paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9902147 | This easily follows from REF. |
math/9902147 | The inclusion MATH of the first equality is obvious, and the inclusion MATH of the second equality follows from REF -REF . To prove the inclusion MATH of the first equality, by REF it is enough to prove that MATH for all MATH. This obviously holds if we prove MATH for every such a MATH since the inclusion MATH of the s... |
math/9902147 | This follows from REF . |
math/9902147 | This follows from REF -REF . |
math/9902147 | The canonical map MATH is a linear isomorphism by REF and the exactness of REF. Moreover it is obviously continuous. Then it is also an homeomorphism because MATH is a NAME topological vector space of finite dimension. |
math/9902147 | By the commutativity of the diagram MATH where all maps are canonical, the result follows directly from REF -REF . |
math/9902147 | Consider the compositions MATH where the first map of each composition is canonical, and the second one is canonically induced by the projections MATH. The first composition is an isomorphism of topological vector spaces by REF -REF , and we know that the second composition is a continuous linear isomorphism REF . Then... |
math/9902147 | This is a direct consequence of REF . |
math/9902147 | This follows from REF since each MATH induces isomorphisms MATH . |
math/9902147 | From REF we get a (non-canonical) isomorphism MATH . Then let MATH be the subspace of MATH that corresponds to MATH by REF. Then REF easily follows from REF , and REF is obvious; in fact, MATH corresponds to MATH by REF. |
math/9902147 | We have MATH since MATH is closed in MATH, yielding MATH because MATH. So MATH as topological vector spaces because all spaces involved are closed subspaces of MATH (see for instance CITE). Now the result follows easily. |
math/9902147 | The space MATH is closed in MATH since MATH is a closed operator, and thus so is its subspace MATH. Hence MATH as NAME spaces, where MATH is the orthogonal complement of MATH in MATH; in particular MATH is closed in MATH too. Obviously, MATH . On the other hand we clearly have MATH and thus MATH and MATH are respective... |
math/9902147 | This follows from REF since we clearly have MATH as topological vector spaces. |
math/9902147 | This follows directly from REF . |
math/9902147 | Let MATH. The restriction MATH induces a homomorphism MATH . We clearly have MATH . So MATH induces a continuous linear isomorphism MATH . Observe that MATH is a NAME topological vector space by REF , and thus MATH and MATH induce norms on MATH that will be also denoted by MATH and MATH, respectively. By REF , the homo... |
math/9902147 | Since MATH is of finite dimension, there is some constant MATH, depending on MATH, so that MATH . Because MATH, any finite dimensional subspace MATH is contained in the sum of finite dimensional subspaces MATH, MATH. Therefore we can assume MATH is contained in some MATH with MATH. Then, for MATH and MATH, we have MATH... |
math/9902147 | The result is proved by induction on MATH. First, the case MATH is trivial. Second, the kernel and the image of MATH in MATH respectively are MATH and MATH, whose canonical projections in MATH are MATH yielding the canonical isomorphism REF for MATH. Since the isomorphism MATH is canonically defined, it corresponds to ... |
math/9902147 | Observe that the first part of the statement means that we have an orthogonal decomposition MATH . Again the result follows by induction on MATH. We have an orthogonal decomposition MATH by REF . Thus the isomorphism MATH is induced by the orthogonal projection MATH onto MATH. On the other hand, the kernel and image of... |
math/9902147 | Write MATH with MATH and MATH. Then MATH by REF and because MATH. |
math/9902147 | Since the image of MATH is closed and equal to its kernel, the hypothesis implies the existence of a sequence MATH such that MATH strongly in MATH. On the other hand we have MATH on MATH, and thus the operator MATH is bounded because MATH and MATH are bounded operators in MATH by REF . Therefore MATH strongly in MATH. ... |
math/9902147 | In this case we have MATH and MATH, which satisfy MATH strongly in MATH. Since MATH we get MATH strongly in MATH. But MATH by REF , and thus we get MATH strongly in MATH. Now observe that MATH, and MATH because MATH and by REF . Then the result follows by REF . |
math/9902147 | First, we check that the assignment MATH, under the conditions in the statement, defines a map MATH - observe that, if such a map is well defined, it is obviously linear - . Suppose there is another MATH and another sequence MATH such that MATH strongly in MATH. Then the sequence MATH satisfies MATH strongly in MATH. T... |
