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math/9906209 | Let MATH. Since MATH is a rank two vector bundle on MATH, we have MATH where MATH denotes the first NAME class of MATH. By NAME duality we have MATH. Now from the exact sequence MATH we compute MATH, and we are done. |
math/9906209 | The formula for the postulation follows immediately from the exact sequence MATH. To obtain the formula for MATH we compute MATH using NAME duality: MATH . |
math/9906209 | Suppose first MATH is linked to MATH by the complete intersection MATH, and let MATH. Let MATH be the intersection of MATH with the reduced plane MATH. We claim that MATH, MATH and MATH. By CITE page REF the ideal sheaf of MATH in MATH is MATH with its natural embedding in MATH. By REF, applying the functor MATH to the exact sequence MATH we obtain a long exact sequence MATH . As MATH, the kernel of the last map is MATH, and our claim follows. Conversely, suppose that MATH is a curve in MATH containing MATH, and such that MATH. The exact sequence REF tells us that there exists a surface MATH containing MATH but not MATH, whose intersection with MATH is MATH. By what we have just shown, MATH. |
math/9906209 | Use the above proposition with MATH. |
math/9906209 | If MATH, by the previous corollary there is a curve obtained from MATH by an elementary biliaison of negative height, hence MATH is not minimal. If MATH, then by REF we have MATH . Now it follows from REF that MATH is minimal: see REF . |
math/9906209 | If MATH and MATH, MATH is irreducible by REF . To handle the case MATH and MATH, we fix homogeneous coordinates MATH on MATH, so that MATH is an equation for MATH, and we look at the family of curves MATH in MATH defined by the global ideal MATH . For MATH, MATH is in MATH, while MATH belongs to MATH. It now follows from REF that MATH is connected. If MATH, by REF we have set theoretically MATH where MATH . In particular, MATH is irreducible for MATH, and for MATH the theorem is a consequence of the following proposition. |
math/9906209 | Suppose the claim is true when MATH. Then by adding MATH times a plane section (that is, by performimg an elementary biliaison of height MATH on MATH, see REF ) we find that the claim is true for all MATH, and we are done. We now construct a family of curves in MATH specializing to one in MATH. This has already been done by NAME in REF , in the case MATH and MATH, and his construction, generalizes without any major modification (NAME noticed this independently CITE). Here are the details. As above we fix homogeneous coordinates MATH on MATH, so that MATH is an equation for MATH. We let MATH and MATH denote respectively the double line MATH and the multiple line MATH. Let MATH have equations MATH where MATH is a form of degree MATH in MATH. To give a curve MATH in MATH with MATH is by REF the same as giving a morphism MATH whose image in in MATH generates MATH at every point. This amounts to choosing forms MATH and MATH in MATH, of degrees MATH and MATH respectively, and such that MATH and MATH have no common zeros on the line MATH. The corresponding morphism MATH sends MATH to MATH and MATH to MATH. For the corresponding curve MATH we have: MATH . Now we look at the family of curves in MATH obtained by flattening the ideal generated by MATH where MATH . For MATH we obtain a curve MATH as above with MATH, MATH and MATH. To see what happens at MATH, we set MATH . It follows that the ideal of the limit scheme MATH contains the ideal MATH . The saturation of MATH is the ideal MATH . But this is the homogeneous ideal of a curve MATH in the double plane: MATH is the line MATH, MATH has equations MATH, and MATH is defined on MATH by the equation MATH. Hence MATH belongs to MATH. In particular, MATH has the same degree and genus as MATH. So we must have MATH, and this finishes the proof. We remark that in this family the zero dimensional scheme MATH associated to MATH is supported at the point MATH for MATH, and at the point MATH for MATH ! |
math/9906209 | Looking at CITE, CITE one sees that MATH is extremal if and only if it is not planar and MATH. From this we deduce that REF implies REF. Curves MATH in the union of two distinct planes MATH are studied in REF: if MATH has degree MATH, then MATH is either minimal or ACM. From this we see that REF implies REF. Finally, we prove REF implies REF. If MATH is obtained from a plane curve by an elementary biliaison of height one on a quadric surface, then one computes MATH and so MATH is extremal. If MATH is a minimal curve in MATH, then by REF, adding a suitable number (namely, MATH) of plane sections to MATH we obtain a curve linearly equivalent to the disjoint union of two plane curves. Thus the NAME module of MATH is isomorphic to MATH where MATH, MATH is the homogeneous coordinate ring of MATH, MATH, MATH are the equations of MATH and MATH, and MATH,MATH are forms of degrees MATH, MATH respectively, having no common zeros along the line MATH. Hence MATH. |
math/9906209 | If MATH is not ACM, the equivalence of MATH and MATH is the content of REF. The same proof works for the ACM case as well (compare REF ). We claim that MATH implies MATH: suppose MATH is obtained from an extremal curve MATH by an elementary biliaison of height one on a quadric surface MATH. If there is a plane MATH whose intersection with MATH has degree MATH and if MATH contains MATH as a component, then MATH with MATH zero-dimensional contained in a line, and MATH is a plane curve. Otherwise, by REF either MATH is an ACM curve with MATH, or MATH is a divisor of type MATH on the smooth quadric surface MATH. So MATH is either ACM with MATH, or MATH is a divisor of type MATH on MATH. Finally, we show that MATH implies MATH. The special cases are clear, so we may assume that there is a plane MATH such that MATH with MATH zero-dimensional contained in a line, and MATH is a plane curve of degree two. Then there is an exact sequence: MATH from which we deduce that MATH, that MATH is contained in a unique quadric surface and that MATH for MATH. Since the NAME function of a curve of degree MATH contained in a quadric surface is symmetric around MATH, MATH is subextremal. |
math/9906209 | The first statement follows from REF . For MATH, MATH and for MATH the NAME scheme MATH is irreducible REF. For MATH and MATH, MATH has two irreducible components, namely the closure of the family of subextremal curves and the family of extremal curves CITE. Hence it is connected by the first statement. |
math/9906211 | Since the arguments are similar, we will prove only REF . Set MATH, and denote by MATH the closed embedding. Note that MATH gives an identification MATH. Applying NAME adjunction formula thrice, we get the commutative diagram MATH . In other words, the identity of MATH factorizes through MATH. One concludes by using CITE. |
