### Table 5. Arbitrary-edge discrepancies for some patterns.

1996

"... In PAGE 17: ... This process is repeated and the standard deviation of the jitter is gradually reduced. Table5 compares the resulting annealed patterns with several other interesting point patterns. The simulated annealing procedure was computationally expensive and has only been used to nd patterns of up to 64 points.... ..."

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### TABLE 13. Retrieval results for variable-length arbitrary passages.

2001

Cited by 27

### Table 2. Time given is the time per character for arbitrary length strings.

### Table 14: Comparison of document retrieval with variable-length arbitrary passages and recom- mended xed-length arbitrary passages.

2001

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### Table I. Dependence of the relative volume of deformed blob and the number of markers necessary to describe its surface on the maximum allowed edge length in the mesh

in SUMMARY

### Table 1: The mean and standard deviation of an arbitrary run and the covariance of two arbitrary successive runs

"... In PAGE 13: ...For instance, although three examples do not di er much in the mean length and standard deviation of an arbitrary run ( Table1 ) and the occurrences of short runs (Table 2), it is much easier for D-MAP 2 to have longer runs (Table 3). Also the dependence between two arbitrary successive runs of D-MAP 2 is much larger than that of D-MAP 1 (see Cov(R1; R2) in Table 1).... ..."

### Table 2: Comparison of the Models on Arbitrary Boolean and Purely Conjunctive Queries. Conjunctive Arbitrary

2000

"... In PAGE 12: ... The only di erence is that the connective between two attributes was selected as either a disjunction or a conjunction by ipping a fair coin. Table2 compares results on arbitrary and purely conjunctive queries (nQ is the query length, tP , Ct and eP are the average online time, query count and error across 200 runs of the algorithms). The maxent models again enjoy a distinct advantage in accuracy over the independence models.... ..."

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### Table 9: FR-12 collection: document retrieval using xed-length arbitrary passages.

2001

"... In PAGE 23: ... The results for prede ned passages are calculated using the pivoted-cosine measure. FR-12 collection The retrieval e ectiveness for xed-length arbitrary passages, whole-document ranking, and the best prede ned passage type, is shown in Table9 . The column marked as \% quot;... ..."

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### Table 10: TREC-45 collection: document retrieval using xed-length arbitrary passages.

2001

"... In PAGE 25: ... However, for large text collections with documents of more uniform length than those in the FR collections, whole-document ranking with the pivoted-cosine measure is expected to perform reasonably well [16, 38], thus reducing the bene ts of passage retrieval. Results for the TREC-45 collection are summarised in Table10 . There is a marked improve- ment when using xed-length arbitrary passages to rank documents, especially for long queries.... ..."

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### Table 1. Possible solutions for M-channel LPPRFB with lters of arbitrary lengths Li = KiM + .

1997

"... In PAGE 9: ... For odd value of M, using a similar argument, we arrive at the conclusion that the total length of all the lters is an odd multiple of M. p The results from Theorem 1, Theorem 2 and Corollary 1 are summarized in Table1 . S stands for symmetric lters; A stands for antisymmetric lters.... In PAGE 9: ... Type A system has even-length lters with di erent symmetry polarity while Type B system has odd-length lters with same symmetry. This can be con rmed using Table1 . Type A system belongs to the rst row (M = 2, and = 0).... In PAGE 10: ... This calls for LPPUFB (GenLOT) with arbitrary-length lters. From the second entry of Table1 , we know that odd-length even-channel GenLOT does not exist. If all lters have the same odd length and M is even, then is odd and Ki = K.... In PAGE 10: .... Km+1 is invertible. For paraunitary systems, Km+1 is orthogonal. From the rst row of Table1 , there are M2 symmetric and M2 antisymmetric lters in even- length even-channel LPPUFB. It can easily be shown that the same approach in [1] (propagating the pairwise time-reversed property) can be applied here to obtain the following factorization for... In PAGE 12: ... The readers can verify that this form of E0(z) allows the rst polyphases to have one order more than the remaining M ? polyphases. Also, since there must be M2 symmetric and M2 antisymmetric lters according to Table1 , S00 and A00 have the same number of rows. The corresponding coe cient matrix P0 with each lter apos;s impulse response arranging row-wise is: P0 = 1 p2 quot; S00 S01 S01J S00J A00 A01 ?A01J ?A00J # : (17) In order for E0(z) to be paraunitary, P0 has to satisfy the following time-domain constraint [6], with hi(n) being its rows: 1 X n=?1 hj(n) h k(n ? `M) = (`) (j ? k): (18) In matrix notation, it is equivalent to E0(z) e E0(z) = I, i.... ..."

Cited by 2