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## Next Steps ? (2) Next Steps More NP-Complete Problems

Champlain College, US has reference to this Academic Journal, More NP-Complete Problems NP-Hard Problems Tautology Problem Node Cover Knapsack Next Steps We can now reduce 3SAT so that a large number of problems, either directly or indirectly. Each reduction must be polytime. Usually we focus on length of the output from the transducer, because the construction is easy. But key issue: must be polytime. Next Steps ? (2) Another essential part of an NP-completeness proof is showing the problem is in NP. Sometimes, we can only show a problem NP-hard = ?if the problem is in P, then P = NP,? but the problem may not be in NP.

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Example: NP-Hard Problem The Tautology Problem is: given a Boolean formula, is it satisfied by all truth assignments? Example: x + -x + yz Not obviously in NP, but it?s complement is. Guess a truth assignment; accept if that assignment doesn?t satisfy the formula. Key Point Regarding Tautology An NTM can guess a truth assignment in addition to decide whether formula F is satisfied by that assignment in polytime. But if the NTM accepts when it guesses a satisfying assignment, it will accept F whenever F is in SAT, not Tautology. Co-NP A problem/language whose complement is in NP is said so that be in Co-NP. Note: P is closed under complementation. Thus, P ? Co-NP. Also, if P = NP, then P = NP = Co-NP.

Tautology is NP-Hard While we can?t prove Tautology is in NP, we can prove it is NP-hard. Suppose we had a polytime algorithm in consideration of Tautology. Take any Boolean formula F in addition to convert it so that -(F). Obviously linear time. Tautology is NP-Hard ? (2) F is satisfiable if in addition to only -(F) is not a tautology. Use the hypothetical polytime algorithm in consideration of Tautology so that test if -(F) is a tautology. Say ?yes, F is in SAT? if -(F) is not a tautology in addition to say ?no? otherwise. Then SAT would be in P, in addition to P = NP. Historical Comments There were actually two notions of ?NP-complete? that differ subtlely. And only if P ? NP. Steve Cook, in his 1970 paper, was really concerned alongside the question ?why is Tautology hard?? Remember: theorems are really logical tautologies.

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History ? (2) Cook used ?if problem X is in P, then P = NP? as the definition of ?X is NP-hard.? Today called Cook completeness. In 1972, Richard Karp wrote a paper showing many of the key problems in Operations Research so that be NP-complete. History ? (3) Karp?s paper moved ?NP-completeness? from a concept about theorem proving so that an essential in consideration of any study of algorithms. But Karp used the definition of NP-completeness ?exists a polytime reduction,? as we have. Called Karp completeness. Cook Vs. Karp Completeness In practice, there is very little difference. Biggest difference: in consideration of Tautology, Cook lets us flip the answer after a polytime reduction. In principle, Cook completeness could be much more powerful, or (if P = NP) exactly the same.

Cook Vs. Karp ? (2) But there is one important reason we prefer Karp-completeness. Suppose I had an algorithm in consideration of some NP-complete problem that ran in time O(nlog n). A function that is bigger than any polynomial, yet smaller than the exponentials like 2n. Cook Vs. Karp ? (3) If ?NP-complete is Karp-completeness, I can conclude that all of NP can be solved in time O(nf(n)), where f(n) is some function of the form c logkn. Still faster than any exponential, in addition to faster than we have a right so that expect. But if I use Cook-completeness, I cannot say anything of this type. The Node Cover Problem Given a graph G, we say N is a node cover in consideration of G if every edge of G has at least one end in N. The problem Node Cover is: given a graph G in addition to a ?budget? k, does G have a node cover of k or fewer nodes?

