nips nips2002 nips2002-203 knowledge-graph by maker-knowledge-mining

203 nips-2002-Using Tarjan's Red Rule for Fast Dependency Tree Construction


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Author: Dan Pelleg, Andrew W. Moore

Abstract: We focus on the problem of efficient learning of dependency trees. It is well-known that given the pairwise mutual information coefficients, a minimum-weight spanning tree algorithm solves this problem exactly and in polynomial time. However, for large data-sets it is the construction of the correlation matrix that dominates the running time. We have developed a new spanning-tree algorithm which is capable of exploiting partial knowledge about edge weights. The partial knowledge we maintain is a probabilistic confidence interval on the coefficients, which we derive by examining just a small sample of the data. The algorithm is able to flag the need to shrink an interval, which translates to inspection of more data for the particular attribute pair. Experimental results show running time that is near-constant in the number of records, without significant loss in accuracy of the generated trees. Interestingly, our spanning-tree algorithm is based solely on Tarjan’s red-edge rule, which is generally considered a guaranteed recipe for bad performance. 1

Reference: text


Summary: the most important sentenses genereted by tfidf model

sentIndex sentText sentNum sentScore

1 It is well-known that given the pairwise mutual information coefficients, a minimum-weight spanning tree algorithm solves this problem exactly and in polynomial time. [sent-6, score-0.621]

2 However, for large data-sets it is the construction of the correlation matrix that dominates the running time. [sent-7, score-0.147]

3 We have developed a new spanning-tree algorithm which is capable of exploiting partial knowledge about edge weights. [sent-8, score-0.532]

4 The partial knowledge we maintain is a probabilistic confidence interval on the coefficients, which we derive by examining just a small sample of the data. [sent-9, score-0.373]

5 The algorithm is able to flag the need to shrink an interval, which translates to inspection of more data for the particular attribute pair. [sent-10, score-0.352]

6 However, the problem of constructing Bayes’ nets from data remains a hard one, requiring search in a super-exponential space of possible graph structures. [sent-14, score-0.19]

7 Namely, we look at dependency trees, which are belief networks that satisfy the additional constraint that each node has at most one parent. [sent-17, score-0.162]

8 In this simple case it has been shown [2] that finding the tree that maximizes the data likelihood is equivalent to finding a minimumweight spanning tree in the attribute graph, where edge weights are derived from the mutual information of the corresponding attribute pairs. [sent-18, score-1.447]

9 Once the weight matrix is constructed, executing a minimum spanning tree (MST) algo- rithm is fast. [sent-21, score-0.501]

10 The time-consuming part is the population of the weight matrix, which takes time O(Rn2 ) for R records and n attributes. [sent-22, score-0.289]

11 This becomes expensive when considering datasets with hundreds of thousands of records and hundreds of attributes. [sent-23, score-0.382]

12 To overcome this problem, we propose a new way of interleaving the spanning tree construction with the operations needed to compute the mutual information coefficients. [sent-24, score-0.579]

13 This algorithm is capable of using partial knowledge about edge weights and of signaling the need for more accurate information regarding a particular edge. [sent-26, score-0.586]

14 The partial information we maintain is in the form of probabilistic confidence intervals on the edge weights; an interval is derived by looking at a sub-sample of the data for a particular attribute pair. [sent-27, score-0.862]

15 Whenever the algorithm signals that a currently-known interval is too wide, we inspect more data records in order to shrink it. [sent-28, score-0.61]

16 Once the interval is small enough, we may be able to prove that the corresponding edge is not a part of the tree. [sent-29, score-0.505]

17 Whenever such an edge can be eliminated without looking at the full data-set, the work associated with the remainder of the data is saved. [sent-30, score-0.464]

18 We have implemented the algorithm for numeric and categorical data and tested it on real and synthetic data-sets containing hundreds of attributes and millions of records. [sent-32, score-0.6]

19 The resulting trees are, in most cases, of near-identical quality to the ones grown by the naive algorithm. [sent-34, score-0.124]

20 In the context of dependency trees, Meila [11] discusses the discrete case that frequently comes up in text-mining applications, where the attributes are sparse in the sense that only a small fraction of them is true for any record. [sent-38, score-0.343]

21 The number of data records is R, the number of attributes n. [sent-41, score-0.48]

22 We denote by ρxy the correlation coefficient between attributes x and y, and omit the subscript when it is clear from the context. [sent-43, score-0.234]

23 2 A slow minimum-spanning tree algorithm We begin by describing our MST algorithm1 . [sent-44, score-0.35]

24 We then proceed to describe its use in the case where some edge weights are known not exactly, but rather only to lie within a given interval. [sent-46, score-0.391]

25 In the following discussion we assume we are given a complete graph with n nodes, and the task is to find a tree connecting all of its nodes such that the total tree weight (defined to be the sum of the weights of its edges) is minimized. [sent-48, score-0.682]

26 We start with a rule to eliminate edges from consideration for the output tree. [sent-50, score-0.334]

27 Following [5], we state the so-called “red-edge” rule: Theorem 1: The heaviest edge in any cycle in the graph is not part of the minimum 1 To be precise, we will use it as a maximum spanning tree algorithm. [sent-51, score-0.974]

