This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: CSCI 2400  Models of Computation Exam 1 Solutions Problem 1 (10 points) Prove that for any integer n ≥ 4, 2 n ≥ n 2 . Solution: Using mathematical induction, Base Case: n = 4 2 4 = 16 ≥ 4 2 √ Inductive hypothesis: Assume that 2 k ≥ k 2 for some k ≥ 4. Inductive step: We will show that given the above assuption, 2 k +1 ≥ ( k +1) 2 is always true: 2 k +1 ≥ ( k +1) 2 2 k + 2 k ≥ k 2 + 2 k + 1 2 k + 2 k ≥ 2 k + 2 k + 1 2 k ≥ 2 k + 1 ... which can be shown via a second inductive proof: Base Case: n = 4 2 4 = 16 ≥ 2*4 + 1 √ Inductive hypothesis: Assume that 2 k ≥ 2 k + 1 for some k ≥ 4. Inductive step: We will show that given the above assuption, 2 k +1 ≥ 2( k +1) + 1 is always true: 2 k +1 ≥ 2( k +1) + 1 2 k + 2 k ≥ 2 k + 2 + 1 2 k + 2 k ≥ 2 k + 2 2 k ≥ 2 √ 1 Problem 2 (10 points) Prove that for any sets R , S , and T , R uniontext ( S intersectiontext T ) = ( R uniontext S ) intersectiontext ( R uniontext T ) Solution: This is the Distributive Law for sets. Venn diagrams make the statement rather obvious. R uniontext ( S intersectiontext T ), union of all colored sections. ( R uniontext S ) intersectiontext ( R uniontext T ), intersection of dark light sections.) 2 However, for a rigorous proof, we must show that both R uniontext ( S...
View
Full Document
 Spring '08
 CAROTHERS
 Natural number, Formal language, Regular expression, Regular language, Nondeterministic finite state machine

Click to edit the document details