1. 1 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Search Algorithms Winter Semester 2004/2005 17 Jan 2005 12th Lecture Christian Schindelhauer schindel@upb.de

2. Search Algorithms, WS 2004/05 2 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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

3. Search Algorithms, WS 2004/05 3 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 ???

4. Search Algorithms, WS 2004/05 4 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Competitive Analysis  How to evaluate the online solution  Classical 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 Cost Alg (x) –to the best possible offline solution (unknown to the cow) Cost offline (x)  Minimize the competitive ratio  =

5. Search Algorithms, WS 2004/05 5 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Solution of the Cow Fence Problem Deterministic Cow-Path 1.dir  left 2.for i  0 to log n do 3. go 2 i steps to direction dir 4. go 2 i 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.

6. Search Algorithms, WS 2004/05 6 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 = 2 x +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’/2 d’/4...

7. Search Algorithms, WS 2004/05 7 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 ?

8. Search Algorithms, WS 2004/05 8 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer The Spiral Solution for the Shoreline Problem Baeza-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k2

9. Search Algorithms, WS 2004/05 9 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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.

10. Search Algorithms, WS 2004/05 10 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 path Variants 1.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:

11. Search Algorithms, WS 2004/05 11 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Spiral Search Cow Spiral-Search-Cow 1.i  0 2.while bull not found do 3. i  i+1 4. 5. explore f(i) steps of ray (i mod m)+1 6. return to the starting point 7. od Theorem The spiral search cow algorithm has a competitive ratio of Proof: Worst case: bull in depth f(i)+1

12. Search Algorithms, WS 2004/05 12 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer The spiral search cow algorithm has a competitive ratio of Theorem Proof: Worst case: bull in depth f(i)+1 Steps of spiral-search-cow: Competitive ratio:

13. Search Algorithms, WS 2004/05 13 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 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:  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

14. Search Algorithms, WS 2004/05 14 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Deterministic and Probabilistic Competitive Ratio  Deterministic Competitive Analysis –Compare the cost of the solution of an instance x Cost Alg (x) –to the best possible offline solution (unknown to the cow) Cost offline (x)  Minimize the competitive ratio  =  Probabilistic (Randomized) Competitive Analysis –Allow the algorithm to use random input –Compare the cost of the expected solution of an instance x E[Cost Alg (x)] –to the best possible offline solution independent from the random numbers unknown to the algorithm Cost offline (x)  Minimize

15. Search Algorithms, WS 2004/05 15 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Why Randomness Helps - 1:1= 1 3:1= 3 4:2= 2 8:2= 4 9:3= 3 10:4= 2.5 11:5= 2.2 19:3= 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

16. Search Algorithms, WS 2004/05 16 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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:= min r>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..

17. Search Algorithms, WS 2004/05 17 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Probabilism versus Determinism m Randomized Competitive Ratio of Smart-Cow Optimal Deterministic Ratio of Spiral-Search-Cow 24.59112...9 37.73232...14.5 410.84181...19.96296... 513.94159...25.41406... 617.03709...30.85984... 720.13033...36.30277...

18. Search Algorithms, WS 2004/05 18 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 geometry  Problem: –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

19. Search Algorithms, WS 2004/05 19 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Lower Bound for the Wall Problem [Papadimitriou, Yannakakis, 1991] Theorem No deterministic algorithm can achieve a competitive ratio of o(n 1/2 ) 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 algorithm is n x n/2 = n 2 /2 –There exists a horizontal line –which is at most n 3/2 steps upwards –and hits at most n 1/2 rectangles –If such a line would not exist then the total area covered by the rectange would be larger than n 2 –Hence, the offline solution is at most n 3/2 –This leads to the ratio n 1/2

20. Search Algorithms, WS 2004/05 20 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Upper Bounds for the Wall Problem Theorem [Blum, Raghavan, Schieber 1991] The Wall problem can be solved with the sweep algorithm with competitive ratio O(n 1/2 ) Theorem [Fiat, Karloff, Rosen, Berman, Blum, Saks 1996] There is a O(n 4/9 log n) competitive randomized algorithm for the wall problem.

21. Search Algorithms, WS 2004/05 21 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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 W f = 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 n 1/2  indicated by T = W/n 1/2 –If it is too expensive perform at most n 1/2 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 n 1/2

22. Search Algorithms, WS 2004/05 22 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Case 4 Case 3 Case 2 The Wall Algorithm 1.dir  downwards 2.count  1 3.W  n 4.T  n 1/2 5.while wall not reached do (phase starts) 6. walk to the right to the next obstacle O 7. If the distance to the nearest corner of O is at most T then 8. perform full sweep 9. else if O spans the entire window then 10. Go to the nearest corner 11. W  2 W 12. T  W/n 1/2 13. count  1 14. reverse dir 15. else if the corner of O in direction dir is inside the window then 16. go to the corner in direction dir of O else 17. count  count+1 18. reverse dir 19. if count > n 1/2 then 20. W  2 W 21. T  W/n 1/2 22. count  1 23. fi 24. fi 25. od Case 1 TT TT W dir TT TT TT TT

23. Search Algorithms, WS 2004/05 23 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer The Wall Algorithm needs O(W f n 1/2 ) steps where W f is the final window size  We bound the number of steps in each phase by c W n 1/2 for a constant c –Then the over-all number of steps is O(W f n 1/2 )  Case 1: –Each full sweep costs T = W/n 1/2 –The number of full sweeps is bounded by n This leads to O(W n 1/2 ) steps  Case 2: –ends a phase without moving, no cost  Case 3 and 4: –each sweep costs at most n vertical steps –The number of sweeps is bounded O(n 1/2 ) by the count mechanism This leads to O(W n 1/2 ) steps TT 1st sweep 2nd sweep 3rd sweep

24. Search Algorithms, WS 2004/05 24 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer The shortest path has at least length  (W f )  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 n 1/2 –After each sweep we have collected a number of rectangles that obstruct each path by at least T=W/ n 1/2 –So all paths inside this window have minimum length n 1/2 W/ n 1/2 = W TT W TT TT

25. Search Algorithms, WS 2004/05 25 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer The Competitive Ratio of the Wall Algorithm  The Wall Algorithm needs O(W f n 1/2 ) steps where W f is the final window size  The shortest path has at least length  (W f )  The competitive ratio is  = O(W f n 1/2 )  (W f ) = O(n 1/2 )

26. Search Algorithms, WS 2004/05 26 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer 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(n 1/2 ) Theorem Fiat, Bar-Eli, Berman, Yan [94] 1.The room problem can be solved with competitive ratio of O(log n) 2.There is no better algorithm.

27. Search Algorithms, WS 2004/05 27 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer s-t-Problem  Instance: –A set of non-overlapping oriented possible unbounded rectangles multiples of the unit size –Starting point s –Target t  Known –coordinates of s and t are known –barriers in distance 1  Problem: –Minimize the path from s to t  Theorem –There is a O(n 1/2 ) competitive algorithm for the s-t-problem

28. 28 HEINZ NIXDORF INSTITUTE University of Paderborn Algorithms and Complexity Christian Schindelhauer Thanks for your attention End of 12th lecture Next lecture:Mo 24 Jan 2005, 11.15 am, FU 116 Next exercise class:Mo 17 Jan 2005, 1.15 pm, F0.530 or We 19 Jan 2005, 1.00 pm, E2.316