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EC 504 functions
创建日期：2024-05-21 17:15:14
所属分类：数学mathematics
标签： EC 504
functions

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EC 504 Midterm Pratice

Instructions: Put your name and BU ID on the first page. This exam is closed book
and no notes – except for 2 pages of personal notes to be passed in the end. No use
of laptops or internet. You have one hour and fifty minutes. Do the easy ones first.
1. (20 pts) Answer True or False to each of the questions below. Each question is worth 2 points.
Answer true only if it is true as stated, with no additional assumptions. No explanation is
needed, but any explanation is likely to earn partial credit, and no explanation will not earn
any credit if the answer is wrong.
(a) N2e−1/ ln(N) ∈ Θ(Neln(N)).
Solution: False: N2e−1/ ln(N) ∈ Θ(N2), since 1/ln(N) → 0 it does not change the
Θ(N2) factor.
(b) If fi(n) ∈ O(n2), then
∑n
i=1 fi(n) ∈ O(n3)
Solution: True. There are n terms, each of which is O(n2), so you can bound this by
a constant times n3, making it O(n3).
(c) The following tree is an AVL binary search tree – explain.
11
/ \
9 23
/ \ / \
8 12 16 50
/ \ / \
1 10 10 80
Solution: False. The key 12 must be the right of 11 and the second 10 to the left of
11.
(d) The second smallest element in a binary min-heap with all elements with distinct values
will always be a child of the root.
Solution: True. The children of the root must be smaller than any of their children.
(e) The set Θ(f(n)) is identical to the set O(f(n))∩Ω(f(n)) (Note: ∩ means intersection
of two sets.)
Solution: False. It is identical to Θ((f(n)) ∩ Ω(f(n)) but not the larger set O(f(n))
that include slower functions. (e.g. like the relation ≤).
(f) It is possible to formulate Quick Sort to be worst case O(N logN) including the cost
of picking a suitable pivot.
Solution: True. You pick the pivot to be them medium using the ()(N) algorirthm
that sorts N/5 columns recursive find the medium for N/5 in the middle row.
1
(g) Consider the array [2, 20, 3, 21, 90, 4,23, 22]. This array has elements in order of a
min-heap.
Solution: True. All path for the root are in decreasing order.
2
/ \
20 3
/ \ / \
21 90 4 23
/
22
(h) A min-Heap (or priority queue) is a null tree or a root node with a minimum value
plus a left and a right min-Heap subtrees.
Solution: True. This is the correct recursive defintion of a min-Heap.
(i) The maximum subsequence sum algorithm finds i and j that maximize
∑j
k=i a[k] for
an array a[N ] of positive and negative integers. The fastest one is a recursive algorithm
with complexity Θ(N logN).
Solution: False. As I demonstrated in class, qx the fastest one, which is a sort of
investors scan, is Θ(N) – suprising!
(j) NN ∈ O( eln(N !) )
Solution: True. The Sterling’s approxmation of the factorial: N ! =
√
2piNNNe−M(1+
O(1/N)) shows that is goes slower than N ! so N ! = O(NN) but not the reverse.
2. (10 pts) Consider the binary tree illustrated below.
*
/ \
+ -
/ \ / \
* 8 - 10
/ \ / \
+ 2 8 *
/ \ / \
6 7 2 3
(a) List the node values in the order in which they would be found in an in-order traversal.
2
(b) List nodes in pre-order.
(c) List the nodes in post=order.
(d) Which order is uses on stack to do the arithmetic? . (HINT the stack order is like
depth first search on this tree!) Evaluate the arithmetic using stack
Solution:
In-order
6 + 7 ∗ 2 + 8 ∗ 8 − 2 ∗ 3 − 10 (1)
pre-order
∗ + ∗ + 6 7 2 8 − − 8 ∗ 2 3 10 (2)
post-order
6 7 + 2 ∗ 8 + 8 2 3 ∗ − − 10 − ∗ (3)
Evaluation of arithmatic. Easiest way is bottom up on the tree (e.g. in-order) to get Result
is −272. The stack has two solutions.
You get −272 if you remeber that we pushed left and then right so for a minus sign is
pop y pop x =⇒ x− y.
You could assume pop x pop y - =⇒ y − x which gives 408. I accept this rule altough
it is inconsistent with the definition of post-order. This could be corrected by exchangeing
left and righ in post order. Hard becuase Minus sign hard because ther are not symmetric.
3
3. (20 pts) Consider the following min-heap.
