91Ó°ÊÓ

The RUDRATA PATH problem NP complete hard problem.

A backtracking algorithm requires a test that looks at a subproblem and quickly declares one of three outcomes:

1. Failure: the subproblem has no solution.

2. Success: a solution to the subproblem is found.

3. Uncertainty.

In the case of SAT, this test declares failure if there is an empty clause, success if there are no clauses and uncertainty otherwise.

With the right test, expand, and choose routines, backtracking can be remarkably effective in practice. The backtracking algorithm we showed for SAT is the basis of many successful satisfiability programs.

a).

A polynomial-time reduction which given any graph $G$ and integer any $C>0$ produces an instance $I(G,C)$ of the TSP such that:

1)If$G$ has a Rudrata path, then$OPT\left(I(G,C)\right)=n,$ the number of vertices in $G$.

2)If $G$ has no Rudrata path, then $OPT\left(I(G,C)\right)â‰\yen n+C.$

This means that even an approximate solution to TSP would enable us to solve RUDRATA PATH.

An approximation algorithm for TSP and let denote its approximation ratio. From any instanceof RUDRATA PATH, we will create an instance $I(G,C)$ of TSP using the specific $constantC=n\mathrm{Î\pm }A.$

In case

(i), it must output a tour of length at most $\mathrm{Î\pm }A\text{ â¶ÄŠ}OPT\left(I(G,C)\right)=n\mathrm{Î\pm }A,$whereas in case

(ii) it must output a tour of length at least $OPT\left(I(G,C)\right)>n\mathrm{Î\pm }A$ .

Thus we can figure out whether$G$has a Rudrata path! Here is the resulting procedure:

Given any graph$G$:

Compute $I(G,C)(withC=n·\mathrm{Î\pm }A)$ and run algorithm $A$ on it ,

If the resulting tour has $length≤n\mathrm{Î\pm }A:$

Conclude that $G$ has a Rudrata path ;

Else: conclude that $G$ has no Rudrata path;

This tells us whether or not $G$ has a Rudrata path;

By calling the procedure a polynomial number of times, if TSP has a polynomial-time approximation algorithm, then there is a polynomial algorithm for the NP-complete RUDRATA PATH problem.

So, unless $P=NP$ , there cannot exist an efficient approximation algorithm for the TSP.

b).

Since the new vertex $x$ has only two neighbors, $sandt,$ any Rudrata cycle in $G$ must consecutively traverse the edges $\{t,x\}and\{x,s\}.$

The rest of the cycle then traverses every other vertex en route from$sandt,$ . Thus deleting the two edges $\{t,x\}and\{x,s\}.$from the Rudrata cycle gives a Rudrata path from $sandt,$in the original graph $G$.

In the case of SAT, this test declares failure if there is an empty clause, success if there are no clauses, and uncertainty otherwise.

With the right test, expand, and choose routines, backtracking can be remarkably effective in practice. The backtracking algorithm we showed for SAT is the basis of many successful satisfiability programs.

As from this instance of RUDRATA CYCLE has a solution.

c).

In this case we must show that the original instance of RUDRATA $(s,t)-PATH$ cannot have a solution either. It is usually easier to prove the contra positive, that is, to show that if there is a Rudrata $(s,t)-PATH$ in $G$, then there is also a Rudrata cycle in . But this is easy: just add the two edges $\{t,x\}and\{x,s\}.$to the Rudrata path to close the cycle.

And here all the crucial but typically easy to check, is that the pre- and postprocessing functions take time polynomial in the size of the instance $(G,s,t).$It is also possible to

go in the other direction and reduce RUDRATA CYCLE to RUDRATA $(s,t)-PATH$

Together, these reductions demonstrate that the two Rudrata variants are in essence the same problem that their descriptions are almost the same.

An approximation algorithm $A$ for TSP and let $\mathrm{Î\pm }A$ denote its approximation ratio. From any instance$G$of RUDRATA PATH, we will create an instance $I(G,C)$ of TSP using the specific$constantC=n\mathrm{Î\pm }A.$

But most of the other reductions are between pairs of problems that, on the face of it, look quite different. they are essentially the same, our reductions will have to cleverly translate between them.