From: Jim Ferry <jferry@delete_this_field.uiuc.edu>
Subject: Infinite Island Optima
Date: Thu, 18 Mar 1999 18:26:56 -0600
Newsgroups: alt.math.recreational,rec.puzzles,sci.math
Keywords: path of given width having minimal length

You are placed (uniformly) at random on an infinite strip of
width w.  The edges are invisible to you.  What path should you
walk to reach an edge if you want to minimize:

1)  The worst-case distance?

2)  The average distance?

I'll address (1).  A special case of (2) (for paths consisting
of two line segments) was solved in the Infinite Island thread
by Jonah (abitpuzzled@my-dejanews.com ).

(1) is equivalent to finding a path of *width* w with minimal
length.  The width of a set is the infimum of the separation
of two parallel lines bounding it.  (Note that this technical
sense of *width* contradicts the expression "an infinite strip
of width w" for finite w.)

(N.B.:  I posted and deleted an incorrect version which stated
that (1) is equivalent to finding the shortest path starting
at the center of a circle of radius w that makes the circle
opaque.  This would be right if you had to cross both edges
without know whether you've crossed them.  I gave a solution
with length (1 + sqrt(3) + 7 pi/6) w, which I guessed was
optimal.)

Problem (1) was solved in

"The Shortest Planar Arc of Width 1," by Ani Adhikari and Jim
Pitman, American Mathematical Monthly (96:4), April 1989,
pp. 309-327.

They call their minimal arc the *caliper*.  To construct it,
consider the arc c(d) formed as follows (1<=d<=2/sqrt(3)):
               
A'---F----B'   In the rectangle on the left |AA'| = 1 and
|   /     |    |AB| = d.  Let F be the midpoint of A'B'.
|  /      |    Draw a circle of radius 1 centered at B.
| /       |    Now draw the shortest path from A to F
|/        |    outside the circle.  c(d) is the union of
A---------B    this arc and its reflection (from F to B).

Let l(d) be the length of c(d).

l(d) = 2 sqrt(d^2-1) + d + (pi - 2 ArcSec(d) - 4 ArcTan(d/2)).

l(d) achieves a minimum at d = d*, where d* is the square root
of the positive root of 3 z^3 + 36 z^2 + 16 z - 64 = 0.

d*    = 1.043590109594984753811841771287022733354889696934037897
l(d*) = 2.278291641440437543661492570717476309150597432749699552

This is slightly better than using two of the sides of an
equilateral triangle (which is c(dt) for dt = 2/sqrt(3)):

dt    = 1.1547
l(dt) = 2.3094

So there you have it.  You can reach an edge by walking at
most 2.2783 w.  The caliper solution may not be unique, though.
The article proves it is optimal over all arcs, but uniquely
so only for *convex* arcs (an arc AB is convex if connecting A
and B with a line segment makes it the boundary of a convex set).

|             Jim Ferry              | Center for Simulation  |
+------------------------------------+  of Advanced Rockets   |
| http://www.uiuc.edu/ph/www/jferry/ +------------------------+
| jferry@expunge_this_field.uiuc.edu | University of Illinois |