paper stringlengths 9 16 | proof stringlengths 0 131k |
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cs/0010018 | We answer a query simply by repeatedly following the pointer MATH for the subinterval MATH that contains the query point. For each block found via this chain of pointers, we look up the value MATH and compare the priorities of the intervals found in this way. Each successive block in the chain corresponds to an interva... |
cs/0010018 | To insert or delete an interval, we create a new copy of each block containing one of the endpoint intervals. By the same argument used to bound query time, there are at most MATH such blocks. For each copied block, we update the priority queues corresponding to subintervals containing the updated interval, copy pointe... |
cs/0010018 | Form a set of intervals MATH with priority MATH for MATH. The maximum priority interval containing MATH has as its left endpoint the predecessor of MATH. Thus, we can use a static version of the data structure described in REF (with MATH) to solve this problem. |
cs/0010018 | We consider a left-right sweep of the rectangles by a vertical line; for each position of the sweep line we maintain a dynamic set of intervals formed by the intersections of the rectangles with the sweep line. This intersection changes only when the sweep line crosses the left or right boundary of a rectangle; at the ... |
cs/0010018 | We consider the minimum spanning tree verification formulation of the problem, and consider the non-tree edges in sorted order by weight. Our algorithm finds the replacements for each path edge in a certain order; when a path edge's replacement is found we reduce the size of the graph by contracting that edge. This con... |
cs/0010018 | We first partition the space MATH into horizontal stripes, according to the maximum-priority horizontal input stripe covering each point in the space; essentially this is just the lower envelope computation of REF . Let MATH denote the minimum priority occurring in this partition. Similarly, we partition the space into... |
cs/0010018 | If the line is horizontal (vertical), the number of cells cut by the line at most doubles at every even (odd) level of the MATH-tree construction, and remains unchanged at every odd (even) level. The result follows from the MATH bound on the number of levels in the tree. |
cs/0010018 | The bound on the number of crossed cells follows immediately from REF . The parent of a maximal covered cell must be crossed, and each crossed cell can have at most one maximal covered child, so the number of maximal covered cells is also MATH. |
cs/0010018 | We build a MATH-tree of the rectangle vertices this can be done in time MATH and perform a depth first traversal of the tree. As we traverse the tree, we maintain at each cell of the traversal the following information: CASE: The maximum priority of a rectangle covering the cell, and one or (if they exist) two rectangl... |
cs/0010019 | Let MATH be a length function and let MATH be a MATH-ensemble. We define the binary relation: MATH . Clearly, this relation is polynomial-time recognizable, since MATH can be computed in polynomial time. Also, the relation is evasive (with respect to MATH) since for every MATH there is at most one MATH satisfying MATH,... |
cs/0010019 | The intuition is that since MATH is evasive, it is infeasible for the forger to find a message MATH so that MATH. Thus, a forgery of the modified scheme must be due to REF , which yields a breaking of the original scheme. Formally, let MATH be an adversary who mounts an adaptive chosen message attack on MATH, and whose... |
cs/0010019 | When we use an ensemble MATH to implement the random oracle in the scheme MATH, we obtain the following real scheme (which we denote MATH): CASE: Uniformly pick MATH, set MATH, MATH, and output MATH. CASE: Output MATH. CASE: Output MATH. Consider now what happens when we use the ensemble MATH to implement the the schem... |
cs/0010019 | Since MATH is evasive, then from REF it follows that MATH is secure in the Random Oracle Model. On the other hand, suppose that one tries to replace the random oracle in the scheme by an ensemble MATH (where MATH be the index in the enumeration). An adversary, given a seed MATH of a function in MATH can then set MATH a... |
cs/0010019 | Below we describe such a signature scheme. For this construction we use the following ingredients. CASE: MATH is a signature scheme, operating in the Random Oracle Model, that is existentially unforgeable under a chosen message attack. CASE: A fixed (and easily computable) parsing rule which interpret messages as tripl... |
