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Subject: Data Structure Midterm (20070514)
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<H3 align=3Dcenter>Data Structure Midterm (20070514)</H3>
<HR>

<OL>
  <P>
  <LI>(5%) Show that 6*2<SUP>n</SUP>+n<SUP>2</SUP> =3D O(2<SUP>n</SUP>). =
Please=20
  use proof by induction if you need to prove some inequality. <FONT=20
  color=3Dgreen><BR>Answer: One way to do this is to prove that=20
  6*2<SUP>n</SUP>+n<SUP>2</SUP>=A1=D87*2<SUP>n</SUP> for n=A1=D94, or =
equivalently,=20
  n<SUP>2</SUP>=A1=D82<SUP>n</SUP> for n=A1=D94. We shall use proof by =
induction.=20
  <OL>
    <LI>Induction base: When n=3D4, n<SUP>2</SUP>=3D2<SUP>n</SUP>=3D16 =
holds.=20
    <LI>Induction hypothesis: When n=3Dk, =
k<SUP>2</SUP>=A1=D82<SUP>k</SUP> holds.=20
    <LI>Induction step: When n=3Dk+1, we have (k+1)<SUP>2</SUP> =3D=20
    k<SUP>2</SUP>+2k+1 =A1=D8 k<SUP>2</SUP>+k<SUP>2</SUP> =3D =
2=A1=D1k<SUP>2</SUP> =A1=D8=20
    2=A1=D12<SUP>k</SUP> =3D 2<SUP>k+1</SUP> </LI></OL></FONT>
  <P></P>
  <LI>(10%) The failure function for KMP (Knuth-Moris-Pratt) algorithm =
is shown=20
  next: <XMP>void fail(char *pat){
	int n=3Dstrlen(pat);
	failure[0]=3D-1;
	for (j=3D1; j<n; j++){
		i=3Dfailure[j-1];
		while ((pat[j]!=3Dpat[i+1]) && (i>=3D0))
			i=3Dfailure[i];
		if (pat[j]=3D=3Dpat[i+1])
			failure[j]=3Di+1;
		else
			failure[j]=3D-1;
	}
}
</XMP>Compute the failure function for each of the following patterns:=20
  <OL type=3Da>
    <LI>(5%) abcadabcab=20
    <LI>(5%) abcxxabcxxyyabcxxabcxy </LI></OL><BR><FONT =
color=3Dgreen>Answer:=20
  <OL>
    <LI>-1 -1 -1 0 -1 0 1 2 3 1=20
    <LI>-1 -1 -1 -1 -1 0 1 2 3 4 -1 -1 0 1 2 3 4 5 6 7 8 -1 =
</LI></OL></FONT><!--
<p><li>(10%) Given a multi-dimensional array =
A[u<sub>0</sub>][u<sub>1</sub>]=A1K[u<sub>n-1</sub>], where its starting =
address is =A3\. Derive the equation to compute the address of =
A[i<sub>0</sub>][i<sub>1</sub>]=A1K[i<sub>n-1</sub>].
-->
  <P></P>
  <LI>(10%) Suppose we have the following declaration for a linked list: =
<XMP>typedef struct list_node *list_pointer;
typedef struct list_node {
	char data;
	list_pointer link;
};
</XMP>Write a C function <I>invert(list_pointer lead)</I> that can =
invert the=20
  list pointed to by lead and return the pointer to the inverted list. =
(Note=20
  that your function should not have more than 15 lines.) <BR><FONT=20
  color=3Dgreen>Answer: <XMP>list_pointer invert(list_pointer lead){
	list_pointer middle=3DNULL, trail;
	while (lead){
		trail=3Dmiddle;
		middle=3Dlead;
		lead=3Dlead->link;
		middle->link=3Dtrail;
	}
	return(middle);
}
	</XMP></FONT>
  <P></P>
  <LI>(30%) Give answers to the following short questions about trees. =
You do=20
  not need to prove them.=20
  <OL type=3Da>
    <LI>(3%) What is the maximum number of nodes on level i of a binary =
tree?=20
    <BR><FONT color=3Dgreen>Answer: 2<SUP>i-1</SUP> </FONT>
    <LI>(3%) What is the maximum number of nodes in a binary tree of =
depth=20
    <I>k</I>? <BR><FONT color=3Dgreen>Answer: 2<SUP>k</SUP>-1 </FONT>
    <LI>(3%) For any nonempty binary tree T, Let n<SUB>0</SUB> be the =
number of=20
