paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0101028 | This lemma follows from the fact that by REF , MATH is bounded. |
cs/0101028 | Assume for contradiction that the claim does not hold. Then there exist a constant MATH and a subsequence MATH of MATH such that MATH for all MATH. Therefore, MATH which equals MATH from the assumption that MATH has a finite competitive ratio. |
cs/0101030 | For any given MATH, MATH is maximized when MATH is a binary tree formed by attaching MATH leaves to a path of MATH vertices. The proof is by induction. CASE: For MATH, the theorem trivially holds. Now assume MATH. CASE: Hypothesis. For every positive integer MATH, the theorem holds. CASE: Step. Let MATH be the smallest integer such that MATH is empty after MATH rakes. Then, at the end of the MATH-th rake, MATH is a path MATH. Let MATH be the subtrees of MATH rooted at vertices in MATH. Let MATH be the number of leaves in MATH. Note that MATH . Since MATH and MATH is homeomorphic, by the induction hypothesis, MATH . Since MATH, MATH . Because MATH is homeomorphic, each MATH has at least one child in MATH. Since MATH, MATH. Then, MATH cannot be a leaf in MATH and thus has at least two children in MATH. Consequently, MATH. Next, note that for all MATH, MATH . With this inequality and the fact that MATH, we can combine the terms in the right-hand side summation of REF to obtain the following inequality. MATH where MATH and MATH. For any given MATH, the summation in REF is maximized when MATH and MATH. Therefore, MATH . The right-hand side of REF is maximized when MATH. This gives the desired bound and finishes the induction proof. |
cs/0101030 | Straightforward. |
cs/0101030 | The ideas are to preprocess MATH for answering queries of lowest common ancestors CITE and to reconstruct subtrees from appropriate tree traversal numberings CITE. |
cs/0101030 | Follows from REF at the end of REF. |
cs/0101030 | Since MATH and MATH have the same label set, all MATH are nonempty. To compute the output MATH, there are two cases depending on whether MATH is empty or nonempty. These cases are computed by REF . The correctness of NAME is then determined by that of REF . These steps can be verified using REF . As for the time complexity, these steps take MATH time using radix sort to evaluate MATH. REF uses REF and takes MATH time. REF take MATH time using tree traversal and radix sort. As discussed in REF preprocesses the input of its NAME calls to maintain their integer indexing assumptions. We reindex the labels and vertices of MATH and MATH and pass the new indices to the calls. We also partition NAME 's MATH-size array to allocate a segment of size MATH to the call with input MATH. Since the total input size of the calls is MATH, this preprocessing takes MATH time in an obvious manner. After this preprocessing, the running time of REF dominates that of NAME. The stated time bounds follow from REF and the fact that MATH is not longer than MATH and the degrees of MATH are at most MATH. |
cs/0101030 | The MATH at REF is a tube system. The heads of the tubes in MATH become children of the tubes in future MATH. The vertices MATH at REF are either leaves of MATH or heads of the tubes in previous MATH. These properties ensure the correctness of the rake-based recursion. The remaining correctness proof uses REF to verify the correctness of REF are straightforward and take MATH time. REF take MATH time using radix sort to access MATH and MATH. At REF , to maintain the integer indexing assumptions for the call to NAME, we simply pass to NAME the indices of MATH and MATH and the whole array of NAME. REF has the same time complexity as NAME. The desired time bounds follow from REF . |
cs/0101030 | The proof is similar to that of REF . The time bounds follow from REF . |
cs/0101030 | The proof is similar to that of REF . The time bounds follow from REF . |
cs/0101030 | The desired minimal forms can be computed by the two steps below: CASE: Sort the pairs in the given condensed forms for MATH into a sequence in the increasing order of the first components of these pairs. CASE: Go through this sequence to delete all unnecessary pairs to obtain the minimal condensed form of MATH. We can use radix sort to implement REF in MATH time for all MATH. REF can be easily implemented in MATH time for all MATH. |
cs/0101030 | The proof of REF uses radix sort in an obvious manner. To prove REF , we assume without loss of generality that MATH is a power of two. The input queries can be evaluated by the following three stages within the desired time bound. For each regular interval MATH, let MATH be the joint of MATH. We use REF MATH times to compute the minimal condensed forms of all MATH. The total size of these forms is MATH. This stage takes MATH time. For each irregular input query MATH, we partition MATH into MATH regular subintervals MATH. Then, the value of MATH is the maximum of those of MATH. These regular queries are point queries for MATH. Together with the given MATH regular queries, we have now generated MATH point queries for all MATH. This stage takes MATH time. We use REF and the minimal condensed forms of MATH to evaluate the points queries generated at Once the values of these point queries are obtained, we can easily compute the values of the input queries. This stage takes MATH time. |
cs/0101030 | Straightforward. |
cs/0101030 | REF follows from the definitions of MATH and MATH. The proofs of REF are similar to REF in the proof of REF below. As for REF , by the definition of MATH, MATH and MATH exist. To show MATH is nonintersecting, we consider the following four cases. The proofs of their symmetric cases are similar to theirs and are omitted for brevity. CASE: MATH is the ceiling of a nonintersecting leaf MATH. Since MATH and MATH are in MATH, MATH and MATH. Then because MATH is nonintersecting, so is MATH. CASE: MATH is the MATH-diagonal of a nonintersecting leaf MATH (or symmetrically, MATH is the MATH-diagonal of a nonintersecting leaf MATH). Since MATH is the floor of MATH, MATH and thus MATH. Let MATH be the smallest index such that MATH and MATH is intersecting. There are two subcases. CASE: MATH does not exist. Then, MATH is nonintersecting and therefore MATH is nonintersecting. CASE: MATH exists. Let MATH be a regular path that contains MATH and is of the same length as MATH. Note that MATH and MATH. There are two subcases. Case REFb REF : MATH. Then MATH is the ceiling of MATH. Since MATH is nonintersecting, it is the ceiling of a nonintersecting leaf in MATH which is a descendant of MATH. Therefore, Case REFb REF is reduced to REF . Case REFb REF : MATH. By the construction of MATH, MATH and thus MATH. By the choice of MATH, MATH is nonintersecting and so is MATH. CASE: MATH is the MATH-diagonal of an intersecting leaf MATH (or symmetrically, MATH is the MATH-diagonal of an intersecting leaf MATH). Since MATH, MATH and MATH. Let MATH be the smallest index such that MATH and MATH is intersecting. There are two subcases. CASE: MATH does not exist. Then, MATH is nonintersecting and therefore MATH is nonintersecting. CASE: MATH exists. Then, MATH and MATH. By the choice of MATH, MATH is nonintersecting. Thus, MATH is nonintersecting. CASE: MATH is the floor of a leaf MATH, which may or may not be intersecting. Let MATH be the lowest ancestor of MATH in MATH such that MATH is not the floor of MATH. This ancestor exists because MATH. There are two subcases. CASE: MATH and MATH. Then, MATH is a subpath of MATH and MATH. Also, MATH. Thus, MATH is the MATH-diagonal of MATH. By the construction of MATH, MATH is the MATH-diagonal of a leaf which is either MATH itself or its descendant. Depending on whether this leaf is nonintersecting or intersecting, Case REFa is reduced to REF or REF. CASE: MATH and MATH. There are two subcases. Case REFb REF : MATH and MATH. Note that MATH, MATH, and MATH is the ceiling of MATH. Since MATH is nonintersecting, MATH is the ceiling of a nonintersecting leaf in MATH which is MATH itself or a descendant. This reduces Case REFb REF to REF . Case REFb REF : MATH and MATH (or symmetrically, MATH and MATH). Note that MATH, MATH, and MATH is the MATH-diagonal of MATH. Then, MATH is the MATH-diagonal of a leaf which is MATH itself or a descendant. Depending on whether this leaf is nonintersecting or intersecting, Case REFb REF is reduced to REF or REF. |