cs/0010019 | In this proof we use the same notations as in the proof of REF. Let MATH be an encryption scheme that is semantically secure in the Random Oracle Model, and we modify it to get another scheme MATH. The key generation algorithm remains unchanged, and the encryption and decryption algorithms utilize a random oracle MATH,... |
cs/0010019 | The proof of REF is a straightforward generalization of the proof of REF. Actually, we need to consider two cases: the case MATH and the case MATH. In the first case, we proceed as in the proof of REF (except that we define MATH). In the second case, for every ensemble MATH, we define the relation MATH . We show that M... |
cs/0010019 | We assume, for simplicity that MATH (and so MATH and MATH). Given MATH as stated, we again adapt the proof of REF. This time, using MATH, we define the relation MATH . Notice that in this definition we have MATH, and also MATH, so this relation is indeed MATH-restricted. Again, it is easy to see that MATH is polynomial... |
cs/0010019 | For simplicity, we consider first the case MATH. Let MATH be a MATH-ensemble. Adapting the proof of REF, we define the relation MATH (Notice that since MATH, the MATH's are indeed in the range of the function MATH.) Clearly, this relation is polynomial-time recognizable. To see that this relation is evasive, notice tha... |
cs/0010019 | Let MATH. For every seed MATH, we define MATH so that MATH equals MATH if MATH is the smallest integer such that MATH. In case MATH holds for all MATH's, we define MATH arbitrarily. Let MATH, and MATH (S stands for ``Small image"). Since MATH is evasive, it is infeasible to find a MATH not in MATH. Thus, for every prob... |
cs/0010021 | Follows immediately from REF . |
cs/0010021 | Follows immediately from REF . |
cs/0010021 | Follows immediately from REF . |
cs/0010021 | Follows immediately from REF . |
cs/0010021 | We use REF to convert the price history MATH and the strategy set MATH into a system of linear constraints MATH and MATH, with the next day's price change MATH determined by MATH for some MATH. Since the values MATH are computable in time polynomial in MATH, this conversion takes time polynomial in MATH. Then, MATH. Si... |
cs/0010021 | Let MATH represent the output of the MATH-th NOR gate, where MATH. Without loss of generality we assume that gate MATH is the output gate. The variables MATH and MATH are dummies to allow for a zero right-hand-side in MATH; our first two constraints are MATH and MATH. Suppose gate MATH has inputs MATH and MATH. The NOR... |
cs/0010021 | For each MATH, let MATH be the constraint MATH. To turn these inequalities into equations, we add slack variables to soak up any excess over MATH, with some additional care taken to ensure that there is a unique assignment to the slack variables for each setting of the variables MATH. We will use the following MATH var... |
cs/0010021 | First of all, MATH because a MATH machine is a CPP machine that happens not to have any abstaining paths. For the inverse direction, represent each abstaining path of a CPP machine by a pair consisting of one accepting and one rejecting path, and each accepting or rejecting path by two accepting or rejecting paths. The... |
cs/0010021 | To show MATH, replace each rejecting path of a MATH machine with an abstaining path in a MATH machine. For the inverse direction, replace each abstaining path of the MATH machine with a rejecting path in the MATH machine. |
cs/0010021 | Immediate from REF and the definition of BCPP and CPP. |
cs/0010021 | Let MATH be a deterministic implementation a MATH machine, where MATH and each MATH supplies a witness for the MATH-th oracle query. We will show that the language MATH accepted by MATH is in MATH. To simplify the presentation, we assume that each oracle query is a Boolean formula with a fixed number MATH of variables,... |
cs/0010021 | Let membership in MATH be computed by MATH. Assume MATH (the case MATH is symmetric). Consider MATH independent executions of MATH with input MATH; call the random variables representing their outputs MATH. Because the executions are independent, for any MATH vector of values MATH. So conditioning on no abstentions, MA... |
cs/0010021 | First we show that bounded market prediction is a member of BCPP. Given a market, construct a noncommittal NAME machine MATH whose input is the price history and strategies, and whose random inputs supply the settings for the population variables MATH. Let MATH abstain if the price history is inconsistent with the inpu... |
