Let an represent the number of bit strings of length n containing two consecutive 1s. What is the recurrence relation for an?
(A) an–2 + an–1 + 2n–2
(B) an–2 + 2an–1 + 2n–2
(C) 2an–2 + an–1 + 2n–2
(D) 2an–2 + 2an–1 + 2n–2
Answer: (A)
Explanation: Simple Solution
One way to solve this is to try for small values and rule out options.
a0 = 0 a1 = 0 a2 = 1 ["11"] a3 = 3 ["011", "110", "111"] a4 = 8 ["0011", "0110", "0111", "1101", "1011", "1100", "1110", "1111"]
If we check for a3, we can see that only A and C satisfy the value. Among (A) and (C), only (A) satisfies for a4.
Another Solution (With Proof)
A string of length n (n >= 2) can be formed by following 4 prefixes 1) 11 followed by a string of length n-2 2) 00 followed by a string of length n-2 3) 01 followed by a string of length n-2 4) 10 followed by a string of length n-2 Number 1 has already two consecutive 1's so number of binary strings beginning with number 3 is 2n-2 as remaining n-2 bits can have any value. Number 2 has two 0's so remaining n-2 bits must have two consecutive 1's. Therefore number of binary strings that can be formed by number 2 is an-2. Number 3 and Number 4 together form all strings of length n-1 and two consecutive 1's.