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Many-one reductions are a special case and stronger form of Turing reductions. With many-one reductions, the oracle (that is, our solution for B) can be invoked only once at the end, and the answer cannot be modified. This means that if we want to show that problem A can be reduced to problem B, we can use our solution for B only once in our solution for A, unlike in Turing reduction, where we can use our solution for B as many times as needed while solving A.
This means that many-one reductions map instances of one problem to instances of another, while Turing reductions compute the solution to one problem, assuming the other problem is easy to solve. The many-one reduction is more effective at separating problems into distinct complexity classes. However, the increased restrictions on many-one reductions make them more difficult to find.

A set B is called many-one complete, or simply m-complete, iff B is recursively enumerable and every recursively enumerable set A is m-reducible to B.

Many-one reductions are often subjected to resource restrictions, for example that the reduction function is computable in polynomial time or logarithmic space; see polynomial-time reduction and log-space reduction for details.
We say that a class C of languages (or a subset of the power set of the natural numbers) is closed under many-one reducibility if there exists no reduction from a language in C to a language outside C.
If a class is closed under many-one reducibility, then many-one reduction can be used to show that a problem is in C by reducing a problem in C to it. Many-one reductions are valuable because most well-studied complexity classes are closed under some type of many-one reducibility, including P, NP, L, NL, co-NP, PSPACE, EXP, and many others. These classes are not closed under arbitrary many-one reductions, however.

A set is many-one reducible to the halting problem if and only if it is recursively enumerable. This says that with regards to many-one reducibility, the halting problem is the most complicated of all recursively enumerable problems. Thus the halting problem is r.e. complete. Note that it is not the only r.e. complete problem.

The specialized halting problem for an individual Turing machine T (i.e., the set of inputs for which T eventually halts) is many-one complete iff T is a universal Turing machine. Emil Post showed that there exist recursively enumerable sets that are neither decidable nor m-complete, and hence that there exist nonuniversal Turing machines whose individual halting problems are nevertheless undecidable.

https://de.wikipedia.org/wiki/Turinggrad#Struktur_der_rekursiv_aufz%C3%A4hlbaren_Turinggrade

https://cs.stackexchange.com/questions/2353/is-computational-power-of-neural-networks-related-to-the-activation-function/2356#2356
https://de.wikipedia.org/wiki/Flei%C3%9Figer_Biber
https://doc.qt.io/qt-5/statemachine-api.html#