 # 2 Problems With Nondeterministic Algorithm and 3-CNF Formula

## Question

1.Describe a Non deterministic polynomial time algorithm for deciding if a simple graph G=(V,E) has a Hamiltonian cycle, i.e Hamilton path for which the begin and end vertices are adjacent.

2. Let C be a 3-CNF formula. A≠-assignment to C is a truth assignment that satisfies C, but in such a way that every clause of C has at least one literal set to true, but also has one literal set to false. Prove that, if a is a≠-assignment for a set of clauses, then so it is its inversion a ̅, where the inversion of an assignment is the assignment that is obtained by inverting each assignment value of a (i.e. 1 to 0, and 0 to 1).

## Solution Preview

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We assume that as input we have the graph G encoded as an adjacency list in a binary notation.
1) We also assume its vertices are counted from 1 to n.
The non-deterministic algorithm will must first call a method to determine a sequence of n+1 numbers from 1 to n.
Then we put the algorithm to verify that each number from the range [1..n] appears only once in the sequence. This can be done simpler by sorting the sequence. The only constraint involves the first and last numbers, which must be the same....
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