math/9906211 | Setting MATH, it follows from CITE or CITE (see also CITE for the case of a single operator) that MATH. Since MATH is a flabby sheaf, one has MATH . The result then follows from REF . |
math/9906211 | Applying REF with MATH, we get MATH . It follows by REF that MATH. Since MATH coincides with MATH outside of the zero section, the fact that MATH is MATH-hyperbolic for any MATH implies that MATH is MATH-hyperbolic. It then follows from CITE or CITE that MATH . |
math/9906211 | Apply REF with MATH, and note that the solution complex MATH is represented by the complex of flabby sheaves MATH . |
math/9906211 | Set MATH. Since MATH is an epimorphism, we have an isomorphism MATH, and a short exact sequence MATH . Applying the functor MATH, we get the long exact cohomology sequence MATH . REF implies MATH for any MATH, and the proof is complete. |
math/9906211 | Since REF - REF are open in MATH, we may find an open neighborhood MATH of MATH in MATH such that REF holds, REF holds in MATH, and moreover MATH for any MATH. Let MATH, and MATH be a MATH-function on MATH as in REF . Consider the morphism of exact sequences, where the vertical arrows are induced by MATH . Consider the stalk-wise analog of REF for MATH. By definition of the micro-support, we are left to prove that MATH and MATH are isomorphisms. This follows from the following considerations. REF states that MATH is non-characteristic for MATH, and by NAME 's theorem this implies that MATH is injective. By REF , MATH is surjective. Moreover, REF MATH says that MATH is surjective, while REF MATH reads MATH. |
math/9906211 | If MATH is admissible, then MATH satisfies REF. If MATH and MATH are two real finite dimensional vector spaces, we identify MATH to MATH by MATH. Then, if MATH and MATH are two cones with MATH, MATH, one has MATH. In particular, since MATH for any MATH, one has MATH. This last set contains MATH by REF . Using the estimate REF follows. Remark that if MATH is an open convex cone in MATH and MATH for MATH, then MATH. By REF , for each MATH there exists MATH with MATH. Then MATH, and REF follows. The proof of REF is similar. |
math/9906211 | CASE: By REF, it is enough to show that MATH. Let MATH, and MATH. There exist a local chart MATH at MATH, and an open conic neighborhood MATH of MATH in MATH, such that MATH. In view of REF, we shall prove that MATH . Since MATH, if MATH is a MATH-path and MATH, then MATH. Let MATH and MATH with MATH. Since the segment of straight line from MATH to MATH is a MATH-path, MATH. CASE: Let us prove that MATH. Let MATH, and MATH. Take a local chart MATH at MATH and an open conic neighborhood MATH of MATH in MATH, such that MATH. Let MATH be an open neighborhood of MATH. By REF, for any MATH, MATH, and MATH, with MATH, and MATH, we have to show that MATH. By definition, MATH if and only if there exist sequences MATH, MATH, with MATH (that is, there is a MATH-path from MATH to MATH). We may assume MATH, MATH. Since MATH, the segment of straight line from MATH to MATH is a MATH-path. Composing the MATH-paths above, we get MATH, which implies MATH as requested. The proof that MATH is similar. |
math/9906211 | REF implies that MATH. REF implies that MATH is compact for any compact subset MATH of MATH. Finally, MATH by REF . |
quant-ph/9906089 | NAME 's inequality and REF give MATH as well as MATH where MATH is the usual operator norm. Thus, MATH for all MATH, which establishes the claim. |
quant-ph/9906089 | Using the product rule and MATH to differentiate MATH gives MATH . |
quant-ph/9906089 | Setting MATH . During the iteration MATH . Hence, setting MATH and noting that MATH we obtain MATH . Setting MATH where MATH denotes the time-ordered exponential, the formal solution of REF is (according to the previous lemma) given by MATH . Observing that MATH and thus MATH we arrive at MATH and thus the total variation from MATH to MATH is MATH . Since MATH is uniformly bounded, MATH is also uniformly bounded for all MATH and thus the sequence MATH is uniformly bounded. MATH for any MATH implies furthermore that MATH is an increasing sequence. Hence, MATH . Consequently MATH . |
quant-ph/9906090 | MATH . |
quant-ph/9906090 | From REF by NAME and NAME, we have only to show that MATH . Let MATH be an arbitrary test which satisfies MATH. From REF , we have MATH and hence, MATH . By taking the minimum, we obtain MATH . Now, let MATH, then MATH and MATH for sufficiently large MATH. Thus, REF yields MATH and hence, MATH . Since MATH is arbitrary, the theorem has been proved. |
quant-ph/9906090 | For all MATH, there exists MATH such that MATH from REF . Putting MATH in REF , we have MATH and hence, MATH . Since MATH is arbitrary, REF has been proved. To show REF , suppose that MATH firstly (see REF ), and MATH attains the maximum in the equation MATH . Then, taking REF into account, MATH is represented parametrically as MATH . By using REF to eliminate MATH from REF , we have MATH . On the other hand, let MATH then we have MATH . To examine the sign of MATH, put MATH, and we have MATH, which indicates that the sign of MATH changes at most once. Therefore MATH takes its maximum at MATH if and only if MATH . This is nothing but REF , and hence, we obtain MATH. In the other cases, it is clear that MATH and MATH . |
quant-ph/9906101 | We omit the easy proof that REF holds in any OML. For MATH that it holds in any OL is apparent from the definitions. For MATH it fails in the non-orthomodular ortholattice from REF but does not fail either in OREF or in WOML, non-OML lattices from REF , and REF. |
quant-ph/9906101 | We give here the proof only for MATH. Others are completely analogous. Let us write the premise MATH as MATH. Hence, MATH and according to CITE MATH. This, together with the other consequence of the premise: MATH, yields MATH CITE, what proves the statement. As for the vice versa part, all four implications fail in OREF which means that they must be orthomodular. |
quant-ph/9906101 | For REF : This is well known and we omit the proof. For REF : See CITE. For REF : MATH and MATH, so MATH; MATH. For REF : From NAME and NAME 's law we have MATH, so MATH; from hypothesis and NAME 's we have MATH, so MATH; from NAME 's we have MATH, so MATH, so MATH. For REF : From hypothesis and REF we have MATH, so MATH. For REF : MATH, so MATH. For REF : We omit the easy verifications. |
quant-ph/9906101 | For REF : Immediate from REF, LREF, LREF and definitions. For REF : Immediate from REF , using REF . For REF : Using REF , MATH; so MATH; so from hypothesis and REF we have MATH; so from REF we conclude MATH. For REF : Immediate from REF . For REF : Immediate from REF . For REF : From REF we have MATH and MATH; so from REF we have MATH and MATH; so from REF we have MATH. |
quant-ph/9906101 | REF and the remark after REF, which applies to any OL in which REF holds. |