    leave nodes and n<SUB>2</SUB> the number of nodes of degree 2. What =
the=20
    relationship between n<SUB>0</SUB> and n<SUB>2</SUB>? <BR><FONT=20
    color=3Dgreen>Answer: n<SUB>0</SUB> =3D n<SUB>2</SUB>+1 </FONT>
    <LI>(3%) What is the depth of a complete binary tree with <I>n</I> =
nodes?=20
    <BR><FONT color=3Dgreen>Answer: floor(log<SUB>2</SUB>n + 1) or=20
    ceil(log<SUB>2</SUB>(n+1)) </FONT>
    <LI>(3%) What are the four 4 methods for tree traversal? Which one =
of these=20
    methods needs the use of a queue? <BR><FONT color=3Dgreen>Answer: =
Preorder,=20
    inorder, postorder, level order. The method of level-order traversal =

    requires the use of a queue. </FONT>
    <LI>(3%) How many null links do we have in a <I>n</I>-node binary =
tree of=20
    linked representation? <BR><FONT color=3Dgreen>Answer: n+1 </FONT>
    <LI>(3%) Use the left child-right sibling representation to convert =
the=20
    following tree into a binary tree.=20
    <CENTER><IMG height=3D200=20
    =
src=3D"http://localhost/jang/courses/cs2351/exam/graph2.png"></CENTER><BR=
><FONT=20
    color=3Dgreen>Answer:=20
    <CENTER><IMG height=3D300=20
    =
src=3D"http://localhost/jang/courses/cs2351/exam/graph2-solution.png"></C=
ENTER></FONT>
    <LI>(3%) In order to access the tree nodes quickly, the binary tree =
in the=20
    previous question is stored level-by-levey in a one-dimensional =
array=20
    A[1..n]. What are the indices of nodes g, j are? <BR><FONT=20
    color=3Dgreen>Answer: g ---&gt; 21, j ---&gt; 171 </FONT>
    <LI>(3%) Draw the binary tree if the preorder sequence is [1, 2, 3, =
4, 5, 6,=20
    7] and the inorder sequence is [2, 4, 3, 1, 6, 5, 7]. <BR><FONT=20
    color=3Dgreen>Answer:=20
    <CENTER><IMG height=3D200=20
    =
src=3D"http://localhost/jang/courses/cs2351/exam/pre-inorder.png"></CENTE=
R></FONT>
    <LI>(3%) Draw the binary tree if the postorder sequence is [7, 4, 2, =
5, 6,=20
    3, 1] and the inorder sequence is [7, 4, 2, 1, 5, 3, 6]. <BR><FONT=20
    color=3Dgreen>Answer:=20
    <CENTER><IMG height=3D200=20
    =
src=3D"http://localhost/jang/courses/cs2351/exam/post-inorder.png"></CENT=
ER></FONT></LI></OL>
  <P></P>
  <LI>(10%) Please use a stack to indicate how to convert the=20
  a/((a+b)*c-d)*e-a*c into a postfix form (You should draw the stack =
status=20
  after each step). <BR><FONT color=3Dgreen>Answer: (In the following =
step, the=20
  bottom of the stack is the first element.)=20
  <TABLE border=3D1>
    <TBODY>
    <TR>
      <TH>Token
      <TH>Stack
      <TH>Output=20
    <TR>
      <TD>a=20
      <TD>&nbsp;=20
      <TD>a=20
    <TR>
      <TD>/=20
      <TD>&nbsp;/=20
      <TD>a=20
    <TR>
      <TD>(=20
      <TD>&nbsp;/(=20
      <TD>a=20
    <TR>
      <TD>(=20
      <TD>&nbsp;/((=20
      <TD>a=20
    <TR>
      <TD>a=20
      <TD>&nbsp;/((=20
      <TD>aa=20
    <TR>
      <TD>+=20
      <TD>&nbsp;/((+=20
      <TD>aa=20
    <TR>
      <TD>b=20
      <TD>&nbsp;/((+=20
      <TD>aab=20
    <TR>
      <TD>)=20
      <TD>&nbsp;/(=20
      <TD>aab+=20
    <TR>
      <TD>*=20
      <TD>&nbsp;/(*=20
      <TD>aab+=20
    <TR>
      <TD>c=20
      <TD>&nbsp;/(*=20
