Search Algorithms Winter Semester 2004/2005 17 Jan 2005 12th Lecture

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HEINZ NIXDORF INSTITUTEUniversity of Paderborn

Algorithms and ComplexityChristian Schindelhauer

Search AlgorithmsWinter Semester 2004/2005

17 Jan 200512th Lecture

Christian Schindelhauer

[email protected]

Search Algorithms, WS 2004/05 2



Spatial Searching

Prolog: Searching with some help

Searching with total Uncertainty Nearsighted Search

– The Cow Path Problem

– The Concept of Competitive Analysis

– Deterministic Solution

– Finding a Shoreline

– Probabilistic Solution

– The Wall Problem

Farsighted Search

– The Watchman Problem

– How to Learn your Environment




The Cow-Path Problem

Given–A near-sighted cow–A fence with a gate–The cow does not know the direction

Task–Find the exit as fast as possible

???




Competitive Analysis

How to evaluate the online solutionClassical approach:

– Worst-case time• This is always n for a fence of length n

– Average case• This is not better

Competitive Analysis– Compare the cost of the solution of an

instance x• CostAlg(x)

– to the best possible offline solution (unknown to the cow)

• Costoffline(x)Minimize the competitive ratio

=

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benötigt.

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Solution of the Cow Fence Problem

Deterministic Cow-Path

1. dir left

2. for i 0 to log n do

3. go 2i steps to direction dir

4. go 2i steps back to the origin

5. revert direction dir

6. od

Theorem [Baeza-Yates, Culberson, Rawlins, 1993]

The deterministic Cow-Path algorithm has a competitive ratio of 9.

This competitive ratio is optimal.





Exit

Performance of the Cow-Path Algorithm

Performance of the best (offline) strategy: d

– where d is the shortest way to the exit Worst case of the Cow-Path Algorithm

– d = 2x+1– Let d’=d-1

Number of steps before finding the exit:

1+1+2+2+4+4+...+d’/2+d’/2+d’+d’+2d’+2d’+d’+1 = 9 d’-1 = 9 d - 10

d’2d’d’

2d’d’+1

d’/2d’/4 ...





The Shoreline Problem

Problem description A boat is lost in a half ocean with a linear

shoreline No compass on board No sight because of dense fog The distance to the shoreline is unknown

Task Find the coast as fast as possible

?




The Spiral Solution for the Shoreline ProblemBaeza-Yates, Culberson, Rawlins, 1993

Solution: Use logarithmic spiral obeying

– where r is the polar radius from the starting point

– and is the polar angle

Numerical optimization leads to a competitive optimal ratio for k=1.250...

The shoreline problem can be solved using the logarithmic spiral method with competitive ratio 13.81...

1

k2




Searching for a point in a Grid

Problem:– Find a spot in a grid without knowing

the coordinates– (finding the restaurant in New York

without policemen) Solution:

– Use a spiral covering all points in Manhattan distance 1,2,3,4,...

Theorem [Baeza-Yates, Culberson, Rawlins, 1993]

– Using the spiral method this problem can be solved with competitive ratio 2d, where d is the Hamming distance between start and target.

– This competitive ratio is optimal.

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benötigt.




Search for a point in m Concurrent Rays

Problem: A robot is at the meeting point of m rays It has to find a point on one of the rays Find the shortest pathVariants1. Variant: the distance n is known:

Then a (2m-1) competitive (deterministic) algorithm optimally solves the case

2. Variant: distance is not known

Visit in round i, i+m, i+2m, .. ray i– no other ordering can improve the ratio

Perform in each ray test– such that ray 1+(i mod m) is searched f(i)

steps deep. Observe for all i>m for all reasonable

algorithms:


benötigt.





Spiral Search Cow

Spiral-Search-Cow1. i 02. while bull not found do3. i i+14.

5. explore f(i) steps of ray (i mod m)+16. return to the starting point7. od

TheoremThe spiral search cow algorithm has a competitive ratio of

Proof:Worst case: bull in depth f(i)+1


benötigt.





