Is the empty problem (or its complement) Karp reducible to any problem in NP?












3












$begingroup$


I'm currently following a course on Complexity Theory, and whilst studying, I came across a rather counterintuitive statement:



If $textbf{P}=textbf{NP}$, the following holds:



For every $A in textbf{NP}$, there is a $B in textbf{NP}$ such that $A leq B$ (where $leq$ means Karp reducible).



However, I do not understand how this applies to the empty problem $emptyset$, and it's complement $Sigma^*$, as these only have no-instances and yes-instances, respectively.



Are there other problems in NP such that these two are reducible to them?






complexity-theory reductions







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  • 1




    $begingroup$
    You don't even need to assume $P=NP$ for this. Just take $A=B$.
    $endgroup$
    – Tom van der Zanden
    16 hours ago










  • $begingroup$
    Hey Tom, I think that the statement from the course meant that $textbf{A} neq textbf{B}$. Otherwise, it is indeed an irrelevant requirement.
    $endgroup$
    – R. dV
    16 hours ago








  • 1




    $begingroup$
    @R.dV It's irrelevant even if you assume $Aneq B$.
    $endgroup$
    – David Richerby
    14 hours ago










  • $begingroup$
    @R.dV Adding to David Richerby's comment: $B neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A setminus { w }$, where $w in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A cup { w }$, where $w notin A$ and $A neq Sigma^ast setminus { w }$).
    $endgroup$
    – dkaeae
    12 hours ago


















3












$begingroup$


I'm currently following a course on Complexity Theory, and whilst studying, I came across a rather counterintuitive statement:



If $textbf{P}=textbf{NP}$, the following holds:



For every $A in textbf{NP}$, there is a $B in textbf{NP}$ such that $A leq B$ (where $leq$ means Karp reducible).



However, I do not understand how this applies to the empty problem $emptyset$, and it's complement $Sigma^*$, as these only have no-instances and yes-instances, respectively.



Are there other problems in NP such that these two are reducible to them?






complexity-theory reductions







share|cite|improve this question









New contributor




R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 1




    $begingroup$
    You don't even need to assume $P=NP$ for this. Just take $A=B$.
    $endgroup$
    – Tom van der Zanden
    16 hours ago










  • $begingroup$
    Hey Tom, I think that the statement from the course meant that $textbf{A} neq textbf{B}$. Otherwise, it is indeed an irrelevant requirement.
    $endgroup$
    – R. dV
    16 hours ago








  • 1




    $begingroup$
    @R.dV It's irrelevant even if you assume $Aneq B$.
    $endgroup$
    – David Richerby
    14 hours ago










  • $begingroup$
    @R.dV Adding to David Richerby's comment: $B neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A setminus { w }$, where $w in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A cup { w }$, where $w notin A$ and $A neq Sigma^ast setminus { w }$).
    $endgroup$
    – dkaeae
    12 hours ago
















3












3








3





$begingroup$


I'm currently following a course on Complexity Theory, and whilst studying, I came across a rather counterintuitive statement:



If $textbf{P}=textbf{NP}$, the following holds:



For every $A in textbf{NP}$, there is a $B in textbf{NP}$ such that $A leq B$ (where $leq$ means Karp reducible).



However, I do not understand how this applies to the empty problem $emptyset$, and it's complement $Sigma^*$, as these only have no-instances and yes-instances, respectively.



Are there other problems in NP such that these two are reducible to them?






complexity-theory reductions







share|cite|improve this question









New contributor




R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I'm currently following a course on Complexity Theory, and whilst studying, I came across a rather counterintuitive statement:



If $textbf{P}=textbf{NP}$, the following holds:



For every $A in textbf{NP}$, there is a $B in textbf{NP}$ such that $A leq B$ (where $leq$ means Karp reducible).



However, I do not understand how this applies to the empty problem $emptyset$, and it's complement $Sigma^*$, as these only have no-instances and yes-instances, respectively.



