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- Path: sparky!uunet!pipex!bnr.co.uk!uknet!acorn!armltd!dseal
- From: dseal@armltd.co.uk (David Seal)
- Newsgroups: rec.puzzles
- Subject: Re: The Worm and the Blanket Problem
- Message-ID: <11007@armltd.uucp>
- Date: 22 Dec 92 16:46:11 GMT
- References: <1992Dec21.010532.14305@thunder.mcrcim.mcgill.edu>
- Sender: dseal@armltd.uucp
- Distribution: rec
- Organization: A.R.M. Ltd, Swaffham Bulbeck, Cambs, UK
- Lines: 105
-
- 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...
-