paper stringlengths 9 16 | proof stringlengths 0 131k |
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quant-ph/0005018 | Fix MATH. Apply REF with MATH and let MATH be the value from the theorem. Let MATH be the string for whose quantum NAME complexity we want to give an upper bound. By REF , we get that the length of the encoding is what was given in the statement of the theorem. By simulating the decoding algorithm to a precision of MAT... |
quant-ph/0005055 | A simple solution to this problem is to embed the classical heuristic MATH into the function used in the algorithm NAME. Let MATH and MATH, so that MATH. By REF , for each function MATH, we have an expected running time in MATH. Let MATH denote the probability that MATH occurs. Then MATH, and we have that the expected ... |
quant-ph/0005055 | For MATH, using standard trigonometric identities, we obtain MATH . The inequality follows directly. |
quant-ph/0005055 | MATH . |
quant-ph/0005055 | Clearly MATH thus using REF we directly obtain the first part of the theorem. We use this fact to prove the next part of the theorem. MATH . For the last part, we use the fact that for MATH, the given expression attains its minimum at MATH in the range MATH. MATH . |
quant-ph/0005055 | After REF , by REF , we have state MATH . After REF , ignoring global phase, we have MATH and after applying MATH we have MATH . We then apply MATH to the first register and measure it in the computational basis. The rest of the proof follows from REF . Tracing out the second register in the eigenvector basis, we see t... |
quant-ph/0005055 | When MATH, the analysis is straightforward. For MATH, let MATH denote MATH and MATH. From REF we have that the probability that REF outputs Count-MATH for MATH is MATH . The previous inequalities are obtained by using the fact that MATH for any MATH and MATH, which can be readily seen by considering the NAME expansion ... |
quant-ph/0005055 | Apply REF with MATH. For each MATH, with probability greater than MATH, outcome MATH satisfies MATH, in which case we also have that MATH. Thus, with probability greater than MATH, we have MATH . Suppose this is the case. Then by REF , with probability at least MATH, MATH and consequently MATH . Hence, with probability... |
quant-ph/0005055 | To find MATH, we run REF to REF of algorithm Basic NAME and then set MATH. A proof analogous to that of REF gives that CASE: MATH with probability at least MATH, and CASE: the expected value of MATH is in MATH. This requires a number of evaluations of MATH which is in MATH , and thus, the expected number of evaluations... |
quant-ph/0005106 | From the definition of the binary entropy function, we have MATH . Using the expansion MATH for MATH, and simplifying, we get MATH which is the claimed bound. |
quant-ph/0005106 | We start with the special case of MATH. By REF , there is a measurement MATH on MATH that realizes the trace norm distance MATH between MATH and MATH. Using NAME 's strategy (see, for example, CITE), the resulting distributions can be distinguished with probability MATH. Let MATH denote the classical random variable ho... |
quant-ph/0005106 | Let MATH. We have: MATH . By REF , MATH, and by REF we have MATH. Thus, MATH. |
quant-ph/0005106 | By REF , there exists a purification MATH of MATH such that MATH. Since MATH and MATH have the same reduced density matrix in MATH, by REF , there is a (local) unitary transformation MATH on MATH such that MATH. Moreover, by REF we have MATH . By REF , MATH, so MATH . This, when combined with the earlier bound on the t... |
quant-ph/0005106 | It is enough to show the lower bound for the two cases when the protocol starts either with MATH or with the other player. Let MATH be the player to start. Note that if we set MATH to a fixed value, say MATH, then we get an instance of MATH. So MATH. But MATH, so the bound of REF applies. Let player MATH be the one to ... |
quant-ph/0005106 | We prove the theorem by induction on MATH. The case MATH is handled by REF . Suppose the theorem holds for MATH. We prove by contradiction that it holds for MATH as well. If MATH, then by REF there is a MATH message protocol for MATH with the wrong player starting, with error MATH, and with the same communication compl... |
quant-ph/0005106 | Let MATH denote the message sent by NAME. For a prefix MATH of length MATH, let MATH be the encoding which is prepared by first fixing MATH and then choosing MATH at random and sending the state MATH. Its density matrix is given by MATH . On the one hand, MATH, the number of qubits in MATH. On the other hand, for MATH,... |
