| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 83.1 | | ORPHAN::BRETT | | Thu Jun 28 1984 09:26 | 7 |
|
None - the first guy divided into 4 squares, the first being 1X1 KM, the
remainder being 0X0 KM - and got lynched by the other three, who then
agreed their father was barmy, split it equally and lived happily ever
after.
/Bevin
|
| 83.2 | | TURTLE::GILBERT | | Thu Jun 28 1984 16:51 | 15 |
| The first son takes a piece that is 1x3/4, figuring that each of his three
brothers will get a piece of size 1/4x1/3, which is the same SHAPE as his own.
The following table summarizes the divisions.
Son takes leaving share
First 1.0 x .75 3 * .333333 x .25 .75
Second .967707 x .25 2 * .0161464 x .25 .241927
Third .245757 x .0322928 1 * .00424333 x .0322928 .007936
Forth .00424333 x .0322928 .000137
Thus, the last son gets a plot of land that's roughly 4.25 meters by 32 meters.
Just enough for a fine driveway, or to raise a bumper crop of spaghetti.
- Gilbert
|
| 83.4 | new solution | HERON::BUCHANAN | fragmentary blue | Mon Oct 31 1988 08:57 | 81 |
| < Note 83.3 by HERON::BUCHANAN "fragmentary blue" >
-< New solution >-
>A man with four sons left his 1 KM square estate in his will with the
>following stipulations: The eldest son was to divide the property into
>four pieces of the same SHAPE and take one for himself. Then the second
>son was to divide the remainder into three pieces of the same shape and
>take one for himself. The third son was to take what remained, divide it
>into two pieces of the same shape, and take one for himself. Assuming these
>three were both greedy and clever, how much was left for the fourth son?
>(The shapes at each division need not be the same as for previous
>divisions. Degenerate solutions involving empty 'shapes' are not acceptable.)
Seems to me Lynn that if the eldest is *particularly* sneaky, he can
get *almost* the whole inheritance, without running into the degeneracy clause
mentioned in the last paragraph.
Key: use the notion of self-similarity.
Let d be less than �km, and observe the following map:
<---d--->
+-----+-+---------------+-+-----+ ^
| +-+ +-+ | |
| 2 | | 3 | d
|-+ +-+ +-+ +-+ |
+-+---+-+ +-+---+-+ v
| |
| |
| |
| 1 |
| |
| |
| |
| +-+---+-+
| +-+ +-+
| | 4 |
| +-+ |
+-----------------------+-+-----+
It's an infinite regress: everytime you get a square, you break off
three corners (it's always obvious which) and give them to the guy who's
adjacent.
Basically, the eldest son divides the field in the following way. He
takes portion 1. The NW square is then divided between brother 1 and brother
2, with the main portion going to brother 2, and the SW, NW and NE subsquares
being divided between brother 1 and brother 2, with the main portions going
to brother 1, and nine sub-subsquares being divided... Similarly, the NE
square is divided between brothers 1 and 3, and the SE square is divided
between brothers 1 and 4.
Observe:
(1) Each brother receives the same shape.
(2) Each piece is connected.
(3) The residue after brother 1 has removed his share is already disconnected
into three pieces, so brothers 2 and 3 have no say, and receive the same as
brother 4.
(4) The area that brother 1 receives is:
1 - 3d� + 9d^4 -... =
1 / (1 +3d�)
d can be made arbitrarily small, so the share of the eldest brother
can be made arbitrarily close to 1.
Andrew
|
| 83.5 | Nice idea! | AKQJ10::YARBROUGH | I prefer Pi | Mon Oct 31 1988 10:02 | 23 |
| Hmmm. I have trouble with the fractal that gets generated in the vicinity
of the smallest squares below, e.g. -+
|
<---d---> +-----------+
v
+-----+-+---------------+-+-----+ ^
| +-+ +-+ | |
| 2 | | 3 | d
|-+ +-+ +-+ +-+ |
+-+---+-+ +-+---+-+ v
| |
| |
| |
| 1 |
especially since a different fractal gets generated for each value of d.
Although that is certainly an ingenious dissection, I think I'd disqualify
it on the grounds that it not acheivable *legally* - the lawyers would be
employed forever trying to get a good measurement! Also, the assumption
that son 1 is greedy implies that he would reduce d to 0, in violation of
my specification that degenerate dissections are disallowed.
Lynn Yarbrough
|
| 83.6 | | HERON::BUCHANAN | fragmentary blue | Mon Oct 31 1988 11:41 | 18 |
| I believe you are confused about several things.
(1) The fact that the fractal is different for different values
of d is irrelevant. For any one d in (0,�), a solution exists. It's then
a matter of greed which d the eldest brother selects.
(2) Your objection about the lawyers is silly.
(3) Your objection about convergence of d to 0 is illogical. For
*any* solution (even .2) is dominated by a greedier one for some value of d.
What you're saying therefore is that *no* solution is acceptable.
What interests me is this: is this excellent problem showing the
merits of self-similarity your own invention? I ask because it is very
felicitous that the number of sons is 4, and not 5 for instance. With 5
I can't figure out how to solve the problem given the *connectivity*
requirement that you were careful to demand. (at least, that's how I
interpreted the word piece).
|
| 83.7 | Gotta keep a sense of humor | AKQJ10::YARBROUGH | I prefer Pi | Mon Oct 31 1988 12:55 | 36 |
| > I believe you are confused about several things.