Are there other problems in NP such that these two are reducible to them?






complexity-theory reductions




complexity-theory reductions






share|cite|improve this question









New contributor




R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 16 hours ago









dkaeae

2,3421922




2,3421922






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R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 16 hours ago









R. dVR. dV

184




184




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R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





R. dV is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.








  • 1




    $begingroup$
    You don't even need to assume $P=NP$ for this. Just take $A=B$.
    $endgroup$
    – Tom van der Zanden
    16 hours ago










  • $begingroup$
    Hey Tom, I think that the statement from the course meant that $textbf{A} neq textbf{B}$. Otherwise, it is indeed an irrelevant requirement.
    $endgroup$
    – R. dV
    16 hours ago








  • 1




    $begingroup$
    @R.dV It's irrelevant even if you assume $Aneq B$.
    $endgroup$
    – David Richerby
    14 hours ago










  • $begingroup$
    @R.dV Adding to David Richerby's comment: $B neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A setminus { w }$, where $w in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A cup { w }$, where $w notin A$ and $A neq Sigma^ast setminus { w }$).
    $endgroup$
    – dkaeae
    12 hours ago
















  • 1




    $begingroup$
    You don't even need to assume $P=NP$ for this. Just take $A=B$.
    $endgroup$
    – Tom van der Zanden
    16 hours ago










  • $begingroup$
    Hey Tom, I think that the statement from the course meant that $textbf{A} neq textbf{B}$. Otherwise, it is indeed an irrelevant requirement.
    $endgroup$
    – R. dV
    16 hours ago








  • 1




    $begingroup$
    @R.dV It's irrelevant even if you assume $Aneq B$.
    $endgroup$
    – David Richerby
    14 hours ago










  • $begingroup$
    @R.dV Adding to David Richerby's comment: $B neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A setminus { w }$, where $w in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A cup { w }$, where $w notin A$ and $A neq Sigma^ast setminus { w }$).
    $endgroup$
    – dkaeae
    12 hours ago










1




1




$begingroup$
You don't even need to assume $P=NP$ for this. Just take $A=B$.
$endgroup$
– Tom van der Zanden
16 hours ago




$begingroup$
You don't even need to assume $P=NP$ for this. Just take $A=B$.
$endgroup$
– Tom van der Zanden
16 hours ago












$begingroup$
Hey Tom, I think that the statement from the course meant that $textbf{A} neq textbf{B}$. Otherwise, it is indeed an irrelevant requirement.
$endgroup$
– R. dV
16 hours ago






$begingroup$
Hey Tom, I think that the statement from the course meant that $textbf{A} neq textbf{B}$. Otherwise, it is indeed an irrelevant requirement.
$endgroup$
– R. dV
16 hours ago






1




1




$begingroup$
@R.dV It's irrelevant even if you assume $Aneq B$.
$endgroup$
– David Richerby
14 hours ago




$begingroup$
@R.dV It's irrelevant even if you assume $Aneq B$.
$endgroup$
– David Richerby
14 hours ago












$begingroup$
@R.dV Adding to David Richerby's comment: $B neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A setminus { w }$, where $w in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A cup { w }$, where $w notin A$ and $A neq Sigma^ast setminus { w }$).
$endgroup$
– dkaeae
12 hours ago






$begingroup$
@R.dV Adding to David Richerby's comment: $B neq A$ is irrelevant since you could just take an arbitrary $A$ and set $B$ to $A$ minus a finite set (e.g., $A setminus { w }$, where $w in A$ and $|A| > 1$) or $A$ plus a finite set not in $A$ (e.g., $A cup { w }$, where $w notin A$ and $A neq Sigma^ast setminus { w }$).
$endgroup$
– dkaeae
12 hours ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

Of course there is.



Just take any non-trivial language $L$ (i.e., $L neq varnothing$ and $L neq Sigma^ast$). Then there are concrete words $x in L$ and $y notin L$.



