# 1 © 2012 Prof. Dr. Franz J. Brandenburg Ranking Problems with Incomplete Information Fixed Parameter Tractability of Distance Problems Franz J. Brandenburg.

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1 © 2012 Prof. Dr. Franz J. Brandenburg Ranking Problems with Incomplete Information Fixed Parameter Tractability of Distance Problems Franz J. Brandenburg University of Passau, Germany

2 © 2012 Prof. Dr. Franz J. Brandenburg Survey the problem the motivation the solution new open problems

3 © 2012 Prof. Dr. Franz J. Brandenburg Similarity of Permutations Definition: Given two total orders or permutations  and  on {1,2,....,n}. How can we measure their dissimilarity? 1) count mismatches --> Kendall-tau 2) count moves --> Spearman footrule 3) others (Hamming, count exchanges of x‘s and y‘s) rotate the middle swap the extremes

4 © 2012 Prof. Dr. Franz J. Brandenburg Kendall tau Distance Definition: Given two total orders or permutations  and  on {1,2,....,n}. The Kendall-tau (or Kemeny) distance K( ) is K( ,  ) = #{(x, y) | x < y and (  (x)–  (x))(  (y)–  (y)) < 0} = # disagreements: x <  y  and y <  x, = # the „dirty“ pairs = # inversions (swaps) to transform  into . = bubbleSort distance  rotate the middle, K( ) = 2 swap the extremes, K( ) = 6

5 © 2012 Prof. Dr. Franz J. Brandenburg Kendall-tau Kendall-tau distance named after Maurice Kendall in 1938 (statistics) invented by Gustav Fechner in 1897 Kendall-tau distance = two-layer crossing problem There is an O(n logn) algorithm to compute - the number of crossings of n lines - the Kendall-tau distance of two total orders Open problem Is there an O(n) algorithm? Compute the inversion numbers (D.E. Knuth 1968) inv(i) = #{ j | j > i and j left of i} Updates in O(log n) in a search tree.

6 © 2012 Prof. Dr. Franz J. Brandenburg Spearman-Footrule distance Spearman-footrule distance or Spearman's rho named after C. Spearman (1904) (correlation ranking in statistics) compute displacements of two permutations move an element by k units the L 1 - vector norm F( ,  ) = ∑ i |(  (i)–  (i)| Lemma For total orders the Spearman footrule distance can be computed in O(n) value 8 value 2

7 © 2012 Prof. Dr. Franz J. Brandenburg Diaconis-Graham Theorem The Diaconis-Graham inquality (1977) K( ,  ) ≤ F( ,  ) ≤ 2K( ,  ) each crossing/mismatch/swap induces a displacement each displacement is repaired by two crossings.

8 © 2012 Prof. Dr. Franz J. Brandenburg Incomplete Information total order x < y or y < x for every pair of candidates x and y tiesx ~ y x and y are equivalent (an equivalence relation) I don't care for x and y bucket orders with equivalent items in a bucket and a total order for the buckets partial order x ? y x any y are unrelated „apples and oranges“ ? is not transitive interval orders, hierarchical orders (trees) contradictory relations with cycles (from Lullus, 1299)

9 © 2012 Prof. Dr. Franz J. Brandenburg Generalization Given: a set of candidates X = {x 1,...,x n } or simply {1,...,n} a partial order π on X, and X is partially ordered by > say x > y if x has a higher ranking if there is a preference for x properties: transitive: x > y and y > z  x > z partial: many pairs are unrelated Example: The Australian Open 2012 Djokovic Nadal Murray Federer a partial order imposes Djokovic beats Federer Murray and Nadal / Federer are incomparable

10 © 2012 Prof. Dr. Franz J. Brandenburg Given: A set of candidates X and a partial order π Representation of π: a DAG (directed acyclic graph) with transitive edges vertices X = {1,...,n} directed edges x ---> y if x > y in π the DAG displays only the generating edges, the transitive reduction e.g. 1 and 8 are unrelated, and 8 > 2, 8 > 3, 8 > 5, 2 and 3 unrelated Partial Orders 1 8 3 2 5 4 6 7

11 © 2012 Prof. Dr. Franz J. Brandenburg Extensions How shall we compare two partial orders?... via their sets of extensions Ext(π) = {total orders  |  does not disagree with π} π(i) < π(i)  (i) <  (i) Ext(π) = {any order obtained from the DAG of π by topological sorting} Example: Ext(π = Ø} = all permutations Ext(π) = {all shuffles from left (8,2,3,5), (8,3,2,5) and right (1, 7,4,6), (1,4,7,6), (1,4,6,7)} 1 8 3 2 5 4 6 7

