I ran into a question on previous Mid-Exam. anyone could clarify me?
Problem A: Given a Complete Weighted Graph G, find a Hamiltonian Tour with minimum weight.
Problem B: Given a Complete Weighted Graph G and Real Number R, Is G has a Hamiltonian Tour with weight at most R?
Suppose there is a machine that solves B. with how many times call of B (each time G and Real number R are given), We Can solve problem A with that machine? suppose the sum of Edges in G up to M.
1) We cannot do this, because there is uncountable state.
2) O(|E|) times
3) O(lg m) times
4) because A is NP-Hard, This is cannot be done.
This is not an answer, I just needed more space than in the comment to follow my discussion with Xoff
I'm just showing here that if you know the minimal weight $R_m$ of an Hamiltonian cycle, you can effectively build one of weight $R_m$ using $|E|$ calls to B.
I assume here that all the weight are positive.
I denote here $B(G,w,R)$ the call of $B$ on $G$ with weight function $w$ and bound $R$.
Let $\{e_1,...,e_n\}$ be the edges of the graph $G$.
Consider the weight function $w_1$ such that $w_1(e_1)=w(e_1)+R_m$ and $w_1(e_i)=w(e_i)$ for $i\neq 1$.
If $B(G,w_1,R_m)$ is false we mark $e_1$.
More generally we define $w_k$ as $w_k(e_k)=w_{k-1}(e_k)+R_m$ and $w_k(e_i)=w(e_i)$ if $e_i$ is marked and for $w_k(e_i)=w_{k-1}(e_i)$ otherwise.
Again if $B(G,w_k,R_m)$ is false we mark $e_k$.
At the end all the edges marked belong to a cycle of weight $R_m$.