Consider an operator $+$ defined on $P(N)$ as follows: $A + B = \{2x\mid x \in A\}\cup \{2x + 1\mid x \in B\}$

Show that both $A$ and $B$ are Turing-reducible to $A+B$

I am kind of confused about this notion of turing reducibility thing.
Does it mean that There is a turing machine which converts an input from $A$ to an element of $B$
on the tape. There is a term called oracle that also pops up. So if we have an oracle for $B$
does it mean that there is a decider for $B$? What's an intuitive workable definition of oracle?


The definition of Turing reducible, is that there exists a Turing reduction - an algorithm for $A+B$ that uses an oracle to $A$.

To understand this, you must first understand oracle machines. Those machines, are like turing machines, but can ask an "oracle" whether any $x$ is in $A$ or not, and would get an answer in $O(1)$, regardless of what $A$ is (doesn't even matter if it is not decidable!)

So, a turing reduction basically means that if you know for any $x$ whether its in $A$ or not in $A$, then you can decide whether a given $x$ is in $A+B$. Formally, this statement can be written as ${(A+B)}^A\in R$.

So basically the question asks you to show that ${(A+B)}^A,{(A+B)}^B\in R$

  • $\begingroup$ So when asked to show that A is turing reducible to A+B. Does that amount to that if given y we could determine if it is in A+B. by using the oracle for A ? So if y/2 is in A then y is in A+B certainly. But in case y is of the form 2x+1 where x is in B. In that case we don't have an oracle for B to determine this. I am kind of confused here how to complete the argument $\endgroup$ – Amit wadhwa Mar 31 at 10:54
  • $\begingroup$ If I understand correctly there is no need to run a turing machine with states and symbols here. And when you talk about oracle for A it just means there is a decider for A $\endgroup$ – Amit wadhwa Mar 31 at 10:55

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