quant-ph/0005106 | By the definition of mutual information, and using REF , MATH . Moreover, from REF , MATH which proves the claim. |
quant-ph/0005106 | For concreteness, we assume that MATH is even, so that MATH is NAME. Let MATH be a protocol that solves MATH with respect to MATH with MATH message qubits, error MATH, and MATH messages starting with NAME. We would like to concentrate on inputs where MATH is fixed to a particular value in MATH. This would give rise to ... |
quant-ph/0005106 | First consider the case when MATH is fixed to some MATH, but the rest of the inputs are as in MATH. In protocol MATH . NAME applies a unitary transformation MATH on his qubits and computes MATH in register MATH (for the message) and MATH (for NAME 's ancilla and input). In MATH the message computation is slightly diffe... |
quant-ph/0005106 | Protocol MATH solves an instance of MATH. NAME is given an input MATH and NAME is given an input MATH. The protocol proceeds as follows. NAME and NAME first reduce the problem to a MATH instance taken from the distribution MATH for a random MATH. To do that, NAME picks MATH at random, sets MATH and sends it to NAME; NA... |
quant-ph/0005106 | Note that MATH is the same as the mutual information MATH when MATH is run on the uniform distribution on MATH. So we prove the claim for the latter. For any MATH, MATH. Therefore by REF (compare REF) we have MATH . As the first message MATH contains only MATH qubits, we have MATH. |
cond-mat/0006367 | We describe a proof which uses knowledge about NAME functions. Let MATH be a partition, that is, a nonincreasing sequence of nonnegative integers. Then the NAME function MATH is defined by (see CITE or CITE) MATH . It is well-known that a combinatorial description of NAME functions may be given in terms of (semistandar... |
cond-mat/0006367 | Using the correspondence between stars and tableaux described in the proof of REF , we see that we must count tableaux with entries at most MATH having at most MATH columns. This enumeration problem (actually the corresponding ``MATH-enumeration" problem) is known under the name NAME - NAME conjecture, and was first pr... |
cond-mat/0006367 | We know that the number of stars with MATH branches of length MATH is given by the product formula MATH . Therefore, proving REF amounts to appropriately rewriting REF and then applying NAME 's formula. For convenience, let us introduce the notations MATH and MATH. Then the product REF can be rewritten as follows, MATH... |
cond-mat/0006367 | The situation is more difficult here, as we do not have a nice closed product formula (such as REF) for MATH-friendly stars. For simplicity, we treat the case of even MATH only, the case of odd MATH being completely analogous. Consider a MATH-friendly star with MATH branches of length MATH. It consists of a family MATH... |
cond-mat/0006367 | The first assertion follows immediately from REF since the number of MATH-friendly stars is bounded below by the number of ``genuine" stars, and is bounded above by the number of MATH-friendly stars in the TK model. So, explicitly, we may choose MATH and MATH . The second assertion can be proved as follows. Clearly, fo... |
cond-mat/0006367 | CASE: By first principles. As in CITE, we could directly use the main theorem of non-intersecting lattice paths (see REF ), to write the number of stars in question in the form MATH where MATH denotes the set of all lattice paths from MATH to MATH that do not go below the MATH-axis. By the ``reflection principle" (see ... |
cond-mat/0006367 | Using the correspondence between stars restricted by a wall and symplectic tableaux described in the second proof of REF , we see that we want to count symplectic tableaux with entries at most MATH having at most MATH rows and at most MATH columns. This problem was encountered before by Proctor CITE. (He was actually i... |
cond-mat/0006367 | We know that the number of stars with MATH branches of length MATH which do not go below the MATH-axis, and whose end points have MATH-coordinates at least MATH, MATH, is given by the product formula MATH . Application of NAME 's formula yields REF after a short calculation. |
cond-mat/0006367 | Again, the situation is more difficult here, as we do not have a nice closed product formula (such as REF) for MATH-friendly stars. We shall follow very closely the line of arguments of the proof of REF . Again, for simplicity, we treat the case of even MATH and MATH only, other cases being completely analogous. As in ... |