I'm confused about a LOT of things.
> (1) The fact that the fractal is different for different values
>of d is irrelevant. For any one d in (0,�), a solution exists. It's then
>a matter of greed which d the eldest brother selects.
As I said, I have trouble with this. That doesn't mean I'm confused about
it. I just can't visualize all the solutions, or appreciate yet how they
ultimately differ from one another.
> (2) Your objection about the lawyers is silly.
Tell that to a lawyer. Most of them would give you a different opinion.
> (3) Your objection about convergence of d to 0 is illogical. For
>*any* solution (even .2) is dominated by a greedier one for some value of d.
>What you're saying therefore is that *no* solution is acceptable.
Not at all. Peter's solution in .2 is what I intended, although it had
occured to me that other solutions might be possible. That solution does
not permit of greedier variants as far as I can determine.
> What interests me is this: is this excellent problem showing the
>merits of self-similarity your own invention? I ask because it is very
>felicitous that the number of sons is 4, and not 5 for instance. With 5
>I can't figure out how to solve the problem given the *connectivity*
>requirement that you were careful to demand. (at least, that's how I
>interpreted the word piece).
The problem is original. I picked 4 sons more or less at random. The
solution in .2 extends easily to 5 sons or more, as the remaining piece in
each case is a rectangle.
Lynn Yarbrough
|
| 83.8 | peace! | HERON::BUCHANAN | fragmentary blue | Mon Oct 31 1988 18:36 | 42 |
| > -< Gotta keep a sense of humor >-
Yes, sorry. Blame my missile guidance system. It thought it
detected inbound NIH, and launched all missiles. By the time the captain
got to the bridge, it was all over bar the congressional hearings.
> The problem is original. I picked 4 sons more or less at random. The
> solution in .2 extends easily to 5 sons or more, as the remaining piece in
> each case is a rectangle.
Congratulations: why not send it in to Math Monthly? But if you really *do*
wish to exclude my solution, then you'd better rephrase the question carefully.
But my interpretation of the puzzle is the one which the readers would get the
biggest buzz from. I think you've invented a *classic* and memorable puzzle.
And you raise a good point: what about the area of the perimeter?
Is it a finite proportion of the whole? And how is the perimeter allocated
between the sons, so that "shape" is the same for all sons?
The areas for the sons add up to unity:
1 = 1/(1+3d�) + 3*(d�/(1+d�))
The allocation of the perimeter is determined by the allocation of the
outside perimeter of the square. This forces us to say that the piece
allocated to brother 1, the eldest, is:
( The closed square, [0,1] x [0,1]
\ The 3 closed squares, side of size d )
union ( The 9 closed squares, side of size d�
\ The 27 closed squares, side of size d�)
...
Thus the self-similarity extends to the perimeter, which is not the
case for the four-rectangle solution in .2. There are I think an infinite
number of points, which are not allocated to any of the four sons in the
recursive solution. One of these would be, assuming the original field to
inhabit the upper right quarter of the Cartesian plane, ( d/(1+1), 1).
I must think about the shape that these avoided points have.
Andrew
|
| 83.9 | | CLT::GILBERT | Multiple inheritence happens | Mon Oct 31 1988 20:06 | 18 |
| I like it! Note that the same approach generalizes to an arbitrary
number of sons. The following illustrates the case for 3 sons:
+---------+---------+---------+---------+---------+
| | | | | |
| | | | | |
| | +-+ +-+ | | +-+ +-+ | |
| +-+ +-+ +-+ +-+ +-+ +-+ |
| |
| |
| |
| |
| |
| |
| |
| |
| |
+-------------------------------------------------+
|
| 83.10 | | HERON::BUCHANAN | fragmentary blue | Tue Nov 01 1988 05:15 | 18 |
| > Note that the same approach generalizes to an arbitrary
> number of sons. The following illustrates the case for 3 sons:
Yes, I was being needlessly self-constrained. Any number of brothers
can be handled. And if I rotate my
sub-squares by forty-five degrees, (ie they look like diamonds)
then I can stick them like polyps to the walls of the field, and each wall can
have diamonds attached to it, without running into difficulties of non-
connectivity.
Also, I needn't have all the subsquares the same size.
Further, I when it comes to embedding sub-diamonds into each of the
diamonds, then I can choose for each diamond which of the four faces is to
be considered the North face (ie that corresponding to the North face of the
original field.
Andrew
|
| 83.11 | fluff -> lawyer | HERON::BUCHANAN | fragmentary blue | Wed Nov 02 1988 08:52 | 23 |
| With regard to the "10% for the lawyers" variant suggested in the next
section (which I'm not sure I understand, but seems reminscent of the extra
camel will puzzle), if I were the eldest brother, I would offer the lawyer
rather the Cantor's dust: the fractal crumbs remaining after the dissection of
the field.
For the example solution with four brothers, where three corners were
reserved for the juniors, with square side d, use the formula:
N * d^D = 1
where N is the number of pieces. ( = 3 here)
d is the scaling factor. ( < � here)
and D is the fractal dimension.
so D = ln(3) / ln(1/d). Thus D can range up to ln(3)/ln(2), but can be made
arbitrarily small.
The area occupied by this fluff is the limit of d^[n*(2-D)] as n ->
infinity. Since D < 2 (indeed D < 1), the area is nearly 0, whatever value of
d is chosen.
Andrew
|