The spiral search cow algorithm has a competitive ratio of

TheoremProof:Worst case: bull in depth f(i)+1Steps of spiral-search-cow:

Competitive ratio:


benötigt.




Theorem: The Spiral-Search-Cow is optimal up to a constant term o(1).

Proof: Visit in round i, i+m, i+2m, .. ray i

–no other ordering can improve the ratioPerform in each ray test

–such that ray 1+(i mod m) is searched f(i) steps deep.

Observe for all i>m for all reasonable algorithms:

Compute the competitive ratio by

Consider the constant c upper bounding

Some (involved) analysis shows that this constant is minimal for

which matches exactly the behavior of the spiral search cow




Deterministic and Probabilistic Competitive Ratio

Deterministic Competitive Analysis– Compare the cost of the solution of an

instance x• CostAlg(x)

– to the best possible offline solution (unknown to the cow)

• Costoffline(x)Minimize the competitive ratio

=Zur Anzeige wird der QuickTime™

Dekompressor „TIFF (Unkomprimiert)“ benötigt.

Probabilistic (Randomized) Competitive Analysis

– Allow the algorithm to use random input

– Compare the cost of the expected solution of an instance x

• E[CostAlg(x)]– to the best possible offline solution

• independent from the random numbers

• unknown to the algorithm• Costoffline(x)

Minimize





Why Randomness Helps

-1:1=

13:1=

34:2=

28:2=

49:3=

310:4=

2.511:5=

2.219:3=6.3..

20:4=5

21:6=3.5

22:7=3.1..

23:8=2.8.. ...

42:6=7

43:7=6.1..

11:5=2.2

• Expected probabilistic competitive ratio: 6• Optimal deterministic ratio: 9





Smart Cow [Kao, Reif, Tate]

5. repeat

6. Explore path (p) up to distance d

7. d d r

8. p p mod m +1

9. until target found

Theorem

For any r>1 Smart Cow has a competitive ratio of

Let c:= minr>1 (1+r)/ln r

and let r* be r minimizing this term

Theorem

Smart Cow is the optimizes the randomized competitive ratio of the cow and the fence problem for r=r* with ratio 1+c = 4.59112..




Probabilism versus Determinism

mRandomized

Competitive Ratio of Smart-Cow

Optimal Deterministic Ratio of Spiral-

Search-Cow

2 4.59112... 9

3 7.73232... 14.5

4 10.84181... 19.96296...

5 13.94159... 25.41406...

6 17.03709... 30.85984...

7 20.13033... 36.30277...




The Wall Problem

Instance:– A set of non-overlapping oriented

rectangles multiples of the unit size in a d x d - square

– Player is nearsighted– If we hit a wall we immediately know

its geometryProblem:

– Minimize the path to an infinite line parallel to the rectangles

Question:– What is the competitive ratio?– i.e. ratio of found path / shortest path




Lower Bound for the Wall Problem[Papadimitriou, Yannakakis, 1991]

TheoremNo deterministic algorithm can achieve a competitive ratio of o(n1/2)

Proof– Place n obstacles of size 1xn into

the deterministic path of the player– such that the middle of each

obstacle – Then, the length of the path of the

algorithmis n x n/2 = n2/2

– There exists a horizontal line – which is at most n3/2 steps

upwards– and hits at most n1/2 rectangles

– If such a line would not exist then the total area covered by the rectange would be larger than n2

– Hence, the offline solution is at most n3/2

– This leads to the ratio n1/2




Upper Bounds for the Wall Problem

Theorem [Blum, Raghavan, Schieber 1991]The Wall problem can be solved with thesweep algorithm with competitive ratio O(n1/2)

Theorem [Fiat, Karloff, Rosen, Berman, Blum, Saks 1996]There is a O(n4/9log n) competitive randomized algorithm for the wall problem.