1
/ \
10 2
/ \ / \
11 12 17 3
/ \ / \ / \ / \
15 18 201 13 155 20 4 5
/ \ /
17 16 21
(a) Show the min-heap which results after inserting the element 7 in the heap. (Indicate
the sequence of steps with arrows.) Then in this new heap show the steps required to
delete element 1 and then delete 11. Draw the final min-heap.
(b) Re-arrange the original min-heap (repeated below) into a max-heap by a “bottom up”
O(N) algorithm. Describe the steps level by level.
1
/ \
10 2
/ \ / \
11 12 17 3
/ \ / \ / \ / \
15 18 201 13 155 20 4 5
/ \ /
17 16 21
Solution:
Start with next to bottom row of 8
1
/ \
10 2
/ \ / \
11 12 17 3
/ \ / \ / \ / \
17 21 201 13 155 20 4 5 <---- Start with row of 8
/ \ /
15 16 18
4
1/ \
10 2
/ \ / \
21 201 155 5 <---- Next row of 4
/ \ / \ / \ / \
17 18 12 13 17 20 4 3
/ \ /
15 16 11
1
/ \
201 155 <---- Next row of 2
/ \ / \
21 13 20 5
/ \ / \ / \ / \
17 18 12 10 17 2 4 3
/ \ /
15 16 11
201 <---- Final Root node
/ \
21 155
/ \ / \
18 13 20 5
/ \ / \ / \ / \
17 11 12 10 17 2 4 3
/ \ /
15 16 1
(c) Using this example for a min-heap with the size N = 18 and a[0] = N. Describe
in a few word an an Θ(N logN) algorithm to sort in place the array elements
a[1], a[2], · · · , a[N ] in descending order (i.e. a[1] > a[2] > · · · > a[N ] )
Solution: a[0] store the size of heap starting with a[0] = N The algorithm is iterative.
You delete the min in a[1] restoring the heap and copying a[a[0]]] = a[1] reducing
the size by by 1 (a[0] = a[0] − 1) until the heap is empty (a[0] = 0) Now the array
a[1], a[2], ..a[N ] is sorted array in descending order. Restorig the heap each time is
(log(N) so the alogritym is O(NlogN)
(d) Solve the recursion relation for the Heap, T (H) = 2T (H − 1) + c0H to show that
“bottom up” scales like T (H) = Ω(N) (HINT: Use N ' 2H)
Solution:
5
Substitute the guess in the equation that is suggested: T (H) = c 2H
c 2H ' c 2 2H−1 + c0H = c 2H + c0H ' c 2H (4)
So the leading behavior is T [H] = c 2H ∈ Ω(H).
6
4. (20 pts) This problem is to construct step by step BST and AVL trees given the following
list of N = 12 elements.
1 2 12 7 3 6 30 36 37 2 40 51
(a) Insert them sequentially into a BST (Binary Search Tree).
Solution: The BST (depth of a node is the numbe of hop form the root. height is the
maximun number of hops to the deapest node.
1 h=7, d = 0
\
2 h=6, d = 1
/ \
h=0, d = 2 2 12 h=5, d = 2
/ \
h=2, d = 3 7 30 h=4, d = 3
/ \
h= 1,d = 4 3 36 h=3, d = 4
\ \
h=0, d = 5 6 37 h=2, d = 5
\
40 h=1, d = 6
\
51 h=0, d = 7
(b) Insert them sequentially into an empty AVL tree, restoring the AVL property after
each insertion. Show the AVL tree which results after each insertion and name the
type of rotation (RR or LL zig-zig or versus RL or LR zig-zag).
Solution: Final AVL form; See next page for sequence of steps!
30 h = 3 d = 0
/ \
3 36 h = 2 d = 1
/ \ / \
2 7 3 48 h = 1 d = 2 (excep 3 is h =0)
/ \ \ / \
1 2 6 37 54 h = 0 d = 3 (excep 3 is h =0)
(c) Give the total height TH(N) and the total depth TD(N) for both the resulting BST
and AVL.
(d) Compare the sum TH(N)+TD(N) for the BST and AVL. For each compare these values
with HN where H is the height of each tree.
7
Solution: BST:
TH(N) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 1 + 2 = 31
and
TD(N) = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 2 + 3 + 4 + 5 = 42
So TH(N) + TD(N) = 73 < HN = 7 ∗ 12 = 84
AVL:
TH(N = 3 + 2 ∗ 2 + 3 ∗ 1 = 10
and
TD(N) = 2 ∗ 1 + 4 ∗ 2 + 5 ∗ 3 = 25
So‘ TH(N) + TD(N) = 35 < HN = 3 ∗ 12 = 36
Note a perfect tree of N = 15 and height H = 3 has
TH(N) = 3 + 2 ∗ 2 + 4 ∗ 1 = 11
and
TD(N) = 2 ∗ 1 + 4 ∗ 2 + 8 ∗ 3 = 34
so TH(N) + TD(N) = 45 = HN = 3 ∗ 15 = 45
I thought this is the only one that saturates the upper bound but a counter
example is to remove one leaf NH → NH −H and that leaf remove d = H
form the sum of TD(N)!