math/9902147 | Consider the following subspaces of MATH: MATH . First, observe that each MATH is the orthogonal complement of MATH in MATH, and thus MATH is the orthogonal complement of MATH in MATH. So the inclusion MATH induces an isomorphism of topological vector spaces MATH whose inverse is induced by the orthogonal projection MA... |
math/9902147 | By the expression MATH it is enough to consider the following three cases. First, assume MATH and let MATH. We clearly have MATH yielding MATH by REF , and because MATH and MATH. Second, suppose MATH and write MATH with MATH . We clearly have MATH yielding MATH by REF . Third, assume MATH, which is contained in MATH. T... |
math/9902147 | Write MATH with MATH, MATH and MATH . Observe that MATH. Since MATH and MATH on MATH, we have MATH. Thus there is some MATH with MATH because MATH. Then MATH satisfies MATH and moreover MATH by REF . Therefore we can assume MATH, and thus MATH. With this assumption, it follows that MATH and MATH by REF , yielding MATH.... |
math/9902147 | By REF, MATH is essentially self-adjoint in MATH. Then, by using for example, REF - REF, so is MATH because MATH is a bounded self-adjoint operator on MATH. But MATH is equal to MATH in MATH and vanishes in its orthogonal complement. Hence MATH is essentially self-adjoint. |
math/9902147 | We have MATH . But MATH on MATH. Then the result follows by REF -REF . |
math/9902147 | REF implies that each inclusion MATH is a compact operator, and MATH. Then the result follows by REF . |
math/9902147 | Observe that the strong convergence MATH in MATH just means the strong convergence MATH. Write MATH with MATH and MATH, according to REF . Then MATH strongly in MATH by REF , yielding MATH strongly in MATH by REF . Moreover MATH by REF and because any linear map MATH is continuous with respect to the MATH norms since M... |
math/9902147 | We can assume that MATH is oriented by using the two fold covering of orientations with standard arguments. Then it is easy to check that the NAME star operator, MATH, satisfies MATH, and MATH on MATH for each integer MATH. Then the result follows from REF . |
math/9902147 | Consider the operators MATH given by REF . Since MATH are of order zero, there is some MATH such that MATH. Because MATH is symmetric, we get MATH for all MATH, yielding MATH . Similarly we get MATH . Therefore, from REF we get MATH for some MATH and all MATH small enough. |
math/9902147 | In the case MATH, REF just means MATH. Therefore MATH by REF . Hence MATH and MATH is uniformly bounded since both MATH and MATH are positive operators. It follows that MATH is uniformly bounded in MATH. Therefore some subsequence of MATH is weakly convergent in MATH (and thus strongly convergent in MATH) to some MATH.... |
math/9902147 | First, we can assume the metric is bundle-like by REF. So we can apply the results of this section. If we had a strict inequality MATH in REF for some MATH, by the isomorphism MATH there are sequences MATH and MATH such that MATH, MATH, and MATH . But then we get a contradiction by REF . So inequality MATH holds in REF... |
math/9902148 | When MATH is a surface, the theorem is proved in CITE. Hence, in what follows we will assume that MATH. Furthermore, as is shown in CITE, the number of closed characteristics is at least MATH. Since MATH when the bundle MATH is trivial, we will assume from now on that MATH is not trivial. We need to recall some details... |
math/9902148 | Throughout the proof we will keep the notations and conventions of REF. Using the symplectic neighborhood theorem, let us identify a neighborhood of MATH in MATH with a neighborhood MATH of the zero section in the total space of the normal bundle MATH so that the symplectic structure on MATH is linear and equal to MATH... |
math/9902148 | The normal bundle MATH to MATH is canonically isomorphic to MATH. Furthermore, let us identify the tangent and cotangent bundles to MATH by means of MATH. Then MATH for all MATH. The compatibility condition of the corollary is equivalent to that for every MATH the eigenvalues of the metric on MATH with respect to MATH ... |
math/9902148 | Note first that to prove the theorem for MATH it suffices to prove the theorem for any form in the cohomology class MATH. Indeed, the fiberwise shift by a one-form MATH sends bounded sets to bounded sets and transforms the twisted symplectic form MATH into the form MATH. Hence without loss of generality, we may assume ... |
