1. Binary Search on a Tree Shay Mozes (Brown University)
Krzysztof Onak (MIT)
Oren Weimann (MIT)

4. Locating differences – Binary Search on a Tree? ?
Directories checksums = 4

5. Locating differences – Binary Search on a Tree? ?
Directories checksums = 5

6. Search Startegy search strategy = decision tree
Optimal strategy = shallowest decision tree a
(a,c)?
(a,b)?
(c,e)? a c
(a,d)? b d
(c,f)?
(e,g)? e f g b d c e f
Binary search – tree is a line.
Decision tree – complete binary tree g

7. Searching in trees and posets
Related Work
Tree edge ranking
Searching in trees and posets
IRV [Disc. Appl. Math. 1991]: 2-approx. in O(nlogn)
TGS [Algorithmica 1995]: O(n3logn)
LY [SODA 1998]: O(n)
BFN [SODA 1997]: O(n4log3n)
LN [ENDM 2001]: 2-approx. in O(nlogn)
OP [FOCS 2006]: O(n3)
This paper: O(n)
Applications:
Ours: filesystem sync, bug detection
Ranking: VLSI design, matrix factorization
BFN – Ben Asher Farchi Newman
LN – Laber Nogueira
OP = Onak Parys
IRV – Iyer, Ratliff, Vijayan
TGS – de la Torre, Greenlaw, Schaffer
LY – Lam, Yue

8. Strategy Function f : E → {1,2,3,…}
If f(e1 ) = f(e2 ), then on the path from e1 to e2 there is e3 with f(e3 ) > f(e1 ) = f(e2 )
strategy function bounded by k
decision tree of height k a
(a,c)?
(a,b)?
(c,e)? a c
(a,d)? b d
(c,f)?
(e,g)? e f g 3 2 1
Intuition: f(e) = k, at most k more queries to go. b d c e f g

9. Solution Approach Given tree
Find strategy function with the least maximum
Convert into a decision tree c a d f e g b
(a,c)?
(a,b)?
(c,e)? a c
(a,d)? b d
(c,f)?
(e,g)? e f g 3 2 1

10. Visibility
e is visible from vertex v if on path from v to e there is no value greater than e
Lexicographic order on visibility sequences e.g.: (3,2,1) > (3,1)
3,1 a 1 3
3,2,1 d c 2 1 e f 1 g

11. Bottom-Up Approach Given visibility sequences at children of v
Extend to minimal visibility sequence at v
Theorem [OP, de la Torre et al.]: Minimal extensions accumulate to an optimal solution
Given strategy functions at children of v v 2 3 3 2 c d f e g 2 1 2 1 1

12. Valid Extension . . . An extension assigns all f(ei)’s v f(e1)? f(e2)?
f(ek)? s1 s2 sk 4 1 4 1 1
. . .

13. Valid Extension . . . An extension assigns all f(ei)’s f(ei)= f(ej) v3 3
f(ek)? s1 s2 sk 4 1 4 1 1
. . .

14. Valid Extension . . . An extension assigns all f(ei)’s f(ei)= f(ej)
f(ei) is not in si v 4
f(e2)?
f(ek)? s1 s2 sk 4 1 4 1 1
. . .

15. Valid Extension . . . An extension assigns all f(ei)’s f(ei)= f(ej)
f(ei) is not in si
f(ei) is in sj f(ej) > f(ei) v 3
f(e2)? 4 s1 s2 sk 4 1 4 1 1
. . .

16. Valid Extension . . . An extension assigns all f(ei)’s f(ei)= f(ej)
f(ei) is not in si
f(ei) is in sj f(ej) > f(ei)
u is in si and sj max{f(ei), f(ei)} > u v 2 3
f(ek)? s1 s2 sk 4 1 4 1 1
. . .

17. Algorithm Outline . . . v f(e1)? f(e2)? f(ek)? s1 s2 sk free values 14 1 1 1
. . .
free values 1 2 3 4 5 6

18. Algorithm Outline . . . set u = max{si} u v f(e1)? f(e2)? f(ek)? s1 s2sk 4 1 1 1
. . .
u is most problematic u
free values 1 2 3 4 5 6

19. Algorithm Outline . . . set u = max{si} while not all edges assigned uv
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . . u
free values 1 2 3 4 5 6

20. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u v
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . .
- u is not problematic cause appears only once, but is not free to use again
- Next largest u that actually appears in some s_i u
free values 1 2 3 4 5 6

21. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u v
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . . u
free values 1 2 3 5 6

22. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise: v
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . . u
free values 1 2 3 5 6

23. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u v
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . .
w is smallest value that can resolve u w u w
free values 1 2 3 5 6

24. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w v
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . .
Its best to assign w to the visibility sequence s.t w makes many values in it disappear –
-NOT NECCESARILY ONE OF THE EDGES THAT u APPEARS IN w u w
free values 1 2 3 5 6 Sj

25. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free v
f(e1)?
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . . w u w
free values 1 3 5 6 Sj

26. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w v 2
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . . w u w
free values 1 3 5 6 Sj

27. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free v 2
f(e2)?
f(ek)? s1 s2 sk 4 1 1 1
. . .
No such values since no integer between 1 and 2 w u w
free values 1 3 5 6 Sj

28. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2
f(e2)?
f(ek)? s1 s2 sk 4 1 1
. . . w u w
free values 1 3 5 6 Sj

29. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2
f(e2)?
f(ek)? s1 s2 sk 4 1 1
. . . u
free values 1 3 5 6

30. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2
f(e2)?
f(ek)? s1 s2 sk 4 1 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 1 3 5 6 Sj

31. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2
f(e2)?
f(ek)? s1 s2 sk 4 1 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 1 5 6 Sj 31

32. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2 3
f(ek)? s1 s2 sk 4 1 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 1 5 6 Sj 32

33. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2 3
f(ek)? s1 s2 sk 4 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 1 5 6 Sj 33

34. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj v 2 3
f(ek)? s1 s2 sk 4 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u
free values 1 5 6 34

35. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 2 3
f(ek)? s1 s2 sk 4 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u
free values 5 6 35

36. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 2 3
f(ek)? w s1 s2 sk 4 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 5 6 Sj 36

37. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 2 3
f(ek)? w s1 s2 sk 4 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 6 Sj 37

38. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w (free previous value)
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 5 3
f(ek)? w s1 s2 sk 4 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 2 6 Sj 38

39. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w (free previous value)
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 5 3
f(ek)? w s1 s2 sk 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u w
free values 2 4 6 Sj 39

40. Algorithm Outline . . . set u = max{si} while not all edges assigned u
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w (free previous value)
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 5 3
f(ek)? s1 s2 sk 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! u
free values 2 4 6 40

41. Algorithm Outline . . . set u = max{si} while not all edges assigned w
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w (free previous value)
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 5 3
f(ek)? s1 s2 sk 1
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! w u w
free values 2 4 6 Sj 41

42. That’s it! Algorithm Outline . . . set u = max{si}
while not all edges assigned
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w (free previous value)
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v 5 3 2 s1 s2 sk
. . .
Here you can see that the maximal sequence w.r.t w is one that does not contain u ! w u w
free values 4 6 Sj 42

43. That’s it! Algorithm Outline set u = max{si}
while not all edges assigned
if u appears once mark u as not free, move to next largest u
otherwise:
w = smallest free value > u
Sj = any maximal sequence w.r.t w
mark w as not free
set current f(ej) = w (free previous value)
mark all Sj values between u and w as free
remove all values < w from Sj
and u = 0 v s1 s1 5 3 2

44. Running Time v s1 s2 sk 4 1 1 1
. . .

45. Running Time
Easy in O(|S1| + |S2| +…+ |Sk|) per vertex O(n2) total v s1 s2 sk 4 1 1 1
But we can actually do it in |s_1|+…+|s_k| yielding n^2 time complexity !
. . . 45

46. Running Time
Easy in O(|S1| + |S2| +…+ |Sk|) per vertex O(n2) total
But, |S1| + |S2| +…+ |Sk| is not a lower bound! v s1 s2 sk 4 1 1 1
But we can actually do it in |s_1|+…+|s_k| yielding n^2 time complexity !
. . . 46

47. Running Time
Easy in O(|S1| + |S2| +…+ |Sk|) per vertex O(n2) total
But, |S1| + |S2| +…+ |Sk| is not a lower bound!
in many cases, the largest values of the largest visibility sequence are unchanged at v itself v s1 s2 sk 4 1 1 1
But we can actually do it in |s_1|+…+|s_k| yielding n^2 time complexity !
. . . 47

48. Σ Linear Running Time . . . q(v) = |S2| +…+ |Sk| k(v) = #v’s children
t(v) = largest value that appears in Sv but not in S1
Theorem: an extension can be computed in time O( k(v)+q(v)+t(v) )
Theorem: k(v)+q(v)+t(v) = O(n)
Differs from [LY98]:
different data structures
No bit tricks
O(n) decision tree construction v Σ v s1 s2 sk 4 1 1 1
Different than Lam and Yue’s solution.
Many details omitted
. . .

49. From Strategy Function to Decision Tree in O(n) Time

50. From Strategy Function to Decision Tree in O(n) Time
(a,c)? 2 1 a c 3 b d c
(a,b)?
(c,e)? 2 1 a b c e e f 1
(a,d)? b
(c,f)?
(e,g)? a d c f e g g a d c f e g

51. From Strategy Function to Decision Tree in O(n) Time
(a,c)?
For all edges e
let s = visibility sequence at bottom(e)
if s contains no values smaller than f(e)
set bottom(e) as the solution when the query on e returns bottom(e)
else, let v1 <...< vk < f(e) in s, let ei be the edge vi is assigned to
set ek as the solution when the query on e returns bottom(e)
for every 1≤ i <k set ei as the solution when the query on ei+1 returns top(ei+1)
set top(e1) as the solution when the query on e1 returns top(e1) 2 1 a c 3 b d c
(a,b)?
(c,e)? 2 1 a b c e e f 1
(a,d)? b
(c,f)?
(e,g)? a d c f e g g a d c f e g

52. Thank You!