To reduce $varnothing$ to $L$, simply map everything to $y$. Then the input is in $varnothing$ (which is false) if and only if $y in L$ (which is also false). Hence, the reduction is correct.



For $Sigma^ast$, do the same but use $x$ instead.




As a note: I assume you are puzzled about $A$ being reduced to $B$. Obviously, in the statement you cite $B$ should at the very least be a non-trivial set (and it seems $textbf{P} = textbf{NP}$ is redundant, as Tom van der Zanden notes in the comments; in fact, the statement is rather fishy, see David Richerby's answer); note you cannot reduce non-trivial sets to $varnothing$ or $Sigma^ast$ (and you cannot reduce either to one another, as David Richerby points out in the comments).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
    $endgroup$
    – R. dV
    16 hours ago












  • $begingroup$
    You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
    $endgroup$
    – dkaeae
    15 hours ago












  • $begingroup$
    Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
    $endgroup$
    – R. dV
    14 hours ago










  • $begingroup$
    Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
    $endgroup$
    – David Richerby
    14 hours ago



















1












$begingroup$

The statement is basically vacuous. Every language is reducible to itself (the reduction is the identity function), so you can just take $B=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
    $endgroup$
    – R. dV
    13 hours ago










  • $begingroup$
    @R.dV OK but that's your assumption and it's not included in the question you say you came across.
    $endgroup$
    – David Richerby
    13 hours ago










  • $begingroup$
    Or $A$ union a finite set if $A$ is a singleton.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
    $endgroup$
    – R. dV
    10 hours ago












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2 Answers
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active

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votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

Of course there is.



Just take any non-trivial language $L$ (i.e., $L neq varnothing$ and $L neq Sigma^ast$). Then there are concrete words $x in L$ and $y notin L$.



To reduce $varnothing$ to $L$, simply map everything to $y$. Then the input is in $varnothing$ (which is false) if and only if $y in L$ (which is also false). Hence, the reduction is correct.



For $Sigma^ast$, do the same but use $x$ instead.




As a note: I assume you are puzzled about $A$ being reduced to $B$. Obviously, in the statement you cite $B$ should at the very least be a non-trivial set (and it seems $textbf{P} = textbf{NP}$ is redundant, as Tom van der Zanden notes in the comments; in fact, the statement is rather fishy, see David Richerby's answer); note you cannot reduce non-trivial sets to $varnothing$ or $Sigma^ast$ (and you cannot reduce either to one another, as David Richerby points out in the comments).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
    $endgroup$
    – R. dV
    16 hours ago












  • $begingroup$
    You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
    $endgroup$
    – dkaeae
    15 hours ago












  • $begingroup$
    Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
    $endgroup$
    – R. dV
    14 hours ago










  • $begingroup$
    Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
    $endgroup$
    – David Richerby
    14 hours ago
















3












$begingroup$

Of course there is.



Just take any non-trivial language $L$ (i.e., $L neq varnothing$ and $L neq Sigma^ast$). Then there are concrete words $x in L$ and $y notin L$.



To reduce $varnothing$ to $L$, simply map everything to $y$. Then the input is in $varnothing$ (which is false) if and only if $y in L$ (which is also false). Hence, the reduction is correct.



For $Sigma^ast$, do the same but use $x$ instead.




As a note: I assume you are puzzled about $A$ being reduced to $B$. Obviously, in the statement you cite $B$ should at the very least be a non-trivial set (and it seems $textbf{P} = textbf{NP}$ is redundant, as Tom van der Zanden notes in the comments; in fact, the statement is rather fishy, see David Richerby's answer); note you cannot reduce non-trivial sets to $varnothing$ or $Sigma^ast$ (and you cannot reduce either to one another, as David Richerby points out in the comments).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
    $endgroup$
    – R. dV
    16 hours ago












  • $begingroup$
    You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
    $endgroup$
    – dkaeae
    15 hours ago












  • $begingroup$
    Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
    $endgroup$
    – R. dV
    14 hours ago










  • $begingroup$
    Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
    $endgroup$
    – David Richerby
    14 hours ago














3












3








3





$begingroup$

Of course there is.



