Computational Geometry Lecture Online Navigation

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Computational GeometryLecture ? Online
Navigation

• Jaeha Lee
• Konkuk Univ.

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Search, exploration, and navigation

• Many problems in AI and Algorithms can be modeled
  as searching (known, unknown) environment in
  order to reach a (known, unknown) target.
• Two factors
• Generality What kind of environment can you
  handle?
• What is your performance measure?
• (There is a tradeoff between generality and
  strength of performance measure.)

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Examples (????)

• Searching a known environment for target t at
  unknown location
• Minimize maxttime-to-reach Traveling Salesman
  Problem
• Given a prior P, minimize EPtime-to-reach
  Traveling Deliveryman Problem
• Minimize maxttime-to-reach /shortest path
  Online navigation
• Looking for a hole in a fence

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Searching for a hole in a fence

• Minimize maxttime-to-reach Go to one end and
  then to the other
• Given a prior P, minimize EPtime-to-reach Use
  dynamic programming. For each interval a,b and
  each endpoint a and b, what is the best way to
  visit the rest of the line?
• Minimize maxttime-to-reach /shortest path
  Optimum deterministic is
• 1 2 4 8 16
• competitive ratio of 9.

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Finding your way without a map

• You are in the (2-dimensional) plane and must
  move from s to t (known) through a scene full of
  obstacles.
• You learn about an obstacle only when you bump
  it. tactile
• How much longer do you walk than an offline
  algorithm that has a map?

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Today

• Unit square obstacles
• Rectangular obstacles

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At least 50 worse than the offline

• At each step, place a unit square in front of the
  online algorithm. If the distance from s to t is
  n gt 1, any online algorithm will walk at least
  1.5n.

t
s
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• We will argue that in such a scene, there is a
  path around obstacles of length at most n3?n,
  that could have been found if you had a map.
• Thus,
• 1.5n/(n3?n) ? 1.5

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The shortest path

• Consider a rectangle of height ?n, and of
  horizontal extent from s to t.
• Imagine sliding the horizontal line down the
  rectangle. Since we put down a total of n unit
  squares, we can slide the horizontal line to some
  position L where it only cuts ?n of these unit
  squares.

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L
t
s

• Offline strategy walk vertically from s to L.
  Walk rightwards along L, taking the shorter route
  around any obstacles encountered.
• Total walk is ?nn?n?n.

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Obvious strategy

• Simply walk right whenever you encounter an
  obstacle, walk around it.
• This is within a factor of 2 of Opt. Why?
• Anyway, online walk is between 1.5 and 2 of Opt.

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Axis-parallel rectangular obstacles

• Infinite wall problem
• Starting point at (0,0) and the target is the
  line xn.
• Room Problem
• Given a room of 2n2n, starting point at (0,0),
  and a target at (n,n).

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At least ?n times the offline path

• At each step, place a ?n1 rectangle in front of
  the online algorithm. If the distance from s to t
  is n gt 1, any online algorithm will walk at least
  (?n/21)n.

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The shortest path

• Consider a rectangle of height n, and of
  horizontal extent from s to t
• Imagine sliding the horizontal line down the
  rectangle. Since we put down a total of n
  rectangles, we can slide the horizontal line to
  some position L where it only cuts ?n of these
  rectangles.

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• Offline strategy walk vertically from s to L.
  Walk rightwards along L, taking the shorter route
  around any obstacles encountered.
• Total walk is nnn.
• Thus online/offline gt ? (?n).

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Online algorithm matching the lower bound

• Idea
• imagine a greedy algorithm
• that goes around the nearer corner whenever it
  meets an obstacle.
• but, unbounded ratio.

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O(n?n)-cost Algorithm idea

• Window of size W (Wn initially)
• ?n sweeps between north and south within the
  window and then double W
• Analysis
• For fixed window size W, online cost is O(W?n)
  and optimal cost (that crosses online path ?n
  times) is ?(W).
• If Wf is final window size, online cost is
  O(Wf?n) and optimal cost is ?(Wf). Thus ratio is
  O(?n) .

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Algorithm more precisely

• Wn, ?W/?n , sweep-dir south
• If nearest corner is nearer than ?, robot goes
  around and back along the same y-coordinate.

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Algorithm more precisely

• If obstacle extends past both sides of window,
  double W, ?, and sweep-dirsouth.
• Otherwise, robot goes in sweep-dir, until it
  meets the corner or window side. In the former
  case, robot moves east. In the latter case, it
  flips sweep-dir.

-W/2
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Analysis more precisely

• Wf size of final window
• ?f final threshold
• Theorem 1. Online cost is O(Wf?n)
• Proof.
• Since window size is doubled, it suffices to
  show theorem for Wf.
• -horizontal segments lt n O(Wf?n)
• -vertical segments by Rule 1 lt2n?f
• -vertical segments by Rule 3 O(Wf?n)
• one sweep distance is O(Wf)
• at most ?n1 sweeps O(Wf?n)

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Analysis more precisely

• Theorem 2. Shortest path length is ?(Wf).
• Proof.
• P shortest path
• - if Ps maximum y-value is gt Wf/2, OK!
• - online path Q while window size is Wf/2 uses
  ?n sweeps.
• Q crosses P at least ?n/2 times.
• For one crossing, optimal vertical path is at
  least ?f/2.
• Thus shortest path length is ?(Wf).

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Room Problem

• Start from a corner of 2n2n room.
• Room contains rectilinear rectangular obstacles.
• Goal is to reach t at the center of the room.
• Note Going from t to s optimally is easy. Greedy
  (South,West) path.
• Note Brute-force exploration takes traversal of
  O(n2) in distance.

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O(n?n)-cost Algorithm idea

• Go to temporary target t at (?n, ?n) from start
• Make greedy (north, east) path.
• Make greedy (south, east) path.
• We have paid cost O(n) and removed the O(?nn)
  area from consideration.

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Analysis more precisely

• Maintain invariant of a known monotone
  obstacle-free path from (n,y0) to (x0,n).
• Begin with x0y00, path along room walls.
• At each iteration, increase x0 and y0 with at
  least ?n with only cost O(n).

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Improvement to n1o(1)

• Apply recursively,
• Let T(n) be cost when d(s,t)n.
• Use t(x0m, y0m).
• Pay only T(m)O(n) to increase x0 and y0 by m.
• Repeat at most 2n/m times.
• T(n) lt 2n/mT(m)O(n)
• So
• T(n) lt n1c/?log n
• Smoother tradeoff get O(nlogn)

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Quiz

• Why is Room Problem easier than Wall Problem?

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Concluding remarks

• Wall Problems
• Deterministic O(?n), tight
• Randomized upper bound O(n4/9)
• Randomized lower bound.
• ?(loglog n) (big gap!!)
• Room Problems
• Ratio of O(log n)
• Improving performance with several trips
• Ratio of O(?n/k) with k trips.