12 © 2012 Prof. Dr. Franz J. Brandenburg topSort An extension of π is a topological sorting topsort: do { get any source x (no incoming edges); print x; delete x; while (there are vertices) } Ext(π) = the set of all topsort runs; all possibilities for "any" Theorem (Brightwell, Winkler, ACM STOC 1991) Computing |Ext(P)| is #P complete. – breadth first8,1, 2, 3, 7, 4, 5, 6 – heap (min-heap1, 4, 6, 7, 8, 2, 3, 5 – best1, 8, 2, 3, 4, 5, 6, 7 1 8 3 2 5 4 6 7

13 © 2012 Prof. Dr. Franz J. Brandenburg Distance Measures Given: a partial order π and a total (partial) order  What is their distance? nearest neighbor distance K NN (π,  ) = min {K( ,  ) |  is an extension of π} Interpretation: the positive view, there is some extension of π at distance ≤ k Hausdorff (farthest neighbor) distance K FN (π,  ) = max {K( ,  ) |  is an extension of π} Interpretation: the negative view all extensions are within distance ≤ k breadth first 8,1, 2, 3, 7, 4, 5, 6 K(, id) = 10 heap (min-heap) 1,4, 6, 7, 8, 2, 3, 5 K(,id) = 11 best 1, 8, 2, 3, 7, 4, 5, 6 K(, id) = 6

14 © 2012 Prof. Dr. Franz J. Brandenburg Distance Measures nearest neighbors = closest red-blue pair or min min farthest neighbors = closest red-blue pair or max max Hausdorff distance = max {min distance{red,blue}} center distance = min {max distance{red,blue}}

15 © 2012 Prof. Dr. Franz J. Brandenburg Measures Hausdorff distance (Felix Hausdorff 1968-1942) is a metric. nearest neighbor, farthest neighbor, center distance are not ! since d(X,Y) = 0 does not imply X=Y and no triangle inequality In R 2, for points p = (x,y) all four distances are in  (n log n).

16 © 2012 Prof. Dr. Franz J. Brandenburg for Partial Orders 1 2 3 4 5 6 7 8 8 1 2 3 7 4 5 6 (breadth first) Kendall-tau = 10 Spearman = 18 1 2 3 4 5 6 7 8 1 4 6 7 8 2 3 5 (min-heap) Kendall-tau = 11 Spearman = 22 1 8 3 2 5 4 6 7 1 2 3 4 5 6 7 8 1 8 2 3 4 5 6 7 (opt) Kendall-tau = 6 Spearman = 12

17 © 2012 Prof. Dr. Franz J. Brandenburg Applications ranking problems –in sport: who is the champion ranking –in metasearch aggregate data from several search engines top-k lists –voring systems

18 © 2012 Prof. Dr. Franz J. Brandenburg Sport Who was the best Formula 1 driver 2011? Sebastian Vettel, Ger392 pts Jenson Button, Eng270 pts Mark Webber, Aus258 pts Schema: weighted Borda scores, sum points (25,18,...,1) Who is the best tennis player? The best possible ranking? –the winner of the Australian open? but Djokovic did not play Federer? –the aggregate winner of the for Grand Slams? evaluate the data from four trees: incomplete data The ranking list is by weighted Borda scores An alternative: US-sports phase 1: scores phase 2: finals

19 © 2012 Prof. Dr. Franz J. Brandenburg Meta Search meta search machines (incl. google) aggregate the rankings (results) of many single searchers searcher_1 = (2,4,3,5,1,...) searcher_2 = (4,1,3,5,2,...) searcher_3 = (5,3,2,1,4,...) e.g. for hotels, flights, rental cars,... but in practice many providers do give you the best offer. They make the \$\$\$ How do they aggregate? How do they compute the ranking? the top - k list? the first page?

20 © 2012 Prof. Dr. Franz J. Brandenburg Rank Aggregation The rank aggregation problem: Given: a collection of total orders / permutations over n elements (  1,  2,...,  m ) Problem: the best compromise (Kemeny score) a permutation or a linear order  * such that distance(  *,  1,  2,...,  m ) = ∑ i distance(  *,  i )  MIN treat every voter as fair as possible (Biedl, B., Deng, Disc. Math 2009) max i {distance (  *,  i )}  MIN As a decision problem: Given k: Is there a  * with distance(  *,  1,  2,...,  m ) ≤ k ?