Just take any non-trivial language $L$ (i.e., $L neq varnothing$ and $L neq Sigma^ast$). Then there are concrete words $x in L$ and $y notin L$.



To reduce $varnothing$ to $L$, simply map everything to $y$. Then the input is in $varnothing$ (which is false) if and only if $y in L$ (which is also false). Hence, the reduction is correct.



For $Sigma^ast$, do the same but use $x$ instead.




As a note: I assume you are puzzled about $A$ being reduced to $B$. Obviously, in the statement you cite $B$ should at the very least be a non-trivial set (and it seems $textbf{P} = textbf{NP}$ is redundant, as Tom van der Zanden notes in the comments; in fact, the statement is rather fishy, see David Richerby's answer); note you cannot reduce non-trivial sets to $varnothing$ or $Sigma^ast$ (and you cannot reduce either to one another, as David Richerby points out in the comments).






share|cite|improve this answer











$endgroup$



Of course there is.



Just take any non-trivial language $L$ (i.e., $L neq varnothing$ and $L neq Sigma^ast$). Then there are concrete words $x in L$ and $y notin L$.



To reduce $varnothing$ to $L$, simply map everything to $y$. Then the input is in $varnothing$ (which is false) if and only if $y in L$ (which is also false). Hence, the reduction is correct.



For $Sigma^ast$, do the same but use $x$ instead.




As a note: I assume you are puzzled about $A$ being reduced to $B$. Obviously, in the statement you cite $B$ should at the very least be a non-trivial set (and it seems $textbf{P} = textbf{NP}$ is redundant, as Tom van der Zanden notes in the comments; in fact, the statement is rather fishy, see David Richerby's answer); note you cannot reduce non-trivial sets to $varnothing$ or $Sigma^ast$ (and you cannot reduce either to one another, as David Richerby points out in the comments).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 12 hours ago

























answered 16 hours ago









dkaeaedkaeae

2,3421922




2,3421922












  • $begingroup$
    Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
    $endgroup$
    – R. dV
    16 hours ago












  • $begingroup$
    You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
    $endgroup$
    – dkaeae
    15 hours ago












  • $begingroup$
    Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
    $endgroup$
    – R. dV
    14 hours ago










  • $begingroup$
    Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
    $endgroup$
    – David Richerby
    14 hours ago


















  • $begingroup$
    Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
    $endgroup$
    – R. dV
    16 hours ago












  • $begingroup$
    You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
    $endgroup$
    – dkaeae
    15 hours ago












  • $begingroup$
    Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
    $endgroup$
    – R. dV
    14 hours ago










  • $begingroup$
    Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
    $endgroup$
    – David Richerby
    14 hours ago
















$begingroup$
Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
$endgroup$
– R. dV
16 hours ago






$begingroup$
Hey @dkaeae, thanks for responding. I thought about something like this, but as far as I knew, this wasn't correct, since we're not mapping the yes-instances from L (so, x $in$ L) to anything in $emptyset$, and vice versa. I don't see how we're mapping yes instances from f(I) to yes-instances of I, as there are no yes-instances of I. I learned that for any karp reduction A $leq$ to be correct, you need two things: For I $in$ as a yes-instance, f(i) $in$ B should be a yes-instance, and vice versa. How could we then map f(i) yes instances to yes instances of I, if I has none?
$endgroup$
– R. dV
16 hours ago














$begingroup$
You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
$endgroup$
– dkaeae
15 hours ago






$begingroup$
You seem to have it the other way around. If you are reducing $varnothing$ to $L$, then you should be mapping yes-instances of $varnothing$ to yes-instances of $L$ and no-instances to no-instances of $L$. There are no yes-instances of $varnothing$; hence, everything is a no-instance and we can afford mapping everything to the same no-instance $y notin L$.
$endgroup$
– dkaeae
15 hours ago














$begingroup$
Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
$endgroup$
– R. dV
14 hours ago




