Kata pengantar singkat. Masalahnya adalah NP-hard.
Sketsa bukti. Kami mengabaikan kendala | F i | ≪ n = | U | dalam masalah yang diposting, karena, untuk setiap instance ( F , U , k ) dari masalah, instance ( F ′ = F n , U ′ = U n , k ) diperoleh dengan mengambil gabungan dari n salinan independen dari ( F , U , k ) (di mana i|Fi|≪n=|U|(F,U,k)(F′=Fn,U′=Un,k)n(F,U,k)i salinan F menggunakan iFi th menyalin dari U sebagai set basis) adalah setara, dan memenuhi kendala (memiliki | F ' i | ≤ n « n 2 = | U ' | ).U|F′i|≤n≪n2=|U′|
Kami memberikan pengurangan dari 3-SAT. Untuk presentasi, pada tahap pertama reduksi, kami mengabaikan kendala e i ∈ F i dalam masalah yang diposting. Pada tahap kedua kami menjelaskan bagaimana memenuhi kendala tersebut sambil mempertahankan kebenaran reduksi.ei∈Fi
Tahap pertama. Perbaiki formula 3-SAT ϕ . Asumsikan WLOG bahwa setiap klausa memiliki tepat tiga literal (masing-masing menggunakan variabel yang berbeda). Hasilkan contoh berikut ( F , U , k ) dari masalah yang diposting, dengan k = 3 .ϕ(F,U,k)k=3
Biarkan n menjadi jumlah variabel dalam ϕ . Ada 3 n + 1 elemen dalam U : satu elemen t (untuk "benar"), dan, untuk setiap variabel x i di φ , tiga elemen x i , ¯ x i , dan f i (untuk "false").nϕ3n+1Utxiϕxix¯¯¯ifi
For each element in UU there is a singleton set containing just that element in FF. Any solution CC therefore includes each of these sets, which contribute their total size 3n+13n+1 to the cost of CC.
In addition, for each variable xixi in ϕϕ there is a "variable" set {xi,¯xi,fi,t}{xi,x¯¯¯i,fi,t} in FF. For each clause in ϕϕ there is a "clause" set in FF consisting of the literals in the clause, and tt. For example, the clause x1∧¯x2∧x3x1∧x¯¯¯2∧x3 yields the set {x1,¯x2,x3,t}{x1,x¯¯¯2,x3,t} in FF.
Claim 1. The reduction is correct: ϕϕ is satisfiable iff some solution CC has cost ∑j|Cj|=5n+1∑j|Cj|=5n+1.
(only if) Suppose ϕϕ is satisfiable. Construct a solution CC consisting of the 3n+13n+1 singleton sets, plus, for each variable xixi, the pair consisting of the true literal and tt. (E.g., {¯xi,t}{x¯¯¯i,t} if xixi is false.) The cost of CC is then 5n+15n+1.
Each variable set {xi,¯xi,fi,t}{xi,x¯¯¯i,fi,t} is the union of three sets: the pair consisting of the true literal and tt, plus two singleton sets, one for each of the other two elements. (E.g., {¯xi,t},{xi},{fi}{x¯¯¯i,t},{xi},{fi}.)
Each clause set (e.g. {x1,¯x2,x3,t}{x1,x¯¯¯2,x3,t}) is the union of three sets: a pair consisting of tt and a true literal, plus two singleton sets, one for each of the other two literals. (E.g., {x1,t},{¯x2},{x3}{x1,t},{x¯¯¯2},{x3}.)
(if) Suppose there is a solution CC of size 5n+15n+1. The solution must contain the 3n+13n+1 singleton sets, plus other sets of total size 2n2n.
Consider first the nn "variable" sets, each of the form {xi,¯xi,fi,t}{xi,x¯¯¯i,fi,t}. The set is the disjoint union of at most three sets in CC. Without loss of generality, it is the disjoint union of two singletons and a pair (otherwise, splitting sets in CC achieves this without increasing the cost). Denote the pair PiPi. The pairs PiPi and PjPj for different variables xixi and xjxj are distinct, because PiPi contains xixi, ¯xix¯¯¯i, or fifi but PjPj does not. Hence, the sum of the sizes of these pairs is 2n2n. So these pairs are the only non-singleton sets in the solution.
Next consider the "clause" sets, e.g, {xi,¯xj,xk,t}{xi,x¯¯¯j,xk,t}. Each such set must be the union of at most three sets in CC, that is, up to two singleton sets and at least one pair PiPi, PjPj, or PkPk. By inspection of the pairs and the clause set, it must be the union of two singletons and one pair, and that pair must be of the form {xi,t}{xi,t} or {¯xj,t}{x¯¯¯j,t} (a literal and tt).
Hence, the following assignment satisfies ϕϕ: assign true to each variable xixi such that Pi={xi,t}Pi={xi,t}, assign false to each variable xixi such that Pi={¯xi,t}Pi={x¯¯¯i,t}, and assign the remaining variables arbitrarily.
Stage 2. The instance (F,U,k=3)(F,U,k=3) produced above does not satisfy the constraint ei∈Fi stated in the problem description. Fix that shortcoming as follows. Order the sets Fi and elements ei in U so that each singleton set corresponds to its element ei. Let m be the number of clauses in ϕ, so |F|=1+4n+m and |U|=1+3n.
Let (F′,U′,k′=4) denote the instance obtained as follows. Let A be a set of 2n+2m new artificial elements, two for each non-singleton set in F. Let U′=U∪A. Let F′ contain the singleton sets from F, plus, for each non-singleton set Fi in F, two sets Fi∪{ai,a′i} and {ai,a′i}, where ai and a′i are two elements in A chosen uniquely for Fi. Now |F′|=|U′|=1+5n+2m and (with the proper ordering of F′ and U′) the constraint e′i∈F′i is met for each set F′i.
To finish, note that (F′,U′,k′=4) has a solution of cost |A|+5n+1 iff the original instance (F,U,k=3) has a solution of cost 5n+1.
(if) Given any solution C of cost 5n+1 for (F,U,k=3), adding the n+m sets {ai,a′i} (one for each non-singleton Fi, so these partition A) to C gives a solution to (F′,U′,k′=4) of cost |A|+cost(C)=|A|+5n+1.
(only if) Consider any solution C′ for (F′,U′,k=4) of cost |A|+5n+1. Consider any pair of non-singleton sets Fi∪{ai,a′i} and {ai,a′i} in F′. Each is the disjoint union of at most 4 sets in C′. By a local-exchange argument, one of these sets is {ai,a′i} and the rest don't contain ai or a′i --- otherwise this property can be achieved by a local modification to the sets, without increasing the cost... (lack of detail here is why I'm calling this a proof sketch). So removing the {ai,a′i} sets from C′ gives a solution C for (F,U,k=3) of cost 5n+1. ⋄