Difference between revisions of "GATE2003 q54"

(→‎{{Template:Author|Arjun Suresh|{{arjunweb}} }})
 
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Define languages L0 and L1 as follows :  
 
Define languages L0 and L1 as follows :  
  
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$L0 = \{< M, w, 0 > |$ $M$ halts on $w\} $
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$L_0 = \{< M, w, 0 > |$ $M$ halts on $w\} $
  
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$L1 = \{< M, w, 1 > |$ $M$ does not halts on$ w\}$  
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$L_1 = \{< M, w, 1 > |$ $M$ does not halts on $w\}$  
  
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Here $< M, w, i >$ is a triplet, whose first component. $M$ is an encoding of a Turing  
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Here $< M, w, i >$ is a triplet, whose first component $M$ is an encoding of a Turing  
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Machine, second component,$ w$, is a string, and third component, $i$, is a bit.  
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Machine, second component $ w$ is a string, and third component $i$ is a bit.  
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Let $L = L0 ∪ L1$. Which of the following is true ?  
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Let $L = L_0 ∪ L_1$. Which of the following is true ?  
 
   
 
   
 
(A) $L$ is recursively enumerable, but is not  
 
(A) $L$ is recursively enumerable, but is not  
  
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(B) $L$ is recursively enumerable, but$ L'$ is not  
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(B) $L$ is recursively enumerable, but $ L'$ is not  
  
 
(C) Both $L$ and $L'$ are recursive   
 
(C) Both $L$ and $L'$ are recursive   
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'''(D) Neither $L$ nor $L'$ is recursively enumerable '''
 
'''(D) Neither $L$ nor $L'$ is recursively enumerable '''
  
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===Solution===
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==={{Template:Author|Arjun Suresh|{{arjunweb}} }}===
 +
 
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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). 
  
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Both $L$ and $L'$ are undecidable. Because halting problem can be solved with both $L$ and $L'$.  
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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 $Lʼ$.  
  
−
Halting problem can be stated as follows: A machine $M$ and a word $w$ are given. You have to tell, if $M$ halts on $w$.  
+
Hence, neither $L$ nor $Lʼ$ is recursively enumerable.
 +
<!--
 +
===Alternate Solution===
  
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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.
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$L_0$ 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 $L_0$.)
  
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Similarly, we can also show that halting problem can be solved with $L'$.  
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$L_1$ 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.
  
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Hence, neither $L$ nor $L'$ is recursively enumerable.
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$L_1$ can be reduced to $L_0'$, and hence $L_0'$ also is not recursively enumerable.
  
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===Alternate Solution===
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$L_1'$ can be reduced to $L_0$, and hence $L_1'$ is recursively enumerable.
  
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$L0$ 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 $L0$.
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Now, $L$ = $L_0 U L_1$
  
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$L1$ 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 $L1$ and also give $w$ to $M$. Either $M$ accepts $w$, or the Turing machine for $L1$ accepts $<M,w,1>$. In either case we have solved halting problem. Hence, $L1$ is not recursively enumerable.
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= <math>RE</math> U not <math>RE</math>
  
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$L1$ can be reduced to $L0'$, and hence $L0'$ also is not recursively enumerable.
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= not <math>RE</math>
  
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$L1'$ can be reduced to $L0$, and hence $L1'$ is recursively enumerable.
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$Lʼ = (L_0 \cup L_1)'$
  
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Now, $L$ = $L0 U L1$
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$= L_0' ∩ L_1'$
  
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= re U not re
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$=$ not <math>RE</math> $∩$ <math>RE</math>
  
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= not re
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$=$ not <math>RE</math>
  
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$L' = (L0 \cup L1)'$
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-->
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{{Template:FBD}}
  
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$= L0' ∩ L1'$
 
  
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$=$ not re $∩$ re
 
  
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$=$ not re
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[[Category: GATE2003]]
 +
[[Category: Automata  questions from GATE]]

Latest revision as of 19:40, 10 September 2014

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

Solution by Arjun Suresh

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.



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Define languages L0 and L1 as follows :

$L0 = \{< M, w, 0 > |$ $M$ halts on $w\} $

$L1 = \{< M, w, 1 > |$ $M$ 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 $L = L0 ∪ L1$. 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

Solution[edit]

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.

Alternate Solution[edit]

$L0$ 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 $L0$.

$L1$ 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 $L1$ and also give $w$ to $M$. Either $M$ accepts $w$, or the Turing machine for $L1$ accepts $<M,w,1>$. In either case we have solved halting problem. Hence, $L1$ is not recursively enumerable.

$L1$ can be reduced to $L0'$, and hence $L0'$ also is not recursively enumerable.

$L1'$ can be reduced to $L0$, and hence $L1'$ is recursively enumerable.

Now, $L$ = $L0 U L1$

= re U not re

= not re

$L' = (L0 \cup L1)'$

$= L0' ∩ L1'$

$=$ not re $∩$ re

$=$ not re