quant-ph/9906101 | CASE: We have shown that REF holds in a WOML. On the other hand, it is easy to prove (using the NAME theorem for example) that MATH holds in an OML; thus by REF it also holds in OL + REF . In other words, the WOML we have defined here is equivalent to the WOML of CITE, and LREF are interchangeable as the WOM law added to an OL. CASE: It is easy to prove either direction of REF from the other using only LREF - LREF. In the proof of REF , we used only REF along with LREF - LREF. Thus in an OL, REF follows from REF, and REF follows from REF . |
quant-ph/9906101 | We extend the proof of REF using the completeness proof for unary quantum logic (for example, CITE) where MATH are the ortholattice mappings for the hypotheses of a deduction. |
quant-ph/9906101 | Since MATH implies MATH, by REF we have that MATH implies MATH in any OML. By REF this also holds in any WOML. On the other hand, assume LREF holds. If MATH then MATH by REF , so MATH by REF, so MATH, so MATH by REF , so LREF holds by REF b. |
quant-ph/9906101 | For LREF: Immediate from REF b. For LREF: We have, using LREF - LREF, MATH, so MATH and MATH. Hence MATH, which becomes the left-hand side of REF after substituting MATH for MATH and MATH for MATH then applying LREF. |
quant-ph/9906101 | Any axiom MATH-REF - MATH-REF is true in any model WOML. Let us put MATH and MATH and let us verify for example MATH-REF for MATH. It maps to MATH. By REF, LREF, LREF, and LREF we get MATH which is true by definition. MATH-REF - MATH-REF we prove analogously. We also have to verify that the set of formulas true in a model MATH is closed under the rules of inference: MATH-RREF. MATH-RREF maps to REF. MATH-RREF maps to MATH which according to REF holds in any WOML since MATH holds in any OML. MATH-RREF mappings we verify analogously. |
quant-ph/9906101 | As we have shown above, MATH is an equivalence relation. In order to be a relation of congruence, the relation of equivalence must be compatible with the operations MATH and MATH. MATH is nothing but MATH-RREF and MATH is MATH-RREF. |
quant-ph/9906101 | That LREF - LREF hold in MATH is obvious. MATH-RREF gives LREF which together with LREF - LREF gives LREF according to REF . |
quant-ph/9906101 | The theorem is a direct consequence of REF and rule MATH-RREF. |
quant-ph/9906101 | Let MATH-REF and MATH-RREF hold for MATH. We form the NAME algebra MATH for this logic using MATH and REF formulated for this MATH. Let us further assume that the so obtained MATH is not orthomodular. But by REF from MATH which must hold in such MATH we obtain the orthomodularity, that is, the contradiction. |
quant-ph/9906101 | Since MATH, the premise MATH yields MATH by REF , wherefrom by MATH-RREF we get MATH. |
quant-ph/9906101 | Since MATH is a model of MATH, to be true for MATH in MATH means: MATH. Hence, by the previous lemma, we get: MATH. |
quant-ph/9906101 | Let us first prove that MATH is an equivalence relation. MATH and MATH are obvious. The proof of the transitivity runs as follows. MATH . Since all the WOML axioms and rules hold in OREF, the last metaconjunction in line REF reduces to MATH by transitivity. Hence the conclusion MATH by definition. In order to be a relation of congruence, the relation of equivalence must be compatible with the operations MATH and MATH. The proofs of the compatibilities run as follows. MATH . In these proofs we used MATH-RREF and MATH-RREF and the corresponding lattice mappings in OREF. |
quant-ph/9906101 | Since all the WOML axioms and rules hold in OREF the proof follows from the proof of REF . |
quant-ph/9906101 | We assume MATH contains at least two propositional variables (or ``primitive" or ``starting" wffs). We pick an evaluation MATH that maps two of them, MATH and MATH, to distinct nodes MATH and MATH of REF that are neither REF nor REF such that MATH [that is, MATH and MATH are on the same side of hexagon OREF ]. From the structure of REF we obtain MATH and MATH. Therefore MATH, that is, MATH. This falsifies MATH. Therefore MATH, providing a counterexample to the OM law for MATH. |
quant-ph/9906101 | MATH is equivalent to MATH and therefore to MATH . |
quant-ph/9906101 | Right to left metaimplication in the line REF holds because all deductions of QL are sound in WOML, and OREF is a WOML. |
quant-ph/9906101 | In MATH is equivalent to MATH. Therefore in WDL MATH is equivalent to MATH. Therefore, since REF holds, REF gives the required result. |
quant-ph/9906101 | The proof actually follows from the proof of REF . We only have to prove that the rules MATH and MATH do hold in MATH. But this is well known. (For example, rules REF on p. REF.) |
cs/9907001 | Consider selecting MATH in the following way: choose a random permutation on the edges of MATH, and let MATH be the first MATH edges in the permutation. Let MATH be the MATH-st edge in the permutation. Then since MATH is equally likely to be any remaining edge, the expected number of edges that take part in a violated constraint is just MATH times the probability that MATH takes part in a violated constraint. But this can only happen if MATH is one of the at most MATH edges involved in the optimal base for MATH. Since this subset is just the first MATH edges in the permutation, and any permutation of this subset is equally likely, this probability is at most MATH. |
cs/9907001 | We use the linear time randomized minimum spanning tree algorithm of CITE, and let MATH denote the set of edges that are in the MST and not in MATH. Note that MATH has exactly as many edges as are in MATH and not in the MST; since each edge in the latter set takes part in a violated swap constraint, the expectation of MATH is MATH by REF . Then it is easy to see that, if tree edge MATH takes part in any violated constraints, at least one must be the constraint corresponding to swap MATH, where MATH is the minimum weight edge in MATH forming a swap with MATH. To find this minimum weight swap for each tree edge, we contract MATH as follows. While MATH has a degree-one vertex that is not adjacent to any edge in MATH, we remove it and its incident edge; that edge can not take part in any swaps with MATH. While MATH has a degree-two vertex that is not adjacent to any edge in MATH, we remove it and merge its two incident edges into a single edge; these two edges share the same minimum swap edge. After this contraction process, the contracted tree MATH has MATH vertices with degree less than three, and therefore MATH total vertices. We apply NAME 's nonlinear minimum spanning tree verification algorithm to this contracted tree to find the best swap in MATH for each contracted tree edge. We then undo the contraction process and propagate the best swap information to the original tree edges. Finally, once we have computed the best swap MATH for each tree edge MATH, we simply compute MATH and MATH and compare the two weights to determine whether this swap leads to a violated constraint. |
