Chapter 9, Problem 3CP

Given the dependency diagram of attributes C1,C2,C3,C4,C5 in a table shown in the following figure, the primary key attributes are underlined.   Make a database with multiple tables from attributes as shown above that are in 3NF, showing PK, non-key attributes, and FK for each table? Assume the tables are already in 1NF. Hint: 3 tables will result after deducing 1NF -> 2NF -> 3NF]

1. Using one of the method described in class and/or textbook (Section 9.1) convert the following regular expression into a state transition diagram: (0+ 10*1)* (01 + 10) Indicate in your answer how did you arrive at the result as follows: Write down all the state transition diagrams that you constructed for all the subexpressions and clearly indicate which diagram corresponds to which expression. Do not simplify any state transition diagram. 2. Consider the following state transition diagram over Σ = {a,b}: b A a a C b B a a b D За a Using the method described in class and in the textbook (Section 9.2) convert the diagram into an equivalent regular expression. Include all the intermediate steps in your answer. 3. Are the languages L1, L2, and L3 below over the alphabet Σ = {a, b, c} regular or non-regular? Justify your answer carefully. (a) L₁ = {a¹b2jc²i : i ≥ 0, j > 2} (b) L₂ = L₁n {akbm c³p: k,m,p≥ 0} (c) L3 = {a²ib²j+1 : i,j ≥ 0}^{akbm c³p : k,m,p ≥ 0}

(1 point) By dragging statements from the left column to the right column below, give a proof by induction of the following statement: an = = 9" - 1 is a solution to the recurrence relation an = 9an-18 with ao = : 0. The correct proof will use 8 of the statements below. Statements to choose from: Note that a₁ = 9a0 + 8. Now assume that P(n) is true for all n ≥ 0. Your Proof: Put chosen statements in order in this column and press the Submit Answers button. Let P(n) be the predicate, "a = 9″ – 1". απ = 90 − 1 = Note that Let P(n) be the predicate, "an 9" - 1 is a solution to the recurrence relation an = 9an-1 +8 with ao = 0." - Now assume that P(k + 1) is true. Thus P(k) is true for all k. Thus P(k+1) is true. Then ak+1 = 9ak +8, so P(k + 1) is true. = 1 − 1 = 0, as required. Then = 9k — 1. ak Now assume that P(k) is true for an arbitrary integer k ≥ 1. By the recurrence relation, we have ak+1 = ak+1 = = 9ak + 8 = 9(9k − 1) + 8 This simplifies to 9k+19+8 = 9k+1 − 1 Then 9k+1 − 1 = 9(9*…