

Seven Puzzles You Think You Must Not Have Heard Correctly - mhb
http://www.math.dartmouth.edu/~pw/solutions.pdf

======
anigbrowl
Fun paper, but I got pissed off with the first one because it depended on
something not stated in the problem; in brief it requires multiple
interdependent participants to adhere to a pre-agreed algorithm for searching
a data structure whose topology they cannot know in advance.

The solution is interesting and worthy, but the problem as written is poorly
stated. I hit the right idea while I was thinking about it but then rejected
it because it seemed like an unjustified assumption. More than one of these
seem to depend on the vagueness or incompleteness of the problem specification
resulting in a meta-solution rather than a formal one.

~~~
RiderOfGiraffes
I don't understand your objection ...

    
    
        > ... it depended on something not stated in the
        > problem; in brief it requires multiple interdependent
        > participants to adhere to a pre-agreed algorithm
        > for searching a data structure whose topology
        > they cannot know in advance.
    

It clearly says:

    
    
        The prisoners have a chance to plot their strategy in advance
    

That seems pretty clear: the prisoners must pre-agree a strategy to do what
they're going to do. Not adhereing to that seems to follow naturally.

What am I missing?

~~~
anigbrowl
Quite correct. But when I was imagining it, I began thinking of different ways
to 'line up' the boxes and it's not clear to me from the problem statement
that the layout is predictable in advance. See below for why (I think) a
circular arrangement would defeat the strategy. Lines can be curved, and
curves may be closed.

~~~
req2
Lines are a special type of curve - a straight curve. This precise and simple
mathematical definition precludes such confusion.

~~~
anigbrowl
Can't argue with that. I guess I'm thinking of those puzzles that say things
like 'can you draw a line that passes through all these points', and when you
can't, the answer turns out to be that it's not a straight line. I'll leave my
wrong arguments up for educational purposes :-)

------
Dove
In the case of the first, I was able to get some pretty good odds -- but
nowhere near what the problem claims. The solution is brilliant, but requires
a little more abstract algebra and combinatorics than I was able to lay my
hands on quickly.

The second problem is surprisingly difficult, and more annoyingly, the 2D case
offers no guidance for the 3D case. Even more annoyingly, the solution, while
obviously correct, is unenlightening. I hate unenlightening proofs.

The third problem suffers from the abstract problem statement and unclear
mechanic; I had to read the solution to understand the problem, which is too
bad. It's a wonderful problem. Fortunately, _understanding_ the solution
wasn't exactly a whole lot simpler than solving the problem might be, so I had
that as compensation. I offer the following (simpler) restatement in the hope
that it will allow someone else to enjoy the problem:

There is a town where each member has on his forehead a blue or red dot. If on
any given day he figures out which it is, he dies in his sleep that night. On
one particular morning, five of them--all with blue dots, unbeknownst to them
--are standing on a street corner when a stranger walks by. "Well," he says,
"I see more than one blue dot is out this fine morning!" Prove that all five
die in their sleep before the week is over.

(If you can do that, the generalization is obvious -- so stating the problem
in a general way only serves to obscure the terms.)

The rest of the problems rapidly lost my interest . . . so I'd say they were
well arranged.

