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Given a graph G = (V, E), construct the following instance of the TSP: the set of cities is the same as V , and the distance between cities u and v is 1 if {u, v} is an edge of G and 1 + otherwise, for some > 1 to be determined The budget of the TSP instance is equal to the number of nodes, |V | It is easy to see that if G has a Rudrata cycle, then the same cycle is also a tour within the budget of the TSP instance; and that conversely, if G has no Rudrata cycle, then there is no solution: the cheapest possible TSP tour has cost at least n + (it must use at least one edge of length 1 + , and the total length of all n 1 others is at least n 1) Thus R UDRATA CYCLE reduces to TSP In this reduction, we introduced the parameter because by varying it, we can obtain two interesting results If = 1, then all distances are either 1 or 2, and so this instance of the TSP satis es the triangle inequality: if i, j, k are cities, then d ij + djk dik (proof: a + b c holds for any numbers 1 a, b, c 2) This is a special case of the TSP which is of practical importance and which, as we shall see in 9, is in a certain sense easier, because it can be ef ciently approximated If on the other hand is large, then the resulting instance of the TSP may not satisfy the triangle inequality, but has another important property: either it has a solution of cost n or less, or all its solutions have cost at least n + (which now can be arbitrarily larger than n) There can be nothing in between! As we shall see in 9, this important gap property implies that, unless P = NP, no approximation algorithm is possible
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We have reduced SAT to the various search problems in Figure 87 Now we come full circle and argue that all these problems and in fact all problems in NP reduce to SAT In particular, we shall show that all problems in NP can be reduced to a generalization of SAT which we call CIRCUIT SAT In CIRCUIT SAT we are given a (Boolean) circuit (see Figure 813, and recall Section 77), a dag whose vertices are gates of ve different types:
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gates and OR gates have indegree 2 gates have indegree 1
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Known input gates have no incoming edges and are labeled false or true Unknown input gates have no incoming edges and are labeled One of the sinks of the dag is designated as the output gate Given an assignment of values to the unknown inputs, we can evaluate the gates of the circuit in topological order, using the rules of Boolean logic (such as false true = true), until we obtain the value at the output gate This is the value of the circuit for the particular assignment to the inputs For instance, the circuit in Figure813 evaluates to false under the assignment true, false, true (from left to right) C IRCUIT SAT is then the following search problem: Given a circuit, nd a truth assignment for the unknown inputs such that the output gate evaluates to true, or report that no such assignment exists For example, if presented with the circuit in Figure 813 we could have 255
Figure 813 An instance of
CIRCUIT SAT
output
true
returned the assignment (false, true, true) because, if we substitute these values to the unknown inputs (from left to right), the output becomes true C IRCUIT SAT is a generalization of SAT To see why, notice that SAT asks for a satisfying truth assignment for a circuit that has this simple structure: a bunch of AND gates at the top join the clauses, and the result of this big AND is the output Each clause is the OR of its literals And each literal is either an unknown input gate or the NOT of one There are no known input gates Going in the other direction, CIRCUIT SAT can also be reduced to SAT Here is how we can rewrite any circuit in conjunctive normal form (the AND of clauses): for each gate g in the circuit we create a variable g, and we model the effect of the gate using a few clauses:
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