math/9902151 | Let MATH be a globular group. We introduce the map MATH defined by MATH and the map MATH defined by MATH. The arrow MATH is certainly a morphism of globular groups from MATH to MATH. In fact, if MATH is an arrow of MATH and if MATH is an element of MATH, then MATH. Therefore the diagram MATH commutes. Now we have to ve... |
math/9902151 | Let MATH be a globular group. For any MATH, let MATH be a free abelian group and MATH an epimorphism of abelian groups. Then the epimorphisms MATH for all MATH induce a natural transformation MATH which is certainly itself an epimorphism. Left adjoint functors and coproduct preserve projective objects CITE. Hence the c... |
math/9902151 | If MATH is a globular group, we introduce the complex of abelian groups MATH defined as follows : MATH and for MATH, MATH, with the differential map MATH if MATH and MATH if MATH with MATH. We have MATH. Let MATH be a natural number and let MATH be a free abelian group. If MATH, let us prove that MATH. Let MATH be a cy... |
math/9902151 | It turns out that MATH is a chain retraction of MATH. If MATH, then MATH. And for MATH, with MATH, we have actually : MATH . |
math/9902151 | If MATH is in MATH, then all MATH are non degenerate in the cubical nerve. But MATH is non degenerate if and only if MATH is MATH-dimensional. Hence the necessity of the condition. Conversely assume that MATH. Then there exists MATH between MATH and MATH such that MATH with MATH and some MATH, ., MATH and with MATH. Th... |
math/9902151 | Obvious. |
math/9902151 | The underlying idea of the proof is as follows. If one wants to define a MATH-functor from MATH to a MATH-category MATH, it suffices to construct MATH-functors from the MATH-faces of MATH to MATH which coincide on the intersection of their definition domains and to fill correctly the interior of MATH. We have necessari... |
math/9902151 | The construction of MATH is exactly the same as the one of connections on the cubical singular nerve of a topological space. Thus there is nothing to verify in the axioms of cubical complex with connections except the relations mixing the two families of degeneracies MATH and MATH : all other axioms are already verifie... |
math/9902151 | The axioms of simplicial sets are immediate consequences of the axioms of cubical set with connections. |
math/9902151 | The proof is quite simple. If MATH is a MATH-dimensional globular cycle, then MATH because of the definition of MATH. And a MATH-dimensional globular boundary MATH is mapped to MATH. |
math/9902151 | Let MATH. Then the natural map MATH from MATH to MATH which sends MATH to MATH corresponds by NAME to a MATH-functor MATH from MATH to MATH, where MATH (respectively, MATH) is the free MATH-category generated by a MATH-morphism MATH (respectively, a MATH-morphism MATH). Set MATH with still the convention MATH and MATH.... |
math/9902151 | Let MATH and let MATH. We have to compare MATH and MATH modulo elements of MATH. We get MATH . Now let us treat the case MATH. Let MATH. We immediately see that MATH and MATH are equal. |
math/9902151 | It is due to the fact that for MATH, the MATH-th homology group of the quotient chain complex MATH is the MATH-th homology group of the normalized chain complex associated to the corner simplicial nerve of MATH. |
math/9902151 | One has MATH since MATH. If MATH, then MATH. Now suppose that MATH and MATH. Then MATH. |
math/9902151 | Suppose that MATH. Define MATH like this REF CASE: if MATH and if MATH, then MATH CASE: if MATH and if MATH, then MATH. If MATH is MATH-dimensional, then MATH . If MATH is of dimension MATH greater than MATH, then either MATH and in this case MATH since MATH. Or MATH and in that case MATH . Since MATH is MATH-dimension... |
math/9902151 | For any MATH of dimension MATH, there exists MATH such that MATH. By convention, we take MATH whenever MATH. However MATH and in the same way, we have MATH. Therefore MATH and MATH . |
math/9902151 | This relation is obviously reflexible and symmetric. It remains to prove the transitivity. Let us consider the following diagram in MATH : MATH and suppose that MATH, MATH, MATH and MATH. Then MATH and MATH. |
math/9902151 | If MATH, it is trivial. Suppose that MATH. Let MATH be the only MATH-functor from MATH to MATH such that MATH and let MATH be the unique functor from MATH to MATH such that MATH. Then MATH so MATH and MATH are homotopic as MATH-functors. Now consider MATH and MATH. Set MATH . First suppose that MATH. Then we have MATH ... |
