Simon S
• Member for 9 years, 1 month
• Last seen more than a month ago
• Leipzig, Germany

Counter-example: $L_1 = \{a^n b^n c^m \mid m,n \ge 1 \}$ $L_2 = \{a^m b^n c^n \mid m,n \ge 1 \}$ $L = (L_1 \cdot L_2^*) \cup \epsilon$ is context-free. Any nonempty word $x \in L^\#$ has a prefix $... View answer Accepted answer 7 votes A language is decidable, iff it is semi-decidable and its complement is semi-decidable. Moreover, a language is recursive-enumerable iff it is semi-decidable and thus you can find an unrestricted ... View answer 3 votes Try using visual designers like this or turingkara-en, which also contains exercises. Keep in mind that the TM is meant to be a theoretical, minimalistic model to simplify proofs. Programming on the ... View answer Accepted answer 7 votes To get the CPL by this definition you first take the average distance from a certain vertex to any other vertex: $$d_v = {{\sum_{v \ne w} d(v,w)} \over {|V(G)| - 1}}$$ After doing so for every vertex ... View answer 2 votes Try constructing a reduced ordered binary decision diagram (ROBDD) as described here. The reducing algorithm removes redundant nodes, and reuses equal subtrees in the decision tree, such that every ... View answer 5 votes Yes$G\$ has a unique MST (assuming it is not a multigraph). Take a look at Kruskal's algorithm: the edges are ordered ascending by weight and then added to the MST unless there is a cycle. After ...