cot5310F07SampleMidterm1Key

cot5310F07SampleMidterm1Key - COT 5310 Fall 2006 Midterm#1...

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Unformatted text preview: COT 5310 Fall 2006 Midterm#1 Sample Name: KEY 16 1 . Choosing from among (REC) recursive , (RE) re non-recursive , (NR) non-re , categorize each of the sets in a) through d). Justify your answer by showing some minimal quantification of some known recursive predicate or by another clear and convincing short argument. a.) { <f,x>| ∀ (t>x) STP(x,f,t) } REC Justification: This is equivalent to STP(x, f, x+1). That’s because STP(x, f, s) implies that STP(x, f, t), where t ≥ s. b.) { f | ∀ x f(x) is a prime number } NR Justification: This is ∀ x ∃ t [ STP(x, f, t) && isPrime(Value(x, f, t) ) ] Some of you thought it was REC because primality is recursive, but that’s not enough since we are looking at all x, and divergence is necessary. c.) { f | ∃ x f(x) ↓ } RE Justification: This is ∃ <t,x> [ STP(x, f, t) ]. It’s just L ne . d.) { x | ∀ f f(x) ↑ } REC Justification: No x has this property. An easy way to see this is by f(x) = x (a base primitive recursive function). This shows that all x are included in the domain of some function. 15 2 . Let set A be recursive, B be re non-recursive and C be non-re. Choosing from among (REC) recursive , (RE) re non-recursive , (NR) non-re , categorize the set D in each of a) through e) by listing all possible categories. No justification is required. a.) D = ~A ∪ B REC, RE b.) D = A ∩ ~C REC, RE, NR c.) D = B – A REC, RE d.) D = C ∪ ~C REC e.) A ⊆ D ⊆ B REC, RE, NR 7 3. Using the definition that S is semi-decidable iff S is the domain of some procedure g S (partial recursive function), prove that if both S and its complement...
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This note was uploaded on 07/14/2011 for the course COT 5310 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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cot5310F07SampleMidterm1Key - COT 5310 Fall 2006 Midterm#1...

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