cs/0101030 | REF - REF are proved below. The proof of NAME REF is similar to those of REF . For all distinct intersecting pairs MATH and MATH, the leaf labels shared by the subtrees MATH where MATH and the subtrees MATH where MATH are different from the shared labels for MATH and MATH. REF then follows from the fact that MATH and MATH share MATH leaf labels. REF . On each level of MATH, for all distinct pairs MATH and MATH, MATH. Thus, each level has at most MATH nonleaf pairs. Consequently, from the second level downwards, each level has at most MATH pairs. These two statements then follows from REF and the fact that the pairs specified in these two statements are within the top MATH levels of MATH. |
cs/0101030 | The proofs of REF are detailed below. The proof of REF is similar to that of REF and is omitted. REF is obvious. REF follows from the first three statements and the fact that if two sets regularly partition MATH, then so does their union. CASE: Note that MATH and MATH. The pair MATH can be the ceiling, the MATH-diagonal, the MATH-diagonal, or the floor of some leaf MATH. These four cases are discussed below. CASE: MATH is the ceiling. Then MATH and MATH. Since MATH and both MATH and MATH are regular, MATH. Since the length of MATH is at most MATH, so is the length of MATH. Thus MATH and MATH is a descendant of MATH. This contradicts the assumption that MATH and this case cannot exist. CASE: MATH is the MATH-diagonal. Then MATH and MATH. As in REF is a descendant of MATH. Thus, there exists a leaf MATH that is a descendant of MATH. Because MATH is the ceiling of this leaf, the existence of this leaf contradicts the assumption that MATH and this case cannot exist. CASE: MATH is the MATH-diagonal. Then, MATH and MATH. As in REF . Since MATH, MATH. Then MATH is a descendant of MATH and MATH. CASE: MATH is the floor. Then, MATH and MATH. As in REF is a descendant of MATH. Thus, there is a leaf MATH which is a descendant of MATH. Since MATH is this leaf's MATH-diagonal, it is in MATH. CASE: Note that MATH consists of the following three types of pairs: CASE: the ceiling, diagonals and floor of a leaf MATH where MATH. CASE: the MATH-diagonal and floor of MATH where MATH is of length MATH. CASE: the pairs in MATH where MATH and MATH is of length MATH. By REF , only the pairs of the first two types may be MATH-irregular. This statement then follows from REF . |
cs/0101030 | This lemma follows from the fact that MATH, MATH and the new labels of MATH and MATH are different from one another and the labels of MATH and MATH. |
cs/0101030 | To form maximum agreement subtrees of MATH and MATH, there are three cases. CASE: MATH accounts for matching MATH to MATH. CASE: MATH accounts for matching MATH to a proper descendant of MATH. CASE: MATH accounts for matching MATH to a proper descendant of MATH. |
cs/0101030 | This lemma follows from REF with a finer case analysis for the cases in the proof of REF . |
cs/0101030 | This lemma follows from REF and is obtained by iterative applications of REF . The following properties are used. Since MATH is nonintersecting, for MATH and MATH, CASE: MATH CASE: MATH CASE: MATH CASE: MATH . |
cs/0101030 | The proof is similar to that of REF and follows from REF . |
cs/0101030 | REF follows from REF . Below we only prove REF for MATH; REF for MATH is similarly proved. We first compute a condensed form MATH for each MATH as follows: CASE: For all MATH, compute MATH and MATH. CASE: For all MATH where MATH is nonempty, do the following steps: CASE: MATH. CASE: Compute all tuples MATH where MATH, MATH, MATH, and MATH. CASE: Find the smallest MATH such that some MATH is obtained at REF . CASE: If there is only one MATH, then add to MATH the pair MATH where MATH. CASE: For all MATH where MATH is nonempty and MATH is empty, do the following steps: CASE: Compute MATH, MATH and MATH where MATH is the root of MATH, MATH, MATH and MATH. CASE: MATH, where MATH. CASE: For all MATH where MATH is empty, MATH. The correctness proof of this algorithm has three cases. CASE: MATH is nonempty. Let MATH. Let MATH. Then, for all MATH and all MATH, MATH and by REF , MATH. There are two subcases for MATH. CASE: REF finds two or more MATH. Then MATH, MATH, and for all MATH, MATH is empty and MATH. CASE: REF finds only one MATH. Then MATH and MATH. For all MATH and MATH. For all MATH, MATH is empty and MATH. Thus, the MATH of REF is a condensed form of MATH for REF . CASE: MATH is nonempty and MATH is empty. This case is similar to Case REFb, and REF computes a correct condensed form MATH for this case. CASE: MATH is empty. This case is obvious, and REF correctly computes a condensed form MATH of MATH for this case. The total size of all MATH is at most that of the MATH mappings of MATH and MATH, which is the desired MATH. REF takes MATH time using REF . The other steps can be implemented in MATH time in straightforward manners using radix sort and tree traversal. As discussed in REF, the MATH mappings are evaluated by radix sort. Once the forms MATH are obtained, we can in MATH time radix sort the pairs in all MATH and then delete all unnecessary pairs to obtain the desired minimal condensed forms. |
cs/0101030 | The value of MATH is that of the point query MATH for MATH, and the value of MATH is that of the interval query MATH. By REF , there are MATH such terms required for the pairs in MATH. Given the results of REF of NAME, we can determine all such terms and the corresponding queries in MATH time. By REF , only MATH of these queries are not MATH-regular. By REF , we can evaluate these queries in MATH time. The terms MATH and MATH are similarly evaluated is MATH time. The analysis for these terms is easier because MATH and it does not involve the notion of MATH-regularity. |
cs/0101030 | To prove REF , we consider the graphs MATH on which the desired terms MATH are defined. Let MATH be the subgraph of MATH constructed by removing all zero-weight edges and all resulting isolated vertices. The edges of MATH are computed as follows: CASE: For all MATH, compute MATH and MATH. CASE: For all MATH is nonempty, do the following steps: CASE: If MATH is nonempty, compute all tuples MATH where MATH, MATH, MATH and MATH. CASE: If MATH is empty, compute the tuple MATH where MATH is the root of MATH, MATH, MATH and MATH. This algorithm captures all the nonzero-weight MATH. At REF , MATH and by REF MATH. Thus, the first two components of the obtained tuples form the edges of all desired MATH and the third components are the weights of these edges. We use REF to implement REF in MATH time. We can implement REF in MATH time using radix sort and tree traversal. Note that REF uses radix sort to evaluate MATH mappings. With the tuples MATH obtained, we use radix sort to construct all desired MATH in MATH time. Let MATH and MATH be the numbers of edges and vertices in MATH, respectively. Since an edge weighs at most MATH, we can compute MATH in MATH and alternatively in MATH time CITE. REF then follows from the fact that MATH, MATH, and by REF the sum of all MATH is at most MATH. To prove REF , we similarly process the bipartite graphs on which the desired terms MATH are defined. The key difference from the third type is that in addition to some of the edges in MATH, we need certain nonzero-weight MATH for MATH and MATH for MATH. Since these edges are required only for intersecting MATH, by REF , MATH such edges are needed. We use REF to compute the weights of these edges in MATH time. Due to these edges, the total time complexity for the fourth type is MATH times that for the third type. |
