Arjun Suresh (talk | contribs) (→{{Template:Author|Arjun Suresh|{{arjunweb}} }}) |
Starry Starr (talk | contribs) |
||
| (One intermediate revision by one other user not shown) | |||
| Line 1: | Line 1: | ||
| − | =='''1.''' If | + | =='''1.''' If $L$ and $L'$ are both recursively enumerable, then $L$ is recursive. Why?== |
==={{Template:Author|Arjun Suresh|{{arjunweb}} }}=== | ==={{Template:Author|Arjun Suresh|{{arjunweb}} }}=== | ||
| − | Given | + | Given $L$ is $RE$. So there is a $TM$, which accepts and halts for all words in $L$. |
| − | Now, if | + | Now, if $L'$ is $RE$, then there is a $TM$, which accepts and halts for all words not in $L$. |
| − | So, if a word is given (either from | + | So, if a word is given (either from $L$ or not from $L$), give it to both those $TM$s. If it is from $L$, the first $TM$ will halt and we say it belongs to $L$. If it is not from $L$, the second one will halt and we say it doesn't belong to $L$. Thus, $L$ becomes recursive. |
<!-- | <!-- | ||
=='''2.''' CYCLE(L) ={xy | yx is in L,L is regular } == | =='''2.''' CYCLE(L) ={xy | yx is in L,L is regular } == | ||
| Line 55: | Line 55: | ||
| − | [[Category: | + | [[Category: Theory of Computation]] |
Given $L$ is $RE$. So there is a $TM$, which accepts and halts for all words in $L$. Now, if $L'$ is $RE$, then there is a $TM$, which accepts and halts for all words not in $L$. So, if a word is given (either from $L$ or not from $L$), give it to both those $TM$s. If it is from $L$, the first $TM$ will halt and we say it belongs to $L$. If it is not from $L$, the second one will halt and we say it doesn't belong to $L$. Thus, $L$ becomes recursive.
Given <math>L</math> is $RE$. So there is a $TM$, which accepts and halts for all words in <math>L</math>. Now, if <math>L'</math> is $RE$, then there is a $TM$, which accepts and halts for all words not in <math>L</math>. So, if a word is given (either from <math>L</math> or not from <math>L</math>), give it to both those $TM$s. If it is from $L$, the first $TM$ will halt and we say it belongs to $L$. If it is not from $L$, the second one will halt and we say it doesn't belong to <math>L</math>. Thus, <math>L</math> becomes recursive.
GATECSE