8
5. (10 POINTS) Consider the following recursion relation.
T (N) = 3T (N/2) +Nk
(a) Substitute T (N) = c0N
γ into the homogeneous equation (i.e. dropping Nk) and de-
termine γ and for what values of k does the exact solution satisfy T (N) = Θ(Nγ).
Solution: The homogeneous equation (i.e. drop the last term has solution by substi-
tution
c0 N
γ = c0 3 (N/2)
γ =⇒ 1 = 3/2γ or γ = log(3)/ log(2) (5)
Base 2 is nice since log2(2) = 1 or γ = log2(3)
This is gives a leading solution T (N) = Θ(Nγ) if k < γ
(b) Now assume that k = γ and find the exact solution in the form: T (N) = c0N
γ +
c1N
γ log2(N). (HINT: First show that c1 = 1. Then Determine c0 in terms of T (1) by
setting N = 1. )
Solution: Note a homogeneous solution never determines the multiplicative const c0.
For this we must find a fulls solution. The substution gives
c0 N
γ + c1N
γ log2(N) = c0 3 (N/2)
γ + c13(N/2)
γ log2(N/2) +N
γ
= c0 3 (N/2)
γ + c1(N)
γ[log2(N)− log2(2)] +Nγ (6)
Obviously the first (homegeneous term) still cancels. We could have ignore this piece
for now.
But to cancel the additive term we need to use both 3/2γ = 1 and again log2(2) = 1,
to get c1 = 1. The setting N = 1 we finally fix c0 by T (1) = c0 1
γ + c11
γ log2(1) = c0.
Actually you can do this step first if you prefer.
9
6. (20 pts) You are interested in compression. Given a file with characters, you want to find
the binary code which satisfies the prefix property (no conflicts) and which minimizes the
number of bits required. As an example, consider an alphabet with 8 symbols, with relative
weights (frequency) of appearance in an average text file 1 give below:
alphabet:| V | A | M | P | I | R | E | S |
weights:| 4 | 40 | 16 | 7 | 30 | 18 | 75 | 8 |
(a) Determine the Huffman code by constructing a tree with minimum external path
length:
∑8
i=1widi. (Arrange tree with smaller weights to the left.) Solution:
V P S M R I A E
4 7 8 16 18 30 40 75
27
40
11
1
Minimize ext length = ∑i wi di
4 V
34
74
16 M
198
49
19
18 R
75 E
7 P
8 S
30A
1
1
1
1
1
1
0
0
0
0
0
Sort Wi =
0
123
I
0
(b) Identity the code for each letter and list the number of bits for each letter and compute
the average number of bits per symbol in this code. Is it less than 3? (You can leave the
answer as a fraction since getting the decimal value is difficult without a calculator.)
Solution:
(c) Give an example of weights for these 8 symbols that would saturate 3 bits per letter.
What would the Huffman tree look like? Is a Huffman code tree always a full tree?
Solution:
1vampire: Origin mid 18th cent: from French, from Hungarian vampir, perhaps form Turkish uber “witch”.
In European folklore, a monster with long pointed canine teeth!
10
RESULTING CODE: AVERAGE BITS/CHAR =508/198 = 2.565
Letter Weight Code Bits Count
! E 75 0 1 75
! A 40 101 3 120
! I 30 111 3 90
! M 16 1000 4 64
! R 18 1001 4 72
! S 8 1100 4 32
! V 4 11010 5 20
! P 67 11011 5 35
Total: 198 508
28
External length = ∑i wi di
It is perfect binary tree. One can have each have weight 10 for example
( 80 )
/ \
(40) (40)
/ \ / \
(20) (20) (20) (20)
/ \ / \ / \ / \
V A M P I R E S
Actually the weight don’t have to exactly equal. Try it with weights:
100, 101, 102, 103, 104, 105, 106, 107
It is still perfect tree1
11
7. (10 POINTS EXTRA CREDIT) Given an array of N-distinct positive integers a[N ] and a
second array of N-distinct positive integers b[N ] to construct the sum:
N−1∑
i=0
a[i] ∗ b[i] (7)
(a) Assume that a[k] is sorted in ascending order. Give an O(N logN) algorithm that
maximizing the sum S by permuting the values of b[k]. Solution: To maximize the
sum you sort the array b[k] into ascending order by any Θ(NlogN) algorithm such
as merg sort. This is an example of every pair {i, j} maximizing the scalar product
a[i]b[i] + a[j]b[j] since a[i] < a[j] and b[i] < b[j]. Any exchang of i, j reduces this
contribution the sum.
