proof that specific disjoint sets are recursively enumerable, but don't lie in a decidable set and its...
let's call a set $A subseteq mathbb{N}$ recursively enumerable if it's "partial characteristic function" $tilde{chi}_A$ is computable, whereby $tilde{chi}_A$ is defined as: $tilde{chi}_A$:= 1, if $x in A$ and $tilde{chi}_A$ is undefined otherwise.
Now for $i = 0,1$ let $P_i$ be the set of all programmes p that terminate for input p and have output i. It's clear that $P_0 cap P_1 = emptyset$.
I'd be glad if you could help me with showing that the sets $P_0$ and $P_1$ are recursively enumerable, but that there's no decidable set E with $P_0 subseteq E$, $P_1 subseteq mathbb{N}backslash E$.
Looking forward to your suggestions!
computability decidability
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
asked Dec 26 at 2:03
Studentu
988
add a comment |
let's call a set $A subseteq mathbb{N}$ recursively enumerable if it's "partial characteristic function" $tilde{chi}_A$ is computable, whereby $tilde{chi}_A$ is defined as: $tilde{chi}_A$:= 1, if $x in A$ and $tilde{chi}_A$ is undefined otherwise.
Now for $i = 0,1$ let $P_i$ be the set of all programmes p that terminate for input p and have output i. It's clear that $P_0 cap P_1 = emptyset$.
I'd be glad if you could help me with showing that the sets $P_0$ and $P_1$ are recursively enumerable, but that there's no decidable set E with $P_0 subseteq E$, $P_1 subseteq mathbb{N}backslash E$.
Looking forward to your suggestions!
computability decidability
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
asked Dec 26 at 2:03
Studentu
988
add a comment |
let's call a set $A subseteq mathbb{N}$ recursively enumerable if it's "partial characteristic function" $tilde{chi}_A$ is computable, whereby $tilde{chi}_A$ is defined as: $tilde{chi}_A$:= 1, if $x in A$ and $tilde{chi}_A$ is undefined otherwise.
Now for $i = 0,1$ let $P_i$ be the set of all programmes p that terminate for input p and have output i. It's clear that $P_0 cap P_1 = emptyset$.
I'd be glad if you could help me with showing that the sets $P_0$ and $P_1$ are recursively enumerable, but that there's no decidable set E with $P_0 subseteq E$, $P_1 subseteq mathbb{N}backslash E$.
Looking forward to your suggestions!
computability decidability
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
asked Dec 26 at 2:03
Studentu
988
let's call a set $A subseteq mathbb{N}$ recursively enumerable if it's "partial characteristic function" $tilde{chi}_A$ is computable, whereby $tilde{chi}_A$ is defined as: $tilde{chi}_A$:= 1, if $x in A$ and $tilde{chi}_A$ is undefined otherwise.
Now for $i = 0,1$ let $P_i$ be the set of all programmes p that terminate for input p and have output i. It's clear that $P_0 cap P_1 = emptyset$.
I'd be glad if you could help me with showing that the sets $P_0$ and $P_1$ are recursively enumerable, but that there's no decidable set E with $P_0 subseteq E$, $P_1 subseteq mathbb{N}backslash E$.
Looking forward to your suggestions!
computability decidability
computability decidability
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
asked Dec 26 at 2:03
Studentu
988
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
asked Dec 26 at 2:03
Studentu
988
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
edited 2 days ago
Andrés E. Caicedo
64.7k8158246
64.7k8158246
asked Dec 26 at 2:03
Studentu
988
asked Dec 26 at 2:03
Studentu
988
asked Dec 26 at 2:03
Studentu
988
988
add a comment |
add a comment |
1 Answer
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The partial characteristic function of $P_0$ is computable by the algorithm: "On input $p$, run program $p$ with input $p$; if and when it terminates with output $0$, print $1$." $P_1$ is recursively enumerable similarly.
