Answer
a. **\(f(\text{"aba"}) = 0,\quad f(\text{"bbab"}) = 2,\quad f(\text{"b"}) = 0\).**
**Range of \(f\)** is \(\{0,1,2,3,\dots\}\).
b. **\(g(\text{"aba"}) = \text{"aba"},\quad g(\text{"bbab"}) = \text{"babb"},\quad g(\text{"b"}) = \text{"b"}\).**
Work Step by Step
Below is a step-by-step solution addressing both parts (a) and (b).
---
## (a) Definition of \(f\) and its values
**Definition of \(f\):**
For any string \(s\) of \(a\)s and \(b\)s,
\[
f(s) =
\begin{cases}
\text{the number of \(b\)s to the left of the left-most \(a\) in \(s\),} & \text{if \(s\) contains at least one \(a\),} \\
0, & \text{if \(s\) contains no \(a\)s.}
\end{cases}
\]
1. **Compute \(f(\text{"aba"})\):**
- The string is \(\text{"aba"}\).
- The left-most \(a\) is in the first position, so there are **0** \(b\)s to the left of it.
- Thus, \(f(\text{"aba"}) = 0\).
2. **Compute \(f(\text{"bbab"})\):**
- The string is \(\text{"bbab"}\).
- The left-most \(a\) appears in the third position (the substring is \(b,b,a,b\)).
- There are **2** \(b\)s to the left of that \(a\).
- Thus, \(f(\text{"bbab"}) = 2\).
3. **Compute \(f(\text{"b"})\):**
- The string is \(\text{"b"}\).
- There are **no** \(a\)s at all.
- By definition, \(f(s) = 0\) if \(s\) has no \(a\)s.
- Thus, \(f(\text{"b"}) = 0\).
### Range of \(f\)
- If a string has no \(a\)s, \(f\) outputs 0.
- If a string has at least one \(a\), say the first \(a\) occurs after \(n\) \(b\)s, then \(f\) outputs \(n\).
- Since \(n\) can be any nonnegative integer (we can prepend as many \(b\)s as we like before the first \(a\)), the **range of \(f\)** is all nonnegative integers:
\[
\{0,1,2,3,\dots\}.
\]
---
## (b) Definition of \(g\) and its values
**Definition of \(g\):**
For any string \(s\) of \(a\)s and \(b\)s, let \(g(s)\) be the **reverse** of \(s\). Formally, if
\[
s = s_1 s_2 \dots s_k
\]
where each \(s_i \in \{a,b\}\), then
\[
g(s) = s_k \dots s_2 s_1.
\]
1. **Compute \(g(\text{"aba"})\):**
- The reverse of \(\text{"aba"}\) is \(\text{"aba"}\).
- Hence, \(g(\text{"aba"}) = \text{"aba"}\).
2. **Compute \(g(\text{"bbab"})\):**
- The string \(\text{"bbab"}\) reversed is \(\text{"babb"}\).
- Hence, \(g(\text{"bbab"}) = \text{"babb"}\).
3. **Compute \(g(\text{"b"})\):**
- The string \(\text{"b"}\) reversed is still \(\text{"b"}\).
- Hence, \(g(\text{"b"}) = \text{"b"}\).