cs/0101030 | The correctness of NAME follows from REF - REF. As for the time complexity, REF is obvious and takes MATH time. By computing MATH, we can compute the sets MATH and MATH. Since the leaf labels of MATH and MATH are from MATH, each level of MATH can be computed in MATH time. Since MATH has MATH levels, MATH and MATH can be computed in MATH time. With these two sets obtained, we can compute all the desired sets in MATH time. Thus, REF takes MATH time. REF takes MATH time using radix sort. The time complexity of REF dominates that of NAME. This step uses REF and takes MATH time or alternatively MATH time. REF spends MATH time using radix sort to create pointers from the pairs in MATH to appropriate predecessors. REF then takes MATH time per pair in MATH and MATH time in total. REF takes MATH time. It uses radix sort to access the desired MATH values and evaluate the input mappings. It also uses REF to compute all MATH and MATH. |
cs/0101030 | By REF , the algorithms in REF compute MATH within the desired time bounds. With straightforward modifications, these algorithms can compute a maximum agreement subtree within the same time bounds. |
cs/0101030 | Given MATH and MATH, we construct two binary evolutionary trees MATH and MATH as follows. Let MATH and MATH be two distinct symbols different from all MATH and MATH. Next, we construct two paths MATH and MATH. MATH is formed by making MATH the root, attaching MATH to MATH as a leaf, and attaching MATH and MATH to MATH as leaves. Symmetrically, MATH is formed by making MATH the root, attaching MATH to MATH, and attaching MATH and MATH to MATH. The lemma follows from the straightforward one-to-one onto correspondence between the longest common subsequences of MATH and MATH and the maximum agreement subtrees of MATH and MATH. |
cs/0101031 | REF is straightforward. To prove REF , let MATH be a maximum weight matching in MATH. Consider the edges in MATH. They form a set MATH of alternating paths and cycles. Since MATH is matched by MATH but not by MATH, MATH is of degree one in MATH. Let MATH be the alternating path in MATH with MATH as an endpoint. Let MATH be the matching obtained by transforming MATH only with MATH. Since MATH is not matched by MATH, MATH is a matching in MATH. MATH can be obtained by further transforming MATH with the remaining alternating paths and cycles in MATH. The net change induced by each of these alternating paths and cycles is non-positive; otherwise, such a path or cycle can improve MATH and we obtain a contradiction. Therefore, MATH, that is, both MATH and MATH are maximum weight matchings in MATH. |
cs/0101031 | CASE: Consider a simple cycle MATH in MATH. Since MATH has no outgoing edges, no MATH equals MATH. By the definition of MATH, MATH is also an alternating cycle in MATH. Therefore, MATH is the net change induced by transforming MATH with MATH. Since MATH is a maximum weight matching in MATH, this net change is non-positive. CASE: Consider an alternating path MATH in MATH starting from MATH. In MATH, MATH is also a simple path. If MATH, then MATH is not matched by MATH, and MATH contains the edge MATH. If MATH, then MATH is matched by MATH, and MATH again contains the edge MATH. Therefore, MATH contains the simple path MATH. The weight of MATH is MATH. The reverse direction of the statement is straightforward. CASE: This statement follows from REF . |
cs/0101031 | By symmetry and REF , we compute MATH for all MATH as follows. CASE: Compute a maximum weight matching MATH of MATH. CASE: Construct MATH as above and find the weights of its longest paths to MATH. CASE: For each MATH, if MATH is matched by MATH, then MATH is the sum of MATH and the weight of the longest path from MATH to MATH in MATH; otherwise, MATH. REF takes MATH time. At REF , constructing MATH takes MATH time, and the single-destination longest paths problem takes MATH time CITE. REF takes MATH time. Thus, the total time is MATH. |
cs/0101031 | See REF to REF. |
cs/0101031 | It follows directly from the definition. |
cs/0101031 | We adapt NAME and NAME 's rooted subtree algorithm CITE to compute MATH. Details are given in REF. |
cs/0101031 | It follows directly from the definition of MATH. |
cs/0101031 | See REF. |
cs/0101031 | Let MATH be the computation time of MATH. The computation is divided into two cases. REF takes MATH time. For REF , a set of subproblems MATH are generated. As to be shown in REF, the time to prepare all these subproblems is also MATH. These subproblems, except possibly one, are each of size less than MATH. For the exceptional subproblem, say, MATH, its computation is again divided into two cases. One case takes MATH time. For the other case, another set of subproblems are generated in MATH time. This time every such subproblem has size less than MATH. Let MATH be the set of all the subproblems generated in both steps. The total size of the subproblems in MATH is at most MATH, and MATH . It follows that MATH. |
cs/0101031 | We adapt preprocessing techniques for on-line product queries in CITE. |
cs/0101031 | This lemma follows from the rooted subtree algorithm and related data structures in CITE. |
cs/0101031 | Observe that MATH if in some maximum agreement subtree of MATH and MATH, the root of MATH is mapped to some node MATH in MATH. On the other hand, MATH for some MATH if in some maximum agreement subtree of MATH and MATH, the root of MATH is not mapped to any node MATH in MATH. |
cs/0101031 | Consider the bipartite graph MATH in which a node MATH is connected to MATH if and only if MATH. This edge is given a weight of MATH. Then, every MATH is a subgraph of MATH. Let MATH be a maximum weight matching of MATH. Observe that if a MATH is not adjacent to any edge in MATH, then MATH is also a maximum weight matching of MATH. Since MATH contains at most MATH edges, there are at least MATH trees MATH not adjacent to any edge in MATH and the corresponding MATH have the same value. |
cs/0101031 | The fact that MATH follows from the construction of MATH. Note that MATH contains MATH edges. All edges in MATH, except MATH and MATH, can be found using MATH time. The weight of these edges can be found in MATH time using REF . To identify MATH and MATH, note that at most MATH instances of MATH are connected to some non-max-child of MATH. All other MATH are partitioned into at most MATH intervals. For each interval, say MATH, by REF , the corresponding MATH which attains the maximum in the set MATH can be found in MATH time. Thus, by scanning all the MATH intervals, MATH can be found in MATH time. MATH can be found similarly. |
cs/0101031 | Let MATH. Let MATH be a path starting from the root of MATH such that every next node is the max-child of its predecessor. Then MATH. Let MATH denote the set of subtrees attached to some node on MATH. The subtrees in MATH are label-disjoint and each has size at most MATH. Thus, MATH . |
