The Double SAT problem asks whether a given satisfiability problem has at least two different satisfying assignments. For example, the problem {{V1, V2}, {V1, V2}, {V1, V2}} is satisfiable, but has only one solution (v₁ = F, v₂ = T). In contrast, {{v₁, V2}, {V1, V2}} has exactly two solutions. Show that Double-SAT is NP-hard.
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- I really need help with this problem, I don't understand itConsider the generalization of the rickety bridge crossing puzzle in which we have n > 1 people whose bridge crossing times are t₁, t2, ..., tn. You may assume that t₁ ≤ t₂ < ... < tn. All the other conditions of the problem remain the same: all begin on the same side of the bridge, at most two people at the time can cross the bridge (and they move with the speed of the slower of the two) and they must carry with them the only flashlight the group has. Design a greedy algorithm for this problem and find how long it will take to cross the bridge by using this algorithm. Find an instance with the smallest number of people for which your greedy algorithm does not yield a minimum crossing time.Give English language translations of the following wff if M(x): x is a man W(x): x is a woman i: Ivan p: Peter W(x, y): x works for y (∃x)(W(x) ∧ (∀y)(M(y) → [W(x, y)]')) which one of this answer choices a. No woman works for any man. b. Every woman works for some man. c. Some woman does not work for any man. d. Every woman works for every man.
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