28 The two are interchangeable, requiring just a reversal of the edge weight comparison operator. [sent-52, score-0.41]

29 While |L| > n − 1 do: ¯ Pick an arbitrary edge e ∈ L \ T . [sent-57, score-0.337]

30 Let e be the heaviest edge on the path in T between the endpoints of e. [sent-58, score-0.504]

31 L contains the set of edges that have been proven to not be in the MST and so L contains the set of edges that still have some chance of being in the MST. [sent-63, score-0.298]

32 Traditionally, MST algorithms use this rule in conjunction with a greedy “blue-edge” rule, which chooses edges for inclusion in the tree. [sent-66, score-0.213]

33 In contrast, we will repeatedly use the red-edge rule until all but n − 1 edges have been eliminated. [sent-67, score-0.213]

34 The proof this results in a minimum-spanning tree follows from [5]. [sent-68, score-0.27]

35 Denote by L the set of edges that have already been ¯ eliminated, and let L = E \ L. [sent-70, score-0.149]

36 As a way to guide our search for edges to eliminate we maintain the following invariant: ¯ Invariant 2: At any point there is a spanning tree T , which is composed of edges in L. [sent-71, score-0.956]

37 ¯ In each step, we arbitrarily choose some edge e in L \ T and try to eliminate it using the red-edge rule. [sent-72, score-0.458]

38 It is clear we only need to compare e with the heaviest edge in P . [sent-75, score-0.459]

39 If e is heavier, we can eliminate it by the red-edge rule. [sent-76, score-0.121]

40 However, if it is lighter, then we can eliminate the tree edge by the same rule. [sent-77, score-0.728]

41 We do so and add e to the tree to preserve Invariant 2. [sent-78, score-0.27]

42 The MIST algorithm can be applied directly to a graph where the edge weights are known exactly. [sent-80, score-0.522]

43 And like many other MST algorithms, it can also be used in the case where just the relative order of the edge weights is given. [sent-81, score-0.436]

44 Now imagine a different setup, where edge weights are not given, and instead an oracle exists, who knows the exact values of the edge weights. [sent-82, score-0.847]

45 In this setup, MIST is still suited for finding a spanning tree while minimizing the number of queries issued. [sent-85, score-0.464]

46 Otherwise, it just ignores the “if” clause altogether and iterates (possibly with a different edge e). [sent-88, score-0.337]

47 For the moment, this setup may seem contrived, but in Section 4, we go back to the MIST algorithm and put it in a context very similar to the one described here. [sent-89, score-0.117]

48 3 Probabilistic bounds on mutual information We now concentrate once again on the specific problem of determining the mutual information between a pair of attributes. [sent-90, score-0.154]

49 We show how to compute it given the complete data, and how to derive probabilistic confidence intervals for it, given just a sample of the data. [sent-91, score-0.167]

50 As shown in [12], the mutual information for two jointly Gaussian numeric attributes X and Y is: 1 I(X; Y ) = − ln(1 − ρ2 ) 2 R ((xi −¯)(yi −¯)) x y i=1 where the correlation coefficient ρ = ρXY = σX σY ˆ2 ˆ2 being the sample means and variances for attributes X and Y . [sent-92, score-0.646]

51 This is a sufficient condition for the use of |ρ| as the edge weight in a MST algorithm. [sent-94, score-0.374]

52 Consequently, the sample correlation can be used in a straightforward manner when the complete data is available. [sent-95, score-0.139]

53 We are trying to estimate i=1 xi · yi given the partial r sum i=1 xi · yi for some r < R. [sent-98, score-0.25]

54 We thus can derive a two-sided confidence interval for i Zi = i xi · yi with probability 1 − δ for some user-specified δ, typically 1%. [sent-103, score-0.302]

55 Given this interval, computing an interval for ρ is straightforward. [sent-104, score-0.168]

56 4 The full algorithm As we argued, the MIST algorithm is capable of using partial information about edge weights. [sent-106, score-0.612]

57 We have also shown how to derive confidence intervals on edge weights. [sent-107, score-0.445]

58 We initialize the tree T in the following heuristic way: first we take a small sub-sample of the data, and derive point estimates for the edge weights from it. [sent-110, score-0.739]

59 Then we feed the point estimates to a MST algorithm and obtain a tree T . [sent-111, score-0.35]

60 When we come to compare edge weights, we generally need to deal with two intervals. [sent-112, score-0.337]

61 We apply this logic to all comparisons, where the goal is to determine the heaviest path edge e and to compare it to the candidate e. [sent-114, score-0.571]

62 If we are lucky enough that all of these comparisons are conclusive, the amount of work we save is related to how much data was used in computing the confidence intervals — the rest of the data for the attribute-pair that is represented by the eliminated edge can be ignored. [sent-115, score-0.595]

