segment.

This exercise is obviously not a mathematical proof, but does serve to make
the proof feel a bit more intuitive. I'd love to construct such a "puzzle"
myself and try it out on a bunch of different contours/tracks.

------
kmill
I spent some time trying to formulate this as an intersection of two regular
expressions (the language being composed of sequences of u and d -- given a
sequence which goes to the top, you can figure out at least one path which
gets you there, and it doesn't matter which you choose)... though it seems
much harder to deal with than just saying "the graph is connected because it
has exactly two vertices of odd degree."

~~~
Jtsummers
Instead of a sequence of u and d, try modeling it as:

Let's go with their ultimate "alphabet", all the pairs based on altitude.
(A,Z), ... (M,M). Note all valid transitions. In fact, you have the exact same
graph as they do since every transition is reversible. (A,Z) is the starting
state, (M,M) is the final state. What you now want is the shortest matching
string: AZ.CX.DY.EW.DU.MM. There ought to be only one in this case.

But there are actually 3 possible machines that can be constructed and need to
be analyzed if you want to go with this method. C < W, C = W, C > W. The case
of C=W is trivial, no intermediate points are introduced. C < W is the case
given above and in the paper. C > W is the same as C < W under a trivially
constructed mapping so it's already solved as well, so ultimately only 2 cases
need be considered given symmetry.

I think the problem of marking it as a sequence of u and d, you'd need a
metric. Some notion of how high or low each peak and valley actually are. Now,
this could be discretized based on relative altitudes. Lowest is 0, next is 1,
next is 2, etc. So you can ignore precision. Still greatly complicates the
matter.

------
Chinjut
The author takes care to make the assumption that A and Z are _global_ minima
and M is a _global_ maximum. But, in reading the proof, it seems not necessary
to make such strong assumptions; would it not work just as well for A and Z to
be merely local minima and M to be a local maximum? (For that matter, it
should work just as well for A, Z, and M to all merely be local extrema of
either sort, so long as A and Z match)

~~~
skj
If A or Z aren't global minima, there might be a spot between A and M that is
lower than all those between Z and M, and therefore no path.

~~~
Chinjut
Yes, I realize now my error: I was reasoning in the degree calculations for
the relevant graph as if "local minimum" was synonymous with "valley", and
similarly for "local maximum" and "peak", but of course A and Z are examples
of local minima that are not typical valleys (having escape routes on only 1
instead of 2 sides), and a similar point applies to M (construed as
impassible).

In fairness, the article itself is glib about this important point (it does
not discuss the degree calculation for (A, X), (X, Z), (M, X), and (X, M) in
general, only for (A, Z) and (M, M) in particular).

------
marlag
Ok, so they DID make it to the top, great! Now, how does this help me get
better at graph theory?

Edit: That being said, I mean no disrespect to this... 20 year old article.
Hope I didn't hurt your feelings, Article. Just trying to find an entry point
into this domain.

~~~
yequalsx
Assuming the question is serious I'll attempt an answer. It shows an example
of how using graph theory one can solve a problem that seemingly at first has
nothing to do with graphs. The proof relies on graph theoretic concepts and
understanding the proof helps one get a firmer grasp on these concepts. This
particular example and this particular experience may not seem helpful but
done often enough you'll end up being good at graph theory.

~~~
marlag
Thx! And I was being serious. For me, one problem of gaining understanding of
the domain is, how do I identify that a problem is indeed a graph problem?

~~~
Jtsummers
Practice, and read about similar puzzles/problems. Track down reprints/scans
of Martin Gardner's _Mathematical Games_ and similar publications.

In general, for identifying if something is a graph problem:

1) Through some mapping, (almost?) everything can be transformed into a graph
problem of some sort. Now, that's a bit too big a set and ignores the real
question.

2) How do I identify that a problem is _practically_ solved with graphs?

Some heuristics (note, these are heuristics, particularly useful for getting
started, but not at all absolutes):

Is it discrete? In the given problem from the link we have a finite number of
extreme points (ends of each line segment), but an infinite number of points
between. Fortunately, thanks to the problem constraint, there are only a few
non-endpoints that we are concerned with. So we can safely ignore the infinity
of possibilities by only examining this finite set.

Are there transitions between these states that are easily modeled as edges?
In the given puzzle, absolutely. And its symmetric so an undirected graph is
suitable (sometimes directed graphs would be more suitable or the only
applicable solution).

After this, proving various things (like that the generalized statement that
all properly constructed puzzles have a solution) will require learning some
of the basic properties of graphs, and how to restate the premises (such as
constraints on altitude and movement) in a graph-theoretic form. It's easy to
intuit the proof now that you have a simplified, but complete, model of the
problem, but harder to state it with mathematical certitude. That part really
will just require more practice and exposure.

At some point you'll develop sufficient vocabulary in the field that you may
not be able to prove it straight off, but you'll know what and where to look
for the elements you need to construct a proof like they have.

~~~
tzs
> Practice, and read about similar puzzles/problems. Track down reprints/scans
> of Martin Gardner's Mathematical Games and similar publications.

The entire set of the 15 books that collected his "Mathematical Games" columns
from Scientific American are available in PDF form on a CD-ROM:

[http://www.amazon.com/Martin-Gardners-Mathematical-Games-
Gar...](http://www.amazon.com/Martin-Gardners-Mathematical-Games-
Gardner/dp/0883855453)