cs/9907001 | The problem is solved by the algorithm above. In each iteration the expected size of the set added to MATH is MATH, so the total size of MATH is MATH. In each iteration we add one more member of the optimal base to MATH, so the algorithm terminates with the correct solution. The steps in which we find the optimal parameter setting for MATH and MATH can be performed by applying REF ; since MATH has MATH edges, the time for these steps is MATH. The step in which we find the edges that take part in a violated constraint can be performed in linear expected time by REF . |
cs/9907001 | The first property follows immediately from REF , since each set of the partition contributes MATH paths, there are MATH sets at the bottom level of the partition, and the number of sets decreases at least geometrically at each level. Similarly, the second property follows, since an edge can belong to MATH paths per level and there are MATH levels. Finally, to prove the third property, let MATH be an arbitrary path in MATH. We describe a procedure for decomposing MATH into few paths MATH. More generally, suppose we have a path MATH contained in a set MATH at some level of a multi-level decomposition (note that the whole tree is the set at the highest level of the partition). Then MATH can be decomposed into at most MATH sets at the next level of the partition; MATH has endpoints in at most two of these sets, and may pass completely through some other sets. Therefore, MATH can be decomposed into the union of two smaller paths in the sets containing its endpoints, together with a single path MATH connecting those two sets. By repeating this decomposition recursively at each level of the tree, we obtain a decomposition into at most two paths per level, or MATH paths overall. |
cs/9907001 | We use the algorithm described above, setting MATH. Therefore, the total size of the sets MATH and MATH (and the total time to find these sets and to perform each iteration) is MATH. Since MATH is constant, there are MATH iterations, and the total time is MATH. |
cs/9907001 | We define a polyhedron MATH by linear inequalities MATH where MATH denotes the weight of a subgraph for the given point MATH, MATH can be any suitable subgraph, and MATH is an additional parameter. To avoid problems with unboundedness, we can also introduce additional normalizing inequalities MATH. Clearly, there exists a point MATH with MATH in MATH if and only if MATH gives a feasible solution to the inverse parametric optimization problem. Although there can be exponentially many inequalities, we can easily define an oracle that either terminates the entire algorithm successfully or acts as a strong separation oracle: to test a point MATH, simply compute the optimal subgraph MATH for the weights defined by MATH. If MATH, we have solved the problem. If MATH, the point is feasible. Otherwise, return the halfspace MATH. Therefore, we can apply the ellipsoid method to find the point maximizing MATH on MATH. If the method returns a point with MATH or terminates early with MATH, we must have solved the problem, otherwise the problem must be infeasible. |
cs/9907011 | This lemma follows from NAME theory in algebra CITE. Let MATH be the field of rational numbers. For each MATH, let MATH. Let MATH be the field generated by MATH over MATH. Also, let MATH be the field generated by MATH over MATH. By induction, MATH, the dimension of MATH over MATH is MATH, and the dimension of MATH over MATH is MATH. Thus, MATH is not a root of any nonzero single variate polynomial over MATH that has a degree less than MATH. Since MATH, by induction, MATH. The lemma is proved at MATH. |
cs/9907011 | For each combination of the bits MATH, MATH is called a conjugate. By the Prime Number Theorem CITE, MATH and thus MATH. Then, since MATH has at most MATH monomials, each conjugate's absolute value is at most MATH. Let MATH. Let MATH be the number of the conjugates that are less than MATH. Let MATH be the number of the other conjugates. Let MATH be the product of all the conjugates. By REF , MATH, and by algebra CITE, MATH is an integer. Thus, MATH and MATH. Hence, MATH; that is, MATH with the desired probability. We next show that if MATH, then MATH. Since MATH, MATH. So approximating MATH reduces each monomial term's absolute value in MATH by at most MATH. Thus, MATH. |
cs/9907011 | This theorem follows from REF immediately. |
cs/9907011 | The proof is straightforward based on the following key facts. There are at most MATH primes MATH, which can be efficiently found via the Prime Number Theorem. Each MATH has at most MATH bits and can be efficiently computed by, say, NAME 's method. |
cs/9907011 | For each MATH as described in REF, let MATH. Then, MATH, where MATH ranges over all the perfect matchings of MATH. It suffices to prove that for distinct perfect matchings MATH and MATH, the monomials MATH and MATH differ by at least one MATH. Let MATH be the subgraph of MATH induced by MATH. MATH is a set of vertex-disjoint cycles. Since MATH, MATH contains at least one cycle MATH. Let MATH be the acyclic digraph obtained from MATH by replacing each edge MATH with the arc MATH. MATH contains two outgoing arcs MATH and MATH of some vertex MATH. So there is an indeterminate MATH used in arc labels for vertex MATH, whose degree is REF in one of MATH and MATH but is REF in the other. Hence, the degree of MATH is REF in one of MATH and MATH but is REF in the other, which makes MATH and MATH distinct as desired. |
cs/9907011 | We separate the total complexity of REF into that for computing MATH and that for all the other computation. For the latter, the running time is dominated by that of REF ; the bit length by that of the entries in MATH at REF ; and the processor count by that of setting up MATH. |
cs/9907011 | We modify REF of the above implementation as follows. CASE: Compute MATH as above. CASE: For each MATH-th entry of MATH, we multiply it with MATH in MATH parallel arithmetic steps using MATH processors. Let MATH be the resulting matrix; note that MATH and each entry of MATH is an integer of at most MATH bits. CASE: Let MATH. Let MATH; note that MATH. We uniformly and independently choose MATH random positive integers MATH using MATH random bits in MATH steps on a single processor. For each chosen MATH, we first compute MATH in MATH parallel arithmetic steps using MATH processors; and then compute MATH instead of MATH. CASE: Output ``MATH has a perfect matching" if and only if some MATH is nonzero. By REF , if MATH, then some chosen MATH does not divide MATH with probability at least MATH. Thus, the overall error probability is at most MATH. We separate the total complexity of REF into that for computing MATH and that for all the other computation. As with REF , the running time of the latter remains dominated by that of REF ; the bit length by that of the entries in MATH at REF ; and the processor count by that of setting up MATH. |
