If $\alpha$ is a bit string, let $C(\alpha)$ be the maximum number of consecutive 0’s in $\alpha$ . [Examples: $C(10010) = 2, C(00110001) = 3$ .] Let $S_n$ be the number of n-bit strings $\alpha$ with $C(\alpha) \leq 2$ . Develop a recurrence relation for $S_1, S_2, . . . .$

Solution

Verified

Step 1

1 of 2

Given:

$\alpha$ is a bit string

$C(\alpha)$ =maximum number of consecutive 0's in $\alpha$

$S_n$ number of $n$ -bit strings $\alpha$ with $C(\alpha)\leq 2$

When $\alpha$ is an $n$ -bit string with at most 2 consecutive zeros and $n\geq 3$ , then $\alpha$ either ends with 1 or with 10 or with 100 (as $\alpha$ has at most two consecutive zeros).

When $\alpha$ ends with 1, then there are $S_{n-1}$ possible strings for the remaining $n-1$ digits and thus there are $S_{n-1}$ strings $\alpha$ that end with 1.

When $\alpha$ ends with 10, then there are $S_{n-2}$ possible strings for the remaining $n-2$ digits and thus there are $S_{n-2}$ strings $\alpha$ that end with 10.

When $\alpha$ ends with 100, then there are $S_{n-3}$ possible strings for the remaining $n-3$ digits and thus there are $S_{n-3}$ strings $\alpha$ that end with 100.