Assessment

The following question was given to ChatGPT.

However, it was later learned that this should not be provable

Determine what's correct, what's incorrect, give a counterexample that shows it

cannot be proved, and assign a letter grade: A$,$B$,$C$,$D$,$F$.$

Rainbow strings are special strings which are made out of only letters from the list

$R,O,Y,G,B,$I,$V.$ They are defined recursively as follows:

i$)$ Any letter from that list is a Rainbow string.

ii$)$ If $S_1$ and $S_2$ are Rainbow strings, where $S_1$ ends with letter $x$ and $S_2$ starts

with letter $y$ where $x$ and $y$ are off by one place $($e$.$g $x$ is $R$ and $y$ is $O.$ Or $x$ is

$V$ and $y$ is $R),$ then $S$ which is $S_1$ concatenated with $S_2$ is a Rainbow string.

Prove by structural induction that len$\frac{S}{2}$ is the maximum number of instances

of letter $Bs$ in the Rainbow string $S.$

$(9)$ To prove that the maximum number of instances of the letter $\text{'}$ B $\text{'}$ in a Rainbow string $R$ is $\frac{\text{l}\text{e}\text{n}(R)}{2},$

we will use structural induction.

Base Case

Case $1$:

For any single letter from the set $R,O,Y,G,B,$I,$V,$ which is a Rainbow string:

If $R$ is a single letter $l,$ where lin$\{R,O,Y,G,B,$I,$V\}$ :

len$(R)=1$

The maximum number of $\text{'}B\text{'}$s in this string is $1$ if $l=B,$ otherwise $0.$

So$,$ the base case holds as:

$\frac{\text{l}\text{e}\text{n}(R)}{2}=\frac{1}{2}$~~$0\mathrm{.}5$

This is consistent with having at most $1\text{'}$ B $\text{'}$ in a single$-$character Rainbow string.

Inductive Step

Assume that for two Rainbow strings $R_1$ and $R_2,$ the maximum number of instances of $\text{'}$ B $\text{'}$ in each

is at most $\frac{\text{l}\text{e}\text{n}(R_1)}{2}$ and $\frac{\text{l}\text{e}\text{n}(R_2)}{2},$ respectively.

Consider the concatenation of these two Rainbow strings to form a new Rainbow string $R$ :

$R=R_1+R_2$

We know from the recursive definition that this concatenation is valid only if the last character of

$R_1$ and the first character of $R_2$ are off by one place in the sequence $R,O,Y,G,B,$I,$V.$

Validating the Inductive Hypothesis

Given the inductive hypothesis:

Let len $(R_1)=n_1$ and len$(R_2)=n_2.$

The maximum number of $\text{'}B\text{'}$s in $R_1$ is $\frac{m_1}{2}.$

The maximum number of $\text{'}B\text{'}s$ in $R_2$ is $\frac{n_2}{2}.$

We need to show that the maximum number of $\text{'}$B$\text{'}$s in $R$ is:

$\frac{\text{l}\text{e}\text{n}(R)}{2}=\frac{n_1+n_2}{2}$

Concatenation Analysis

Since $R=R_1+R_2,$ the length of $R$ is:

len$(R)=n_1+n_2$

By the inductive hypothesis, the maximum number of $\text{'}$B$\text{'}$s in $R_1$ is $\frac{n_1}{2}$ and in $R_2$ is $\frac{n_2}{2}.$

Combination of Maximums

The maximum number of $\text{'}B\text{'}$s in $R$ can be at most the sum of the maximum number of $\text{'}$B$\text{'}$s in $R_1$

and $R_2$ :

MaxBs in $R\frac{n_1}{2}+\frac{n_2}{2}=\frac{n_1+n_2}{2}$

Therefore, by structural induction, we have shown that the maximum number of instances of the

letter $\text{'}$ B $\text{'}$ in any Rainbow string $R$ is $\frac{\text{l}\text{e}\text{n}(R)}{2}.$

This completes the proof.