cs/9907011 | The running time of REF is dominated by those of REF . The error probability follows from REF . |
cs/9907015 | By symmetry. we only prove the first statement. MATH, where MATH. Assume to the contrary that MATH and MATH. Then MATH. We swap MATH with MATH. Let MATH. Now MATH becomes the parent of MATH and MATH. This rearrangement of nodes does not affect the value of node MATH, and the costs of MATH, MATH, and MATH remain unchanged. Let MATH be the new subtree with root MATH. Let MATH be the entire new tree resulted from the swapping. Since MATH and MATH have the opposite signs, MATH. Hence, MATH. Thus, MATH, contradicting the optimality of MATH. This completes the proof. |
cs/9907015 | Straightforward. |
cs/9907015 | Since MATH, MATH and thus MATH. Next, MATH. This complete the proof of the first statement. The second statement follows from the fact that MATH if and only if MATH. |
cs/9907015 | CASE: Note that MATH. Thus, MATH. Since MATH, MATH. Hence, MATH, that is, MATH. CASE: Since MATH, we have MATH. Also, since MATH and MATH, we have MATH. Next, for each MATH, we have MATH. Then by the triangular inequality and REF , MATH. CASE: By REF , MATH. Thus MATH. Then, since MATH, MATH. |
cs/9907015 | Assume to the contrary that a type-REF node MATH is added to a node MATH in the form of MATH with MATH. Then MATH for some MATH. Let MATH be the parent of MATH and MATH. Since MATH, MATH cannot be the root of MATH. Let MATH be the sibling of MATH. Let MATH be the parent of MATH and MATH. Let MATH be the root of MATH. Let MATH be the path from MATH to MATH in MATH. Let MATH be the number of nodes on MATH. Since MATH has MATH internal nodes, MATH. We rearrange MATH to obtain a new tree MATH as follows. First, we replace MATH with MATH; that is, MATH now has subtrees MATH and MATH. Let MATH be the remaining tree; that is, MATH is MATH after removing MATH. Next, we create MATH such that its root has subtrees MATH and MATH. This tree rearrangement eliminates the cost MATH from MATH but may result in a new cost in the form of MATH on each node of MATH. The total of these extra costs, denoted by MATH, is at most MATH. Then, MATH, contradicting the optimality of MATH. This completes the proof. |
cs/9907015 | CASE: Assume that the statement is untrue. Then, since all negative leaves have values MATH, some negative internal node MATH has an absolute value greater than MATH and two negative children MATH and MATH. Since MATH, some MATH has a positive sibling MATH. We pick such a MATH at the lowest possible level of MATH. Let MATH be the parent of MATH and MATH. By REF , MATH. Then MATH. Since all positive leaves have values less than MATH, MATH is an internal node with two children MATH and MATH. Since MATH, MATH, and MATH, by REF , MATH must have a positive child and a negative child. Without loss of generality, let MATH be positive and MATH be negative. Then MATH. Since MATH is at the lowest possible level, MATH, for otherwise we could find a MATH at a lower level under MATH. We swap MATH with MATH. Let MATH be the new subtree rooted MATH. Let MATH. Since MATH and MATH, we have MATH. Since MATH, we have MATH. Let MATH. Then, MATH, which contradicts the optimality of MATH because the costs of the internal nodes not mentioned above remain unchanged. CASE: Assume that this statement is false. Then, since all positive leaves have values less than MATH, some internal node MATH has a value at least MATH as well as two positive children. Since MATH, some such MATH has a negative sibling MATH. By REF , MATH. Hence MATH, contradicting REF . |
cs/9907015 | CASE: By REF , MATH. Thus, MATH with MATH. To rule out MATH by contradiction, assume MATH with MATH. Since by REF all positive leaves have values less than MATH, MATH is an internal node. By REF , MATH has two children MATH and MATH. Since MATH, MATH is not the root and by REF , MATH has a negative sibling MATH. By REF , MATH. Let MATH be the parent of MATH and MATH. Then MATH. We swap MATH with MATH. Let MATH be the parent of MATH and MATH. Now MATH is the parent of MATH and MATH. Let MATH be the new subtree rooted at MATH after the swapping. Since MATH remains the same, MATH. If MATH, then MATH; otherwise, MATH and thus MATH. In either case, MATH, contradicting the optimality of MATH. CASE: The proof is similar to that of REF . By REF , MATH with MATH. To rule out MATH by contradiction, assume MATH with MATH. By REF , MATH has a positive sibling MATH and two children MATH and MATH. Let MATH be the parent of MATH and MATH. Then MATH. We swap MATH with MATH. Let MATH be the parent of MATH and MATH. Now MATH is the parent of MATH and MATH. Let MATH be the new subtree rooted at MATH after the swapping. Since MATH is the same, MATH. If MATH, then MATH; otherwise, MATH and thus MATH. So MATH, contradicting the optimality of MATH. |
cs/9907015 | Assume to the contrary that MATH is not a leaf. By REF , MATH has two children MATH and MATH. By REF , MATH has two children MATH and MATH, contradicting REF . |
cs/9907015 | To prove the lemma by contradiction, by REF , we assume MATH. Let MATH and MATH be the two children of MATH. Let MATH be the sibling of MATH; by REF , MATH or MATH. Let MATH be the parent of MATH and MATH. Then MATH. By REF , there are two cases based on the values of MATH and MATH with the symmetric cases omitted. CASE: MATH and MATH. Swap MATH with MATH. Let MATH be the new parent of MATH and MATH. Then MATH is the parent of MATH and MATH. Let MATH be the new subtree rooted at MATH. Then MATH. Whether MATH or MATH, we have MATH and thus MATH, which contradicts the optimality of MATH. CASE: MATH and MATH. There are two subcases based on MATH. CASE: MATH. We swap MATH with MATH. Let MATH be the new parent of MATH and MATH. Then MATH. CASE: MATH. We swap MATH with MATH. Let MATH be the new parent of MATH and MATH. By REF , both MATH and MATH are leaves, and thus by REF , MATH. Therefore, MATH. Therefore, in either subcase of REF the swapping results in an addition tree over MATH with smaller cost than MATH, reaching a contradiction. |