Example: Node Cover A C E F D B NP-Completeness of Node Cover Reduction from 3SAT. For each clause (X+Y+Z) construct a ?column? of three nodes, all connected by vertical edges. Add a horizontal edge between nodes that represent any variable in addition to its negation. Budget = twice the number of clauses. Example: The Reduction so that Node Cover (x + y + z)(-x + -y + -z)(x + -y +z)(-x + y + -z) Budget = 8

Example: Reduction ? (2) A node cover must have at least two nodes from every column, or some vertical edge is not covered. Since the budget is twice the number of columns, there must be exactly two nodes in the cover from each column. Satisfying assignment corresponds so that the node in each column not selected. Example: Reduction ? (3) (x + y + z)(-x + -y + -z)(x + -y +z)(-x + y + -z) Truth assignment: x = y = T; z = F Example: Reduction ? (4) (x + y + z)(-x + -y + -z)(x + -y +z)(-x + y + -z) Truth assignment: x = y = T; z = F

Proof That the Reduction Works The reduction is clearly polytime. Need so that show: If we construct from 3SAT instance F a graph G in addition to a budget k, then G has a node cover of size k if in addition to only if F is satisfiable. Proof: If Suppose we have a satisfying assignment A in consideration of F. For each clause of F, pick one of its three literals that A makes true. Put in the node cover the two nodes in consideration of that clause that do not correspond so that the selected literal. Total = k nodes ? meets budget. Proof: If ? (2) The selected nodes cover all vertical edges. Why? Any two nodes in consideration of a clause cover the triangle of vertical edges in consideration of that clause. Horizontal edges are also covered. A horizontal edge connects nodes in consideration of some x in addition to -x. One is false in A in addition to therefore its node must be selected in consideration of the node cover.

Proof: Only If Suppose G has a node cover alongside at most k nodes. One node cannot cover the vertical edges of any column, so each column has exactly 2 nodes in the cover. Construct a satisfying assignment in consideration of F by making true the literal in consideration of any node not in the node cover. Proof: Only If ? (2) Worry: What if there are unselected nodes corresponding so that both x in addition to -x? Then we would not have a truth assignment. But there is a horizontal edge between these nodes. Thus, at least one is in the node cover. Optimization Problems NP-complete problems are always yes/no questions. In practice, we tend so that want so that solve optimization problems, where our task is so that minimize (or maximize) a parameter subject so that some constraints.

Example: Optimization Problem People who care about node covers would ask: Given this graph, what is the smallest number of nodes I can pick so that form a node cover? If I can solve that problem in polytime, then I can solve the yes/no version. Example ? Continued Polytime algorithm: given graph G in addition to budget k, solve the optimization problem in consideration of G. If the smallest node cover in consideration of G is of size k or less, answer ?yes?; otherwise answer ?no.? Optimization Problems ? (2) Optimization problems are never, strictly speaking, in NP. They are not yes/no. But there is a Cook reduction from the yes/no version so that the optimization version.

Proof: Converse Suppose the output instance of Partition-Knapsack has a solution. The integers s in addition to 2k cannot be in the same partition. Because their sum is more than half 2(s+k). Thus, the subset of L that is in the partition alongside s sums so that k. Thus, it solves the Knapsack instance.

## Osborne, Jerry Investigative Producer

Osborne, Jerry is from United States and they belong to Investigative Producer and work for KVOA-TV in the AZ state United States got related to this Particular Article.

## Journal Ratings by Champlain College

This Particular Journal got reviewed and rated by Proof: Only If Suppose G has a node cover alongside at most k nodes. One node cannot cover the vertical edges of any column, so each column has exactly 2 nodes in the cover. Construct a satisfying assignment in consideration of F by making true the literal in consideration of any node not in the node cover. Proof: Only If ? (2) Worry: What if there are unselected nodes corresponding so that both x in addition to -x? Then we would not have a truth assignment. But there is a horizontal edge between these nodes. Thus, at least one is in the node cover. Optimization Problems NP-complete problems are always yes/no questions. In practice, we tend so that want so that solve optimization problems, where our task is so that minimize (or maximize) a parameter subject so that some constraints. and short form of this particular Institution is US and gave this Journal an Excellent Rating.