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GATE | GATE-CS-2015 (Set 1) | Question 65
  • Last Updated : 20 Feb, 2015

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.

Quiz of this Question

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