cs/0101031 | Note that MATH are almost identical to the subproblems considered in REF in that all the MATH have exactly one shrunk leaf each. Using exactly the same approach, we can compute the auxiliary information in MATH, MATH, MATH. |
cs/0101031 | By REF , MATH and MATH can be computed in time MATH and afterwards, for each node MATH, MATH and MATH can be retrieved in MATH time. For each node MATH, MATH is the auxiliary information stored at MATH in MATH and can be retrieved in MATH time. |
cs/0101031 | Let MATH be the set of labels of the atomic leaves of MATH. An internal node MATH can be either an auxiliary node, a compressed node, or a node of MATH. If MATH, then MATH. Thus, MATH and MATH. If MATH is a compressed node, then we need to compute MATH and MATH. Recall that MATH represents some tree path MATH of MATH, where MATH is the closest to the root, that is, MATH. Thus, MATH, MATH, and MATH. Thus, MATH time is sufficient for finding the auxiliary information stored at every internal node of MATH. |
cs/0101031 | If MATH is a compressed leaf in MATH, MATH's parent MATH must not be an auxiliary node. Depending on whether MATH is a compressed node, we have two cases. CASE: MATH is not a compressed node. We must compute MATH. Note that MATH is also in MATH. When MATH is constructed from MATH, some of the subtrees of MATH attached to MATH are replaced by MATH and no longer exist in MATH. Let MATH be these subtrees. Observe that both MATH and MATH represent the same set of subtrees in MATH. Thus, CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. These four values can be found in MATH time. Since MATH are subtrees attached to MATH in MATH, MATH is at most the degree of MATH in MATH. Moreover, the sum of the degrees of all internal nodes of MATH is MATH. Therefore, MATH time suffices to compute the auxiliary information for all the compressed leaves in MATH whose parents are not compressed node. CASE: MATH is a compressed node. We need to compute MATH, MATH, MATH, MATH, MATH and MATH. Note that MATH is compressed from a tree path MATH in MATH where MATH is the closest to the root. Moreover, MATH is compressed from the subtrees hanging between MATH and MATH. For every MATH, let MATH be the set of subtrees of MATH attached to MATH that are compressed into MATH. Both MATH and the subtrees in MATH represent the same set of subtrees in MATH. The auxiliary information stored at MATH can be expressed as follows. CASE: MATH. CASE: MATH for some MATH. CASE: MATH for some MATH. CASE: MATH. CASE: MATH. CASE: MATH. These values can be found in MATH time, where MATH is the degree of MATH in MATH. Thus, the auxiliary information for every compressed leaf of MATH, whose parents are compressed nodes, can be computed in MATH time. |
cs/0101031 | The proof of this lemma is the same as that of REF . |
cs/0101031 | Let MATH and MATH be the two shrunk leaves of MATH. Assume that MATH is also a shrunk leaf in MATH and MATH represents MATH, that is, MATH represents the subtree MATH of MATH. Let MATH be the subtree obtained by connecting MATH and MATH. By the same argument as in REF , the auxiliary information in MATH can be computed based on the values MATH, MATH and MATH for all MATH. The value MATH can be found in MATH. The values MATH and MATH for all MATH can be retrieved in MATH time after MATH and MATH are computed in MATH time based on REF . Then the auxiliary information in MATH can be computed in MATH time. |
cs/0101031 | We adopt the framework of NAME and NAME 's algorithm CITE, which is essentially a sparsified dynamic programming based on the following formula. For any internal nodes MATH of MATH and MATH of MATH, MATH where MATH denotes the maximum size of all the agreement subtrees of MATH and MATH in which MATH is mapped to MATH. Our algorithm differs from NAME and NAME 's algorithm in the way how each individual MATH is computed. When MATH and MATH are ordinary evolutionary trees, each MATH is found by computing a maximum weight matching of some bipartite graph, and it takes MATH time to compute MATH. Below, we show that when MATH and MATH have compressed and shrunk leaves, each MATH can be found either in constant time or by computing at most two maximum weight bipartite matchings of similar graphs but with edge weights bounded by MATH instead of MATH. Thus, we can compute MATH using the same sparsified dynamic programming in CITE; as a by-product, we can afterwards retrieve MATH for any node MATH of MATH in MATH time. The enlarged upper bound of edge weights increases the time complexity to MATH, though. In the rest of this section, we show how each MATH is computed. First, we consider the case when MATH is a leaf. The following case analysis shows that MATH time suffices to compute MATH. CASE: MATH is an atomic leaf. If MATH contains a leaf with the same label as that of MATH, then MATH; otherwise it equals zero. CASE: MATH is an auxiliary leaf. Then, MATH. CASE: MATH is a compressed leaf. By definition, MATH can only be mapped to MATH, MATH or the least common ancestor MATH of MATH and MATH. If MATH has no shrunk leaves, then MATH. If MATH has only one shrunk leaf, say MATH, then MATH. If MATH has two shrunk leaves, MATH must also contain MATH and MATH. Next, we consider the case when MATH is an internal node. Assume that MATH is an atomic leaf. Then MATH if MATH contains a leaf with the same label as that of MATH, and zero otherwise. If MATH is a shrunk leaf, say MATH, then MATH. It remains to consider the case when MATH is an internal node. Due to the nature of dynamic programming, we only need to compute MATH, then we can apply REF to compute MATH. We further divide our discussion into the following three cases. CASE: MATH is an auxiliary internal node. In such case, MATH has only two children, one of them is an auxiliary leaf. From definition, an auxiliary leaf will not be mapped to any node in any agreement subtree of MATH and MATH; thus, there is no agreement subtree in which MATH is mapped to MATH and MATH. CASE: MATH is an ordinary internal node. As in CITE, we first construct the bipartite graph defined as follows: Let MATH and MATH be the set of children of MATH and MATH, respectively; define MATH to be the bipartite graph formed by the edges MATH with MATH, and MATH is given a weight MATH. If none of MATH's children is an auxiliary leaf, then MATH. Otherwise, let MATH be the child of MATH which is an auxiliary. In this case, MATH also has a compressed child MATH. Other than MATH and MATH, no other child of MATH is a compressed and auxiliary leaf. On the other hand, consider the rooted subtrees MATH rooted at the children MATH of MATH. If the shrunk leaves appear together in one of such subtrees, then by definition, MATH cannot be mapped to any shrunk leaf in any agreement subtree of MATH and MATH, and MATH. If the shrunk leaves appear in two different subtrees rooted at two children MATH and MATH of MATH, then MATH . CASE: MATH is a compressed internal node. By definition of a compressed node, the structure of MATH is very restrictive - MATH has exactly three children MATH, MATH, and an auxiliary internal node MATH; MATH has two children, an auxiliary leaf MATH and an uncompressed internal node MATH; see REF . To find MATH, we note that there are only six possible ways on how the MATH are mapped to MATH and MATH. We consider each of these cases and MATH is the maximum of the values found. We only discuss the case where MATH and MATH are mapped to MATH and MATH, respectively. The other cases can be handled similarly. Let MATH be the path between MATH and MATH. Let MATH denote the set of subtrees hanged on MATH. The size of the largest agreement subtrees of MATH and MATH in which MATH and MATH are mapped to MATH and MATH, respectively, equals MATH . Note that using the technique in CITE, we can precompute MATH for all MATH in MATH time. Afterwards, REF can be found in constant time. |