cond-mat/0006367 | We know that the number of watermelons with MATH branches of length MATH and with deviation MATH (where MATH) is given by REF. Therefore our task is to estimate the sum MATH . We follow the standard way of carrying out such estimations, as described in CITE. The dominant terms in the sum on the right-hand side are thos... |
cond-mat/0006367 | The first step is the same as in the proof of REF . We transform MATH-friendly watermelons into families of non-intersecting lattice paths by shifting the MATH-th path up by MATH units. Thus, MATH-friendly watermelons with MATH branches of length MATH and deviation MATH are in bijection with families MATH of non-inters... |
cond-mat/0006367 | The number of watermelons with MATH branches of length MATH and with deviation MATH (where MATH) which do not go below the MATH-axis is given by REF. We wish to sum this expression over all MATH with MATH and then approximate it. This is done completely analogously to the proof of REF . The only difference is that here... |
cond-mat/0006367 | As we have already seen, in the context of MATH-friendly models the situation is more difficult, because we do not have a nice closed product formula (such as REF) for the number of MATH-friendly watermelons. The first step is the same as in the proof of REF . We transform MATH-friendly watermelon into families of noni... |
cond-mat/0006367 | We first prove REF for MATH. The NAME summation theorem (see CITE, CITE) says that MATH for suitable functions MATH. It is for example valid for continuous, absolutely integrable functions MATH of bounded variation. The choice of MATH in REF gives REF for MATH upon little manipulation. From now on let MATH. In REF we c... |
cs/0006008 | It is easy to see that from the time process MATH becomes active, it performs each unit of work at most once, partial checkpoints each subchunk at most once (and hence performs at most MATH partial checkpoints), and full checkpoints every chunk at most once (and hence performs at most MATH full checkpoints). Each parti... |
cs/0006008 | REF is immediate from REF and the definition of MATH. We prove REF simultaneously. To do so, we need a careful way of counting the total number of messages sent and the total amount of work done. A given unit of work may be performed a number of times. If it is performed more than once, say by processes MATH, we say th... |
cs/0006008 | REF follows from the fact that in each useful round, the chain either performs work, or checkpoints to some group MATH the fact that a subchunk MATH was performed, or checkpoints the fact that group MATH was informed that chunk MATH was performed. The discussion above shows that no unit of work is repeated and hence th... |
cs/0006008 | The proof is straightforward. We start with REF . In the calculations below, we use ``(MATH)" to denote the value REF if MATH and REF otherwise. Similarly, MATH denotes REF if MATH, and REF otherwise. Recall that MATH denotes MATH mod MATH. MATH . If MATH, then MATH and REF follows. (In the first equality we replaced M... |
cs/0006008 | We first show that if MATH is in MATH and becomes active at round MATH with MATH, then there are at most MATH useless rounds in MATH. We proceed by induction on MATH. If MATH, the result is trivial. If MATH, then MATH is MATH's successor for some MATH in the activation chain and MATH received its last message from MATH... |
cs/0006008 | Fix an execution MATH of Protocol MATH. The proof proceeds by induction on the round MATH. The base case of MATH holds trivially since only process REF is active then. Assume the claim for MATH, and we will show it for MATH. If MATH, the claim holds trivially. Thus, we can assume MATH. Suppose that the last ordinary me... |
cs/0006008 | REF were argued in the beginning of REF. For REF , let MATH be the last process that is active and consider its activation chain. We want to find the last round MATH in which MATH is active. It follows from REF that the maximal number of useful rounds performed by any chain is MATH. Therefore, applying REF with MATH we... |
cs/0006008 | By assumption, one of the processes is correct, say MATH. At some point process MATH will become active, since once every other process has retired process MATH will not extend its deadline. It is straightforward from inspection of the algorithm that at any time during the execution of the algorithm MATH if and only if... |