63 The price we need to pay here is looking at more data to shrink the confidence intervals. [sent-119, score-0.152]

64 We do this by choosing one edge — either a tree-path edge or the candidate edge — for “promotion”, and doubling the sample size used to compute the sufficient statistics for it. [sent-120, score-1.137]

65 After doing so we try to eliminate again (since we can do this at no additional cost). [sent-121, score-0.121]

66 If we fail to eliminate we iterate, possibly choosing a different candidate edge (and the corresponding tree path) this time. [sent-122, score-0.795]

67 The choice of which edge to promote is heuristic, and depends on the expected success of resolution once the interval has shrunk. [sent-123, score-0.505]

68 Consider the comparison of the path-heaviest edge to an estimate of a candidate edge. [sent-126, score-0.404]

69 The candidate edge’s confidence interval may be very small, and yet still intersect the interval that is the heavy edge’s weight (this would happen if, for example, both attribute-pairs have the same distribution). [sent-127, score-0.493]

70 We may be able to reduce the amount of work by pretending the interval is narrower than it really is. [sent-128, score-0.236]

71 We therefore trim the interval by a constant, parameterized by the user as , before performing the comparison. [sent-129, score-0.168]

72 To construct the correlation matrix from the full data, each of the R records needs to be considered for each of the n attribute pairs. [sent-142, score-0.399]

73 We evaluate the performance of our algorithm 2 by adding the number of records that were actually scanned for all the attribute-pairs, and n dividing the total by R 2 . [sent-143, score-0.332]

74 Figure 2 shows that the amount of data the algorithm examines is a constant that does not depend on the size of the data-set. [sent-147, score-0.246]

75 Note that the reported usage is an average over the number of attributes. [sent-152, score-0.122]

76 However this does not mean that the same amount of data was inspected for every attribute-pair — the algorithm determines how much effort to invest in each edge separately. [sent-153, score-0.557]

77 The running time is plotted against the number of data attributes in Figure 3. [sent-155, score-0.294]

78 A linear relation is clearly seen, meaning that (at least for this particular data-generation scheme) the algorithm is successful in doing work that is proportional to the number of tree edges. [sent-156, score-0.35]

79 For our algorithm the risk is making the wrong decision about which edges to include in the resulting tree. [sent-158, score-0.264]

80 In particular, we can just run the original algorithm on a small sample of the data, and use the generated tree. [sent-161, score-0.139]

81 2e+06 20 records 40 60 80 100 120 140 160 number of attributes Figure 2: Data usage (indicative of absolute running Figure 3: Running time as a function of the number time), in attribute-pair units per attribute. [sent-164, score-0.631]

82 004 data usage Figure 4: Relative log-likelihood vs. [sent-176, score-0.159]

83 To examine this effect we have generated data as above, then ran a 30-fold cross-validation test for the trees our algorithm generated. [sent-181, score-0.252]

84 We also ran a sample-based algorithm on each of the folds. [sent-182, score-0.13]

85 This variant behaves just like the full-data algorithm, but instead examines just the fraction of it that adds up to the total amount of data used by our algorithm. [sent-183, score-0.218]

86 We see that our algorithm outperforms the sample-based algorithm, even though they are both using the same total amount of data. [sent-185, score-0.148]

87 The reason is that using the same amount of data for all edges assumes all attribute-pairs have the same variance. [sent-186, score-0.254]

88 This is in contrast to our algorithm, which determines the amount of data for each edge independently. [sent-187, score-0.442]

89 Apparently for some edges this decision is very easy, requiring just a small sample. [sent-188, score-0.185]

90 The sample-based algorithm would not put more effort into those high-variance edges, eventually making the wrong decision. [sent-190, score-0.185]

91 The baseline (0) is the log-likelihood of the tree grown by the original algorithm using the full data. [sent-193, score-0.389]

92 Keep in mind that the sample-based algorithm has been given an unfair advantage, compared with MIST: it knows how much data it needs to look at. [sent-195, score-0.188]

93 The alternative is to use a fixed amount (specified either as a fraction or as an absolute count), which is likely to be too much or too little. [sent-198, score-0.12]

94 √ × √ × √ × × × × × √ × each training fold, we ran our algorithm, followed by a sample-based algorithm that uses as much data as our algorithm did. [sent-218, score-0.247]

95 Table 1 shows whether the 99% confidence interval for the log-likelihood difference indicates that either of the algorithms outperforms the other. [sent-220, score-0.168]

96 6 Conclusion and future work We have presented an algorithm that applies a “probably approximately correct” approach to dependency-tree construction for numeric and categorical data. [sent-224, score-0.3]

97 Experiments in sets with up to millions of records and hundreds of attributes show it is capable of processing massive data-sets in time that is constant in the number of records, with just a minor loss in output quality. [sent-225, score-0.645]

98 While we have derived formulas for both numeric and categorical data, we currently do not allow both types of attributes to be present in a single network. [sent-230, score-0.373]

99 An accelerated Chow and Liu algorithm: fitting tree distributions to high dimensional sparse data. [sent-285, score-0.27]

100 Using Tarjan’s red rule for fast dependency tree construction. [sent-291, score-0.504]


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