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Language $L_1$ is polynomial time reducible to language $L_2$. Language $L_3$ is polynomial time reducible to language $L_2$, which in turn polynomial time reducible to language $L_4$. Which of the following is/are true?

  1. $\text{ if } L_4 \in P, \text{ then } L_2 \in P$
  2. $\text{ if } L_1 \in P\text{ or } L_3 \in P, \text{ then } L_2 \in P$
  3. $L_1 \in P, \text{ if and only if } L_3 \in P$
  4. $\text{ if } L_4 \in P, \text{ then } L_3 \in P$
  1. II only
  2. III only
  3. I and IV only
  4. I only
in Theory of Computation by Veteran (105k points)
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3 Answers

+34 votes
Best answer
  1. $L_1$ is polynomial time reducible to $L_2$. So, $L_2$ is at least as hard as $L_1$. 
  2. $L_3$ is polynomial time reducible to $L_2$. So, $L_2$ is at least as hard as $L_3$. 
  3. $L_2$ is polynomial time reducible to $L_4$. So, $L_4$ is at least as hard as $L_2$. 

If $L_4$ is in $P, L_3$, $L_2$ and $L_1$ must also be in $P$. So, I and IV are true. 

We can have $L_1$ in $P$ and $L_2$ not in $P$, and none of the given conditions are violated. So, II is false. 

Assume $L_3$ not in $P$. Now, Since $L_2$ must be at least as hard as $L_3$, it must also be not in $P$. But $L_1$ is less harder than $L_1$ as per condition 1, and it can be in $P$ without violating any given conditions. So, III is false.

Hence, C is choice.

More Info: http://gatecse.in/wiki/Some_Reduction_Inferences

by Veteran (431k points)
edited by
0
Precise answer
0
I am not getting that line "But L1 is less harder than L1 as per condition 1" ???
0
Typo error  : But L1 is less harder than "L2" as per condition 1 . Because L2 is atleast as hard as L1 which means L2 can be more bigger problem than L1 or at least equal to L1.
+12 votes
from question we can conclude that ...

L1<L2
L3<L2<L4
now take first statement
I. if L4 belongs P, then L2 belongs P this is true because L2 is polynomial time reducible to language L4
 

II. if L1 belongs P or L3 belongs P, then L2 belongs P not possible because L1, L3 is is polynomial time reducible to language L2, not L2 is polynomial time reducible to language L1,L2.
 

III. L1 belongs P, if and only if L3 belongs P this is not true because looking at L3 we can not define L1 or simply we can say that L1 is not polynomial time reducible to language L3
 

IV. if L4 belongs P, then L1 belongs P and L3 belongs P this is true because we know that L1<L2 & L3<L2 and L2 is polynomial time reducible to language L4 so we can say  L1&L2 also be polynomial time reducible to language L4.
So option C is correct.
by Active (5k points)
0

In the above explanation all the  <  should be changed to \leq

If X is polynomial time reducible to Y, i.e  X \leqp  Y   then X is atmost as hard as Y. Therefore if X belongs to class P then, Y may or may not belong to class P.

0

you are right .....but here above sign (< ) is not mean to the less than i use this just for understanding purpose   .. consider this (L1<L2)  as  L1 is polynomial time reducible to language L2.
 

0

II. if L1 belongs P or L3 belongs P, then L2 belongs P not possible because L1, L3 is is polynomial time reducible to language L2, not L2 is polynomial time reducible to language L1,L2.

It should be then L2 belongs to P may or may not be possible

0

@Anoop Sonkar Please explain part 4

+2 votes
Option C.
by Active (2k points)
Answer:

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