cs/0006008 | It is immediate from the description of the algorithm that all nonretired processes have received a message from MATH by the time it has performed MATH units of work (at level MATH) after round MATH. Thus, we compute an upper bound on the time it takes for MATH to perform MATH units of work starting at round MATH. In t... |
cs/0006008 | The proof is an easy induction on MATH, since when a MATH-th rank process becomes active, it knows about everything its parent knew when it became active, and at least one more piece of work or failure. |
cs/0006008 | The proof is by induction on MATH. The base case, MATH, is straightforward. Let MATH, and assume that all parts of the lemma hold for smaller values of MATH. We prove it for MATH. For REF , observe that by the inductive hypothesis, REF holds at the beginning of round MATH. If no process is active in round MATH, then no... |
cs/0006008 | If process MATH's reduced view is MATH and it does not receive a message within MATH steps, then it becomes active. Each message that MATH receives increases its reduced view. Thus, MATH becomes active in at most MATH rounds. Once it becomes active, arguments similar to those used in REF show that it retires in at most... |
cs/0006008 | We proceed by induction on MATH. The case MATH is vacuous. Assume that MATH and the result holds for MATH. If MATH, then it must be the case that MATH received its message from MATH, MATH, and MATH is the successor of MATH in the cyclic order on MATH, as computed by MATH in round MATH. It is easy to see that the result... |
cs/0006008 | Given MATH, MATH, and an execution MATH of Protocol MATH, we consider the sequence of triples MATH, with one triple in the sequence for every time a process MATH sends an ordinary message reporting a unit of work MATH to a process MATH, listed in the order that the work was performed. We must show that the length of th... |
cs/0006008 | REF implies that the amount of real work units that are performed and reported to MATH is at most MATH. In addition, each of the MATH processes may perform one unit without reporting it (because it retired immediately afterwards). Summing the two, REF follows. For REF implies that each MATH, performs at most MATH repor... |
cs/0006008 | For REF , an easy induction on MATH shows that by the end of phase MATH, no more than MATH units of work remain to be done, and no more than MATH units of work have been done. It follows that at most MATH units of work are done altogether. (We remark that there is nothing special about the factor ``half" in our require... |
cs/0006009 | Given that MATH, we have by REF that MATH iff MATH. Since MATH, this holds iff MATH. Again by REF this is true iff MATH, and we are done. |
cs/0006009 | Let MATH be a correct (joint) protocol for the coordinated attack problem, with MATH being the corresponding system. Consider a ground language consisting of a single fact MATH ``both generals are attacking", let MATH assign a truth value to this formula in the obvious way at each point MATH, and let MATH be the corres... |
cs/0006009 | Fix MATH. Without loss of generality, we can assume MATH. Let MATH be the number of messages received in MATH up to (but not including) time MATH. We show by induction on MATH that if MATH, then MATH iff MATH. We assume that all the runs mentioned in the remainder of the proof have the same initial configuration and th... |
cs/0006009 | Recall that communication between the generals is not guaranteed (that is, it satisfies REF above), and we assume that in the absence of any successful communication neither general will attack. Thus, if we take MATH to be ``both generals are attacking", then MATH does not hold at any point in a run in which no message... |
cs/0006009 | We sketch the proof for MATH; the proof for MATH is analogous. We assume that all runs mentioned in this proof have the same initial configuration and the same clock readings as MATH. If MATH is a run such that MATH holds at some point in MATH, let MATH be the first time in MATH that processor MATH knows MATH. Let MATH... |
cs/0006009 | The proof is analogous to that of REF . Assume that MATH is a joint protocol that guarantees that if either party attacks then they both eventually attack, and let MATH be the corresponding system. Let MATH``At least one of the generals has started attacking". We first show that when either general attacks, then eventu... |
