No, the inverse statement will be not true any more. Intuitively, problem A is reducible to problem B if solutions to B exist and give solutions to A whenever A has solutions. Thus, solving A cannot be harder than solving B. When B belongs to P, A can't belong to NP, because B is harder than A. But on the inverse direction, if A belongs to P, B has no need to be a member of P problems. Actually, B can be a NP problem as a NP problem is definitely harder than a P problem according to the definition of computability.
On Fri, Aug 27, 2010 at 3:16 AM, Avik Mitra <[email protected]> wrote: > An instance of a problem for an algorithm A can be coded in > binary string of 0 and 1. Let us assume that there exists a standard > coding that is valid for this discussion. Then the set of instances > that satisfies the algorithm i.e. meets the criteria will form a > language. If A runs in polynomial-time then the language is said to in > P and set of all such languages constitutes P. > [Description is from "Introduction to Algorithm" by RivestThomas H. > Cormen, Charles E. Leiserson, > Ronald L. Rivest, Clifford Stein, 2nd Edition.] > > Now it can be proved that if L1 and L2 are two languages such > that L1 and can reduces to L2 in polynomial time, then if L2 is in P > then L1 is also in P.[lemma 34.3 of the same book] Now can we prove or > disprove the following? > > If L1 and L2 are two languages such that L1 can be reduced to L2 in > polynomial time, then if L1 is in P then L2 is also in P. > > -- > You received this message because you are subscribed to the Google Groups > "Algorithm Geeks" group. > To post to this group, send email to [email protected]. > To unsubscribe from this group, send email to > [email protected]. > For more options, visit this group at > http://groups.google.com/group/algogeeks?hl=en. > > -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to [email protected]. To unsubscribe from this group, send email to [email protected]. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
