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ioda
3 years ago
7

Two well-known NP-complete problems are 3-SAT and TSP, the traveling salesman problem. The 2-SAT problem is a SAT variant in whi

ch each clause contains at most two literals. 2-SAT is known to have a polynomial-time algorithm. Is each of the following statements true or false?
1. 3-SAT ≤p TSP.2. If P ¹ NP, then 3-SAT ≤p 2-SAT.3. If P ¹ NP, then no NP-complete problem can be solved in polynomial time.
Engineering
1 answer:
Hitman42 [59]3 years ago
7 0

3-SAT ≤p TSP

If P ¹ NP, then no NP-complete problem can be solved in polynomial time.

both the statements are true.

<u>Explanation:</u>

  • 3-SAT ≤p TSP due to any  complete problem of NP to other problem by exits of reductions.
  • If P ¹ NP, then 3-SAT ≤p 2-SAT are the polynomial time algorithm are not for 3-SAT. In P, 2-SAT is found, 3- SAT polynomial time algorithm implies the exit of reductions. 3 SAT does not have polynomial time algorithm when P≠NP.
  • If P ¹ NP, then no NP-complete problem can be solved in polynomial time. because for the NP complete problem individually gets the polynomial time algorithm for the others. It may be in P for all the problems, the implication of latter is P≠NP.
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Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance x!

= cos 75⁰ = \frac{x^!}{10ft}

    x! = 10 * cos 75^{o} = 2.59 ft

calculate distance x

= cos 75⁰ = \frac{x}{30ft}

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * x! + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

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3 years ago
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Explanation:

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