cs/0006009 | Fix a run MATH, time MATH, and formula MATH. Since MATH is MATH-reachable from MATH in the graph corresponding to the complete-history interpretation, there exist points MATH, MATH, , MATH such that MATH, MATH, and for every MATH there is a processor MATH that has the same history at MATH and at MATH. We can now prove ... |
cs/0006009 | Let MATH be a system with temporal imprecision and MATH be a point of MATH. Suppose MATH (otherwise clearly MATH is reachable from MATH). Let MATH be the greatest lower bound of the set MATH. We will show that MATH is reachable from MATH and that MATH. Since MATH is a system with temporal imprecision, there exists a MA... |
cs/0006010 | The first point is true by definition of MATH. Let us focus on the second point. Assume that: MATH . MATH here above is the worst possible assumption with respect of the number of cut elimination steps, necessary to normalize MATH at level MATH, because: CASE: we assume that all the contraction nodes at MATH, that is, ... |
cs/0006012 | Assume a pair of crossing constituents appears in the output of the constituent voting technique. Each of the constituents must have received at least MATH votes from the MATH parsers. Let MATH be the sum of the votes for the assumed constituents. MATH because none of the parsers contains crossing brackets so none of t... |
cs/0006012 | The technique for this proof comes from NAME 's work on multiple sequence alignment, although his goal was to show that a particular biological sequence alignment technique was good under a given goodness measure CITE. The edit distance in question must be symmetric. That is, it must take the same number of edits to tr... |
cs/0006033 | Let MATH be the MGU of MATH and MATH. By REF , we have that MATH and MATH are nicely moded and MATH is input-linear. Thus by REF MATH is nicely moded, and hence MATH is MATH-nicely moded. |
cs/0006033 | Suppose that MATH is MATH-robustly typed. By REF we have MATH. Suppose MATH. Since MATH is obtained by unifying MATH with a head of a clause MATH, and MATH, it follows that MATH. By REF , MATH. Since MATH is MATH-nicely moded, MATH and so MATH. |
cs/0006033 | The proof is by induction on the position MATH in the derivation. The base case MATH is trivial since MATH. Now suppose the result holds for some MATH and MATH exists. By REF , MATH is permutation simply typed. Thus the result follows for MATH by REF . |
cs/0006033 | Since MATH is robustly typed and types are closed under instantiation, there exists a substitution MATH such that MATH, MATH, and MATH is correctly typed. Since MATH is nicely moded, MATH. Since MATH, it follows that MATH and hence MATH is nicely moded. Since MATH is well typed, it follows by REF that MATH is well type... |
cs/0006033 | Let MATH be an infinite delay-respecting derivation of MATH. Assume, for the purpose of deriving a contradiction, that MATH contains only finitely many steps where a non-robust atom is resolved. Then there exists an infinite suffix MATH of MATH containing no steps where a non-robust atom is resolved. Consider the first... |
cs/0006033 | Suppose MATH is MATH-robustly typed (note that MATH exists by REF ). Let MATH be an atom in MATH with all its ancestors in safe positions. By REF , MATH is correctly typed in its input positions and hence selectable. Moreover, since MATH is a safe position, MATH. It follows that if the proper ancestors of MATH are not ... |
cs/0006033 | By REF , MATH, and thus MATH is a vector of constants. Since MATH is already a vector of non-variable terms, it follows that MATH is a vector of constants and thus MATH. Therefore MATH. |
cs/0006033 | The proof is by induction on the length of MATH. Let MATH and MATH. The base case holds by the assumption that MATH is MATH-ground. Now consider some MATH where MATH and MATH exists. By REF , MATH and MATH are permutation simply typed and hence type-consistent in all argument positions. The induction hypothesis is that... |
cs/0006033 | By REF is non-variable in all bound positions, and MATH is a linear vector having flat terms in all bound positions, and variables in all other positions. Thus there is a substitution MATH such that MATH and MATH, which shows REF . Since MATH is a linear vector of variables, there is a substitution MATH such that MATH ... |
