cot6410F10Midterm2Key

# cot6410F10Midterm2Key - COT 6410 Fall 2010 Exam#2 Name KEY...

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COT 6410 Fall 2010 Exam#2 Name: KEY 12 1 . Choosing from among (REC) recursive , (RE) re non-recursive, (coRE) co-re non-recursive , (NRNC) non-re/non-co-re , categorize each of the sets in a) through d). Justify your answer by showing some minimal quantification of some known recursive predicate. a) A = { f | f(f) } t STP(f,f,t) RE b.) B = { f | range(f) is a proper subset of } x <y,t> [STP(f,x,t) Value(f,y,t) ≠ x] NRNC c.) C = { f | f(0) take at least 100 steps to converge, if at all } ~STP(f,0,99) REC d.) D = { f | f diverges everywhere } <x,t> ~STP(f,x,t) coRE 6 2 . Define, compare and contrast the notions of Countable and Recursively Enumerable for some set S . Question 7 actually gives one of these definitions. A set S is countable iff it can be placed in a 1-1 correspondence with a subset of the Natural numbers. That is, S is countable iff there is an injective mapping (not necessarily computable) that associates each element of S with a unique element of . Alternatively, S is countable iff S is empty or there is a surjective mapping (not necessarily computable) that associates each element of with a unique element of S. A set S is recursively enumerable (re) iff it is either empty or there exists a total computable function that effectively maps the Natural numbers onto the set. That is, S is re, non-empty iff there is an algorithm (a total computable function), f, whose domain is and whose range is S. Every re set is countable, but not every countable set is re. The issue is that re requires a mapping that is computable, whereas countable just requires the existence of a mapping; whether or not that mapping is computable is not relevant. Thus, all subsets of

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