(b) Given the result of part (a) give an algorithm that minimizes S in O(N).
Solution: To minimize the result of part (a) one just reverse the order of b[k] with
N/2 swaps of b[i] and b[N − 1− i] for i = 0, 1, · · ·N/2
(c) If you are allowed to permute independently both a[k] and b[k] how many different
solutions will maximize S? Solution: There are N ! solution because give maximun
Smax =
N−1∑
k=0
a[k] ∗ b[k] (8)
as describe in part (a) is unchange by any simultaineous permutation of k (pi(k)
Smax =
N−1∑
k=0
a[pi(k)] ∗ b[pi(k)] (9)
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CIVL2010 ECMT2150 CS 333 CP1406 KIT308 Economics 701 ECON3740 ECE 5759 BFC3340 FINC2012 COMP4500 MATH3871 FIT3080 18-819C Economic ECO202Y5Y PHY100 FINC-780 Econ 100B EC602 IE582 EE450 COMP9517 WRDS 150B ECON 570 LECTURE 2 ECA5101 ECE 4420 ADAD9311 FIT3031 CS 544 MF 703 CS 6140 KIT501 ENGSCI 344 CSC171 IE 332 FINM8007 21S2 COSC 1114 FE545 E471 SIT102 CSC108 MAT301 IAE 101 COP3503 STAT2004 EC 303 CS1026 STAT1203 MATH3171 ISOM5160 IFN657 FIT2014 ACCT2102 DSCI550 STATS 330 ELEC421 COMP3670 QBUS6840 MEES:698B MAST30027 MIET1077 Economics IFT 6758 FIT5122 BUSN5101 CptS 111 HRMT5502 Math2018 PP 312 INFS3200 CZ3005 CSSE1001 SENG 2011 MKTG 553Q BUFN 640 COMP507 ETF5952 DATA 311 ECE4043 BANK3011 finance M3H623544 HW1 INFS 7410 AI6103 STAT3888 INFS3208 ACCT7103 ECOS3013 PSYCH 20B F79MA STAT3006 ENGG952 2X MGSC 7000 ECE4132 MSF 526 MATH 3033 CDS 511 EEEE4120 EEET2465 ECA5103 MATH2070 ELEC3104 BU51019 EEE8147 ENGR20003 COSC 1107 ETF3200 FNCE 435 MSFI 435 ISyE6420 GENG4405 HW ETC5523 IE 7315 MOEC0021 STA 106 ENGG 1300A ENGG 1300 OLET1139 STAT 3503 STAT 206 FEM11149 ELEN 4720 COMP20003 SOLA1070 CS574 C11CS LINB29H3 MAST90083 ECON7200 STAT 610 LINB29 DSM-5 COMP3161 EEEN40010 MATH 4431 K353 STA 3386 IERG4300 COMS W4167 ENG5298 AMME 2302 CS3910 COMPSCI 320 STATS 369 GR5242 COSC 328 CSCI435 MGTS1301 INFS2200 COMP3900 CIS 375 Module AE03 COMP3121 CS260 HAM503 MGMT5050 INT3075 COSC 285 GEOL0029 GEGE2X01 STAT1201 MATH1061 MMF1941H STAT 5060 CS 520 COMP 4320 MSIN0104 000T FINS3648 ECMT2160 ECON4003 Phys 129L CSE 142 CS 3130 MGEB01 EECS 340 Finance (20443/20444) ECMM149 COMP5310 EEEN60180 CSC 111 AT2 CSCI 4430 STAB57 MA/CSC 427 MATH322 D100 CSCI544 CITS3004 CI6227 CSE 565 APCO 1P01 852L1 ELEC S411F COMP3207 STATS330 ECON7530 FIT5149 CSI4104 CMPT 371 COMP201 SEHS4696 FI4003 MATH6182 CG1 WS MATH401 MGT 205 QTM 110 COMP 0137 MACE12201 FIT5210 ECE 544 BISM7206 HUDM 6055 ECON3203 STATS 210 MSCI 570 MATH6173 COMPSCI4039 Accounting COM3503 AOE2024 ECON213 PSYC10004 Control Data 100/200A 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