Deterministic Optimal Solution for the Wall Problem

Techniques– Move upwards, downwards or right

• But never to the left– Algorithm works in phases

• Guess distance n by W• If something fails we proceed with the

next phase• at the end Wf = O(n)

– Use a window to prevent drifting apart• In each phase this window size

doubles– If payable use a full sweep

• Circumvent a rectangle on the shortest route and switch back to the original height

• if it costs at most n1/2

indicated by T = W/n1/2

– If it is too expensive perform at most n1/2

sweeps in the window• A Sweep is a monotone path upwards

or downwards (indicated by dir)• Use a counter (count) that avoids too

many sweeps more than n1/2




Case 4

Case 3

Case 2

The Wall Algorithm

1. dir downwards2. count 13. W n4. T n1/2

5. while wall not reached do (phase starts) 6. walk to the right to the next obstacle O7. If the distance to the nearest corner of O

is at most T then8. perform full sweep9. else if O spans the entire window then10. Go to the nearest corner11. W 2 W12. T W/n1/2

13. count 114. reverse dir15. else if the corner of O in direction dir is inside

the window then16. go to the corner in direction dir of O

else17. count count+118. reverse dir19. if count > n1/2 then20. W 2 W21. T W/n1/2 22. count 123. fi24. fi25. od

Case 1

T

T

W

dir

T

T

dirT

T




The Wall Algorithm needs O(Wf n1/2) stepswhere Wf is the final window size

We bound the number of steps in each phase by c W n1/2 for a constant c

– Then the over-all number of steps is O(Wf n1/2)

Case 1:– Each full sweep costs T = W/n1/2

– The number of full sweeps is bounded by n

• This leads to O(W n1/2) stepsCase 2:

– ends a phase without moving, no costCase 3 and 4:

– each sweep costs at most n vertical steps

– The number of sweeps is bounded O(n1/2) by the count mechanism

• This leads to O(W n1/2) steps

T

1st sweep

2nd sweep

3rd sweep




The shortest path has at least length (Wf)

We need to prove that the increase of window size is justified

Case 2: An obstacle spans the complete window

–Then the shortest path cannot lie within the window and therefore it is at least W/2

Case 4: The number of sweeps is larger than n1/2

–After each sweep we have collected a number of rectangles that obstruct each path by at least T=W/ n1/2

–So all paths inside this window have minimum length n1/2 W/ n1/2 = W

T

W

T

T





The Competitive Ratio of the Wall Algorithm

The Wall Algorithm needs O(Wf n1/2) steps where Wf is the final window size

The shortest path has at least length (Wf)

The competitive ratio is

=Zur Anzeige wird der QuickTime™

Dekompressor „TIFF (Unkomprimiert)“ benötigt.

O(Wf n1/2)

(Wf)

= O(n1/2)




The Room Problem

Instance:

– A set of non-overlapping oriented rectangles multiples of the unit size in a d x d - square

– Starting point in the corner

– Player is nearsighted Problem:

– Minimize the path to the middle of the square

Observation:

– shortest path has length of at most d

Theorem [Blum, Raghavan, Schieber 1991]

The room problem can be solved with competitive ratio of O(n1/2)

Theorem Fiat, Bar-Eli, Berman, Yan [94]

Ø The room problem can be solved with competitive ratio of O(log n)

Ø There is no better algorithm.




s-t-Problem

Instance:

–A set of non-overlapping oriented possible unbounded rectangles multiples of the unit size

–Starting point s

–Target tKnown

–coordinates of s and t are known

–barriers in distance 1 Problem:

–Minimize the path from s to tTheorem

–There is a O(n1/2) competitive algorithm for the s-t-problem

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Thanks for your attentionEnd of 12th lecture

Next lecture: Mo 24 Jan 2005, 11.15 am, FU 116Next exercise class: Mo 17 Jan 2005, 1.15 pm, F0.530 or We 19 Jan 2005, 1.00 pm, E2.316

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Search Algorithms Winter Semester 2004/2005 17 Jan 2005 12th Lecture