Solutions-Test-One-291-W18-R.docx - Test 1 CS291 Discrete Structures II Spring 2018 March 21 Wednesday 4:00pm-5:15pm CLOSE BOOK Your class roll number

# Solutions-Test-One-291-W18-R.docx - Test 1 CS291 Discrete...

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Test 1 CS291: Discrete Structures II Spring 2018 March 21, Wednesday, 4:00pm-5:15pm CLOSE BOOK, Your class roll number __________________ Last Name ______________________ First Name ________________________ There are 7 problems with a total of 100 points. 1 (10 points) Solve the following recurrence relation: a n = 10 a n -1 - 25 a n -2 ; a 1 = 3, a 2 = 20, Solution: Characteristic equation is t 2 – 10 t + 25 = ( t - 5) 2 = 0. It has one multiple root t = 5 with multiplicity 2. Therefore, its total solution is a n = 5 n + n 5 n We use the initial condition to get and . For n = 1, we get 3 = 5 + 5 (1) For n = 2, we get 20 = 25 + 50 (2) Solving (1) and (2), we get = 2 5 and = 1 5 . Therefore, the answer is a n = 2 5 ×5 n + 1 5 n 5 n = 2 5 n -1 + n 5 n -1 1
Two ways to tile a 1x1 board Seven ways to tile a 1x2 board Black one red one blue two red two blue red blue blue red Green white 2 (10 points) Suppose we can only use 5 kinds of tiles to cover a 1 n board: (1) 1 1 size with red color, (2) 1 1 size with blue color, (3) 1 2 size with black color, (4) 1 2 size with green color, (5) 1 2 size with white color. Let W ( n ) be the number of ways to tile the 1 n board, n = 1, 2, …. For example, W (1) = 2, W (2) = 7, as shown by the following figure. Please find a recurrence relation for W ( n ) and solve it. Solution: The recurrence relation is: The initial conditions are W (1) = 2, W (2) = 7, When n > 2, W ( n ) = 2 W ( n -1) + 3 W ( n -2). Its characteristic equation is t 2 – 2 t - 3 = 0. ( t – 3)( t +1) = 0.

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• Fall '18
• Eric Swartz
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