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\begin_body

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
exerc1[01]
\end_layout

\end_inset

In the binary tree (2),
 let 
\family typewriter
INFO(P)
\family default
 denote the letter stored in 
\family typewriter
NODE(P)
\family default
.
 What is 
\begin_inset Formula $\mathtt{INFO(LLINK(RLINK(RLINK(T))))}$
\end_inset

?
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset


\begin_inset Formula $H$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc4[20]
\end_layout

\end_inset

The text defines three basic orders for traversing a binary tree;
 another alternative would be to proceed in three steps as follows:
\end_layout

\begin_layout Enumerate
Visit the root,
\end_layout

\begin_layout Enumerate
traverse the right subtree,
\end_layout

\begin_layout Enumerate
traverse the left subtree,
\end_layout

\begin_layout Standard
using the same rule recursively on all nonempty subtrees.
 Does this new order bear any simple relation to the three orders already discussed?
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

Yes,
 this is the same as postorder but backwards.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc10[20]
\end_layout

\end_inset

What is the largest number of entries that can be in the stack at once,
 during the execution of Algorithm T,
 if the binary tree has 
\begin_inset Formula $n$
\end_inset

 nodes?
 (The answer to this question is very important for storage allocation,
 if the stack is being stored consecutively.)
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\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

There can be up to 
\begin_inset Formula $n$
\end_inset

 nodes in the stack,
 in the case that all the right links are null and the tree is just a linked list using the 
\begin_inset Formula $\mathtt{RLINK}$
\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc13[24]
\end_layout

\end_inset

Design an algorithm analogous to Algorithm T that traverses a binary tree in 
\emph on
preorder
\emph default
,
 and prove that your algorithm is correct.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer
\end_layout

\end_inset


\end_layout

\begin_layout Enumerate
[Initialize.] Set stack 
\family typewriter
A
\family default
 empty,
 and set the link variable 
\begin_inset Formula $\mathtt{P}\gets\mathtt{T}$
\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:23113b"

\end_inset

[
\begin_inset Formula $\mathtt{P}=\Lambda$
\end_inset

?] If 
\begin_inset Formula $\mathtt{P}=\Lambda$
\end_inset

,
 go to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113d"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:23113c"

\end_inset

[Visit 
\begin_inset Formula $\mathtt{P}$
\end_inset

.] Visit 
\begin_inset Formula $\mathtt{NODE(P)}$
\end_inset

.
 Do 
\begin_inset Formula $\mathtt{A}\Leftarrow\mathtt{RLINK(P)}$
\end_inset

,
 
\begin_inset Formula $\mathtt{P}\gets\mathtt{LLINK(P)}$
\end_inset

,
 and return to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113b"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:23113d"

\end_inset

[Visit right hand side.] If 
\family typewriter
A
\family default
 is empty,
 the algorithm terminates.
 Otherwise set 
\begin_inset Formula $\mathtt{P}\Leftarrow\mathtt{A}$
\end_inset

 and return to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113b"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Standard
We now prove by induction on the number of nodes that,
 if we start in step 2,
 we traverse the subtree starting at 
\begin_inset Formula $\mathtt{P}$
\end_inset

 in preorder,
 leave the stack unchanged,
 and proceed to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113d"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

,
 which clearly would show that the algorithm is correct.
\end_layout

\begin_layout Standard
For an empty tree,
 this is obvious.
 For a tree with 
\begin_inset Formula $n$
\end_inset

 nodes,
 that we start with the root is obvious,
 we do that at step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113c"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 with the stack empty.
 After that step,
 setting 
\begin_inset Formula $\mathtt{P}\gets\mathtt{LLINK(P)}$
\end_inset

 and returning to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113b"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 means that we get to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23113d"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

 after traversing the left subtree,
 which by the induction hypothesis would leave us on step 4 with the stack like it was before but adding the node 
\begin_inset Formula $\mathtt{RLINK(P)}$
\end_inset

.
 Then step 4 clearly means to traverse 
\begin_inset Formula $\mathtt{RLINK(P)}$
\end_inset

 and leave 
\family typewriter
A
\family default
 like it was before starting processing 
\family typewriter
P
\family default
.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc16[22]
\end_layout

\end_inset

The diagrams in Fig.
 24 help to provide an intuitive characterization of the position of 
\begin_inset Formula $\mathtt{NODE(Q}\$\mathtt{)}$
\end_inset

 in a binary tree,
 in terms of the structure near 
\begin_inset Formula $\mathtt{NODE(Q)}$
\end_inset

:
 If 
\begin_inset Formula $\mathtt{NODE(Q)}$
\end_inset

 has a nonempty right subtree,
 consider 
\begin_inset Formula $\mathtt{Q}=\$\mathtt{P}$
\end_inset

,
 
\begin_inset Formula $\mathtt{Q}\$=\mathtt{P}$
\end_inset

 in the upper diagrams;
 
