Let's prove the inductive step $($where $T$ is not a single leaf$).$ Call its left

and right subtrees $T_1$ and $T_2.$ Assume that $l(T_1)2*rank(T_1)$ and

$l(T_2)2*rank(T_2)$

The proof breaks into $3$ cases, depending on whether rank$(T_1)=$rank$(T_2)rank(T_1)>$rank$(T_2)l(T)2*rank(T)rank(T_1),$ and prove that $l(T)$

$2*rank(T)rank(T_1),$

rank$(T_1)=$rank$(T_2),or$ rank$(T_1)>$rank$(T_2).$ Let's $do$ the first case for

this question: assume that rank$(T_1),$ and prove that $l(T)$

$2*rank(T)$ Now do the second case $).$ Again, you can assume

$(T_1)2*rank(T_1)$ and $(T_2)2*rank(T_2)$ and your goal is to prove

$t(T)2*rank(T)$ The last case $(rank(T_1)>$rank$(T_2))$ is similar to the first, but with the

roles of $T_1$ and $T_2$ switched. Reproduce it here.Finally, show that rank is not controlled by $.$ Provide an infinite list of

trees where rank is unbounded and $$ is bounded.Consider the following recursive definition of a function $$ from complete

binary trees to rational numbers.

$(T)=\{\begin{array}{ccc}0& ifTis\text{a}single\text{l}\text{e}\text{a}\text{f}& \\ 1+\frac{1}{2}((T_1)+(T_2))& ifT\text{h}\text{a}\text{s}\text{s}\text{u}\text{b}\text{t}\text{r}\text{e}\text{e}\text{s}{T}_1\text{a}\text{n}\text{d}& T_2\end{array}$

You can see that the definition is almost like size, except for that $\frac{1}{2}.($And it's

like the example from class, except the denominator is $2$ instead of $4.)$

Now recall from class the definition of rank of a tree:

rank$(T)=\{\begin{array}{ccc}0& ifTis\text{a}single\text{l}\text{e}\text{a}\text{f}\text{,}\text{e}\text{l}\text{s}\text{e}& \\ \text{m}\text{a}\text{x}& \{rank(T_1),\text{r}\text{a}\text{n}\text{k}(T_2)\}if\text{r}\text{a}\text{n}\text{k}(T_1)\text{r}\text{a}\text{n}\text{k}(T_2),\text{e}\text{l}\text{s}\text{e}& \\ 1+\text{r}\text{a}\text{n}\text{k}& (T_1)if\text{r}\text{a}\text{n}\text{k}(T_1)=\text{r}\text{a}\text{n}\text{k}& (T_2)\end{array}$

$($Notice that in the last line, $1+$rank$(T_1)$ could have just as well been expressed

as $1+$rank$(T_2),$ as this is the case where they are equal.$)$

In this problem set, you will determine which of these quantities controls

the other. First we will show that rank controls $,$ specifically that $l(T)2.$

rank$(T)$ for any tree $T,$ by induction. The base case is easy: if $T$ is a single leaf,

then $(T)=$rank$(T)=0,$ so $(T)2*rank(T).$

Problems start on the next page.

$1.$ Let$$s prove the inductive step $($where T is not a single leaf$).$ Call its left

and right subtrees T$1$ and T$2.$ Assume that $\backslash$iota $($T rank$($T$1)$ and

$\backslash$iota $($T rank$($T$2).$

The proof breaks into $3$ cases, depending on whether rank$($T$1)$ rank$($T$2),$

rank$($T$1)=$ rank$($T$2),$ or rank$($T$1)>$ rank$($T$2).$ Let$$s do the first case for

this question: assume that rank$($T$1)$ rank$($T$2),$ and prove that $\backslash$iota $($T

$2$ rank$($T$).$

$2.$ Now do the second case $($rank$($T$1)=$ rank$($T$2)).$ Again, you can assume

$\backslash$iota $($T rank$($T$1)$ and $\backslash$iota $($T rank$($T$2)$ and your goal is to prove

$\backslash$iota $($T rank$($T$).$

$3.$ The last case $($rank$($T$1)>$ rank$($T$2))$ is similar to the first, but with the

roles of T$1$ and T$2$ switched. Reproduce it here.

$4.$ Finally, show that rank is not controlled by $\backslash$iota $.$ Provide an infinite list of

trees where rank is unbounded and $\backslash$iota is bounded.