cs/0101031 | Straightforward. |
cs/0101031 | We can examine all the entries of MATH whose values differ from MATH in MATH time. By examining all such entries of MATH, we can construct MATH and arrange the integers in MATH in ascending order. Thus, REF takes MATH time. REF takes MATH time. For REF , we first analyze the time required to process one nonempty MATH. Note that MATH for every MATH. Thus, MATH can be computed in MATH time using the result of REF . Summing over all MATH, computing all MATH takes MATH time. Applying REF to the sequence MATH, MATH, MATH, MATH, MATH takes MATH time. In total, it takes MATH time to process one MATH, and REF takes MATH time. Thus, the total time of our preprocessing is MATH. |
cs/0101031 | Let MATH. A crucial step is to find MATH. Without loss of generality, assume MATH. To find MATH, we first find the smallest integer MATH in MATH that is greater than MATH, and the largest integer MATH in MATH that is smaller than MATH. Since MATH is sorted, we can find MATH and MATH in MATH time. If MATH, then MATH; otherwise, MATH is the set of integers between MATH and MATH in MATH. If MATH, then MATH. Because of REF of our preprocessing, we can find MATH in MATH time. If MATH, then MATH and MATH equals the maximum of CASE: MATH, CASE: MATH, CASE: MATH. Note that REF equals the maximum of MATH, which can be computed in MATH time after REF of our preprocessing. Since MATH and MATH, REF enables us to compute REF in MATH time. As a result, MATH can be answered in MATH time. |
cs/0101032 | Let MATH be the table constructed from MATH by also publishing the suppressed cells that are not in the strongly connected components of MATH. By REF , MATH remains a nonzero analytic function of MATH. Also, the connected components of the suppressed graph MATH of MATH are the strongly connected components of MATH. Thus, to prove the lemma, it suffices to prove it for MATH, MATH, and MATH. Let MATH be a fixed bounded feasible assignment of MATH. Let MATH is a bounded feasible assignment of MATH. Since MATH is an analytic invariant of MATH, the function MATH is an analytic invariant of MATH with MATH and its value is zero over MATH. Because MATH contains a nonempty open subset of MATH, MATH is zero over MATH. By REF and the strong connectivity of the connected components of MATH, MATH and MATH is zero over MATH. Thus, it suffices to show that if MATH is connected for all connected components MATH of MATH, then MATH for some MATH. To construct MATH, let MATH. Let MATH be the connected component of MATH that contains MATH. By the connectivity of MATH, there is a vertex-simple path MATH in MATH between the endpoints of MATH. Let MATH be the vertex-simple cycle formed by MATH and MATH. Next, direction-blindly label MATH with MATH labeled MATH. Since MATH is a nonzero power series, MATH for some vector MATH. Note that MATH is not necessarily in MATH. So, let MATH, where MATH is the component of MATH at variable MATH. Then, by REF , MATH. Because MATH is in MATH, MATH appears only in the term MATH in MATH. Thus MATH and MATH have the same component values at the variables in MATH. Since the variables not in MATH do not appear in any expansion of MATH, MATH, proving the lemma. |
cs/0101032 | It is equivalent to show that MATH is not totally protected if and only if MATH contains some edges not in the strongly connected components of MATH or for some strongly connected component MATH of MATH, the graph MATH is not connected. The MATH direction follows from REF . As for the MATH direction, if MATH contains some edges not in the strongly connected components of MATH, then by REF , MATH contains some invariant cells of MATH and thus cannot be totally protected. If for some strongly connected component MATH of MATH, the graph MATH is not connected, then some subset MATH of MATH is a minimal edge cut of MATH. By REF , MATH has a linear invariant MATH with MATH and thus MATH is not totally protected. |
cs/0101032 | This problem can be solved within the desired time bound by means of REF and linear-time algorithms for computing connected components and strongly connected components CITE. |
cs/0101032 | Given an instance MATH and MATH of REF with MATH, the desired instance of REF is the total graph MATH and the suppressed graph MATH of MATH, and MATH itself. This transformation can easily be computed in linear time. There are two directions to show that it reduces REF to REF . Assume that MATH is a desired set for REF . By REF is true. Also, every strongly connected component of MATH is a union of edge-disjoint connected components in MATH and MATH. Therefore, by REF holds. As a result, MATH itself is a desired set for REF . On the other hand, assume that MATH is a desired set for REF . Let MATH be the set of all edges in MATH that are also in the strongly connected components of MATH. By REF and the total protection of MATH in MATH, the connected components of MATH are the strongly connected components of MATH. Thus, MATH holds REF . Next, because a connected component of MATH is included in a strongly connected component of MATH, by REF , MATH also holds REF and thus is a desired set for REF . Given an instance MATH, MATH, and MATH of REF , the desired instance of REF with MATH is MATH itself and the table defined as follows. For each vertex in MATH (respectively, MATH), there is a row (respectively, column). The upper and lower bounds for each cell are REF. For each edge MATH in MATH, its corresponding cell is at the row and column corresponding to its endpoints. The value of that cell is REF (respectively, REF) if MATH is undirected (respectively, directed from MATH to MATH, or directed from MATH to MATH). For each edge MATH in MATH, its corresponding cell is suppressed. Note that the total and suppressed graphs of this table are MATH and MATH themselves. Thus, the remaining proof details for this reduction are essentially the same as for the other reduction. |
cs/0101032 | For each MATH, let MATH be an element in MATH; by the assumption of this lemma, these elements exist. Next, let MATH and MATH; by REF , these two sets exist. Now, let MATH. Note that MATH. Since MATH consists of MATH edges and MATH consists of at most MATH edges, MATH has at most MATH edges. MATH holds REF because MATH consists of the edges in the traversable cycles MATH. REF of MATH follows from the fact that MATH connects MATH, which forms the only connected component of MATH with more than one vertex. |
cs/0101032 | By REF , MATH must contain some edge MATH for each MATH with MATH. By REF , MATH. Now let MATH. To calculate the size of MATH, note that by REF , MATH must also contain MATH and at least one edge leaving MATH for each MATH. Thus MATH. Then MATH because MATH. |
cs/0101032 | The proof uses arguments similar to those in the proof of REF . The strong connectivity properties in REF can be ignored because this section assumes that the total graph of MATH is undirected. The forest structure of MATH follows from its minimality. |