cs/9907015 | By REF , and REF, each MATH can only be added to some MATH or to some MATH. In turn, MATH can only be the sum of MATH and some MATH. In turn, MATH is the sum of some MATH and MATH. Hence, in MATH, MATH leaves in MATH are added in pairs. The sum of each pair is then added to a leaf node MATH. This sum is then added to a leaf node in MATH. This sum is a type-REF node with value MATH, which can only be added to another type-REF node. Let MATH be the three leaves in MATH associated with each MATH and added together as MATH in MATH. The cost of such a subtree is MATH. There are MATH such subtrees MATH. Their total cost is MATH. Hence, MATH. If MATH is not a positive instance of REF, then for any MATH, there is some subtree MATH with MATH. Then, the value of the root MATH of MATH is MATH. Since MATH is a type-REF node, it can only be added to a type-REF node. No matter how the MATH root values MATH and the MATH leaves MATH are added, some node resulting from adding these MATH numbers is nonzero. Hence, MATH. If MATH is a positive instance of REF, let MATH with MATH form a REF-set partition of MATH; that is, MATH is the union of these MATH REF-sets and for each MATH, MATH. Then each REF-set can be added to one MATH and one MATH as MATH, resulting in a node of value zero and contributing no extra cost. Hence, MATH. This completes the proof. |
cs/9907015 | By REF , it suffices to construct a reduction MATH from REF to AT. Let MATH, which is polynomial-time computable. By REF , MATH is a positive instance of AT if and only if MATH is a positive instance of REF. Then, by REF , MATH is a desired reduction. |
cs/9907015 | Let MATH be a leaf in MATH. There are two cases. CASE: MATH is some critical leaf MATH or MATH. Let MATH be the parent of MATH and MATH in MATH for MATH. Then MATH. CASE: MATH is some noncritical leaf MATH. Let MATH be the sibling of MATH in MATH. Let MATH be the parent of MATH and MATH. There are three subcases. CASE: MATH is also a leaf. Since MATH is noncritical, MATH has the same sign as MATH and is also a noncritical leaf. Thus, MATH. CASE: MATH is an internal node with the same sign as MATH. Then MATH. CASE: MATH is an internal node with the opposite sign to MATH. If MATH, then MATH; if MATH, then MATH. So, we always have MATH. Observe that MATH . Simplifying the sum of these two inequalities based on the case analysis, we have MATH as desired. |
cs/9907015 | By case analysis, if MATH and MATH, then MATH. Thus, if MATH, then pairing MATH with MATH returns the minimum MATH. For the case MATH, let MATH be an infinitesimally small positive number. Let MATH be MATH with additional MATH copies of MATH. Then, MATH is the minimum over all possible critical matchings of MATH. Thus, MATH is the minimum over all possible critical matching of MATH. The case MATH is symmetric to the case MATH. Since MATH is sorted, the running time of REF is MATH. |
cs/9907015 | REF both take MATH time. By REF also takes MATH time and thus REF takes MATH time. As for the error analysis, let MATH be the addition tree constructed at REF . Then MATH. Let MATH be the number of levels of MATH. Since MATH is a balanced tree, MATH and thus MATH. By assumption, MATH has at least two numbers with the opposite signs. So there are at most MATH numbers to be added pairwise at REF . Thus, MATH. Next, by REF , since MATH is a minimum critical matching of MATH, we have MATH. In summary, MATH. |
cs/9907015 | If MATH is unsorted (respectively, sorted), then a NAME tree over MATH can be constructed in MATH CITE (respectively, MATH CITE) time. |
cs/9907015 | For an addition tree MATH and a node MATH in MATH, the depth of MATH in MATH, denoted by MATH, is the number of edges on the path from the root of MATH to MATH. Since MATH is a NAME tree over MATH and every MATH is a NAME tree over MATH, there exists some MATH such that for each MATH, its depth in MATH is at least its depth in MATH. Furthermore, in MATH, the depth of each MATH is at least that of MATH. Therefore, MATH . Also note that for MATH, MATH. Hence, MATH . In summary, MATH. Since REF takes MATH time and the others take MATH time, the total running time of REF is as stated. |
cs/9907015 | Follows from REF . |
cs/9907018 | Pick any spanning tree MATH of the graph of a MATH-form. Let MATH be some leaf of MATH. Let MATH denote the unique vertex incident to MATH in MATH, and glue it to MATH. Now perform a depth-first traversal of MATH rooted at MATH, and label newly visited vertices as MATH, each gluing to its parent. The result is a gluing sequence for the polyform, such that only one polygon REF glues to MATH. |
cs/9907018 | Consider any MATH-form MATH of type MATH for MATH. We will prove that MATH rotates into MATH, and hence MATH rotates into all MATH-forms of type MATH. Consider a gluing sequence for MATH from REF . The union of the first MATH copies of MATH is some MATH-form MATH of type MATH. By REF , MATH can be rotated into MATH. Now attach the remaining copies of MATH one by one in the order specified by the gluing sequence. At each step, if we have a rotation of MATH into the first MATH copies of MATH, then attaching the next gives us a rotation of MATH. By induction, we reach a rotation of MATH into the desired MATH-form MATH. Now MATH contains the first copy of MATH in the gluing sequence, call it MATH. The gluing sequence from REF glues only one copy of MATH to MATH. Provided MATH, MATH has already glued a copy of MATH to MATH, and hence no other copies of MATH are glued to MATH. The above construction makes precisely one attachment corresponding to each gluing other than the first MATH gluings; thus, no copies of MATH are ever attached to MATH. |
cs/9907018 | Suppose there were a hinging MATH of five squares that could rotate into every pentomino. We first consider the NAME, which is a MATH rectangle. Because MATH rotates into the NAME, the squares must be hinged one after the other in a chain, ordered by their position in the rectangle. An example without the hinges is given in REF . We next consider the NAME, in which the five squares form a (Greek) cross. Four of the five squares are arms of the NAME. Each of these four has two adjacent vertices that do not touch any other square, and thus cannot have hinges on them. It follows that none of these four squares can have hinges at diagonally opposite vertices. Thus, at most one of the five squares can have hinges at diagonally opposite vertices. REF gives an example of such a hinging of the NAME (there are several). In this example, only square REF, the middle square, has hinges at diagonally opposite vertices. We next consider the NAME, in which three squares are stacked one on top of the other, and the other two squares are to the right and to the left of the top square in the stack. One end of the chain (call it square REF) must be on the bottom of the stack, because it is adjacent to only one other square (which must necessarily be square REF). The top middle square cannot be square REF, for otherwise it would be impossible to connect all the squares in a chain. Thus, in particular, the other end of the chain (square REF) must be either the left or the right square at the top. This argument limits us to the configuration in REF and its mirror image. Suppose that we transform MATH from the NAME to the NAME while leaving square REF in the same orientation. If the NAME lies horizontally, then square REF must rotate MATH clockwise, causing square REF to have hinges at diagonally opposite