Lecture 27: P, NP, Co-NP, NP-Complete, PSPACE
Recall that we write “x is in P if there exists a polynomial time algorithm a(*) such that s is in x if and only a(s) returns yes.” Define x’ = { s | s not in x}. Is x’ in P? Yes, because we can define a’(*) as follows:
if a(s) returns “yes” then return “no” otherwise return “yes”
Let y be in NP. What about y’? Let y’ = {s | s not in y}. To say “y is in NP” means that there exists a polynomial time certifier b(*,*): s is in y if and only if there exists t such that b(s, t) returns “yes,” in polynomial bounded size. So, let’s look at y’:
s in y’
<=> s not in y
<=> ! there exists t such that b(s,t) returns “yes”
<=> for all t, b(s, t) returns “no”
Co-NP = {y’ | y is in NP}. Does NP = Co-NP? We don’t know. If we assume P=NP, though, it follows that P=Co-NP as well. Generally the consensus is that NP does not equal Co-NP. Here is a diagram that might help explain P, NP, Co-NP, and PSPACE:
P-SPACE is defined as {x | x has an algorithm using at most polynomial space}. P is a subset of PSPACE, because you cannot use more than polynomial space in only polynomial time. To say x is in PSPACE implies that x’ is in PSPACE using the same inversion algorithm as above. You can also create the following chain:
x in Co-NP => x’ in NP => x’ in SPACE => x in PSPACE
This entry was posted on Friday, April 1st, 2005 at 6:38 pm and is tagged with polynomial time algorithm, polynomial space, inversion algorithm, np, subset, consensus, lt. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback.
Leave a Reply
Please take time to enjoy the archives: May 2005 (1) April 2005 (11) March 2005 (11) February 2005 (15) January 2005 (7)
Fresh, related resources:
- There were no results.