cs/0101032 | The key idea is that an optimal MATH for REF can be obtained by connecting the vertices of each MATH first with edges in MATH, which can be used for free, next with edges in MATH, and then with edges outside MATH. Let MATH be a set of MATH edges in the complement of MATH that becomes MATH after REF. Then, MATH is a desired output MATH for REF , showing that REF - M REF can indeed reduce REF to REF . REF is the only step that requires more than linear time. It is important to avoid directly computing MATH at REF. Computing these complement graphs takes MATH time if some MATH contains a constant fraction of the vertices in MATH. In such a case, if MATH is sparse, then the time spent on computing MATH alone is far greater than the desired complexity. Instead of this naive approach, REF uses efficient techniques recently developed for complement graph problems CITE and takes the desired MATH time. |
cs/0101033 | The statements are proved separately as follows. CASE: For MATH, let MATH be the collection of the left edges of MATH for MATH. We prove by induction on MATH the claim that MATH is a tree spanning over MATH. Then, since MATH, the claim implies the statement. For the base case MATH, the claim trivially holds. The induction hypothesis is that the claim holds for MATH. The induction step is to prove the claim for MATH. MATH is obtained from MATH by adding the left edge MATH of MATH. By the induction hypothesis, MATH is a tree spanning over MATH. Since MATH is the leftmost neighbor of MATH on MATH, MATH is some MATH with MATH and MATH. Thus, MATH contains MATH, and MATH is a tree spanning over MATH. CASE: The proof is symmetric to that of REF . MATH has MATH vertices and MATH edges. The edges MATH are not in MATH. Thus, since MATH and MATH have MATH edges each, MATH has MATH edges. Then, since MATH is acyclic and does not contain MATH and MATH, MATH is a spanning tree of MATH. |
cs/0101034 | The proof follows from REF . |
cs/0101034 | REF are both in NP. To prove their completeness, by REF , it suffices to reduce REF to REF . Given an instance MATH, MATH, MATH of REF , an instance MATH of REF is constructed as follows: CASE: Let MATH. The vertices MATH correspond to MATH, but MATH corresponds to no MATH. CASE: Let MATH. The vertices MATH correspond to MATH of MATH, but MATH corresponds to no MATH. CASE: Let MATH consist of the following edges: CASE: The edge between MATH and MATH is MATH. CASE: For all MATH with MATH, the edge between MATH and MATH is MATH. CASE: For all MATH with MATH, the edge between MATH and MATH is MATH. CASE: For each MATH and each MATH, if MATH, then the edge between MATH and MATH is MATH; otherwise it is MATH. CASE: Let MATH. CASE: Let MATH. The above construction can be easily computed in polynomial time. The next two claims show that it is indeed a desired reduction from REF to REF . If some MATH with MATH has at least one element in each MATH, then some MATH consists of at most MATH edges such that every connected component of MATH is strongly connected and bridge-free. To prove this claim, observe that for each MATH, some MATH exists. By REF , MATH exists. By REF , MATH exists. Let MATH. Note that MATH consists of MATH edges. MATH consists of at most MATH edges. Thus MATH has at most MATH edges. For all MATH with MATH, the edges MATH form a vertex-simple traversable cycle. Because MATH consists of the edges in these cycles, every connected component of MATH is strongly connected and bridge-free. This finishes the proof of REF . If some MATH consists of at most MATH edges such that every connected component of MATH is strongly connected and bridge-free, then some MATH with MATH has at least one element in each MATH. To prove this claim, observe that for all MATH with MATH, by REF , MATH contains MATH but no edge pointing from MATH. Because every connected component of MATH is strongly connected, MATH contains an edge MATH for some MATH. By REF , MATH. Let MATH. Note that MATH contains MATH but MATH contains no edges pointing from MATH. Because every connected component of MATH is strongly connected, MATH must also contain at least one edge pointing from each vertex in MATH. Thus MATH contains at least MATH edges. Then MATH because MATH. This finishes the proof of REF and thus that of REF . |
cs/0101034 | The proof is similar to that of REF . |
cs/0101034 | Let MATH be a sum minimal invariant of MATH. If MATH consists of an edge not in any strong component of MATH, then MATH is trivially rectangular. Otherwise, by REF MATH is a bipartite minimal cut set of a strong component MATH of MATH. By REF , MATH has two connected components MATH and MATH. Let MATH and MATH be the sets of endpoints of MATH in MATH and MATH, respectively. By the bipartiteness of MATH, without loss of generality the vertices in MATH are rows in MATH and those in MATH are columns. Then MATH is rectangular because MATH consists of the edges between MATH and MATH in MATH. |
cs/0101034 | The directions MATH are straightforward. The direction MATH follows from the fact that by REF MATH are the only factors in the decomposition in REF for a positive invariant MATH with MATH. To prove MATH, note that because MATH is a positive invariant for all MATH, by REF there is a sum minimal invariant MATH with MATH. Since MATH is also rectangular, by REF , MATH for some MATH. Because MATH share no variable, by the minimality of MATH and coefficient comparison MATH equals MATH and thus is a sum minimal invariant. To prove the desired uniqueness of MATH, let MATH be a sum minimal invariant with MATH. By REF , MATH is rectangular. By REF , MATH for some MATH. Because MATH is nonzero, some MATH. Because MATH do not share variables, MATH. Then, MATH by coefficient comparison and the minimality of MATH. |
cs/0101034 | The proof is similar to that of REF . |
cs/0101034 | The proof of the lemma for the positive invariants and that for the general invariants are similar; only the former is detailed here. For the direction MATH , REF follows from the second condition of the definition of MATH being protected. Then, REF follows from REF , and REF . For the direction MATH , by REF , the first condition of MATH being protected is satisfied. The second condition then follows from REF . |
cs/0101034 | If some MATH, then MATH is protected if and only if MATH is protected. Thus without loss of generality assume that no MATH is empty. Then, by REF the protection definitions with respect to the positive, sum, and rectangular invariants are all equivalent. Similarly, by REF , those with respect to the general and unitary invariants are also equivalent. This theorem then follows directly from REF . |
cs/0101034 | This lemma follows from REF . |
cs/0101034 | This theorem follows directly from REF . |
cs/0101034 | REF follows from REF in a straightforward manner using linear-time algorithms for connectivity and strong connectivity CITE. REF follows from REF . The key step is to test the bipartite-MATH-connectivity of MATH within the stated time bound. We first construct two auxiliary graphs MATH and MATH. For each vertex MATH, replace MATH with MATH copies in MATH. For each MATH and each edge MATH in MATH between MATH and a vertex MATH, replace MATH with MATH copies between MATH and the MATH copies of MATH in MATH. MATH is obtained by exchanging MATH and MATH in the construction. Because MATH is connected and each vertex in MATH is duplicated MATH times, MATH has a vertex cut MATH of at most MATH vertices if and only if MATH is a subset of MATH and is a vertex cut of MATH. A symmetrical statement for MATH also holds. Thus MATH is bipartite-MATH-connected if and only if both MATH and MATH are MATH-connected. This corollary then follows from the fact CITE that the MATH-connectivity of a MATH-vertex graph can tested in MATH time if MATH. |
