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Let A ≤m B denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?
- a)If A ≤m B and B is recursive then A is recursive.
- b)If A ≤m B and A is undecidable then B is undecidable.
- c)If A ≤m B and B is recursively enumerable then A is recursively enumerable.
- d)If A ≤m B and B is not recursively enumerable then A is not recursively enumerable.
Correct answer is option 'D'. Can you explain this answer?
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Verified Answer
Let A ≤m B denotes that language A is mapping reducible (also kn...
- A ≤m B means language A is mapping reducible to language B.Thus, A cannot be harder than B. Since, A can be reduced to B, instead of deciding A we can now decide B. So, the first three options are correct.
- As B is not recursively enumerable, it doesn't guarantee A is not recursively enumerable.Thus, if A ≤m B and B is not recursively enumerable then A is not recursively enumerable. Therefore, answer is D is correct
Please comment below if you find anything wrong in the above post.
Most Upvoted Answer
Let A ≤m B denotes that language A is mapping reducible (also kn...
False Statement: If A m B and B is not recursively enumerable then A is not recursively enumerable.
Explanation:
To understand why this statement is false, let us first define what it means for a language to be mapping reducible (many-to-one reducible) to another language.
Mapping Reducibility:
A language A is said to be mapping reducible to another language B (denoted by A m B) if there exists a computable function f such that for every input x:
- If x is in A, then f(x) is in B.
- If x is not in A, then f(x) is not in B.
Now let us examine the options one by one to see why the given statement is false.
a) If A m B and B is recursive then A is recursive.
This statement is true. If A is mapping reducible to B and B is recursive (decidable), then A can also be decided using the same reduction function. This means that if we have an algorithm to decide B, we can use the reduction function to transform any instance of A into an instance of B and use the algorithm to decide A as well.
b) If A m B and A is undecidable then B is undecidable.
This statement is true. If A is mapping reducible to B and A is undecidable, then B must also be undecidable. This is because if we have an algorithm to decide B, we can use the reduction function to transform any instance of A into an instance of B and use the algorithm to decide A as well, which contradicts the fact that A is undecidable.
c) If A m B and B is recursively enumerable then A is recursively enumerable.
This statement is true. If A is mapping reducible to B and B is recursively enumerable, then we can construct a Turing machine that enumerates all the strings in B. We can use the reduction function to map each string in A to a string in B and output it if it is in A. This way, we can effectively enumerate all the strings in A, making A recursively enumerable.
d) If A m B and B is not recursively enumerable then A is not recursively enumerable.
This statement is false. Just because B is not recursively enumerable does not imply that A is also not recursively enumerable. The mapping reducibility does not preserve the property of recursive enumerability. A language can be mapping reducible to a non-recursively enumerable language and still be recursively enumerable.
Therefore, the false statement is option D.
Explanation:
To understand why this statement is false, let us first define what it means for a language to be mapping reducible (many-to-one reducible) to another language.
Mapping Reducibility:
A language A is said to be mapping reducible to another language B (denoted by A m B) if there exists a computable function f such that for every input x:
- If x is in A, then f(x) is in B.
- If x is not in A, then f(x) is not in B.
Now let us examine the options one by one to see why the given statement is false.
a) If A m B and B is recursive then A is recursive.
This statement is true. If A is mapping reducible to B and B is recursive (decidable), then A can also be decided using the same reduction function. This means that if we have an algorithm to decide B, we can use the reduction function to transform any instance of A into an instance of B and use the algorithm to decide A as well.
b) If A m B and A is undecidable then B is undecidable.
This statement is true. If A is mapping reducible to B and A is undecidable, then B must also be undecidable. This is because if we have an algorithm to decide B, we can use the reduction function to transform any instance of A into an instance of B and use the algorithm to decide A as well, which contradicts the fact that A is undecidable.
c) If A m B and B is recursively enumerable then A is recursively enumerable.
This statement is true. If A is mapping reducible to B and B is recursively enumerable, then we can construct a Turing machine that enumerates all the strings in B. We can use the reduction function to map each string in A to a string in B and output it if it is in A. This way, we can effectively enumerate all the strings in A, making A recursively enumerable.
d) If A m B and B is not recursively enumerable then A is not recursively enumerable.
This statement is false. Just because B is not recursively enumerable does not imply that A is also not recursively enumerable. The mapping reducibility does not preserve the property of recursive enumerability. A language can be mapping reducible to a non-recursively enumerable language and still be recursively enumerable.
Therefore, the false statement is option D.
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Let A ≤m B denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?a)If A ≤m B and B is recursive then A is recursive.b)If A ≤m B and A is undecidable then B is undecidable.c)If A ≤m B and B is recursively enumerable then A is recursively enumerable.d)If A ≤m B and B is not recursively enumerable then A is not recursively enumerable.Correct answer is option 'D'. Can you explain this answer?
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