\begin_inset Formula $\mathtt{NODE(Q}\$\mathtt{)}$
\end_inset

 is the 
\begin_inset Quotes eld
\end_inset

leftmost
\begin_inset Quotes erd
\end_inset

 node of that right subtree.
 If 
\family typewriter

\begin_inset Formula $\mathtt{NODE(Q)}$
\end_inset


\family default
 has an empty right subtree,
 consider 
\begin_inset Formula $\mathtt{Q}=\mathtt{P}$
\end_inset

 in the lower diagrams;
 
\begin_inset Formula $\mathtt{NODE(Q}\$\mathtt{)}$
\end_inset

 is located by proceeding upward in the tree until after the first upward step to the right.
\end_layout

\begin_layout Standard
Give a similar 
\begin_inset Quotes eld
\end_inset

intuitive
\begin_inset Quotes erd
\end_inset

 rule for finding the position of 
\begin_inset Formula $\mathtt{NODE(Q}*\mathtt{)}$
\end_inset

 in a binary tree in terms of the structure near 
\family typewriter
NODE(Q)
\family default
.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

If 
\begin_inset Formula $\mathtt{NODE(Q)}$
\end_inset

 has a nonempty left subtree,
 consider 
\begin_inset Formula $\mathtt{Q}=*\mathtt{P}$
\end_inset

 in the upper diagrams;
 then 
\begin_inset Formula $\mathtt{Q}*=\mathtt{P}$
\end_inset

 is the left child of 
\begin_inset Formula $\mathtt{Q}$
\end_inset

.
 Otherwise,
 if 
\begin_inset Formula $\mathtt{NODE(Q)}$
\end_inset

 has a nonempty right subtree,
 
\begin_inset Formula $\mathtt{Q}*$
\end_inset

 is the right child of 
\begin_inset Formula $\mathtt{Q}$
\end_inset

.
 Otherwise both subtrees are empty and 
\begin_inset Formula $\mathtt{Q}*$
\end_inset

 can be found by going upward until finding a node that is a left child (which could be 
\begin_inset Formula $\mathtt{Q}$
\end_inset

 itself) and taking its right sibling.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc17[22]
\end_layout

\end_inset

Give an algorithm analogous to Algorithm S for determining 
\begin_inset Formula $\mathtt{P}*$
\end_inset

 in a threaded binary tree.
 Assume that the tree has a list head as in (8),
 (9),
 and (10).
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

If 
\family typewriter
P
\family default
 points to a node of a threaded binary tree,
 this algorithm sets 
\begin_inset Formula $\mathtt{Q}\gets\mathtt{P}*$
\end_inset

.
\end_layout

\begin_layout Enumerate
[Is the left subtree nonempty?] If 
\begin_inset Formula $\mathtt{LTAG(P)}=1$
\end_inset

,
 set 
\begin_inset Formula $\mathtt{Q}\gets\mathtt{LLINK(P)}$
\end_inset

 and terminate the algorithm.
\end_layout

\begin_layout Enumerate
[Search to the right.] Set 
\begin_inset Formula $\mathtt{Q}\gets\mathtt{RLINK(P)}$
\end_inset

.
 If 
\begin_inset Formula $\mathtt{RTAG(P)}=1$
\end_inset

,
 terminate the algorithm.
 Otherwise set 
\begin_inset Formula $\mathtt{P}\gets\mathtt{Q}$
\end_inset

 and repeat this step.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
exerc28[00]
\end_layout

\end_inset

After Algorithm C has been used to make a copy of a tree,
 is the new binary tree 
\emph on
equivalent
\emph default
 to the original,
 or 
\emph on
similar
\emph default
 to it?
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

It's equivalent (and similar).
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc30[22]
\end_layout

\end_inset

Design an algorithm that threads an unthreaded tree;
 for example,
 it should transform (2) into (10).
 
\emph on
Note:

\emph default
 Always use notations like 
\begin_inset Formula $\mathtt{P}*$
\end_inset

 and 
\begin_inset Formula $\mathtt{P}\$$
\end_inset

 when possible,
 instead of repeating the steps for traversal algorithms like Algorithm T.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

This algorithm threads an unthreaded tree 
\begin_inset Formula $\mathtt{T}$
\end_inset

.
 The fields 
\begin_inset Formula $\mathtt{LTAG}$
\end_inset

 and 
\begin_inset Formula $\mathtt{RTAG}$
\end_inset

 in the nodes of 
\begin_inset Formula $\mathtt{T}$
\end_inset

 are considered to initially contain arbitrary values that we do not need to preserve.
\end_layout

\begin_layout Enumerate
[Initialize list head and variables.] Get 
\begin_inset Formula $\mathtt{P}\Leftarrow\mathtt{AVAIL}$
\end_inset

 and set 
\begin_inset Formula $\mathtt{LTAG(P)}\gets\mathtt{RTAG(P)}\gets0$
\end_inset

,
 
\begin_inset Formula $\mathtt{LLINK(P)}\gets\mathtt{T}$
\end_inset

,
 
\begin_inset Formula $\mathtt{RLINK(P)}\gets\mathtt{P}$
\end_inset

,
 
\begin_inset Formula $\mathtt{T}\gets\mathtt{P}$
\end_inset

,
 and 
\begin_inset Formula $\mathtt{Q}\gets\mathtt{P}\$$
\end_inset

.
 