cs/0101034 | This corollary follows from REF in a straightforward manner using linear-time algorithms for strong connectivity and REF-connectivity CITE. |
cs/0101034 | The proof follows from REF . |
cs/0101034 | REF are both in NP. To prove their completeness for MATH, by REF , it suffices to reduce REF to REF with MATH. Given an instance MATH, MATH, MATH of REF , let MATH be the instance constructed for REF . The next two claims show that this transformation is indeed a desired reduction. If some MATH with MATH has at least one element in each MATH, then some MATH consists of at most MATH edges such that every nonsingleton connected component of MATH is strongly connected and bipartite-REF-connected. To prove this claim, observe that for each MATH, some MATH exists. Let MATH, which exists by REF of the construction of MATH, MATH, and MATH. By REF , MATH exists. Let MATH. Note that MATH consists of MATH edges. MATH consists of at most MATH edges. Thus MATH has at most MATH edges. For all MATH with MATH, the edges MATH form a vertex-simple traversable cycles. These cycles all go through MATH and form the only nonsingleton connected component of MATH. This component is clearly strongly connected and bipartite-REF-connected. This finishes the proof of REF . If some MATH consists of at most MATH edges such that every nonsingleton connected component of MATH is strongly connected and bipartite-REF-connected, then some MATH with MATH has at least one element in each MATH. The proof of this claim is the same as that of REF , and uses only the componentwise strong connectivity of MATH. This finishes the proof of REF . |
cs/0101034 | The proof is similar to that of REF . |
cs/0101034 | The proof follows from REF . |
cs/0101034 | Because a protected table has no invariant cells, each row or column has either no suppressed cell or at least two suppressed cells. It suffices to prove that if MATH holds this condition, then the statements below are equivalent: CASE: Every positive invariant is a positive linear combination of MATH. CASE: Every nonzero sum invariant is a positive linear combination of MATH. CASE: Every nonzero rectangular sum invariant of MATH is a positive linear combination of MATH. CASE: The nonzero linear invariants among MATH are the only sum minimal invariants of MATH. The directions MATH and MATH are straightforward. The direction MATH follows from REF . To prove the direction MATH, note that for each MATH with MATH, MATH is a nonzero sum invariant. By REF there is a sum minimal invariant MATH with MATH. MATH is also rectangular. By REF , MATH where MATH. By coefficient comparison there is some MATH with MATH. Because MATH, MATH. Then MATH because two distinct MATH cannot share more than one cell and each nonempty MATH contains at least two cells. Thus MATH equals MATH and is a sum minimal invariant. To prove the desired uniqueness of MATH, let MATH be a sum minimal invariant with MATH. By REF , MATH is rectangular. By REF , MATH where MATH. By coefficient comparison there is some MATH with MATH. Because MATH, MATH. Then MATH by coefficient comparison and the minimality of MATH. |
cs/0101034 | By REF , it suffices to prove that the following statments are equivalent: CASE: MATH is protected. CASE: The nonzero invariants among MATH are the only sum minimal invariants of MATH. Also each MATH contains either no suppressed cell or at least two suppressed cells. CASE: Each connected component of MATH is strongly connected and bridge-free. Also the nonempty sets among MATH are the only bipartite minimal edge cuts of the strong components of MATH. The equivalence MATH follows from the proof of REF . The equivalence MATH follows from REF . |
cs/0101034 | This is an immediate corollary of REF . |
cs/0101034 | The proof follows from REF . |
cs/0101034 | REF are both in NP. To prove their completeness, by REF , it suffices to reduce REF to REF . Given an instance MATH, MATH, MATH of REF , let MATH be the instance constructed for REF with the modification below: CASE: Rule MATH: Let MATH. This construction can be computed in polynomial time. The next two claims show that it is a desired reduction from REF to REF . If some MATH with MATH has at least one element in each MATH, then some MATH consists of at most MATH edges such that every nonsingleton connected component of MATH is strongly connected and bipartite-complete. To prove this claim, observe that for each MATH, some MATH exists. Let MATH. Let MATH. Let MATH be the set of edges in MATH from MATH to MATH. Let MATH be the set of edges in MATH from MATH to MATH. Let MATH. Note that MATH has at most MATH edges because MATH has at most MATH vertices. For each MATH with MATH, the edge MATH is in MATH by REF of the construction of MATH, MATH, and MATH. Also, MATH, MATH, and MATH are in MATH. These four edges form a vertex-simple traversable cycle. These cycles form the only nonsingleton connected component in MATH. Because these cycles all go through MATH, this component is strongly connected. By the choice of MATH, this component is bipartite-complete. This finishes the proof of REF . If some MATH consists of at most MATH edges such that every nonsingleton connected component of MATH is strongly connected and bipartite-complete, then some MATH with MATH has at least one element in each MATH. To prove this claim, observe that because every connected component of MATH is strongly connected, for each MATH with MATH, the set MATH contains some edges MATH and MATH. Then MATH and MATH exists in MATH by REF of the construction of MATH, MATH, and MATH. Let MATH. Let MATH be the connected component of MATH that contains MATH. Then MATH also contains MATH and MATH. By the completeness of MATH, the set MATH has at least MATH edges. Thus MATH because MATH. This finishes the proof of REF and thus that of REF . |
cs/0101034 | The proof is similar to that of REF . |
cs/0101034 | This theorem follows from REF . |
cs/0101034 | REF are straightforward. REF follows from REF . |
gr-qc/0101107 | By REF , the evaluation of a relativistic spin network is bounded by the evaluation of the network with the edge label MATH for each edge. |
gr-qc/0101107 | Since the function MATH is bounded and is asymptotic to MATH as MATH, for any MATH there exists MATH with MATH . The second part follows because MATH is bounded by REF. |
gr-qc/0101107 | Using notation as in this picture: we need to show MATH. By the law of cosines for hyperbolic trigonometry CITE we have MATH . Adding these equations we obtain MATH . By the triangle inequality we have MATH, so MATH from which the desired result follows. |
gr-qc/0101107 | This is proved by iteration. Suppose MATH and MATH are two points in the set which are the farthest distance apart. Then using REF , we can replace both MATH and MATH by the barycentre of MATH and MATH without increasing the quantity MATH . By iterating this process, all of the points in the set converge to the arithmetic mean MATH. |
gr-qc/0101107 | This is proved by induction on MATH. The induction starts with MATH by REF . Let MATH be a barycentre for the first MATH points. Then by the induction hypothesis MATH . Next we find a barycentre MATH for MATH points at MATH and REF point at MATH. To do this, note that all these points lie on a geodesic, and so the barycentre is constructed by REF . We thus have MATH . |