      <TD>aab+c=20
    <TR>
      <TD>-=20
      <TD>&nbsp;/(-=20
      <TD>aab+c*=20
    <TR>
      <TD>d=20
      <TD>&nbsp;/(-=20
      <TD>aab+c*d=20
    <TR>
      <TD>)=20
      <TD>&nbsp;/=20
      <TD>aab+c*d-=20
    <TR>
      <TD>*=20
      <TD>&nbsp;*=20
      <TD>aab+c*d-/=20
    <TR>
      <TD>e=20
      <TD>&nbsp;*=20
      <TD>aab+c*d-/e=20
    <TR>
      <TD>-=20
      <TD>&nbsp;-=20
      <TD>aab+c*d-/e*=20
    <TR>
      <TD>a=20
      <TD>&nbsp;-=20
      <TD>aab+c*d-/e*a=20
    <TR>
      <TD>*=20
      <TD>&nbsp;-*=20
      <TD>aab+c*d-/e*a=20
    <TR>
      <TD>c=20
      <TD>&nbsp;-*=20
      <TD>aab+c*d-/e*ac=20
    <TR>
      <TD>eos=20
      <TD>&nbsp;=20
      <TD>aab+c*d-/e*ac*- </TR></TBODY></TABLE></FONT>
  <P></P>
  <LI>(10%) Please use a stack to indicate how to evaluate the postfix=20
  expression 62/3-42*+ step by step. (You should draw the stack status =
after=20
  each step.) <BR><FONT color=3Dgreen>Answer: (In the following step, =
the bottom=20
  of the stack is the first element.)=20
  <TABLE border=3D1>
    <TBODY>
    <TR>
      <TH>Token
      <TH>Stack=20
    <TR>
      <TD>6=20
      <TD>&nbsp;6=20
    <TR>
      <TD>2=20
      <TD>&nbsp;6 2=20
    <TR>
      <TD>/=20
      <TD>&nbsp;3=20
    <TR>
      <TD>3=20
      <TD>&nbsp;3 3=20
    <TR>
      <TD>-=20
      <TD>&nbsp;0=20
    <TR>
      <TD>4=20
      <TD>&nbsp;0 4=20
    <TR>
      <TD>2=20
      <TD>&nbsp;0 4 2=20
    <TR>
      <TD>*=20
      <TD>&nbsp;0 8=20
    <TR>
      <TD>+=20
      <TD>&nbsp;8 </TR></TBODY></TABLE></FONT>
  <P></P>
  <LI>(10%) Draw the max-heap tree after the following operations are =
performed=20
  beginning with an empty heap: insert 43, insert 31, insert 68, insert =
24,=20
  delete-max, insert 51, insert 44, insert 53, insert 69, insert 71, =
delete-max.=20
  <BR><FONT color=3Dgreen>Answer:=20
  <CENTER><IMG height=3D350=20
  =
src=3D"http://localhost/jang/courses/cs2351/exam/maxHeap.png"></CENTER></=
FONT>
  <P></P>
  <LI>(15%)=20
  <OL type=3Da>
    <LI>(5%) Let b<SUB>n</SUB> be the number of different permutations=20
    obtainable by passing the numbers 1, 2, 3, ..., n through a stack =
and=20
    deleting in all possible ways. Give the recursive formula for =
b<SUB>n</SUB>.=20
    (Be sure to specify the initial condition for your formula.) =
<BR><FONT=20
    color=3Dgreen>Answer: b<SUB>n</SUB> =3D <FONT=20
    face=3Dsymbol>S</FONT><SUB>i=3D0</SUB><SUP>n-1</SUP> b<SUB>i</SUB>=20
    b<SUB>n-i-1</SUB>, b<SUB>0</SUB>=3D1 </FONT>
    <LI>(5%) Let c<SUB>n</SUB> be the number of different ways to =
compute the=20
    product of n matrices. What is the relationship between =
b<SUB>n</SUB> and=20
    c<SUB>n</SUB>? Why? <BR><FONT color=3Dgreen>Answer: c<SUB>n+1</SUB> =
=3D=20
    b<SUB>n</SUB><BR>(You need to state the reason.) </FONT>
    <LI>(5%) Derive an explicit formula for b<SUB>n</SUB> using the =
method=20
    described in the class. <BR><FONT color=3Dgreen>Answer: =
b<SUB>n</SUB> =3D=20
    C<SUB>n</SUB><SUP>2n</SUP>/(n+1)<BR>(You need to show how you derive =
the=20
    answer.) </FONT></LI></OL></LI></OL>
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