21 © 2012 Prof. Dr. Franz J. Brandenburg Facts Theorem: The rank aggregation problem under the Kendall-tau distance (1) is NP hard - for many voters (Bartoldi, Tovey, Trick, 1989) - for even numbers > 4 (Dwork et al, WWW 2001) by a complex reduction from feedback arc set (small corrections by Biedl, Brandenburg, Deng Disc. Math. 2009) - in the max version (Biedl et al 2009) (2) in O(n) for two voters (you and your boy/girlfriend/husband...)... take any of the two or any inbetween (crossing)

22 © 2012 Prof. Dr. Franz J. Brandenburg Facts (2) (3) Kemeny score (rank aggregation with Kendal tau) is fixed parameter tractable for several parameters Kemeny score number of candidates max and average range of candidate positions (Betzler,Fellows, Guo, Niedermeier, Rosamond, TCS 410(45) 2009) In contrast: (4) rank aggregation under the Spearman footrule distance in O(n 3 ) for any number of voters (and even with local weights) (Dwork et al, 2001) by weighted matching: for i,j = 1,..., n set w i,j = What does it cost to place i at j?... and matching does the rest.

23 © 2012 Prof. Dr. Franz J. Brandenburg Results on Distances Given:a partial order π and a total order id = (1,...,n) nearest neighbor distance K(π, id) a total order  in Ext(π) such that K( ,id) ––> MIN Theorem ( Brandenburg, Gleissner, Hofmeier, Walcom 2012 / J. Comb. Opt. to appear) The nearest neigbor distance problem is NP-hard, Kendall tau: by reduction from one-sided crossing minimization in the version OSCM-4-stars Spearman: by reduction from clique fixed mobile Idea: the lower level is π with 4 elements per point no relations between points, and extra blockers and about n 2 crossings

24 © 2012 Prof. Dr. Franz J. Brandenburg NP-hard: What‘s next Approximation if we cannot solve the problem exactly (if P ≠ NP) can we solve it up to some small error Theorem The nearest neigbor distance problem of a partial and a total order is 2-approximable for the Kendall tau distance... by a reduction to a constraint feedback arc set problem on tournaments and an adaptation/improvement of the 3-approximation of Schalekamp/van Zylen using Quicksort on the feedback arc set problem is 4-approximable for the Spearman footrule distance... using the Diaconis-Graham inequality

25 © 2012 Prof. Dr. Franz J. Brandenburg FPT Given:a partial order π and a total order  over {1,....,n} a parameter k Problem: find an extension  of π in polynomial time such that – distance( ,  ) ≤ k – or show that all extensions of π have distance at least k+1 Theorem (Brandenburg, Hofmeier, Gleißner, Walcom 12, J.Comb. Opt) The distance problems for a partial order π and a total order  – for nearest neighbor Kendall tau distance – for nearest neighbor Spearman footrule distance are fixed parameter tractable with a linear kernel. Transform the problem into a small version of size 2k.

26 © 2012 Prof. Dr. Franz J. Brandenburg Intuition.... the distance problem is NP-hard but Suppose there are 1.000 elements and k=100. Then at most 100 "critical" pairs may cross at least 800 elements are not involved. GOAL: find these "800" elements and remove them solve the problem only on the "critical" pairs - naive by exhaustive search on all ≤ k! extensions of π. TODO: Improve upon the search...x..y... in π...y..x... in 

27 © 2012 Prof. Dr. Franz J. Brandenburg Our Key: a Derivation Given: a partial order π and a total order  over X = {1,....,n} The derivation   (π) of π in direction  is a binary relation over X For two elements x,y letx < y in   (π) by (i) agreement x < y both in π and  (ii) overrule x y in  (iii) takeover x  y in π and x < y in  Example 1 < 4,6,7 by agreement 1 < 2,3,5,8 by takeover 8 < 2,3,5 by overrule 1,4,6,7 < 8 by takeover cycle : 8 < 2 < 4 < 8 (overrule takeover takeover)... and only 8 is involved in eight cycles, excluding 1. 1 8 3 2 5 4 6 7

28 © 2012 Prof. Dr. Franz J. Brandenburg Derivation   (π) Lemma ( Brandenburg, Gleissner, Hofmeier Walcom 2012, J. Comb. Opt.)   (π) is complete, defined for all x,y.   (π) may have cycles. A cycle is made from one overrule and two overtakes. Proof: Completeness is by definition Cycles by a case analysis for (x,y,z) If follows from the transitivity of π and  that an agreement cannot be part of a cycle and that two overrules x < y, y < z imply x < z by the transitivity of a partial order

