High School

Prove that the class BPP is closed under complement, union, and intersection. That is:

- If [tex]L \in \text{BPP}[/tex], then its complement [tex]\Sigma^* - L \in \text{BPP}[/tex].
- If [tex]L_1, L_2 \in \text{BPP}[/tex], then the intersection [tex]L_1 \cap L_2 \in \text{BPP}[/tex].
- If [tex]L_1, L_2 \in \text{BPP}[/tex], then the union [tex]L_1 \cup L_2 \in \text{BPP}[/tex].

Answer :

Final answer:

The class BPP is closed under complement, union, and intersection. To prove this, we show that if a language L is in BPP, then its complement is also in BPP, the intersection of two languages in BPP is also in BPP, and the union of two languages in BPP is also in BPP.

Explanation:

Proof that BPP is closed under complement:

  1. Assume L is in BPP and let M be a probabilistic polynomial-time Turing machine that decides L with bounded error.
  2. We construct a new Turing machine M' that decides the complement of L, Σ* - L.
  3. M' runs M on the input x and accepts if M rejects x, and rejects if M accepts x.
  4. Since M has bounded error, the probability of M' accepting or rejecting is 1. Therefore, M' decides the complement of L, and thus Σ* - L is in BPP.

Proof that BPP is closed under intersection:

  1. Assume L1 and L2 are in BPP and let M1 and M2 be probabilistic polynomial-time Turing machines that decide L1 and L2 with bounded error.
  2. We construct a new Turing machine M' that decides L1 ∩ L2.
  3. M' runs M1 and M2 separately on the input x, and accepts if both M1 and M2 accept x.
  4. Since M1 and M2 have bounded errors, the probability of M' accepting is at least the product of the probabilities of M1 and M2 accepting. Therefore, M' decides L1 ∩ L2, and thus it is in BPP.

Proof that BPP is closed under union:

  1. Assume L1 and L2 are in BPP and let M1 and M2 be probabilistic polynomial-time Turing machines that decide L1 and L2 with bounded error.
  2. We construct a new Turing machine M' that decides L1 ∪ L2.
  3. M' runs M1 and M2 separately on the input x, and accepts if either M1 or M2 accepts x.
  4. Since M1 and M2 have bounded errors, the probability of M' accepting is at least the sum of the probabilities of M1 and M2 accepting. Therefore, M' decides L1 ∪ L2, and thus it is in BPP.

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