cs/0006033 | We show how MATH is computed, where we consider three stages. In the first, MATH and MATH are unified. In the second, the output positions are unified where the bindings go from MATH to MATH. In the third, the output positions are unified where the bindings go from MATH to MATH. REF illustrates which variables are boun... |
cs/0006033 | For simplicity assume that MATH and each clause body do not contain two identical atoms. Let MATH, MATH and MATH be a delay-respecting derivation of MATH. The idea is to construct an NAME MATH of MATH such that whenever MATH uses a clause MATH, then MATH uses the corresponding clause MATH in MATH. It will then turn out... |
cs/0006033 | In this proof, by a MATH-step we mean a MATH-step, for some MATH; likewise we define a MATH-step. By REF , no MATH-step can instantiate any descendant of MATH or MATH. Thus the MATH-steps can be disregarded, and without loss of generality, we assume MATH is empty. Suppose MATH is a delay-respecting derivation for MATH ... |
cs/0006033 | If MATH is an atom using a predicate in MATH such that the set MATH is non-empty and bounded, we define MATH. Thus, for each atom MATH and substitution MATH such that MATH and MATH are defined MATH . To measure the size of a query, we use the multiset containing the level of each atom whose predicate is in MATH. The mu... |
cs/0006033 | Suppose there is an infinite left-based derivation MATH of MATH. Then letting MATH, MATH, we can write MATH where MATH are the queries in MATH where a non-robust atom is selected. By REF , there are infinitely many such queries. We derive a contradiction. By REF , the non-robust atoms in each query in MATH have only an... |
cs/0006046 | An assignment of colors to the original MATH-CSP instance's variables solves the problem if and only if, for each constraint, there is at least one pair MATH in the constraint that does not appear in the coloring. In our transformed problem, we choose one variable per original constraint, with the colors available to t... |
cs/0006046 | Let the two colors allowed at MATH be MATH and MATH. Define MATH to be the set of pairs MATH is a constraint. We then include MATH to our set of constraints. Any pair MATH does not reduce the space of solutions to the original problem since if both MATH and MATH were present in a coloring there would be no possible col... |
cs/0006046 | It is safe to choose the colors MATH and MATH, since these two choices do not conflict with each other nor with anything else in the CSP instance. |
cs/0006046 | Any solution involving MATH can be changed to one involving MATH without violating any additional constraints, so it is safe to remove the option of coloring MATH with color MATH. Once we remove this option, MATH is restricted to two colors, and we can apply REF . |
cs/0006046 | We may safely assign color MATH to MATH and remove it from the instance. |
cs/0006046 | No coloring of the instance can use MATH, so we can restrict MATH to the remaining two colors and apply REF . |
cs/0006046 | If no constraint exists, we can solve the problem immediately. Otherwise choose some constraint MATH. Rename the colors if necessary so that both MATH and MATH have available the same three colors MATH, MATH, and MATH, and so that MATH. Restrict the colorings of MATH and MATH to two colors each in one of four ways, cho... |
cs/0006046 | We perform the reduction above MATH times, taking polynomial time and giving probability at least MATH of finding a correct solution. If we repeat this method until a solution is found, the expected number of repetitions is MATH. |
cs/0006046 | If MATH and MATH are both three-color variables, then the instance can be colored if and only if we can color the instance formed by replacing them with a single four-color variable, in which the four colors are the remaining choices for MATH and MATH other than MATH REF . Thus in this case we can reduce the problem si... |
cs/0006046 | The second constraint for MATH can not involve MATH, or we would be able to apply REF . We choose either to use color MATH or to restrict MATH to avoid that color REF . If we use color MATH, we eliminate choice MATH and another choice on the other neighbor of MATH. If we avoid color MATH, we may safely use color MATH. ... |
cs/0006046 | We assume that the instance has no color choice with only a single constraint, or we could apply one of REF to achieve the given work factor. We say that MATH implies MATH if there are constraints from MATH to every other color choice of MATH. If the target MATH of an implication is not the source of another implicatio... |