cs/0010021 | Similar to the proof of REF . |
cs/0010028 | We prove the lemma by structural induction on the syntax of the underlying program. It uses the concrete semantic rules of REF , the definition of MATH in REF , and the specifications of the abstract operations given in REF. The proof is straightforward due to the close correspondence of the concrete and the abstract s... |
cs/0010028 | The result follows from the definition of TAB in REF , the definition of TCB in REF, and REF . |
cs/0010028 | Let MATH be the concrete semantics of the underlying program. Since sat is pre-consistent, there exists a concrete behavior MATH such that CASE: MATH, and CASE: sat safely approximates MATH. The first condition implies that MATH since TCB is monotonic and MATH. The second condition and REF imply that MATH safely approx... |
cs/0010028 | Assume that sat safely approximate MATH. Let MATH and MATH. It must be proven that-MATH . Assume that the left part of the implication holds. REF implies that MATH . Since sat is a post-fixpoint and Cc is monotonic, MATH and then MATH . |
cs/0010028 | Let us abbreviate MATH by MATH. It is sufficient to prove that, for any MATH and any MATH, MATH . Fix MATH, MATH, and MATH satisfying the left part of the implication. By REF , MATH . Since sat safely approximates every MATH, MATH for all MATH. Finally, since MATH is chained-closed, MATH . |
cs/0010028 | The proof is in three steps. First we construct a sequence MATH of lower-approximations of MATH which is not necessarily a chain; then we modify it to get a chain MATH; finally, we show that MATH. The proof uses the following property of program substitution sequences, whose proof is left to the reader. If MATH, MATH a... |
cs/0010028 | The result is an immediate consequence of REF . |
cs/0010028 | The fact that MATH, for all MATH, is a direct consequence of the definition of the operation MATH. Moreover, the hypotheses on the sequences ensure that MATH; thus the sequence MATH is stationary. |
cs/0010029 | By structural induction on MATH. |
cs/0010029 | By structural induction on MATH. |
cs/0010029 | By structural induction on MATH. |
cs/0010029 | By structural induction on MATH. |
cs/0010029 | For any type MATH, we have MATH, hence by REF , MATH, that is MATH has no solution. For the second proposition, we prove its contrapositive. Suppose MATH has a solution, say MATH. By definition of a maximum and REF , we have MATH. Hence by REF , MATH. By the rules in REF , MATH. Therefore MATH, since otherwise MATH wou... |
cs/0010029 | The proof of the first part is by structural induction. For the base case, suppose MATH. Then by Rule (Var), MATH and hence for some MATH, we have MATH. Thus again by (Var), MATH. Now consider the case MATH where the inductive hypothesis holds for MATH. By REF , there exists a type substitution MATH such that MATH, and... |
cs/0010029 | The proof of the first part is by structural induction. For the base case, suppose MATH. Then by Rule (Var), MATH. If MATH, there is nothing to show. If MATH, then by definition of an ordered substitution, MATH and hence MATH where MATH. Now consider the case MATH where the inductive hypothesis holds for MATH. By REF ,... |
cs/0010029 | We first show that MATH is a well-defined type substitution. Since MATH is linear, MATH and hence MATH is uniquely defined. Moreover, since MATH is idempotent, MATH cannot occur in MATH. Therefore MATH, and hence by the condition on MATH in the statement, MATH. For the inequality MATH and for each MATH such that MATH h... |
cs/0010029 | Consider a non-variable position MATH in MATH. There is exactly one inequality in MATH with MATH as left-hand side. Moreover, MATH is a flat type (declared range type of a function), thus linear, and (because of indexing the parameters in MATH by MATH) has no parameters in common with any other left-hand side of MATH. ... |
cs/0010029 | Termination is proved by remarking that the sum of the sizes of the terms in left-hand sides of inequalities strictly decreases after each application of a rule. By REF the initial system is left-linear and acyclic, and one can easily check that each rule preserves the left-linearity as well as the acyclicity of the sy... |
cs/0010029 | Suppose MATH is computed by the algorithm of REF , and that MATH is the sequence of systems of this computation, that is, MATH is equal to MATH viewed as a substitution. By REF , MATH. In particular, this means that no system MATH REF contains an inequality MATH where MATH and MATH is not a parameter. It is easy to see... |