math/9902151 | Take two homotopic MATH-functors MATH and MATH. Let MATH be a globular MATH-cycle. Then MATH. Therefore MATH. Now take a globular MATH-cycle MATH with MATH. Then MATH. Therefore MATH. |
math/9902151 | We make the proof for MATH. The homology of the non normalized complex associated to a simplicial group is equal to its homotopy REF . Therefore it suffices to find an homotopy between MATH and MATH in MATH for any MATH. We can suppose without loss of generality that there exists a MATH-dimensional morphism MATH of MAT... |
math/9902151 | We give only a sketch of proof. CASE: First of all, we observe that the functor from MATH to the category MATH of sets MATH is representable. We denote by MATH the representing MATH-category. It is equal to the direct limit of the diagram MATH . We denote by MATH and MATH the two canonical embeddings of MATH in MATH re... |
math/9902151 | The induction equations define a fillable MATH-shell as defined in REF . |
math/9902151 | Obvious for MATH and MATH. Let us suppose that MATH and let us proceed by induction on MATH. Since MATH by the previous proposition, then the evaluation map MATH from MATH to MATH is surjective. Now let us prove that MATH and MATH and MATH imply MATH. Since MATH and MATH are in MATH, then one sees immediately that the ... |
math/9902151 | By REF , any MATH-cube MATH of MATH (respectively, of MATH) is determined by its MATH-shell of MATH-faces MATH and by its image in MATH (respectively, MATH). |
math/9902157 | If MATH, then it is obvious that the edges of MATH lie in an essential annulus in MATH. Assume MATH. Let MATH. If the endpoints of MATH appear in this order when one travels around MATH clockwise, say, then those of MATH appear in the same order when one travels around MATH anticlockwise, since MATH and MATH have disti... |
math/9902157 | Let MATH be the disk bounded by MATH in MATH. Then MATH, since MATH is essential in MATH by the assumption on MATH. If both sides of MATH on MATH meet MATH, replace MATH by MATH. Then MATH gives a new NAME torus of MATH satisfying MATH. However this contradicts the choice of MATH, since MATH. Hence all vertices of MATH... |
math/9902157 | Assume for contradiction that the edges of MATH lie in a disk MATH in MATH. Let MATH be the subgraph of MATH consisting of two vertices MATH and MATH along with the edges of MATH. First, suppose that MATH. By the cut-and-paste operation of MATH, it can be assumed that any component in MATH is essential in MATH. Therefo... |
math/9902157 | Let MATH be the label pair of MATH. Assume that the edges of MATH lie in a disk MATH in MATH for contradiction. By the same argument in the proof of REF , we may assume that MATH. If MATH, then MATH gives a punctured lens space in a solid torus, which is impossible. Therefore MATH. Choose an innermost component MATH of... |
math/9902157 | Let MATH, MATH be the label pairs of MATH and MATH, respectively. By REF , the edges of MATH give an essential cycle in MATH after shrinking two fat vertices MATH and MATH to points, and MATH. Then MATH is contained in the solid torus, MATH say, and the union MATH gives a NAME band MATH properly embedded in MATH, after... |
math/9902157 | By REF , MATH and MATH. We remark that MATH. Hence MATH and MATH have the label pair MATH. We may assume that MATH and MATH. Then MATH is a solid torus, and MATH runs twice in the longitudinal direction on MATH by REF . Furthermore, the annulus MATH is parallel to MATH in MATH. Similarly, MATH is a solid torus, and MAT... |
math/9902157 | Assume that there are MATH mutually parallel edges MATH in MATH, numbered successively. We may assume that MATH has the label MATH at one endpoint for MATH, and MATH has the label MATH. By the parity rule, MATH has the label MATH at the other endpoint for some MATH. If MATH, then MATH and MATH form MATH-cycles with dis... |
math/9902157 | Let MATH and MATH be MATH-cycles in MATH with the label pairs MATH, respectively, where MATH. Let MATH be the face of MATH bounded by MATH. By REF , the edges of MATH lie in an essential annulus in MATH, and MATH. If we shrink MATH to its core radially, then MATH gives a NAME band MATH properly embedded in MATH or MATH... |
math/9902157 | We may assume that MATH has the label MATH at one endpoint for MATH. If MATH has the label MATH at the other endpoint, then MATH is a MATH-cycle. Therefore we suppose that MATH has the label MATH at the other endpoint for some MATH. Then MATH and MATH form MATH-cycles with disjoint label pairs. By the same argument as ... |