$begingroup$
Yes, that's true, but we were shown that the contraposition of mapping no instances of I -> f(I) is to map yes-instances of f(I) -> I. This was ofcourse done with problems that have yes and no instances. But given your answers, I see now how it works for non-trivial languages. Thanks again :)
$endgroup$
– R. dV
14 hours ago












$begingroup$
Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
$endgroup$
– David Richerby
14 hours ago




$begingroup$
Re your note at the end, you can't reduce between $emptyset$ and $Sigma^*$, either.
$endgroup$
– David Richerby
14 hours ago











1












$begingroup$

The statement is basically vacuous. Every language is reducible to itself (the reduction is the identity function), so you can just take $B=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
    $endgroup$
    – R. dV
    13 hours ago










  • $begingroup$
    @R.dV OK but that's your assumption and it's not included in the question you say you came across.
    $endgroup$
    – David Richerby
    13 hours ago










  • $begingroup$
    Or $A$ union a finite set if $A$ is a singleton.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
    $endgroup$
    – R. dV
    10 hours ago
















1












$begingroup$

The statement is basically vacuous. Every language is reducible to itself (the reduction is the identity function), so you can just take $B=A$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
    $endgroup$
    – R. dV
    13 hours ago










  • $begingroup$
    @R.dV OK but that's your assumption and it's not included in the question you say you came across.
    $endgroup$
    – David Richerby
    13 hours ago










  • $begingroup$
    Or $A$ union a finite set if $A$ is a singleton.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
    $endgroup$
    – R. dV
    10 hours ago














1












1








1





$begingroup$

The statement is basically vacuous. Every language is reducible to itself (the reduction is the identity function), so you can just take $B=A$.






share|cite|improve this answer









$endgroup$



The statement is basically vacuous. Every language is reducible to itself (the reduction is the identity function), so you can just take $B=A$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 14 hours ago









David RicherbyDavid Richerby

70k15106196




70k15106196












  • $begingroup$
    Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
    $endgroup$
    – R. dV
    13 hours ago










  • $begingroup$
    @R.dV OK but that's your assumption and it's not included in the question you say you came across.
    $endgroup$
    – David Richerby
    13 hours ago










  • $begingroup$
    Or $A$ union a finite set if $A$ is a singleton.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
    $endgroup$
    – R. dV
    10 hours ago


















  • $begingroup$
    Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
    $endgroup$
    – R. dV
    13 hours ago










  • $begingroup$
    @R.dV OK but that's your assumption and it's not included in the question you say you came across.
    $endgroup$
    – David Richerby
    13 hours ago










  • $begingroup$
    Or $A$ union a finite set if $A$ is a singleton.
    $endgroup$
    – David Richerby
    12 hours ago










  • $begingroup$
    Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
    $endgroup$
    – R. dV
    10 hours ago
















$begingroup$
Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
$endgroup$
– R. dV
13 hours ago




$begingroup$
Hi David, Tom already commented this on the question, as I stated there, the assumption I made for this questions is that $textbf{B}neqtextbf{A}$
$endgroup$
– R. dV
13 hours ago












$begingroup$
@R.dV OK but that's your assumption and it's not included in the question you say you came across.
$endgroup$
– David Richerby
13 hours ago




$begingroup$
@R.dV OK but that's your assumption and it's not included in the question you say you came across.
$endgroup$
– David Richerby
13 hours ago












$begingroup$
Or $A$ union a finite set if $A$ is a singleton.
$endgroup$
– David Richerby
12 hours ago




$begingroup$
Or $A$ union a finite set if $A$ is a singleton.
$endgroup$
– David Richerby
12 hours ago












$begingroup$
Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
$endgroup$
– R. dV
10 hours ago




$begingroup$
Yeah that's correct David, I did not get any further context from the statement; but given the difficulty they usually propose, such a trivial answer wouldn't make a lot of sense. Anyway, your last comment does make sense, so thanks :)
$endgroup$
– R. dV
10 hours ago










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