If they were separated by a decidable set $E$ as in the question, let $p$ be a program that outputs $1$ when $pin E$ and outputs $0$ when $pnotin E$, i.e., $p$ computes the characteristic function of $E$. What would $p$ do when fed input $p$? It must eventually halt and output $0$ or $1$, since that's all $p$ ever does on any input. If it outputs $0$, then that means $pin P_0subseteq E$, so, by our choice of $p$, it should, on input $p$, compute $1$. Similarly, if $p$ on input $p$ outputs $1$, then $pin P_1$, so $pnotin E$, and $p$ on input $p$ should output $0$. So we have a contradiction in either case. Therefore, no such program $p$ can exist.
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
add a comment |
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The partial characteristic function of $P_0$ is computable by the algorithm: "On input $p$, run program $p$ with input $p$; if and when it terminates with output $0$, print $1$." $P_1$ is recursively enumerable similarly.
If they were separated by a decidable set $E$ as in the question, let $p$ be a program that outputs $1$ when $pin E$ and outputs $0$ when $pnotin E$, i.e., $p$ computes the characteristic function of $E$. What would $p$ do when fed input $p$? It must eventually halt and output $0$ or $1$, since that's all $p$ ever does on any input. If it outputs $0$, then that means $pin P_0subseteq E$, so, by our choice of $p$, it should, on input $p$, compute $1$. Similarly, if $p$ on input $p$ outputs $1$, then $pin P_1$, so $pnotin E$, and $p$ on input $p$ should output $0$. So we have a contradiction in either case. Therefore, no such program $p$ can exist.
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
add a comment |
The partial characteristic function of $P_0$ is computable by the algorithm: "On input $p$, run program $p$ with input $p$; if and when it terminates with output $0$, print $1$." $P_1$ is recursively enumerable similarly.
If they were separated by a decidable set $E$ as in the question, let $p$ be a program that outputs $1$ when $pin E$ and outputs $0$ when $pnotin E$, i.e., $p$ computes the characteristic function of $E$. What would $p$ do when fed input $p$? It must eventually halt and output $0$ or $1$, since that's all $p$ ever does on any input. If it outputs $0$, then that means $pin P_0subseteq E$, so, by our choice of $p$, it should, on input $p$, compute $1$. Similarly, if $p$ on input $p$ outputs $1$, then $pin P_1$, so $pnotin E$, and $p$ on input $p$ should output $0$. So we have a contradiction in either case. Therefore, no such program $p$ can exist.
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
add a comment |
The partial characteristic function of $P_0$ is computable by the algorithm: "On input $p$, run program $p$ with input $p$; if and when it terminates with output $0$, print $1$." $P_1$ is recursively enumerable similarly.
If they were separated by a decidable set $E$ as in the question, let $p$ be a program that outputs $1$ when $pin E$ and outputs $0$ when $pnotin E$, i.e., $p$ computes the characteristic function of $E$. What would $p$ do when fed input $p$? It must eventually halt and output $0$ or $1$, since that's all $p$ ever does on any input. If it outputs $0$, then that means $pin P_0subseteq E$, so, by our choice of $p$, it should, on input $p$, compute $1$. Similarly, if $p$ on input $p$ outputs $1$, then $pin P_1$, so $pnotin E$, and $p$ on input $p$ should output $0$. So we have a contradiction in either case. Therefore, no such program $p$ can exist.
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
The partial characteristic function of $P_0$ is computable by the algorithm: "On input $p$, run program $p$ with input $p$; if and when it terminates with output $0$, print $1$." $P_1$ is recursively enumerable similarly.
If they were separated by a decidable set $E$ as in the question, let $p$ be a program that outputs $1$ when $pin E$ and outputs $0$ when $pnotin E$, i.e., $p$ computes the characteristic function of $E$. What would $p$ do when fed input $p$? It must eventually halt and output $0$ or $1$, since that's all $p$ ever does on any input. If it outputs $0$, then that means $pin P_0subseteq E$, so, by our choice of $p$, it should, on input $p$, compute $1$. Similarly, if $p$ on input $p$ outputs $1$, then $pin P_1$, so $pnotin E$, and $p$ on input $p$ should output $0$. So we have a contradiction in either case. Therefore, no such program $p$ can exist.
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
answered Dec 26 at 2:32
Andreas Blass
49.2k351106
49.2k351106
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
add a comment |
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
Oh, I understand, it's so much easier than I thought! Thank you very much for the good explanation!
– Studentu
2 days ago
add a comment |
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