vertices. Square REF must rotate MATH clockwise, square REF rotates MATH clockwise, and square REF rotates MATH clockwise, as shown in REF . But this requires that two squares, squares REF, have hinges at diagonally opposite vertices. Because this possibility has been ruled out, we cannot transform MATH from the NAME to the NAME while leaving square REF in the same orientation and having the NAME lie horizontally. Suppose that we transform MATH from the NAME to the NAME while leaving square REF in the same orientation, with the NAME standing vertically. Then square REF must rotate MATH counterclockwise. This would leave both of its hinges adjacent to square REF, as shown in REF . Clearly squares REF cannot be connected in this way. This exhausts all cases for transforming MATH from the NAME to the NAME. Thus the desired hinging MATH does not exist. |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . We can attach a square MATH to the hinging of a polyomino MATH as follows; refer to REF . Let MATH be the triangle in this hinging that shares an edge with MATH. One of its base vertices, say MATH, is also incident to MATH, and it must be hinged to some other triangle MATH. We split MATH into two right isosceles triangles MATH and MATH so that both have a base vertex at MATH. Now we replace MATH's hinge at MATH with a hinge to MATH, and add a hinge from MATH to MATH at MATH. Finally, MATH and MATH are hinged together at their other base vertex. The result is a hinging of MATH triangles rotated into MATH. We can optionally swap MATH and MATH in order to avoid crossings between the hinges. |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . One square in this domino, MATH, has hinges at diagonally opposite vertices just as before, but the other square, MATH, has only one hinge. By symmetry, we can arrange in REF for MATH to be chosen as slippery, and hence all attachments act as in REF . |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . We can attach a regular MATH-gon MATH to the hinging of a MATH-regular MATH as follows; refer to REF . Let MATH be the triangle in this hinging that shares an edge with MATH. Both of its base vertices are also incident to MATH. Let MATH be either one of MATH's base vertices, and suppose that it is hinged to triangle MATH. We split MATH into MATH isosceles triangles MATH, , MATH so that both MATH and MATH have a base vertex at MATH. Now we replace MATH's hinge at MATH with a hinge to MATH, and add a hinge from MATH to MATH at MATH. Finally, MATH and MATH are hinged together at their common base vertex, for all MATH. The result is a hinging of MATH triangles rotated into MATH. We can optionally renumber MATH, , MATH as MATH, , MATH in order to avoid crossings between the hinges. |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . One regular MATH-gon MATH in this MATH-regular has hinges at all its vertices just as before, but the other regular MATH-gon MATH has only two hinges. By symmetry, we can arrange in REF for MATH to be chosen as slippery, and hence all attachments act as in REF . |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . We can attach a regular MATH-gon MATH to the hinging of a MATH-regular MATH similar to the proof of REF ; refer to REF . Let MATH be the piece in this hinging that shares an edge with MATH. Both of the base vertices of one of its constituent triangles are incident to MATH. Let MATH be such a base vertex of MATH that is not joined to a base vertex of another constituent triangle of MATH (which we call lone base vertices), and suppose that it is hinged to piece MATH. We split MATH into MATH pieces MATH, , MATH so that both MATH and MATH have a lone base vertex at MATH. Now we replace MATH's hinge at MATH with a hinge to MATH, and add a hinge from MATH to MATH at MATH. We hinge MATH and MATH together at their common lone base vertex, for all MATH. Finally, if MATH is odd, we choose one of the pieces MATH, , MATH to be a single triangle instead of a double triangle, appropriately so that the single triangles appear periodically in the resulting cycle of pieces, with a period of MATH. The result is the desired hinging of MATH pieces rotated into MATH. We can optionally renumber MATH, , MATH as MATH, , MATH in order to avoid crossings between the hinges. |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . One regular MATH-gon MATH in this MATH-regular has hinges at roughly half of its vertices just as before, but the other regular MATH-gon MATH has only two hinges. By symmetry, we can arrange in REF for MATH to be chosen as slippery, and hence all attachments act as in REF . |
cs/9907018 | Apply REF with MATH. The case MATH is shown in REF . Note that in contrast to all previous dissections, there are no hinges at the vertices of the monabolo. There are, however, hinges at the midpoints of all the edges. We can attach a half-square MATH to the hinging of a polyabolo MATH at these midpoints as shown in REF . There are three cases according to relative orientations of MATH and the incident half-square. But in all cases we obtain the same hinged dissection REF with triangles pointing outward from the cycle, and alternating in orientation along the cycle. |
cs/9907018 | Both can be viewed as a MATH-abolo, by splitting a square in the MATH-omino into four pieces (REF , left) and splitting a square in the MATH-omino into two pieces (REF , right). |
cs/9907018 | There is one detail omitted in the discussion above, so let us go through a formal proof. Apply REF with MATH. (While this lemma was designed for general MATH-forms, it applies equally well to restricted MATH-forms.) The case MATH just takes the decomposition of MATH described above, and hinges it at the midpoints of edges of MATH. Adding a copy of MATH to a restricted polyform of type MATH requires special care. Let MATH denote the copy of MATH to which we want to attach MATH, and let MATH denote the edge of MATH to which MATH will attach. We need to place MATH against MATH such that the rigid motion mapping MATH to MATH and MATH to MATH also maps the pieces of MATH to the pieces of MATH (where ``pieces" refer to the subdivision described above). Certainly there is a rigid motion MATH mapping the pieces of MATH to the pieces of MATH, so we should attach MATH's edge MATH to MATH's edge MATH. The only possible wrinkle is that if MATH has symmetry, in the sense that there is a rigid motion MATH from MATH to itself, then we might instead attempt to attach MATH to MATH. Fortunately, we get to choose the orientation of the subdivision of MATH for the copy we are attaching. We can explore all symmetric versions of the subdivision of MATH and choose the one that places MATH against MATH. |