cs/0006046 | We can assume from REF that each constraint connects MATH to a different variable. Then if we choose to use color MATH, we eliminate MATH and remove a choice from each of its neighbors, either eliminating them or reducing their number of choices from four to three. If we don't use MATH, we eliminate that color only. So... |
cs/0006046 | For convenience suppose that the four-color neighbor is MATH. We can assume MATH has only two constraints, else it would be covered by a previous lemma. Then, if MATH and MATH do not form a triangle with a third (variable,color) pair (REF , left), we choose either to use or avoid color MATH. If we use MATH, we eliminat... |
cs/0006046 | Let MATH be the neighbor with two constraints. Note that (since the previous lemma is assumed not to apply) all neighbors of MATH have only three color choices. First, suppose MATH and MATH are not part of a triangle of constraints (REF , top). Then, if we choose to use color MATH we eliminate four variables, while if ... |
cs/0006046 | Let MATH and MATH be variables in a small component MATH. Then each (variable,color) pair in MATH from variable MATH has exactly one constraint to a distinct (variable,color) pair from variable MATH, so the numbers of pairs from MATH equals the number of pairs from MATH. The assertions that each variable has the same n... |
cs/0006046 | A component with MATH uses up all color choices for all four variables. Thus we may consider these variables in isolation from the rest of the instance, and either color them all (if possible) or determine that the instance is unsolvable. The remaining small components have MATH. Such a component may be drawn with the ... |
cs/0006046 | Choose some arbitrary pair MATH as a starting point, and perform a breadth first search in the graph formed by the pairs and constraints in the component. Let MATH be the first pair reached by this search where MATH is not one of the variables adjacent to MATH, let MATH be the grandparent of MATH in the breadth first s... |
cs/0006046 | Let MATH, MATH, MATH, MATH, and MATH be a witness for the component. Then we distinguish subcases according to how many of the neighbors of MATH are pairs in the witness. CASE: If MATH has a constraint with only one pair in the witness, say MATH, then we choose either to use color MATH or to avoid it. If we use it, we ... |
cs/0006046 | We split into subcases: CASE: Suppose the cycle passes through five consecutive distinct variables, say MATH, MATH, MATH, MATH, and MATH. We can assume that, if any of these five variables has four color choices, then this is true of one of the first four variables. Any coloring that does not use both MATH and MATH can... |
cs/0006046 | We form a bipartite graph, in which the vertices correspond to the variables and components of the instance. We connect a variable to a component by an edge if there is a (variable,color) pair using that variable and belonging to that component. Since each pair in a good three-component or small two-component is connec... |
cs/0006046 | We employ a backtracking (depth first) search in a state space consisting of MATH-CSP instances. At each point in the search, we examine the current state, and attempt to find a set of smaller instances to replace it with, using one of the reduction lemmas above. Such a replacement can always be found in polynomial tim... |
cs/0006046 | Randomly choose a subset of four values for each variable and apply our algorithm to the resulting MATH-CSP problem. Repeat with a new random choice until finding a solvable MATH-CSP instance. The random restriction of a variable has probability MATH of preserving solvability so the expected number of trials is MATH. E... |
cs/0006046 | Let the cycle MATH consist of vertices MATH, MATH, MATH, MATH. We can assume without loss of generality that it has no chords, since otherwise we could find a shorter cycle in MATH; therefore each MATH has a unique neighbor MATH outside the cycle, although the MATH need not be distinct from each other. Note that, if an... |
cs/0006046 | Suppose the subset forms a MATH-vertex tree, and let MATH be a vertex in this tree such that each subtree formed by removing MATH has at most MATH vertices. Then, if MATH is REF-colored, some two of the three neighbors of MATH must be given the same color, so we can split the instance into three smaller instances, each... |
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