cs/0010029 | Since MATH is a minimal matcher, we have MATH . It remains to be shown that there exists a type substitution MATH such that MATH as defined above is an ordered substitution. Let MATH be the solution of MATH corresponding to MATH, and MATH be the solution of MATH corresponding to MATH (see REF ). Note that since MATH is... |
cs/0010029 | By CITE, MATH is nicely moded. Let MATH and MATH be the variable typings used to type MATH and MATH, respectively (in the sense of REF ). Let MATH be the selected atom and MATH. By REF , MATH. Moreover, MATH for some type substitution MATH. Let MATH. Note that since MATH, MATH is a variable typing. By REF , we have MAT... |
cs/0010031 | Consider a recursive call MATH of NAME. Let MATH be the node that is selected to be processed in REF . All of MATH's predecessors in the original graph MATH have either been processed in a previous REF or deleted in some REF . Therefore, the current weight of MATH, MATH, as seen by the recursive call MATH, is just MATH... |
cs/0010031 | We will prove the result for NAME. The full result follows from REF . Clearly, the returned set of nodes MATH is an independent set. By REF , we need only show that MATH is a MATH-approximation with respect to MATH and MATH. We will prove this by induction on the recursion. The base case of the recursion is trivial, si... |
cs/0010031 | NAME computes MATH for each node MATH in time proportional to its indegree, and computes MATH for each node in time proportional to its outdegree, for a total time of MATH. In the case of NAME, a recursive call is made at most once for each node in the graph, and defining MATH and MATH in each call takes time proportio... |
cs/0010031 | See CITE. |
cs/0010031 | For each MATH in MATH, MATH is a clique. |
cs/0010031 | Let MATH be a node of MATH. Let MATH, MATH, MATH, and MATH be the set of all successors of MATH in MATH, MATH, and MATH, respectively. Let MATH be any independent subset of MATH. Then CASE: MATH, and CASE: MATH is an independent subset of MATH, implying MATH. |
cs/0010031 | Let MATH be some independent set of successors of MATH. Under the conditions of the lemma each MATH intersects MATH. But since the MATH do not themselves intersect, each must intersect MATH in a distinct element. Thus there are at most MATH of them. |
cs/0010031 | Let MATH be a tree decomposition of MATH with width MATH. We will use this tree decomposition to construct an ordering of the connected node subsets of MATH, with the property that if MATH then either MATH or MATH intersects some frontier set MATH with at most MATH elements. The full result then follows from REF . Choo... |
cs/0010031 | CITE gives a recursive MATH algorithm for computing tree decompositions of constant-treewidth graphs based on a linear time algorithm for finding approximate separators for small node subsets. Replacing this separator-finding subroutine with the linear time algorithm of CITE gives a MATH time algorithm for computing a ... |
cs/0010031 | Given an undirected MATH-node graph MATH, construct a MATH-node directed acyclic graph MATH by REF directing the edges of MATH in any consistent order, and REF adding a new source node MATH to MATH with edges from MATH to every node in MATH. Let MATH be an independent set in MATH. Then every node in MATH is a successor... |
cs/0010031 | Let MATH be any independent set in MATH. Choose the ordering so that all nodes in MATH precede all nodes not in MATH. Then for any MATH, MATH has no predecessors in the oriented graph and MATH. Let MATH be the independent set computed by the algorithm. If MATH is in MATH but not MATH, it must have a successor MATH in M... |
cs/0010031 | We will prove the result for NAME; by REF the same result holds for NAME. Let MATH order the nodes in order of decreasing weight. Let us show by induction on MATH that if the greedy algorithm chooses a node MATH, then MATH; but if the greedy algorithm does not choose MATH, then MATH. Suppose we are processing some node... |
cs/0010031 | The proof that both algorithms return the same approximation is similar to the proof of REF . The proof of the approximation ratio follows the same structure as the proof of REF . We prove the result for NAME. By REF , we need only show that the returned set of bids MATH is a MATH-approximation with respect to MATH and... |
