exam1 - CSCI 2400 Models of Computation Exam 1 Solutions...

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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...
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  • Spring '08
  • CAROTHERS
  • Natural number, Formal language, Regular expression, Regular language, Nondeterministic finite state machine

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exam1 - CSCI 2400 Models of Computation Exam 1 Solutions...

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