math/9902157 | Since MATH does not contain trivial loops, the unique vertex MATH has valency at most MATH in MATH (see CITE). Therefore the edges of MATH are partitioned into at most MATH families of parallel edges. If there is an edge MATH in MATH with MATH, then MATH for some MATH by REF , that is, MATH contains a NAME bottle. Then... |
math/9902157 | By REF , the interior of a black (white) bigon is disjoint from MATH. Then the proof of CITE remains valid. Remark that a final contradiction comes from the fact that a NAME bottle will be found in a solid torus MATH or MATH. |
math/9902157 | If MATH and MATH are parallel in MATH and have the same edge class label, then they are also parallel in MATH. Then MATH contains a NAME band by CITE, which contradicts the fact that MATH is hyperbolic. |
math/9902157 | Suppose that there are MATH mutually parallel edges. Then there are two bigons with the same color among these MATH parallel edges. By REF , these two bigons have the same pair of edge class labels. This contradicts REF . |
math/9902157 | Let MATH be adjacent parallel edges of MATH. By REF , these three edges have distinct edge class labels. Let MATH be the edge class labels of MATH respectively. Let us denote the endpoints of MATH by MATH for MATH. See REF . MATH . Note that MATH and MATH appear consecutively around the vertex MATH in the order, when t... |
math/9902157 | The unique vertex MATH has valency at most MATH in MATH, and the edges of MATH are partitioned into at most MATH families of parallel edges. Recall that MATH has valency MATH in MATH. By REF , MATH cannot contain MATH mutually parallel edges. If MATH contains MATH mutually parallel edges, then we have MATH by REF . If ... |
math/9902157 | This follows from CITE. |
math/9902157 | If MATH contains two MATH-cycles on disjoint label pairs, then MATH is MATH for some MATH by REF . But this is impossible by REF . |
math/9902157 | The vertex MATH has valency MATH in MATH. Recall that the edges of MATH are partitioned into at most three families of mutually parallel edges. Let MATH be a family of mutually parallel edges in MATH, and suppose that MATH consists of more than MATH edges, MATH numbered consecutively. Note that MATH by REF . We may ass... |
math/9902157 | The vertex MATH has valency MATH in MATH. By REF , MATH consists of three families of mutually parallel edges, each containing exactly MATH edges. Then there is no MATH-cycle in MATH, but there are two NAME cycles MATH and MATH of length three in MATH. Let MATH be the face of MATH bounded by MATH for MATH. We may assum... |
math/9902157 | Since the vertex MATH has valency MATH in MATH, there are more than MATH mutually parallel edges in MATH, which contradicts REF . |
math/9902157 | Take a regular neighborhood MATH of MATH in MATH. Let MATH. Then MATH is an annulus whose boundary defines a link MATH in MATH. Note that MATH is incompressible in the exterior of MATH, and MATH has an unknotting tunnel. Then MATH is a MATH-bridge torus link by CITE. Furthermore, an unknotting tunnel of such a link is ... |
math/9902157 | If there are three mutually parallel edges in MATH, there are two MATH-cycles MATH and MATH whose faces MATH and MATH lie on opposite sides of MATH. Since MATH by REF , we can assume that MATH for MATH by REF . Then we may also assume that MATH and MATH. Note that MATH and MATH are handlebodies of genus two, since the ... |
math/9902157 | Recall that MATH is separating in MATH, and therefore the faces of MATH are partitioned into black and and white ones. This implies that MATH has no parallel edges, since MATH has just three edges. Then there are two NAME cycles MATH and MATH of length three in MATH. Let MATH be the face of MATH bounded by MATH for MAT... |
math/9902157 | MATH has exactly five edges. By REF , these edges of MATH are partitioned into three families, two pairs of parallel edges and one edge which is not parallel to the others. However, this configuration contradicts the fact that the faces of MATH are divided into black and white sides. |
cs/9903009 | Assume that there is a node such that the deviation of its degree MATH from MATH is greater than MATH, that is, MATH. From the lower bound on MATH corresponding to the assumption that MATH is random in MATH, we can estimate an upper bound on MATH, as follows: In a description of MATH given MATH we can indicate which ed... |
cs/9903009 | The only graphs with diameter REF are the complete graphs which can be described in MATH bits, given MATH, and hence are not random. It remains to consider MATH is a MATH-random graph with diameter greater than REF. which contradicts REF from some MATH onwards. It remains to consider MATH is a MATH-random graph with di... |