cs/0010031 | Similar to the proof of REF . The additional MATH term comes from having to apply up to MATH budget constraints to each node; since MATH, this term also covers the cost of reading the MATH from the input and initializing the variables for each subset. |
cs/0010031 | Since each heavy bid consumes more than half a bidder's budget, each bidder can win at most one bid. This is just a simple unweighted budget constraint and can be solved as described in REF for a performance bound of MATH. |
cs/0010031 | This proof uses the same structure and notation as the proof of REF . An upper bound MATH on the MATH-revenue of any feasible set of bids is MATH. With regards to a MATH-maximal set of bids, if MATH cannot be added to the set because the budget constraint MATH will be exceeded, the existing bids in the set must have we... |
cs/0010031 | The sum of the optimal revenues for the heavy and light bids is at least equal to the optimum revenue among all bids. From REF , the better of the two solutions will be within a factor of MATH of the optimum for the general problem. For the running time, observe that decomposing the bids into heavy and light bids takes... |
cs/0010032 | Contained in REF. |
cs/0010032 | Contained in REF. |
cs/0010032 | See REF. |
cs/0010032 | See REF. |
cs/0010032 | First we show that a (minimal) model of one of MATH and MATH is also a model of the other: CASE: The conditional facts in MATH are logical consequences of MATH. By induction on MATH, it follows that MATH is implied by MATH. Therefore, every model of MATH is also a model of MATH. CASE: Next, we prove that every minimal ... |
cs/0010032 | If this were not the case, there were a minimal model MATH of MATH which does not satisfy MATH. Because of MATH and MATH, there must be a model MATH of MATH that is smaller than MATH, but not a model of MATH. Since MATH is smaller than MATH, it assignes the same truth values to the default negation literals. But then i... |
cs/0010032 | CASE: Let MATH be a static expansion of MATH. By the lemma above, MATH. But then it easily follows that MATH is also a static expansion of MATH: It has to satisfy MATH . Since MATH, the formula MATH is anyway contained in the preconditions of MATH, so the union with MATH changes nothing. Assume conversely that MATH is ... |
cs/0010032 | CASE: Consider the interpretation MATH that extends MATH by interpreting MATH as false and all remaining atoms as true. Suppose that it were not a model of MATH, that is, it would violate some disjunction in MATH. Since all atoms that MATH interprets as false do not appear in MATH, and all the remaining atoms except MA... |
cs/0010036 | The proof comes immediately from the proof of CITE . |
cs/0010036 | The result is obvious if MATH. If MATH, it is clear from the definition that a non dual configuration can not belong to a strongly connected component since it does not belong to a circuit. So we just have to prove that the set of dual configurations is a strongly connected component. Let MATH and MATH be two dual conf... |
cs/0010036 | From the shot vector definition, we obtain: MATH . Since the necessary and sufficient condition to apply the rule on position MATH of MATH is MATH, we are going to focus on the difference MATH. MATH which proves the lemma. |
cs/0010036 | Let MATH and MATH be two sequences of transitions from MATH to MATH : MATH . Let us recall that: MATH . Assume that MATH. We have MATH. We are going to show that there exists a path of positive length from MATH to MATH, which is a contradiction. For that, we are going to build step by REF sequence of transitions from M... |
cs/0010036 | In order to show that MATH we can consider two sequences of transitions, one from MATH to MATH and the other from MATH to MATH, and then make a similar proof as the one used in REF . On the other hand, let MATH and MATH be two non dual configurations of MATH such that MATH. Let MATH be a sequence of transitions from MA... |
cs/0010036 | Let us assume that MATH and MATH are incomparable (otherwise MATH and MATH are comparable and the result is obvious). We are first going to prove that MATH is reachable from MATH. For that, we just have to find a configuration MATH such that MATH and MATH. Let MATH be a sequence of transitions from MATH to MATH and let... |
cs/0010036 | Assume that MATH. Let MATH be such that MATH. We are first going to show step by step that there exists a path from MATH to MATH in which player MATH never plays. Let MATH be such that MATH is maximal among the MATH and such that MATH. Since MATH, such MATH exists and MATH. We have MATH, so MATH and MATH. Let MATH be t... |
