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$L_1 = \{< M, w, 1 > |$ $M$ does not halts on $w\}$ | $L_1 = \{< M, w, 1 > |$ $M$ does not halts on $w\}$ | ||
| − | Here $< M, w, i >$ is a triplet, whose first component | + | Here $< M, w, i >$ is a triplet, whose first component $M$ is an encoding of a Turing |
| − | Machine, second component | + | Machine, second component $ w$ is a string, and third component $i$ is a bit. |
| + | |||
Let $L = L_0 ∪ L_1$. Which of the following is true ? | Let $L = L_0 ∪ L_1$. Which of the following is true ? | ||
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'''(D) Neither $L$ nor $L'$ is recursively enumerable ''' | '''(D) Neither $L$ nor $L'$ is recursively enumerable ''' | ||
| − | === | + | ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}=== |
| − | Both $L$ and $ | + | Both $L$ and $Lʼ$ are undecidable and not even semi-decidable (not recursively-enumerable). Because halting problem can be solved with both $L$ and $Lʼ$. |
| − | Halting problem can be stated as follows: A machine $M$ and a word $w$ are given. You have to tell, if $M$ halts on $w$. | + | Halting problem can be stated as follows: A machine $M$ and a word $w$ are given. You have to tell, if $M$ halts on $w$. |
| − | + | ||
| − | So, to solve halting problem $<M,w>$ using $L$, just give $<M,w,0>$ and $<M,w,1>$ to two instances of $T$ which is the Turing machine for $L$. If $T$ accepts the triplet $<M,w,0>$, it means $M$ halts on $w$ => we have solved halting problem. If $T$ accepts the triplet $<M,w,1>$, it means $M$ doesn't halt on $w$ => we have solved halting problem. We know that either $<M,w,0>$ or $<M,w,1>$ is in $L$. So, if $L$ is recursively enumerable, $T$ is bound to stop on at least one of these inputs | + | So, to solve halting problem $<M,w>$ using $L$, just give $<M,w,0>$ and $<M,w,1>$ to two instances of $T$ which is the Turing machine for $L$. If $T$ accepts the triplet $<M,w,0>$, it means $M$ halts on $w$ => we have solved halting problem. If $T$ accepts the triplet $<M,w,1>$, it means $M$ doesn't halt on $w$ => we have solved halting problem. We know that either $<M,w,0>$ or $<M,w,1>$ is in $L$. So, if $L$ is recursively enumerable, $T$ is bound to stop on at least one of these inputs ($TM$ for a recursively enumerable language stops and accepts, when provided with a word in its language). |
| − | Similarly, we can also show that halting problem can be solved with $ | + | Hence, using $L$ we can solve halting problem => $L$ is not recursively enumerable. |
| − | + | Similarly, we can also show that halting problem can be solved with $Lʼ$. | |
| − | |||
| + | Hence, neither $L$ nor $Lʼ$ is recursively enumerable. | ||
| + | <!-- | ||
===Alternate Solution=== | ===Alternate Solution=== | ||
| − | $L_0$ is recursively enumerable. (Given $<M,w,0>$, we can just give $w$ to $M$. If $M$ halts on $w$, $<M,w,0>$ is element of $L_0$. | + | $L_0$ is recursively enumerable. (Given $<M,w,0>$, we can just give $w$ to $M$. If $M$ halts on $w$, $<M,w,0>$ is an element of $L_0$.) |
| − | $L_1$ is not recursively enumerable. Because, halting problem can be solved with it. To decide if a | + | $L_1$ is not recursively enumerable. Because, halting problem can be solved with it. To decide if a Turing machine $M$ accepts a word $w$, just give $<M,w,1>$ to the Turing machine for $L_1$ and also give $w$ to $M$. Either $M$ accepts $w$, or the Turing machine for $L_1$ accepts $<M,w,1>$. In either case we have solved halting problem. Hence, $L_1$ is not recursively enumerable. |
$L_1$ can be reduced to $L_0'$, and hence $L_0'$ also is not recursively enumerable. | $L_1$ can be reduced to $L_0'$, and hence $L_0'$ also is not recursively enumerable. | ||
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Now, $L$ = $L_0 U L_1$ | Now, $L$ = $L_0 U L_1$ | ||
| − | = | + | = <math>RE</math> U not <math>RE</math> |
| − | = not | + | = not <math>RE</math> |
| − | $ | + | $Lʼ = (L_0 \cup L_1)'$ |
$= L_0' ∩ L_1'$ | $= L_0' ∩ L_1'$ | ||
| − | $=$ not | + | $=$ not <math>RE</math> $∩$ <math>RE</math> |
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| − | < | ||
| − | |||
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| − | < | ||
| + | $=$ not <math>RE</math> | ||
| + | --> | ||
| + | {{Template:FBD}} | ||
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[[Category: GATE2003]] | [[Category: GATE2003]] | ||
| − | [[Category: Automata | + | [[Category: Automata questions from GATE]] |
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Define languages L0 and L1 as follows :
$L_0 = \{< M, w, 0 > |$ $M$ halts on $w\} $
$L_1 = \{< M, w, 1 > |$ $M$ does not halts on $w\}$
Here $< M, w, i >$ is a triplet, whose first component $M$ is an encoding of a Turing Machine, second component $ w$ is a string, and third component $i$ is a bit.