gr-qc/0101107 | Define MATH. Using REF we obtain MATH . The summation here, and in the rest of the proof of the lemma, is over values of MATH from MATH to MATH. If we work in spherical coordinates using the barycentre of the points MATH as our origin, this implies MATH . We next break this integral over MATH into two parts, and estimate these separately using two bounds: MATH from REF , and MATH where MATH . We obtain MATH for some constant MATH depending only on MATH and MATH. Finally, the triangle inequality implies MATH for all choices of MATH, so MATH . |
gr-qc/0101107 | Firstly, consider introducing an extra vertex connected to an integrable graph MATH by three or more extra edges. REF shows that doing the integral over the extra vertex first introduces an extra multiplicative factor of MATH in the integral of MATH for the graph MATH. However the function MATH is bounded so the new graph is also integrable. Adding an extra edge does not affect the integrability of a graph because MATH, by REF . Finally, for two integrable graphs joined by identifying one vertex we have that the evaluation of the resulting graph is the product of the evaluations of the two pieces, and so in particular the graph is integrable. This follows from taking the vertex where the two pieces are joined as the vertex which is not integrated over in the definition of the evaluation. |
gr-qc/0101107 | We need to show for any choice of numbers MATH for MATH and a point MATH, the integral MATH converges. First we integrate out MATH using REF , obtaining MATH where MATH. Next integrate over another variable, say MATH. With MATH this gives MATH . By REF we have MATH . To get a good bound on the integral here, we resort to a coordinate system in which two of the coordinates are MATH while the third is the angle MATH between the plane containing MATH and a given plane containing MATH and MATH. The ranges of these coordinates are MATH where we set MATH. Coordinates of this sort can also be defined in Euclidean space, where they are closely akin to prolate spheroidal coordinates CITE, but here all the formulas are a bit different, since we are working in hyperbolic space. The main thing we need is a formula for the volume form in these coordinates, MATH which is proved in the Appendix. Using this formula we can do the integral REF in the MATH coordinate system, obtaining MATH or doing the integral over MATH and MATH and then MATH, MATH for some constants MATH and MATH independent of all the parameters in this problem. We conclude the proof by using this bound on MATH to bound the integral MATH. By REF we have MATH and by REF this gives MATH . The right-hand side is finite so the proof is complete. |
hep-ph/0101101 | We first consider the case of MATH non-singular. In fact, in this case there is an open neighborhood of MATH in MATH where all matrices are non-singular. We can apply the parametrization REF to every matrix in that neighborhood. Using the normality of MATH REF and MATH, we can write (dropping the argument MATH for brevity) MATH and MATH, with MATH, MATH unitary and MATH, MATH diagonal. Define MATH diagonal by MATH. Then MATH and MATH satisfy REF . Assume now MATH has a null eigenvalue with multiplicity MATH. We can choose MATH so that MATH with MATH. Similarly, we can choose MATH so that MATH, with MATH and MATH. We then define MATH and proceed as above. |
hep-th/0101063 | Let MATH be the arc with radius MATH shown in REF . Then MATH . Since MATH, then there exists MATH such that MATH . Thus MATH . Using MATH we obtain for REF MATH . Therefore, as MATH, we obtain MATH . |
hep-th/0101063 | Since MATH, then MATH CITE. A straightforward application of NAME 's inequality shows that MATH. Then, from the above lemma MATH . |
hep-th/0101089 | In the zeroth order in MATH REF implies MATH . This holds provided the boundary REF are compatible. In the MATH-th REF order in MATH REF implies: MATH where the quantity MATH is given by MATH and we have assumed that MATH for MATH. The necessary and sufficient condition for REF to have a solution is MATH. Let us first show explicitly that MATH. Indeed, in view of the zeroth order equation MATH . Then a particular solution for MATH is MATH . The proof of the statement goes further along the standard induction procedure CITE: one can first check that MATH provided MATH satisfy REF for MATH; one then finds: MATH . Finally, one can check that MATH is a unique solution of REF for MATH provided the additional condition MATH is imposed. |
hep-th/0101089 | Let us represent adjoint action MATH of the BRST charge and function MATH as the sum of homogeneous components with respect to MATH: MATH with MATH. In the MATH-th order in MATH REF then becomes MATH where MATH is given by MATH . The consistency condition for REF is MATH. Let us show that MATH provided REF is fulfilled for any MATH and MATH carries nonnegative ghost number. Indeed, in the zeroth order in MATH REF rewrites as MATH . The consistency condition is obviously fulfilled since MATH . The later equality follows because MATH is independent of MATH. Assume that MATH are given for MATH and REF is fulfiled for MATH. Consider then the identity: MATH . One can see that MATH were MATH denote terms of order higher than MATH. In the MATH-th order in MATH . REF then implies: MATH. That MATH allows one to construct solution iteratively: MATH . One can indeed check that MATH since MATH and MATH. The later equality is obvious for MATH; the fact that MATH follows because MATH has a nonnegative ghost number. Thus the first part of the statement is proved. Further, let MATH and MATH satisfy REF. Assume that MATH for any MATH. In the MATH-th order in MATH . REF then implies: MATH . Since MATH and MATH, one can represent MATH as MATH . Thus MATH satisfies REF and coincides with MATH up to to the terms of order higher than MATH. Iteratively applying this procedure one can construct function MATH such that MATH . Finally, let MATH and MATH satisfy REF and additional condition MATH. For MATH one then has: MATH . This implies that MATH because MATH is at least linear in MATH. This proves second item. |
hep-th/0101089 | It is useful to introduce new coordinate functions MATH where MATH are the coefficients of an arbitrary symmetric connection on MATH. The reason is that unlike MATH that have inhomogeneous transformation properties, the coordinate functions MATH transform as the coefficients of a REF-form on MATH. Note, that functions MATH and MATH also form a local coordinate system on MATH and conditions MATH and MATH are obviously equivalent. Any BRST exact function MATH (that is, a function that can be represented as MATH) evidently vanishes when MATH. Conversely, assume that MATH vanishes when MATH and carries nonnegative ghost number. Then it can be represented as MATH where functions MATH and MATH can be taken in the form MATH and MATH respectively. One can also choose MATH and MATH such that they transform as the components of a section of MATH and components of a vector field on MATH respectively. We introduce MATH where MATH is the curvature of MATH. Note that MATH transform as components of a vector field on MATH; in particular, MATH is the globally defined function on MATH. Picking MATH as MATH one can indeed check that MATH . Finally, it follows from REF that there exists function MATH such that MATH. |
hep-th/0101089 | The proof is a direct generalization of that of the analogous statement from CITE. Assume that MATH for MATH and MATH doesn't depend on MATH. Then the solution can be constructed iteratively in the form MATH where MATH and MATH is given by MATH . |
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