Undecidability of: $|w in L| geq 1, L={w in {0,1}^*|a_0·#_0(w)+a_1·#_1(w)- a_1a_0=0}$
$begingroup$
Let $a_0, a_1 in mathbb{N} setminus {0}$ and $L={w in {0,1}^*|a_0·#_0(w)+a_1·#_1(w)- a_1a_0=0}$ . Let's assume problem $P$ that, language of Turing machine accepts at least one word from language $L$. Using membership problem, is $P$ undecidable?
Membership problem of $w in L land L in unrestricted$, is undecidable but is semi-decidable. Prove of semi-decitability is quite easy by simulating of Turing machine $T$, where $L(T)=L$. But how can we show that $P$ is undecidable?
turing-machines decidability
asked Nov 30 '18 at 10:11
nocturnenocturne
687
$endgroup$
add a comment |
$begingroup$
Let $a_0, a_1 in mathbb{N} setminus {0}$ and $L={w in {0,1}^*|a_0·#_0(w)+a_1·#_1(w)- a_1a_0=0}$ . Let's assume problem $P$ that, language of Turing machine accepts at least one word from language $L$. Using membership problem, is $P$ undecidable?
Membership problem of $w in L land L in unrestricted$, is undecidable but is semi-decidable. Prove of semi-decitability is quite easy by simulating of Turing machine $T$, where $L(T)=L$. But how can we show that $P$ is undecidable?
turing-machines decidability
asked Nov 30 '18 at 10:11
nocturnenocturne
687
$endgroup$
$begingroup$
You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L land L$.
$endgroup$
– platty
Nov 30 '18 at 10:12
$begingroup$
I edited the question, is it better?
$endgroup$
– nocturne
Nov 30 '18 at 10:16
$begingroup$
I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable?
$endgroup$
– platty
Nov 30 '18 at 10:16
$begingroup$
Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w in L$ and rejects $w in sum^* setminus L$
$endgroup$
– nocturne
Nov 30 '18 at 10:22
add a comment |
$begingroup$
Let $a_0, a_1 in mathbb{N} setminus {0}$ and $L={w in {0,1}^*|a_0·#_0(w)+a_1·#_1(w)- a_1a_0=0}$ . Let's assume problem $P$ that, language of Turing machine accepts at least one word from language $L$. Using membership problem, is $P$ undecidable?
Membership problem of $w in L land L in unrestricted$, is undecidable but is semi-decidable. Prove of semi-decitability is quite easy by simulating of Turing machine $T$, where $L(T)=L$. But how can we show that $P$ is undecidable?
turing-machines decidability
asked Nov 30 '18 at 10:11
nocturnenocturne
687
$endgroup$
Let $a_0, a_1 in mathbb{N} setminus {0}$ and $L={w in {0,1}^*|a_0·#_0(w)+a_1·#_1(w)- a_1a_0=0}$ . Let's assume problem $P$ that, language of Turing machine accepts at least one word from language $L$. Using membership problem, is $P$ undecidable?
Membership problem of $w in L land L in unrestricted$, is undecidable but is semi-decidable. Prove of semi-decitability is quite easy by simulating of Turing machine $T$, where $L(T)=L$. But how can we show that $P$ is undecidable?
turing-machines decidability
turing-machines decidability
asked Nov 30 '18 at 10:11
nocturnenocturne
687
asked Nov 30 '18 at 10:11
nocturnenocturne
687
edited Nov 30 '18 at 10:14
nocturne
asked Nov 30 '18 at 10:11
nocturnenocturne
687
asked Nov 30 '18 at 10:11
nocturnenocturne
687
asked Nov 30 '18 at 10:11
nocturnenocturne
687
687
$begingroup$
You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L land L$.
$endgroup$
– platty
Nov 30 '18 at 10:12
$begingroup$
I edited the question, is it better?
$endgroup$
– nocturne
Nov 30 '18 at 10:16
$begingroup$
I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable?
$endgroup$
– platty
Nov 30 '18 at 10:16
$begingroup$
Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w in L$ and rejects $w in sum^* setminus L$
$endgroup$
– nocturne
Nov 30 '18 at 10:22
add a comment |
$begingroup$
You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L land L$.
$endgroup$
– platty
Nov 30 '18 at 10:12
$begingroup$
I edited the question, is it better?
$endgroup$
– nocturne
Nov 30 '18 at 10:16
$begingroup$
I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable?
$endgroup$
– platty
Nov 30 '18 at 10:16
$begingroup$
Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w in L$ and rejects $w in sum^* setminus L$
$endgroup$
– nocturne
Nov 30 '18 at 10:22
$begingroup$
You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L land L$.
$endgroup$
– platty
Nov 30 '18 at 10:12
$begingroup$
You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L land L$.
$endgroup$
– platty
Nov 30 '18 at 10:12
$begingroup$
I edited the question, is it better?
$endgroup$
– nocturne
Nov 30 '18 at 10:16
$begingroup$
I edited the question, is it better?
$endgroup$
– nocturne
Nov 30 '18 at 10:16
$begingroup$
I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable?