cs/9907018 | The graph of the hinging structure (in which vertices represent pieces and edges represent hinges) can be any planar graph. First take a spanning tree of that graph, and remove all other hinges. This transformation removes the cycles from the hinging structure while keeping the pieces connected. Now for each (original) piece with only one hinge, we cut it along a polygonal line from that hinge to any other point on the boundary (for example, another vertex), and add a hinge between the two pieces at that point. For each original piece that has at least two hinges, we cut along a tree of line segments that is interior to the piece and has leaves at the hinges, and we replace each original hinge with two ``parallel" hinges. The result is a cyclicly hinged dissection MATH, which can be rotated as MATH can because it simply subdivides and adds more degrees of motion. See REF for an example. If MATH has MATH pieces, then when we reduce the hinging structure to a spanning tree it has MATH hinges. Our construction doubles every original hinge, and adds an additional hinge for every leaf (a piece adjacent to only one original hinge), for a total of at most MATH. There is one piece per hinge in a cyclicly hinged dissection, so the number of pieces in MATH is at most MATH. |
cs/9907018 | Start with the cyclicly hinged dissection from REF . Now we want to add cuts so that there is a hinge at the midpoint of each edge in MATH for all MATH. This can be done as follows. For each MATH, and for each edge MATH of MATH that does not already have a hinge at its midpoint, consider the piece MATH whose boundary contains MATH's midpoint when MATH is rotated into MATH. Refer to REF . Let MATH and MATH denote the paths of MATH's boundary connecting the two hinges incident to MATH, where the paths include their endpoints. Order MATH and MATH so that the midpoint of MATH is on MATH. Pick an arbitrary point MATH on MATH, add a polygonal cut from MATH to the midpoint of MATH, and add a hinge connecting the two pieces at the midpoint of MATH. REF shows the special case in which MATH is chosen to be an endpoint of MATH, that is, a hinge. In this case, the hinge at MATH is assigned to the piece of MATH that is not incident to the other endpoint of MATH, and the hinged dissection remains cyclic. Performing this operation for all choices of MATH and MATH, we obtain a cyclicly hinged dissection MATH that can be repeated MATH times to obtain MATH. See REF for a complete example. Each repetition of MATH can be thought of, in particular, as a subdivision of MATH with hinges at the midpoints. Thus, as we proved in REF , MATH can be rotated into any restricted polyform of type MATH, and this holds for any MATH. We started with the MATH-piece cyclicly hinged dissection from REF , and added at most one piece per midpoint of an edge of a polygon MATH. Therefore, we added MATH pieces, for a total of MATH. |
cs/9907018 | Apply REF with MATH, for both MATH and MATH symmetrically; we will focus on MATH. The case MATH is solved by the given hinged dissection MATH. Now consider attaching a regular MATH-gon MATH to a rotation of MATH into some MATH-regular MATH. Let MATH be the polygon to which we are attaching MATH. Let MATH be a hinge at the midpoint or an endpoint of the edge of MATH to which we are attaching MATH, which is guaranteed to exist because MATH is extendible. Split MATH into two chains, MATH and MATH at MATH. Point MATH is either the midpoint or an endpoint. If it is the midpoint, make cuts in MATH such that the cuts in MATH exhibit MATH-rotational symmetry about MATH. Split MATH at point MATH into MATH and MATH, and splice them together at the original endpoints to give MATH. Then splice MATH into MATH and MATH, and the result is MATH rotated into the desired polyregular MATH. If MATH is an endpoint of the common side of MATH and MATH, make cuts in MATH so that MATH are identical with respect to MATH-rotational symmetry about MATH, where MATH is the interior angle of MATH. We can then form MATH by cutting and splicing as in the previous case. Again MATH rotates into the desired polyregular MATH. |
cs/9907018 | This follows directly from REF and the above construction. |
cs/9907018 | Again this follows directly from REF and the above construction. |
cs/9907024 | Let MATH be the nearest neighbor graph of MATH: that is, the vertices of MATH are the points of MATH, and MATH define an edge of MATH if and only if MATH is the nearest neighbor of MATH (denoted by MATH) or MATH is the nearest neighbor of MATH in MATH. MATH is well known to be a subgraph of MATH, the NAME triangulation of MATH, and to have maximum degree REF. We denote by MATH the degree of MATH in MATH, and by MATH the expectation when MATH is chosen uniformly in MATH. Then we have MATH notice that MATH is a random variable; result holds since MATH and MATH are random subsets of MATH and that the average degree of a vertex in a triangulation is less than REF. But even if MATH is a random point in MATH, the vertex MATH, the nearest neighbor of MATH in MATH, is not uniformly random. MATH . |
cs/9907024 | Let MATH and let MATH be the points of MATH lying in a section of angle MATH having apex MATH sorted by increasing distance to MATH. Clearly, a disk of center MATH passing through MATH REF cannot contain MATH and thus, if MATH, a necessary condition for MATH to be in the disk having as diameter the segment defined by MATH and the nearest neighbor of MATH in MATH is that no point of MATH is in the sample MATH which has probability MATH. Using six sections around MATH to cover the whole plane, and summing over the choice of MATH we get the claimed result. Notice that the disk of center MATH and passing through the nearest neighbor of MATH contain the disk of diameter the line segment defined by these two points, and thus the bound apply also to that circle. |
cs/9907024 | Let MATH be an edge of MATH intersecting segment MATH. If MATH does not exist in MATH, it means that MATH is an internal edge of the region retriangulated when MATH is inserted in MATH. Since MATH is a random point in MATH, the expected number of such edges is REF since it equals the average degree of MATH in MATH minus REF. If MATH exists in MATH, one end-point MATH of MATH must belong to the disk of diameter MATH, denoted MATH, (otherwise any disk through the end-points of MATH must contain MATH or MATH and MATH cannot belong to MATH). The expected number of edges of MATH intersecting MATH is bounded by the sum of the degrees of the vertices in MATH . Notice that REF was established for a fixed MATH and a random MATH which allows to use it inside the sum over MATH. Thus we get a total expected cost for the march bounded by MATH. |
cs/9907024 | All the triangles MATH examined in phase REF have a vertex in the disk of center MATH passing through MATH. Thus we can argue similarly as in REF , denoting MATH the disk of center MATH through MATH: MATH . |
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