Recurrence Relation Problems: Detailed Analysis

Recurrence Relation Problems: Detailed Analysis Overview This document presents solutions to several recurrence relation problems, focusing on deriving recurrence relations, solving for initial conditions, and computing specific values for the recurrence. Detailed stepby-step processes are included for clarity. — Problem 1: Bit Strings with Consecutive Zeros Problem Statement We are tasked to analyze bit strings of length n and find: (a) A recurrence relation for the number of bit strings of length n that contain at least one pair of consecutive 0s. (b) The initial conditions for the recurrence relation. (c) The number of bit strings of length n = 7 that contain at least one pair of consecutive 0s. Solution Part (a): Recurrence Relation Let an represent the number of bit strings of length n that contain at least one pair of consecutive 0s. Complementary Counting: To derive an , we use complementary counting: an = 2n − xn , where xn represents the number of bit strings of length n with no consecutive 0s. Recurrence Relation for xn : To compute xn : • If the bit string starts with 1, the remaining n − 1 bits form any valid string of length n − 1, contributing xn−1 . • If the bit string starts with 0, the next bit must be 1, and the remaining n − 2 bits form any valid string of length n − 2, contributing xn−2 . Thus: xn = xn−1 + xn−2 . Recurrence Relation for an : Substituting xn into an = 2n − xn : an = 2n − (xn−1 + xn−2 ). Analyzing overlapping contributions from shorter strings leads to: an = an−1 + an−2 + 2n−2 . 1 Part (b): Initial Conditions The initial conditions are: • a1 = 0: A bit string of length 1 cannot contain consecutive 0s. • a2 = 1: Only one bit string of length 2 contains consecutive 0s (00). Part (c): Compute a7 To compute a7 , use the recurrence an = an−1 + an−2 + 2n−2 and the initial conditions a1 = 0 and a2 = 1: a3 = a2 + a1 + 23−2 = 1 + 0 + 2 = 3, a4 = a3 + a2 + 24−2 = 3 + 1 + 4 = 8, a5 = a4 + a3 + 25−2 = 8 + 3 + 8 = 19, a6 = a5 + a4 + 26−2 = 19 + 8 + 16 = 43, a7 = a6 + a5 + 27−2 = 43 + 19 + 32 = 94. Final Answer: The number of bit strings of length 7 that contain at least one pair of consecutive 0s is: a7 = 94. — Problem 2: Vending Machine Problem Problem Statement A vending machine dispensing books of stamps accepts only 1coins,1 bills, and 5bills.W earetaskedto : Find a recurrence relation for the number of ways to deposit n dollars into the vending machine, where the order of deposits matters. Determine the initial conditions. Compute the number of ways to deposit n = 10 dollars. Solution Part (a): Recurrence Relation Let an represent the number of ways to deposit n dollars into the vending machine. To compute an : • If the first deposit is 1 (coin or bill), the remaining amount is n − 1, contributing 2an−1 (two ways: coin or bill). • If the first deposit is 5 (bill), the remaining amount is n − 5, contributing an−5 . Thus, the recurrence relation is: an = 2an−1 + an−5 , for n ≥ 5. Part (b): Initial Conditions The initial conditions are: a0 = 1, a1 = 2, a2 = 4, a3 = 8, a4 = 16. 2 Part (c): Compute a10 Using the recurrence an = 2an−1 + an−5 and the initial conditions: a5 = 2a4 + a0 = 2(16) + 1 = 33, a6 = 2a5 + a1 = 2(33) + 2 = 68, a7 = 2a6 + a2 = 2(68) + 4 = 140, a8 = 2a7 + a3 = 2(140) + 8 = 288, a9 = 2a8 + a4 = 2(288) + 16 = 592, a10 = 2a9 + a5 = 2(592) + 33 = 1217. Final Answer: The number of ways to deposit 10is : a10 = 1217. — 3