What is the difference between Turing-reducibility and m-reducibility?
$begingroup$
The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.
We shall write $A ≤_{m} B$
Let be $A, B subset mathbb{N}$
is it true that $A leq_{m} B implies Aleq_{T}B$?
Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.
If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?
logic computability
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
$endgroup$
add a comment |
$begingroup$
The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.
We shall write $A ≤_{m} B$
Let be $A, B subset mathbb{N}$
is it true that $A leq_{m} B implies Aleq_{T}B$?
Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.
If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?
logic computability
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
$endgroup$
add a comment |
$begingroup$
The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.
We shall write $A ≤_{m} B$
Let be $A, B subset mathbb{N}$
is it true that $A leq_{m} B implies Aleq_{T}B$?
Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.
If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?
logic computability
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
$endgroup$
The set $A$ is many-one reducible (m-reducible) to the set $ B $ if there is a
total computable function $f$ such that $x ∈ A $ iff $f(x) ∈ B$ for all $x$.
We shall write $A ≤_{m} B$
Let be $A, B subset mathbb{N}$
is it true that $A leq_{m} B implies Aleq_{T}B$?
Where $Aleq_{T}B$ is that $A$ is Turing-reducible to the set B.
If it is true, why do the mathematicians use more Turing-reducibility than m-reducibility? Can you give me some counterexample of two sets $A,Bsubset mathbb{N}$ where $Aleq_{T}B$ and $neg (Aleq_{m}B)$?
logic computability
logic computability
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
edited Dec 10 '18 at 22:18
Nicolás Castro
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
asked Dec 10 '18 at 22:11
Nicolás CastroNicolás Castro
83
83
add a comment |
add a comment |
1 Answer
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$begingroup$
It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.
An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.
This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)
Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
add a comment |
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$begingroup$
It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.
An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.
This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)
Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
add a comment |
$begingroup$
It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.
An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.
This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)
Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
add a comment |
$begingroup$
It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.
An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.
This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)
Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
$endgroup$
It is true that $le_m$ implies $le_T$. The converse, however, is false, so they are truly different things.
An easy example of this is supplied by the following: any set is Turing equivalent to its complement, but in general a set need not be many-one reducible to its complement. For example, if $C$ is a non-recursive recursively enumerable set, then $C$ is not many-one reducible to its complement.
This helps motivate $le_T$ as opposed to $le_m$: intuitively, a set has exactly as much information as its complement, just "twisted around" (in a very computable way); $le_T$ reflects this but $le_m$ doesn't. (That said, there are also situations where $le_m$ is more natural.)
Incidentally, it's worth noting that there are other reducibilities strictly between $le_m$ and $le_T$, such as truth-table reducibility $le_{tt}$ (with respect to which a set is equivalent to its complement).
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
answered Dec 10 '18 at 22:18
Noah SchweberNoah Schweber
124k10150287
124k10150287
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
add a comment |
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
Do you know the relation between $leq_{c}$ and the others $leq_{T}, leq_{m}$ and $leq_{tt}$? Where $Aleq_{c}B$ iff $ Ain L[B]$.
$endgroup$
– Nicolás Castro
Dec 10 '18 at 22:34
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
$begingroup$
@NicolásCastro $le_c$ is vastly, vastly, vastly broader. Not only does $le_T$ (and hence $le_m,le_{tt},$ etc.) imply $le_c$, it's consistent with ZFC that there is only one $le_c$-degree at all! They're just not on the same playing field at all.
$endgroup$
– Noah Schweber
Dec 10 '18 at 23:23
add a comment |
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