From: dseal@armltd.co.uk (David Seal)
Newsgroups: rec.puzzles
Subject: Re: The Worm and the Blanket Problem
Date: 22 Dec 92 16:46:11 GMT
Distribution: rec
Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK

In article <1992Dec21.010532.14305@thunder.mcrcim.mcgill.edu>
mouse@thunder.mcrcim.mcgill.edu (der Mouse) writes:

>None.  But the blanket gets to move to follow it.  (Blanket motion is
>rigid, purely translation and rotation.)
>
>Blanket motion may or may not be allowed to be discontinuous; all the
>answers I've seen so far are for the discontinuity-allowed case.

One solution I've seen posted so far works with continuous motion of the
blanket, namely the circle of diameter 1. All you need to do is keep its
centre over the centre of the worm, so if the worm moves contnuously, so
does the blanket.

The square of diagonal 1 comes close. This is guaranteed to cover the worm
if you make one diagonal of the blanket cover both endpoints of the worm,
then slide it along until the midpoint of the worm is on the other diagonal.
This position changes continuously as the worm moves continuously, *except*
when the two endpoints of the worm coincide. E.g. consider the worm moving
as follows:

      *                   *                             ^
     / \                  ||                            *|
    /   \       ==>       || (folded in half)     ==>   ||
   /     \                ||                            |*
  *       *               **                            |
                                                        *

I.e. first it folds around its midpoint until the two halves are together,
then it moves to extend one part and shorten the other. The rule above would
place the blanket at an angle of 45 degrees to the horizontal throughout
this motion. However, during the first part of the motion, it would be
centred halfway between the two endpoints of the worm; during the second, it
would be centred on the midpoint of the worm. At the midpoint of the motion,
therefore, the blanket would have to "jump" a distance of 1/2 upwards.

However, I observe that this "jump" seems to be artificial: the worm can be
covered quite easily when its two endpoints are close together. The problem
is that the exact rule we've chosen to position the blanket is
discontinuous, and not necessarily that *any* such rule has to be
discontinuous. This suggests modifications of this rule, whereby the blanket
moves according to the rule if the worm's endpoints are far apart and by
some other rule if they are close together. I don't know quite how to do
this, though: if anyone has any ideas, please post them.

If this can be done, a better continuously-moving blanket might be achieved
by taking the union of a square of diagonal 1-2D and a 2D by 1-2D rectangle
(at an angle of 45 degrees to the square) with semicircular endcaps of
radius D:

                    *
                  / | \
                /   |   \
              /     |     \
     _--*-----      |1/2-D -----*--_
    /   |D          |          D|   \
   | D  |   1/2-D   |   1/2-D   |  D |
   *----*-----------*-----------*----*
   |    |           |           |    |
    \_  |D          |          D|  _/
       -*-----      |1/2-D -----*--
              \     |     /
                \   |   /
                  \ | /
                    *

This would be moved to keep the points at distance D and 1-D along the worm
under the horizontal axis and the point 1/2 the way along the worm on the
vertical axis, at least while the first two points are far apart (a similar
modification to the one above would be used when they were close together to
keep the motion continuous). The central part of the worm would be covered
by the square of diagonal 1-2D, while the two ends would be covered by the
rectangle and its circular caps.

Of course, getting the details right of what happens when the points at a
distance D and 1-D along the worm are close together is the tricky part! If
it can be achieved, though, this gives us a blanket of area:

  PI*D^2 + 2*D*(1-2*D) + 2*(1/2-2*D)^2
   = PI*D^2 + 1/4 + (1/2-2*D)^2        if 0 <= D <= 1/4

  PI*D^2 + 2*D*(1-2*D)
   = PI*D^2 + 1/4 - (1/2-2*D)^2        if 1/4 <= D <= 1/2

If I've done the calculations right, this is minimised at D = 1/(4+PI) =
approximately 0.140025, giving an area of:

  PI/(4+PI)^2 + 1/4 + (PI/(8+2*PI))^2

  = 1/4 + (4*PI + PI^2)/(8+2*PI)^2

  = 1/4 + PI/(16+4*PI)

  = 1/2 - 1/(PI+4)

  = 1/2 - D

i.e. about 0.359975.

Anyone able to fill in the holes in the ideas above?

David Seal
dseal@armltd.co.uk

All opinions are mine only...