29 © 2012 Prof. Dr. Franz J. Brandenburg Cycle Rule Given: a partial order π and a total order  over X = {1,....,n} a parameter k Cycle rule for the reduction: For every element x remove x and keep k if x is not in a cycle, in fact not in a triangle x < y < z < x of   (π) and there is no overrule on x. Example There is no cycle and no overrule on 1. All cycles use 8 -- 2, 8 --3, 8 -- 5 1 8 3 2 5 4 6 7

30 © 2012 Prof. Dr. Franz J. Brandenburg Proof Lemma The cycle rule preserves the Kendall-tau distance. Proof: (sketch) Consider a nearest neighbor of  in Ext(π) with K( ,  ) ≤ k For every x which is removed define pred(x) = {y | y < x in   (π)} succ(x) = {z | x < z in   (π) }. Claim 1:There is an extension π* of π such that y < x for every y  pred(x) and x < z for every z  succ(x), Then x serves as a "separator". otherwise, consider the first y  pred(x) with y < x in   (π) and x < y in π*. Then... by some case analysis...y and its left neighbor can be swapped which contradict to "y is the first„ This needs some more work (see our papers). pred(x) x succ(x) in π* pred(x) x succ(x) in 

31 © 2012 Prof. Dr. Franz J. Brandenburg Proof In the running example, 1 can be removed. Place 1 at the first position in the extension of π. 1 2 3 4 5 6 7 8 1 8 2 3 4 5 6 7 then K( ) = 6 1 8 3 2 5 4 6 7

32 © 2012 Prof. Dr. Franz J. Brandenburg Kernel Lemma The nearest neighbor Kendall tau distance between π and  is ≤ k if after the cycle rule i.e. after the removal of all "separators" x there is an instance with at most 2k elements which has K(π 2k,  2k ) ≤ k. Find this solution by exhaustive search an the 2k! extensions of π 2k. Conclusion The distance problem is FPT. TODO Improve the search for K(π 2k,  2k ) ≤ k.

33 © 2012 Prof. Dr. Franz J. Brandenburg some open problems for parameterized complexity

34 © 2012 Prof. Dr. Franz J. Brandenburg Bucket Orders Theorem (Fagin et al 2006) There are O(n logn) algorithms to compute the distances (Kendall-tau aud Spearman) between bucket orders. with x < y and ties x ~ y The buckets are totally ordered. Theorem The rank aggregation problem is – NP-hard for total orders under Kendall-tau (Dwork et al 2001) – in P for many total orders under Spearman (Dwork et al 2001). – NP-hard for many bucket orders under Spearman (Brandenburg, Gleissner, Hofmeier FAW-AAAI 2011) OPEN: Is it FPT?

35 © 2012 Prof. Dr. Franz J. Brandenburg 1-planarity Definition (G. Ringel, 1965) A graph G is 1-planar if each edge is crossed at most once (by all other edges) Properites an edge coloring black with crossings red x blue a 6-vertex coloring (Borodin 1984) #edges < 4n-8 (Pach, Toth 1997, and others) not closed under edge contraction there are infinitely many minimal non-1-planar graphs (Korzhik, 2007) test is NP-hard (Korzhik, Mohar Graph Drawing 2008, LNCS 5166)

36 © 2012 Prof. Dr. Franz J. Brandenburg 1-planar + Rotation System Definition a rotation system (embedding) of a graph G = (V,E) is the cylic order of the edge (neighbors) of v for each vertex v The crossing pair system of a graph G = (V,E) is G together with all pairs (e,e‘) of crossing edges. Lemma Given a crossing pair system. Test for 1-planarity is in O(n), and there is a straight-line drawing of G on a polynomial size grid. Claim (under work) (Auer, Brandenburg, Gleißner, Reislhuber) Given a rotation system: Test for 1-planarity is NP-hard.... by a reduction from planar 3-SAT

37 © 2012 Prof. Dr. Franz J. Brandenburg Parameterized Complexity Given: a graph G = (V,E) a parameter k Problem: Is G 1-planar with at most k pairs of crossing edges? Given: G with a rotation system and k Problem: Is G 1-planar with at most k pairs of crossing edges? Given: a directed graph G = (V,E) a parameter k Problem: Is G upward 1-planar with at most k pairs of crossing edges? i.e. G has a 1-planar drawing such that all edges are upward (Y-monontone) I need your help! Thank you

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