Arjun Suresh (talk | contribs) |
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===Solution=== | ===Solution=== | ||
| + | Since, <math>f</math> is a polynomial time computable bijection and $f_{-1}$ is also polynomial time computable, $L_1$ and $L_2$ should have the same complexity (isomorphic). This is because, given a problem for $L_1$, we can always do a polynomial time reduction to $L_2$ and vice verse. Hence, the answer is 'C', as in 'A', $L_1$ and $L_2$ can be finite, in 'B', $L_1$ and $L_2$ can be in $P$ and in 'D', $L_1$ and $L_2$ can be recursive. Only, in 'C' there is no intersection for , and hence it can't be true. | ||
| − | + | Alternatively, we can prove 'C' to be false as follows: | |
| − | + | Given $L_2$ is decidable. Now, for a problem in $L_1$, we can have a $TM$, which takes an input x, calculates f(x) in polynomial time, check f(x) is in $L_2$ (this is decidable as $L_2$ is decidable), and if it's then output yes and otherwise outputs no. Thus $L_1$ must also be decidable. | |
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Consider two languages $L_1$ and $L_2$ each on the alphabet $\Sigma$. Let $f : \Sigma → \Sigma$ be a polynomial time computable bijection such that $(\forall x) [ x\in L_1$ iff $f(x) \in L_2]$. Further, let $f^{-1}$ be also polynomial time computable.
Which of the following CANNOT be true ?
(A) <math>L_1</math> $\in P$ and <math>L_2</math> is finite
(B) <math>L_1</math> $\in NP$ and <math>L_2</math> $\in P$
(C) <math>L_1</math> is undecidable and <math>L_2</math> is decidable
(D) <math>L_1</math> is recursively enumerable and <math>L_2</math> is recursive
Since, <math>f</math> is a polynomial time computable bijection and $f_{-1}$ is also polynomial time computable, $L_1$ and $L_2$ should have the same complexity (isomorphic). This is because, given a problem for $L_1$, we can always do a polynomial time reduction to $L_2$ and vice verse. Hence, the answer is 'C', as in 'A', $L_1$ and $L_2$ can be finite, in 'B', $L_1$ and $L_2$ can be in $P$ and in 'D', $L_1$ and $L_2$ can be recursive. Only, in 'C' there is no intersection for , and hence it can't be true.
Alternatively, we can prove 'C' to be false as follows: Given $L_2$ is decidable. Now, for a problem in $L_1$, we can have a $TM$, which takes an input x, calculates f(x) in polynomial time, check f(x) is in $L_2$ (this is decidable as $L_2$ is decidable), and if it's then output yes and otherwise outputs no. Thus $L_1$ must also be decidable.
Consider two languages $L_1$ and $L_2$ each on the alphabet $\Sigma$. Let $f : \Sigma → \Sigma$ be a polynomial time computable bijection such that $(\forall x) [ x\in L_1$ iff $f(x) \in L_2]$. Further, let $f^{-1}$ be also polynomial time computable.
Which of the following CANNOT be true ?
(A) <math>L_1</math> $\in P$ and <math>L_2</math> is finite
(B) <math>L_1</math> $\in NP$ and <math>L_2</math> $\in P$
(C) <math>L_1</math> is undecidable and <math>L_2</math> is decidable
(D) <math>L_1</math> is recursively enumerable and <math>L_2</math> is recursive
A language <math>L</math> is recursively enumerable (partially decidable), iff there is a Turing Machine <math>M</math>, which can enumerate all the strings of <math>L</math>.
So, suppose <math>L_2</math> is decidable. Now we have a <math>TM</math> <math>T</math>, which can enumerate all strings of <math>L_2</math>. For each of these strings <math>w</math>, we can find <math>x</math>, such that <math>f(x) = w</math>, using the function $f^{-1}$. Since $f^{-1}$ is bijective, it ensures that for each of the string <math>w</math> enumerated by <math>T</math>, we are getting a unique $f^{-1}(w)$ (because of one-one property of bijection) and we are guaranteed that we'll generate all strings of $L_1$ (because of "co-domain = range" property of bijection),. i.e.; we are actually enumerating all strings of <math>L_1</math>. Thus, we are getting a <math>TM</math> which is enumerating all strings of <math>L_1</math>, which means <math>L_1</math> should also be recursively enumerable (partially decidable).
Thus, if <math>L_2</math> is decidable, <math>L_1</math> will at-least be partially decidable.
GATECSE