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True or False: If \( T \) is a reduction from \( L_1 \) to \( L_2 \) where \( L_1 \) and \( L_2 \) are subsets of \( E^* \), then:

A. If \( L_1 \) is decidable, then \( L_2 \) is decidable.
B. If \( L_1 \) is undecidable, then \( L_2 \) is undecidable.
C. If \( L_2 \) is decidable, then \( L_1 \) is decidable.
D. If \( L_2 \) is undecidable, then \( L_1 \) is undecidable.

Answer :

a) If L1 is decidable then L2 is decidable: True.

b) If Lj is undecidable then L2 is undecidable: False.

c) If L2 is decidable then L1 is decidable: False.

d) If L2 is undecidable then L1 is undecidable: False.

a) If L1 is decidable then L2 is decidable:

True. A reduction from L1 to L2 means that an algorithm for L2 can be constructed using an algorithm for L1. If L1 is decidable, then there is an algorithm that can determine membership in L1. Using this algorithm and the reduction, we can construct an algorithm to determine membership in L2 as well, making L2 decidable.

b) If Lj is undecidable then L2 is undecidable:

False. The reduction alone does not provide information about the decidability of L2. It is possible for L2 to be decidable even if Lj is undecidable.

c) If L2 is decidable then L1 is decidable:

False. The reduction only guarantees the direction from L1 to L2, not the other way around. The decidability of L1 cannot be determined based solely on the decidability of L2.

d) If L2 is undecidable then L1 is undecidable:

False. The reduction does not provide information about the decidability of L1. L1 can still be undecidable even if L2 is undecidable.

Learn more about the topic of Decidability here:

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