$endgroup$
– platty
Nov 30 '18 at 10:16
$begingroup$
I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable?
$endgroup$
– platty
Nov 30 '18 at 10:16
$begingroup$
Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w in L$ and rejects $w in sum^* setminus L$
$endgroup$
– nocturne
Nov 30 '18 at 10:22
$begingroup$
Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w in L$ and rejects $w in sum^* setminus L$
$endgroup$
– nocturne
Nov 30 '18 at 10:22
add a comment |
1 Answer
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To show that a language over Turing Machine encodings is undecidable, we can use Rice's Theorem, which states that any non-trivial semantic property of Turing Machines is undecidable. Here, our property is "Turing Machine $M$ accepts at least one word from language $L$." We first note that, for every $a_0,a_1 in mathbb{N} setminus {0}$, $L$ is nonempty and does not contain every string. For the former, $1^{a_0} in L$ for any $a_0,a_1$; for the latter, observe that $varepsilon notin L$ for any such $L$.
From here, we just need to show that this property is nontrivial. This requires showing that there is a Turing machine that accepts a language satisfying this property, and one that accepts a language which does not satisfy this property. Turing machines accepting the languages $Sigma^*$ and $emptyset$ suffice, respectively.
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
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$begingroup$
To show that a language over Turing Machine encodings is undecidable, we can use Rice's Theorem, which states that any non-trivial semantic property of Turing Machines is undecidable. Here, our property is "Turing Machine $M$ accepts at least one word from language $L$." We first note that, for every $a_0,a_1 in mathbb{N} setminus {0}$, $L$ is nonempty and does not contain every string. For the former, $1^{a_0} in L$ for any $a_0,a_1$; for the latter, observe that $varepsilon notin L$ for any such $L$.
From here, we just need to show that this property is nontrivial. This requires showing that there is a Turing machine that accepts a language satisfying this property, and one that accepts a language which does not satisfy this property. Turing machines accepting the languages $Sigma^*$ and $emptyset$ suffice, respectively.
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
To show that a language over Turing Machine encodings is undecidable, we can use Rice's Theorem, which states that any non-trivial semantic property of Turing Machines is undecidable. Here, our property is "Turing Machine $M$ accepts at least one word from language $L$." We first note that, for every $a_0,a_1 in mathbb{N} setminus {0}$, $L$ is nonempty and does not contain every string. For the former, $1^{a_0} in L$ for any $a_0,a_1$; for the latter, observe that $varepsilon notin L$ for any such $L$.
From here, we just need to show that this property is nontrivial. This requires showing that there is a Turing machine that accepts a language satisfying this property, and one that accepts a language which does not satisfy this property. Turing machines accepting the languages $Sigma^*$ and $emptyset$ suffice, respectively.
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
$endgroup$
add a comment |
$begingroup$
To show that a language over Turing Machine encodings is undecidable, we can use Rice's Theorem, which states that any non-trivial semantic property of Turing Machines is undecidable. Here, our property is "Turing Machine $M$ accepts at least one word from language $L$." We first note that, for every $a_0,a_1 in mathbb{N} setminus {0}$, $L$ is nonempty and does not contain every string. For the former, $1^{a_0} in L$ for any $a_0,a_1$; for the latter, observe that $varepsilon notin L$ for any such $L$.
From here, we just need to show that this property is nontrivial. This requires showing that there is a Turing machine that accepts a language satisfying this property, and one that accepts a language which does not satisfy this property. Turing machines accepting the languages $Sigma^*$ and $emptyset$ suffice, respectively.
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
$endgroup$
To show that a language over Turing Machine encodings is undecidable, we can use Rice's Theorem, which states that any non-trivial semantic property of Turing Machines is undecidable. Here, our property is "Turing Machine $M$ accepts at least one word from language $L$." We first note that, for every $a_0,a_1 in mathbb{N} setminus {0}$, $L$ is nonempty and does not contain every string. For the former, $1^{a_0} in L$ for any $a_0,a_1$; for the latter, observe that $varepsilon notin L$ for any such $L$.
From here, we just need to show that this property is nontrivial. This requires showing that there is a Turing machine that accepts a language satisfying this property, and one that accepts a language which does not satisfy this property. Turing machines accepting the languages $Sigma^*$ and $emptyset$ suffice, respectively.
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
answered Nov 30 '18 at 10:30
plattyplatty
3,370320
3,370320
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$begingroup$
You don't include a condition in the set-builder notation. What is this supposed to say? It's also unclear what you mean by $L land L$.
$endgroup$
– platty
Nov 30 '18 at 10:12
$begingroup$
I edited the question, is it better?
$endgroup$
– nocturne
Nov 30 '18 at 10:16
$begingroup$
I think so. Is $P$ supposed to be a language of Turing Machine encodings, and the question is whether $P$ is undecidable?
$endgroup$
– platty
Nov 30 '18 at 10:16
$begingroup$
Question is whether the P is undecidable. Problem P is undecidable when is not decidable which means we can construct a Turing machine $T$, where it decides whether it accepts $w in L$ and rejects $w in sum^* setminus L$
$endgroup$
– nocturne
Nov 30 '18 at 10:22