管理精英宣言最新課件

1、A simple Test For the Consecutive Ones PropertyWithout PC-trees!Consecutive 1s Property of matrices Given a (0,1)- matrix M, does there exist a PERMUTATION of the COLUMINS of M such that the 1s in the ROWS are consecutive? 1 2 3 4 1 1 0 0 1 0 0 1 0 1 1 03 2 1 40 1 1 00 0 1 11 1 0 0Consecutive requir

2、ement on the rows Each row i of M can be viewed as a requirement that those columns with a 1 in row j must be consecutive. Booth and Lueker 1976 showed that the consecutive ones property can be tested using P-Q trees in linear time. They process the consecutive requirement in a row by row fashion.P-

3、Q Trees Two types of internal nodes: P-nodes & Q-nodesChildren of a P-node can be “permuted arbitrarily”Children of a Q-node can only be “reversed” QP1234L(T) = all permutations generated by T In the example, L(T) = 1234,1243,4321,3421 Intermediate On-Line OperationsStrictly Overlapping Relationship

4、s Two columns are say, to overlap strictly if they overlap but none is contained in the other. Such a pair of rows implies the following column partition: 1 - - - - - - 1 1 - - - 1 1 - - - - 1 uvIdeal Situation If there is a vertex ordering v1 , v2 ,vm such that each vi strictly overlaps with some v

5、j with j i, then it is trivial to test the consecutive ones propertyPartition Before 1- - - - - - 1 1 - - - 1 1 - - - - 1 After 1 - - - 1 1 - - 1 1 - - 1 1 - - - 1 1 - - - - 1 The General Case (I) Define the graph G on the set of rows whose edge set consists of those strictly overlapping pairs of co

6、lumns. Each connected component of G satisfies the above “ideal situation”. The corresponding submatrices are called prime The matrix satisfies the COP iff each of its prime submatrices doesAn example of the Graph G1234567891016437981052The General Case (II) However, we cannot afford to compute all

7、the edges in G, which could take O(r2 ) time. We shall compute a subset of edges that contain a spanning tree of each connected component. Note that the process of obtaining the component actually decompose the matrix into prime submatricesAn Efficiency NoteThe # of strictly adjacent pairs is |A| |B

8、| . Let a, b bethe least indexed rows in A,B, respectively. To connect A,B, it suffices to make a adjacent to all rows in B and b adjacent to all rows in A.ABabAn Efficiency NoteThe # of strictly adjacent pairs is |A| |B| . Let a, b bethe least indexed rows in A,B, respectively. To connect A,B, it s

9、uffices to make a adjacent to all rows in B and b adjacent to all rows in A.ABabRepresentative Rows vA and vBvv1/21/21 1 1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 Let v be adjacent to both A and B. But, vA and vB are forbidden to be made adjacent to A,BvAvBvAvBvBvAClassifying the neighbors of a row u uBDCAAppe

10、nd A(u),B(u) and D(u) to PT(u).Append uD to PT(w) for all w in C(u) whose index is smaller than Ind(uD)Delete the row u and use an artificial column u to replace the region covered by columns of uAdd edges from u to nodes of PT(u)-FB(u)6145321 10 00 00 00 00 10 0 0 0 00 0 0 1 10 0 1 1 01 1 0 0 00 1

11、1 0 01 1 0 0 0161235640 0 0 00 0 0 10 1 1 00 0 1 1.5 0 1 11 11 00 00 00 0452631632235646453165321 .50 01 00 00 0 0 0 1 1 0 0 1 .5 1 1 35640.506451645321 10 11 100.556465164532.5 .51 00.556Lemma 1 If uj FB(ui)PT(ui), ij, ui and uj are connected in GLemma 2 If one of the ui and uj (ij) is contained in

12、 the other and the containment is changed before iteraion i, ui and uj are connected in G.0.5uiuiujujukuk0The sub-graph G generated by the algorithmG is a spanning sub-graph of G(M) with the same components.Claim 1. G is a subgraph of G(M). If(ui,uj) G(M), (ui,uj) GClaim 2. if(ui, uj) G(M), then ui

13、and uj belong to the same component of G(M) Claim 1G is a subgraph of G(M)ukBukAukuk0.50.5In this case, ui is in FB(uj) and uj is in FB(ui)1. ui and uj are independent originally.2. ui is contained in uj originally. (Lemma 2)Claim 2If(ui, uj) E(G(M), then ui and uj belong to the same component of G.

14、 Let ui,uj be the minimal bad pair. (for all other bad pair (up,uq) either ip or jq) Consider the changing of intersection relationship “intersect” to “contain” (case 1) “intersect” to “independent” (case 2)Case 1 :“intersect” to “contain” ui and uj intersect originally. Let one of the ui and uj be

15、contained in the other after iteration k. Consider the following two subcases:Case 1.1: Both ui and uj overlap uk.Case 1.2: Only one of the ui and uj (say, z) overlaps uk ( The other is named eA)Case 1.1 Both ui and uj overlap ukukukui is connected to uj through ukuiuiujujCase 1.2 one of ui and uj (

16、say, z) overlaps ukzeAzeAukzeAukAuk is connected to z and ukA. We shall verify if ukA is connected to eA.ukCase 1.2 Only one of the ui and uj (said) z overlaps uk Case (i) uka is contained in eA originallyBy lemma 2, uka is connected to eA. Case (ii) uka contains eA originallyzeAukAuk-1(eA) -1(ukA)

17、-1(z)If z is deleted at iteration t (t -1(eA) )zeAukAukt-1(eA) -1(z) -1(utD)eA connects utD. utD connects t. t connects z. Case 1.2 Case (iii) ukA is indepenet eA originally Let ukA overlap eA atfer interation t. ukA is connected to eA via ut Case (iv) ukA intersect eA originally (ukA, eA) becomes t

18、he minimal bad pair. (a contradiction) It concludes that ukA is connected to eA in G such that eA and z is connected in G.Case 2 “intersect” to “independent” ui and uj intersect originally. Let one of the ui and uj become indepedent after iteration k. consider the following two subcases:Case 2.1: Bo

19、th ui and uj overlap uk.Case 2.2: Only one of the ui and uj (said) z intersects uk (The other is named eA)Case 2.1 Both ui and uj overlap ukukukui is connected to uj through uk in GuiuiujujCase 2.2 Only one of the ui and uj (say, z) intersects ukzeAzeAzeAukAuk is connected to z and ukA. We shall ver

20、ify if ukA is connected to eA.ukCase 2.2 Only one of the ui and uj (said) z intersects uk (i) ukA is independent to eA or one is contained in the other originally.Check Claim 1 (ii) ukA intersects eA originally.If ukA is not connected to eA, (ukA ,eA) becomes the minimal bad pair. (a contradiction)L

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