Thanks for an enjoyable waste of an evening!

~~~
cousin_it
The second problem is really good. I just thought up a solution that, though
more complex, seems to me much less mysterious, and the 2D case does shed
guidance on 3D.

One-dimensional case:

    
    
        a1 < a2
    

Two-dimensional case:

    
    
        a1*b1 < a2*b2 (area)
        a1+b1 < a2+b2 (sides projected outwards, triangle inequality)
    

Three-dimensional case:

    
    
        a1*b1*c1 < a2*b2*c2 (volume)
        a1*b1+a1*c1+b1*c1 < a2*b2+a2*c2+b2*c2 (surface area, projected outwards)
        a1+b1+c1 < a2+b2+c2 (our problem)
    

Notice the pattern already? If one N-dimensional box lies inside the other,
the inequality holds for all elementary symmetric polynomials of box
dimensions.

<http://en.wikipedia.org/wiki/Elementary_symmetric_polynomial>

To prove it, we need only notice that the symmetric polynomial of degree M is
proportional to the average volume of the M-dimensional "shadow" of the
N-dimensional box, averaged over all projection directions uniformly. (To
convince yourself of that, work through the two-dimensional case.) On the
other hand, if a box lies inside another box, each "shadow" of the smaller box
lies within the corresponding "shadow" of the bigger box. Done!

------
calambrac
I'm all for thinking outside the box, but the whole point of a puzzle is to
set up a set of constraints and work your way from there. The solution to #1
seems to contradict that spirit - without spoiling it, it's true the
description didn't say they couldn't do what they need to do, but it also
didn't say they couldn't riot, kill all the guards, and just escape to avoid
playing the game altogether...

~~~
dreish
I fail to see how the solution to #1 contradicts the plain meaning of the
constraints. I think it's an ingenious and surprising result.

~~~
anigbrowl
I guess it depends on how you interpret 'lined up'. What if the room is
circular, the table runs around the circumference, and you enter the room
through a trapdoor in the center? Then your entry point to the list is
effectively randomized and on any given trial the algorithm will work no
better than chance.

~~~
req2
I believe the standard interpretation of 'lined up' is 'in a line', as in a
'straight curve', which disallows your otherwise ingenious foil.

~~~
anigbrowl
True. But as soon as I thought of the circle, I said to myself 'ah, the idea
that it's a _straight_ line is exactly the sort of assumption that you gets
you into trouble!' so I threw the whole solution out.

I figured a straight line would be equivalent to a random arrangement with
each of the boxes being numbered, so I assumed the vagueness was designed to
lead one into making a foolish assumption about their ordinal presentation.
Maybe I'm paranoid :-)

------
michael_dorfman
Six of those were mind-blowers, but the solution to one should be obvious to
most hackers.

------
MaysonL
I call BS on Problem 1: it's bait and switch!

The answer to the puzzle #1 is simply totally incorrect: how long the
permutations are has nothing to do with whether your name is in any of the
boxes in the permutation that the prisoners start with your name.

EDIT: After thinking about, and trying a few small {try 4 boxes) examples, I'm
not so sure [i.e. I guess I blew it].

~~~
jibiki
> The answer to the puzzle #1 is simply totally incorrect: how long the
> permutations are has nothing to do with whether your name is in any of the
> boxes in the permutation that the prisoners start with your name.

The answer is correct. Your name has to be in the cycle started with your box,
this follows from the fact that permutations are bijections. It is impossible
to have something like:

    
    
      0->1->2->1
    

(Because then 1 would be in two boxes at once, which is impossible.)

------
tome
If you like the Random Native puzzle, try this one for size!

[http://www.srcf.ucam.org/~te233/maths/puzzles/evenharder.htm...](http://www.srcf.ucam.org/~te233/maths/puzzles/evenharder.html)

------
jongraehl
"post offices and currier services" - <http://en.wikipedia.org/wiki/Currier>

------
muerdeme
Correction to the Roger Federer problem: the setting must be the U.S. Open for
the solution to make any sense.

~~~
tome
I don't get that at all. Why? The scoring at the US Open is not different to
Wimbledon as far as I know.

~~~
muerdeme
The U.S. Open is the only grand slam tournament that uses final set tiebreaks.
Hence the marathon 5th set (16-14) between Roddick and Federer at this year's
Wimbledon.

~~~
andrew1
But the final (i.e. fifth) set is never reached in the solution. At Wimbledon
it is entriely correct that you can win the match in three or four sets by
winning the third or fourth set in a tie-break. There is not a problem with
the solution as stated.

~~~
muerdeme
"Final" means the set that will end the match, not necessarily the fifth set.

~~~
andrew1
It may do elsewhere but it doesn't at Wimbledon.