\emph on
(We'll use 
\begin_inset Formula $\mathtt{Q}$
\end_inset

 as a pointer to the node being considered and 
\begin_inset Formula $\mathtt{P}$
\end_inset

 and 
\begin_inset Formula $\mathtt{R}$
\end_inset

 as pointers to the next and previous nodes,
 respectively.
 We only ever compute the next node in inorder from the last node calculated this way,
 which makes it trivial to use algorithm T as a coroutine.)
\end_layout

\begin_layout Enumerate
\begin_inset CommandInset label
LatexCommand label
name "enu:23122b"

\end_inset

[Are we over?] If 
\begin_inset Formula $\mathtt{RLINK(Q)}=\mathtt{Q}$
\end_inset

,
 the algorithm terminates.
 
\emph on
(This loop only happens in the list head.)
\end_layout

\begin_layout Enumerate
[Step.] Set 
\begin_inset Formula $\mathtt{R}\gets\mathtt{Q}\$$
\end_inset

.
 If 
\begin_inset Formula $\mathtt{LLINK(Q)}=\Lambda$
\end_inset

,
 set 
\begin_inset Formula $\mathtt{LTAG(Q)}\gets0$
\end_inset

 and 
\begin_inset Formula $\mathtt{LLINK(Q)}\gets\mathtt{P}$
\end_inset

,
 otherwise set 
\begin_inset Formula $\mathtt{LTAG(Q)}\gets1$
\end_inset

.
 Similarly,
 if 
\begin_inset Formula $\mathtt{RLINK(Q)}=\Lambda$
\end_inset

,
 set 
\begin_inset Formula $\mathtt{RTAG(Q)}\gets0$
\end_inset

 and 
\begin_inset Formula $\mathtt{RLINK(Q)}\gets\mathtt{R}$
\end_inset

,
 otherwise set 
\begin_inset Formula $\mathtt{RTAG(Q)}\gets1$
\end_inset

.
 Finally set 
\begin_inset Formula $\mathtt{P}\gets\mathtt{Q}$
\end_inset

,
 
\begin_inset Formula $\mathtt{Q}\gets\mathtt{R}$
\end_inset

,
 and return to step 
\begin_inset CommandInset ref
LatexCommand ref
reference "enu:23122b"
plural "false"
caps "false"
noprefix "false"
nolink "false"

\end_inset

.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
rexerc37[24]
\end_layout

\end_inset

(D.
 Ferguson.) If two computer words are necessary to contain two link fields and an 
\begin_inset Formula $\mathtt{INFO}$
\end_inset

 field,
 representation (2) requires 
\begin_inset Formula $2n$
\end_inset

 words of memory for a tree with 
\begin_inset Formula $n$
\end_inset

 nodes.
 Design a representation scheme for binary trees that uses less space,
 assuming that 
\emph on
one
\emph default
 link and an 
\begin_inset Formula $\mathtt{INFO}$
\end_inset

 field will fit in a single computer word.
\end_layout

\begin_layout Standard
\begin_inset ERT
status open

\begin_layout Plain Layout


\backslash
answer 
\end_layout

\end_inset

If we allow for a second null value (call it 
\begin_inset Formula $\lambda$
\end_inset

,
 which could be e.g.
 a pointer to a specific word in the program code or to a sentinel constant,
 or 4001 in MIX,
 or 1 in MMIX),
 then we could implement the solution in the book,
 which is that,
 if 
\begin_inset Formula $\mathtt{LINK(P)}=\Lambda$
\end_inset

,
 then 
\begin_inset Formula $\mathtt{LLINK(P)}=\mathtt{RLINK(P)}=\Lambda$
\end_inset

,
 and otherwise 
\begin_inset Formula $\mathtt{LLINK(P)}=\mathtt{LINK(P)}$
\end_inset

 and 
\begin_inset Formula $\mathtt{RLINK(P)}=\mathtt{LINK(P)}+1$
\end_inset

 (or 
\begin_inset Formula $+\mathtt{sizeof(int)}$
\end_inset

),
 except that if 
\begin_inset Formula $\mathtt{LINK(LINK(P))}=\lambda$
\end_inset

 or 
\begin_inset Formula $\mathtt{LINK(LINK(P)}+1\mathtt{)}=\lambda$
\end_inset

 then 
\begin_inset Formula $\mathtt{LLINK(P)}$
\end_inset

 or 
\begin_inset Formula $\mathtt{RLINK(P)}$
\end_inset

 is respectively 
\begin_inset Formula $\Lambda$
\end_inset

.
\end_layout

\end_body
\end_document