cs/0010036 | Let us consider the dominance ordering on dual configurations, in which a configuration MATH is greater or equal to a configuration MATH if and only if MATH, MATH. The greatest element of this order is clearly MATH and the maximal length of a chain in this order is MATH. Let MATH and MATH be two dual configurations suc... |
cs/0010037 | For any letter MATH consider MATH and MATH. Since MATH is sub-normalised it follows that for each letter MATH, there is MATH such that MATH that is, for each MATH, its greatest lower bound constraint is less or equal than its least upper bound constraint. Now, let MATH be an interpretation such that CASE: MATH and MATH... |
cs/0010037 | Similarly to REF , for any letter MATH, consider MATH and MATH. Since MATH is sub-normalised, it follows that for each letter MATH, there is MATH such that MATH. Now, let MATH be an interpretation such that CASE: MATH and MATH; CASE: MATH, for all letters MATH . It is easily verified that MATH satisfies MATH. |
cs/0010037 | MATH . Assume MATH and suppose to the contrary that there is a MATH such that MATH. Therefore, there is a fuzzy interpretation MATH such that MATH and MATH. Let MATH be the following four-valued interpretation: MATH . We show on induction on the structure of a proposition MATH that MATH satisfies MATH iff MATH. CASE: I... |
cs/0010037 | Since MATH, MATH returns true. Therefore, there is a completed branch MATH. Suppose that for each completed branch MATH, MATH, there is a letter MATH occurring in MATH such that both MATH and MATH. As a consequence, collecting all the MATH expressions in branches MATH, informally MATH is equivalent to MATH and, thus, M... |
cs/0010037 | MATH . Assume MATH and MATH. If MATH then MATH, by REF . Otherwise, MATH implies MATH (as a NNF of MATH is normalised and by REF ). So, let us show that MATH. Suppose to the contrary that MATH. From REF , there is an interpretation MATH such that MATH, MATH and for no letter MATH, MATH. Consider the following fuzzy int... |
cs/0010037 | Assume MATH. NAME all meta-literals in MATH with MATH. Consider a deduction of MATH, which returns false, and let MATH be the deduction tree. As a consequence, all branches MATH in MATH are closed. Let us consider the following substitution, MATH, in each branch MATH. For each meta literal MATH occurring in MATH, MATH ... |
cs/0010037 | From hypothesis, from REF it follows immediately that either MATH or MATH holds. |
cs/0010037 | MATH . Assume MATH and MATH. Consider a DNF MATH of MATH. From MATH it follows that MATH and, thus, for each MATH, MATH, that is, MATH is unsatisfiable. NAME all the meta literals in a NNF of MATH with MATH. Let MATH be all the branches of a deduction of MATH. Consider a branch MATH. Obviously, MATH is closed. Therefor... |
cs/0010037 | MATH . Assume that for both MATH, MATH holds. If MATH then condition MATH is trivially satisfied. Otherwise, assume MATH. From REF , MATH follows. But then, we know that MATH. As a consequence, for MATH no interpretation satisfies MATH. Therefore, since by REF holds for MATH, it follows that for MATH, MATH has to be un... |
cs/0010037 | MATH . Assume that MATH. Suppose to the contrary that MATH such that MATH. Therefore, there is a fuzzy interpretation MATH such that MATH and MATH. But, from the hypothesis MATH follows. Absurd. MATH . Assume that for all MATH, MATH. Suppose to the contrary that MATH. Therefore, there is a fuzzy interpretation MATH suc... |
hep-th/0010227 | In such a case the Hamiltonian REF is invariant under the NAME reflection symmetry MATH and it is always possible to find in the NAME space MATH a complete basis of stationary states with definite U REF symmetry. If the energy level is not degenerate, the corresponding physical state MATH has to be either even or odd u... |
hep-th/0010227 | Since the potential term MATH is non-trivial it gives a non-trivial contribution to the energy of stationary states. The states with lowest energy which are parity even vanish at MATH, where the potential terms attains its maximal value, and cannot have the same energy as parity odd states which vanish at MATH, where t... |
hep-th/0010227 | The basic strategy is similar to that used in the planar rotor. Because of the non-degeneracy of the energy levels, any stationary state must have a definite MATH-parity symmetry with respect to the four points MATH where the circles with holonomies MATH and MATH cross each other. There are four quantum parity symmetry... |
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