cs/9903009 | Given MATH, let MATH be the set of the least MATH nodes directly adjacent to MATH. Assume by way of contradiction that there is a node MATH of MATH that is not directly connected to a node in MATH. We can describe MATH as follows: CASE: A description of this discussion in MATH bits; CASE: A literal description of MATH ... |
cs/9903009 | Let MATH be as in the statement of the theorem. By REF we know that from each node MATH we can route via shortest paths to each node MATH through the MATH directly adjacent nodes of MATH that have the least indexes. By REF , MATH has diameter REF. Once the message has reached node MATH its destination is either node MA... |
cs/9903009 | Let MATH and MATH be as in the statement of the theorem. By REF we know that from each node MATH we can shortest path route to each node MATH through the first MATH directly adjacent nodes MATH of MATH. By REF , MATH has diameter REF. Relabel MATH such that the label of node MATH equals MATH followed by the original la... |
cs/9903009 | Let MATH and MATH be as in the statement of the theorem. By REF we know that from each node MATH we can shortest path route to each node MATH through the first MATH directly adjacent nodes MATH of MATH. By REF , MATH has diameter REF. Consequently, each node in MATH is directly adjacent to some node in MATH. Hence, it ... |
cs/9903009 | Let MATH and MATH be as in the statement of the theorem. By REF , MATH has diameter REF. Therefore the following routing scheme has stretch factor MATH. Let node MATH store a shortest path routing function. All other nodes only store a shortest path to node MATH. To route from a originating node MATH to a target node M... |
cs/9903009 | Let MATH and MATH be as in the statement of the theorem. From REF we know that from each node MATH we can shortest path route to each node MATH through the first MATH directly adjacent nodes of MATH. By REF , MATH has diameter REF. So the local routing function - representable in MATH bits - is to route directly to the... |
cs/9903009 | At each node we can give the neighbors by the positions of REF's in a binary string of length MATH. Since each node has at most MATH neighbours by REF , a permutation of port-assignments to neighbors can have NAME complexity at most MATH CITE. This permutation MATH describes part of the local routing function by for ea... |
cs/9903009 | Since for MATH-random labeled graphs on MATH the node degree of every node is MATH by REF , we can in each source node describe the appropriate outgoing edges (ports) for each destination node by REF's in a binary string of length MATH. For each source node it suffices to store at most MATH such binary strings correspo... |
cs/9903009 | Let MATH be a MATH-random graph. Let MATH be the local routing function of node MATH of MATH, and let MATH be the number of bits used to store MATH. Let MATH be the standard encoding of MATH in MATH bits as in REF . We now give another way to describe MATH using some local routing function MATH. CASE: A description of ... |
cs/9903009 | In the proof of this theorem we need the following combinatorial result. Let MATH and MATH be arbitrary natural numbers such that MATH. Let MATH, for MATH, be natural numbers such that MATH. If MATH, then MATH . By induction on MATH. If MATH, then MATH and clearly MATH if MATH. Supposing the claim holds for MATH and ar... |
cs/9903009 | If the graph cannot be relabeled and the port-assignment cannot be changed, the adversary can set the port-assignment of each node to correspond to a permutation of the destination nodes. Since each node has at least MATH neighbours by REF , such a permutation can have NAME complexity as high as MATH CITE. Because the ... |
cs/9903009 | Consider the graph MATH with MATH nodes depicted in REF . Each node MATH in MATH is connected to MATH and to each of the nodes MATH. Fix a labeling of the nodes MATH with labels from MATH. Then any labeling of the nodes MATH with labels from MATH corresponds to a permutation of MATH and vice versa. Clearly, for any two... |
cs/9903009 | Let MATH be a graph on nodes MATH satisfying REF with MATH. Then we know that MATH satisfies REF . Let MATH be the local routing function of node MATH of MATH, and let MATH be the number of bits used to encode MATH. Let MATH be the standard encoding of MATH in MATH bits as in REF . We now give another way to describe M... |
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