/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} 23E In the NODE-DISJOINT PATHS probl... [FREE SOLUTION] | 91影视

91影视

In the NODE-DISJOINT PATHS problem, the input is an undirected graph in which some vertices have been specially marked: a certain number of 鈥渟ources鈥 s1,s2,,sk and an equal number of 鈥渄estinations鈥 t1,t2,,tk. The goal is to find k node-disjoint paths (that is, paths which have no nodes in common) where the ith path goes from si to ti. Show that this problem is NP-complete.Here is a sequence of progressively stronger hints.

  1. Reduce from 3SAT .
  2. For a 3SAT formula with m clauses and n variables, use k=m+n sources and destinations. Introduce one source/destination pair (sx,tx)for each variable x , and one source/destination pair (sc,tc) for each clause c .
  3. For each 3SAT clause, introducenew intermediate vertices, one for each literal occurring in that clause and one for its complement.

Notice that if the path from sc to tc goes through some intermediate vertex representing, say, an occurrence of variable x, then no other path can go through that vertex. What vertex would you like the other path to be forced to go through instead?

Short Answer

Expert verified

The given problem is NP-Complete.

Step by step solution

01

Explain 3SAT problem.

3SAT problem is the satisfiability problem, that finds the fastest algorithm that shows the given Boolean formula is satisfiable. The problem is satisfiable is the formula results 1 .

02

Show that the given problem is NP-complete

As given, for any clause c=(l1l2l3), set sc to and to respectively. For each variable , is concatenated with all the to , forms the path. All the concatenates to forms another new path.

From the two paths , one must be chosen, that ensures the consistency of variables. If any clause chooses the vertex , then no other clause is allowed to choose the same vertex.

Consider the following example, Consider that the CNF is to be verified as a satisfiable problem. Let , be the NODE-DISJOINT PATH problem as follows:

Therefore, the given problem is NP-complete has been shown by the above example.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Show that if P=NP then the RSA cryptosystem (Section 1.4.2) can be broken in polynomial time.

Determine which of the following problems are NP-complete and which are solvable in polynomial time. In each problem you are given an undirected graph G=(V,E), along with:

(a)A set of nodesLV , and you must find a spanning tree such that its set of leaves includes the set L.

(b)A set of nodes LV, and you must find a spanning tree such that its set of leaves is precisely the set L.

(c)A set of nodesLV , and you must find a spanning tree such that its set of leaves is included in the set L.

(d)An integer k, and you must find a spanning tree withk or fewer leaves.

(e)An integer k, and you must find a spanning tree withk or more leaves.

(f)An integer k, and you must find a spanning tree with exactlyk leaves.

Search versus decision. Suppose you have a procedure which runs in polynomial time and tells you whether or not a graph has a Rudrata path. Show that you can use it to develop a polynomial-time algorithm for RUDRATA PATH (which returns the actual path, if it exists).

We are feeling experimental and want to create a new dish. There are various ingredients we can choose from and we鈥檇 like to use as many of them as possible, but some ingredients don鈥檛 go well with others. If there arepossible ingredients (numbered 1to n), we write down an matrix giving thediscordbetween any pair of ingredients. Thisdiscordis a real number between 0.0and 1.0, where means 鈥渢hey go together perfectly鈥 and 1.0 means 鈥渢hey really don鈥檛 go together.鈥 Here鈥檚 an example matrix when there are five possible ingredients.

In this case, ingredients 2and 3go together pretty well whereas1and5clash badly. Notice that this matrix is necessarily symmetric; and that the diagonal entries are always . 0.0Any set of ingredients incurs apenaltywhich isthe sum of all discord values between pairs of ingredients.For instance, the set of ingredients{1,3,5}incurs a penalty of 0.2+1.0+0.5=1.7

.We want this penalty to be small.

EXPERIMENTAL CUISINE

Input:, nthe number of ingredients to choose from D;,the nn鈥 discord鈥 matrix; some numberp0

OUTPUT:The maximum number of ingredients we can choose with penalty p.

Show that ifEXPERIMENTAL CUISINEis solvable in polynomial time, then so is 3SAT.

On page 266we saw that 3SATremainsNP-complete even when restricted to formulas in which each literal appears at most twice.

(a)Show that if each literal appears at mostonce,then the problem is solvable in polynomial time.

(b)Show that INDEPENDENT SET remains NP-complete even in the special case when all the nodes in the graph have degree at most 4.

See all solutions

Recommended explanations on Computer Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.