Let $L = L_0 ∪ L_1$. Which of the following is true ?
(A) $L$ is recursively enumerable, but is not
(B) $L$ is recursively enumerable, but $ L'$ is not
(C) Both $L$ and $L'$ are recursive
(D) Neither $L$ nor $L'$ is recursively enumerable
Both $L$ and $Lʼ$ are undecidable and not even semi-decidable (not recursively-enumerable). Because halting problem can be solved with both $L$ and $Lʼ$.
Halting problem can be stated as follows: A machine $M$ and a word $w$ are given. You have to tell, if $M$ halts on $w$.
So, to solve halting problem $<M,w>$ using $L$, just give $<M,w,0>$ and $<M,w,1>$ to two instances of $T$ which is the Turing machine for $L$. If $T$ accepts the triplet $<M,w,0>$, it means $M$ halts on $w$ => we have solved halting problem. If $T$ accepts the triplet $<M,w,1>$, it means $M$ doesn't halt on $w$ => we have solved halting problem. We know that either $<M,w,0>$ or $<M,w,1>$ is in $L$. So, if $L$ is recursively enumerable, $T$ is bound to stop on at least one of these inputs ($TM$ for a recursively enumerable language stops and accepts, when provided with a word in its language).
Hence, using $L$ we can solve halting problem => $L$ is not recursively enumerable. Similarly, we can also show that halting problem can be solved with $Lʼ$.
Hence, neither $L$ nor $Lʼ$ is recursively enumerable.
Define languages L0 and L1 as follows :
$L_0 = \{< M, w, 0 > |$ $M$ halts on $w\} $
$L_1 = \{< M, w, 1 > |$ $M$ does not halts on $w\}$
Here $< M, w, i >$ is a triplet, whose first component. $M$ is an encoding of a Turing Machine, second component,$ w$, is a string, and third component, $i$, is a bit. Let $L = L_0 ∪ L_1$. Which of the following is true ?
(A) $L$ is recursively enumerable, but is not
(B) $L$ is recursively enumerable, but $ L'$ is not
(C) Both $L$ and $L'$ are recursive
(D) Neither $L$ nor $L'$ is recursively enumerable
Both $L$ and $L'$ are undecidable. Because halting problem can be solved with both $L$ and $L'$.
Halting problem can be stated as follows: A machine $M$ and a word $w$ are given. You have to tell, if $M$ halts on $w$.
So, to solve halting problem $<M,w>$ using $L$, just give $<M,w,0>$ and $<M,w,1>$ to two instances of $T$ which is the Turing machine for $L$. If $T$ accepts the triplet $<M,w,0>$, it means $M$ halts on $w$ => we have solved halting problem. If $T$ accepts the triplet $<M,w,1>$, it means $M$ doesn't halt on $w$ => we have solved halting problem. We know that either $<M,w,0>$ or $<M,w,1>$ is in $L$. So, if $L$ is recursively enumerable, $T$ is bound to stop on at least one of these inputs. Hence, using $L$ we can solve halting problem => $L$ is not recursively enumerable.
Similarly, we can also show that halting problem can be solved with $L'$.
Hence, neither $L$ nor $L'$ is recursively enumerable.
$L_0$ is recursively enumerable. (Given $<M,w,0>$, we can just give $w$ to $M$. If $M$ halts on $w$, $<M,w,0>$ is element of $L_0$.
$L_1$ is not recursively enumerable. Because, halting problem can be solved with it. To decide if a turing machine $M$ accepts a word $w$, just give $<M,w,1>$ to the Turing machine for $L_1$ and also give $w$ to $M$. Either $M$ accepts $w$, or the Turing machine for $L_1$ accepts $<M,w,1>$. In either case we have solved halting problem. Hence, $L_1$ is not recursively enumerable.
$L_1$ can be reduced to $L_0'$, and hence $L_0'$ also is not recursively enumerable.
$L_1'$ can be reduced to $L_0$, and hence $L_1'$ is recursively enumerable.
Now, $L$ = $L_0 U L_1$
= re U not re
= not re
$L' = (L_0 \cup L_1)'$
$= L_0' ∩ L_1'$
$=$ not re $∩$ re
$=$ not re
GATECSE