MT2 Exam CMPE-371 10.01.2015 (120 min, 100 points)
St. Name, Surname______________________________________ St.Id#_____________
Group___________________________
Instructors
Gr1 - Asst. Prof. Dr. Adnan Acan
Gr2 – Assoc. Prof. Dr. Alexander Chefranov
Note: Use the last sheet for your calculations
Q1 (30 points). For the matrices A1(3,4), A2(4,5), A3(5,4), A4(4,6) with sizes given in the
parenthesis, use Dynamic programming to define minimal number of floating point
multiplications to calculate the product of the matrix chain A1∙A2∙A3∙A4.
a) (5 points). Write out a recursive formula for calculation of the optimal complexity
0, 𝑖 = 𝑗
𝑚𝑖𝑗 = { min (𝑚𝑖𝑘 + 𝑚𝑘+1,𝑗 + 𝑝𝑖−1 𝑝𝑘 𝑝𝑗 )
𝑖≤𝑘<𝑗
b) (5 points). Write out array p=(3,4,5,4,6)
c) (5 points). What two matrices are to be filled? What do they contain? What are the sizes of the
matrices? How they are to be initialized? In what order their elements are to be filled?
Matrices (𝑚𝑖𝑗 ) and (𝑠𝑖𝑗 ), the former matrix contains optimal cost values, and the latter one
contains values of k giving minimum. The matrices have 4 rows and 4 columns. The matrix m is
initialized by zeroes on the main diagonal. Elements of the matrices are filled by diagonals
d) (5 points). Fill in the matrices
m
0
60
120
0
80
0
s
192
176
120
0
1
2
2
3
3
3
I=1,j=2, k=1, m12=m11+m22+p0*p1*p2=0+0+3*4*5=60
I=2, j=3, k=2, m23=m22+m33+p1*p2*p3=0+0+4*5*4=80
I=3,j=4,k=3,m34=m33+m44+p2*p3*p4=0+0+5*4*6=120
I=1,j=3,k=1, m11+m23+p0*p1*p3=0+80+3*4*4=128
K=2,m12+m33+p0*p2*p3=60+0+3*5*4=120
M13=120, k=2
I=2, j=4, k=2, m22+m34+p1*p2*p4=0+120+4*5*6=240
K=3,m23+m44+p1*p3*p4=80+0+4*4*6=176
M24=176,k=3
1
I=1, j=4, k=1, m11+m24+p0*p1*p4=0+176+3*4*6=248
K=2, m12+m34+p0*p2*p4=60+120+3*5*6=270
K=3, m13+m44+p0*p3*p4=120+0+3*4*6=192
M14=192, k3
e) (5 points). What is the minimal computational complexity of the matrix chain?
192
f). (5 points). Show the optimal parenthesization
(A1*A2)*A3)*A4
Q.2. (30 points). Let the knapsack capacity is W=50 kg, and five items are given
Item#
1
2
3
4
Weight (wi,
10
30
20
10
kg)
Value (vi, $)
20
40
60
45
5
30
20
a) (20 points). Use Dynamic programming to solve the 0-1 Knapsack problem given
above
a.1) (4 points). Write out the recursive formula used in the method
𝑏(𝑘, 𝑤) = {
𝑏(𝑘 − 1, 𝑤), 𝑤𝑘 > 𝑤
max⁡(𝑏(𝑘 − 1, 𝑤), 𝑏(𝑘 − 1, 𝑤 − 𝑤𝑘 ) + 𝑣𝑘
a.2) (4 points). Define the size of the matrix to be used in the method. What it shall
contain? How it shall be initialized? In what order its elements are to be filled?
The matrix has 6 rows for 0..5 items, and 6 columns for weights 0,10,20,30,40,50. It
shall contain best solution value b(i,w) for i items and w weight of a knapsack. It shall
be initialized by zeroes in the first row and column. It shall be filled by rows
a.3) (4 points). Fill in the matrix
i\w
0
10
0
0
0
1
0
20
2
0
20
3
0
20
4
0
45
5
0
45
20
0
20
20
60
65
65
30
0
20
40
80
105
105
40
0
20
60
80
125
125
50
0
20
60
100
125
125
2
a.4) (4 points).What is the maximal value (in $) the knapsack can have?
125
a.5) (4 points).What items are in the optimal solution of the 0-1 Knapsack problem?
4,3,1
b) (10 points). Use Greedy algorithm to the Fractional Knapsack problem given above
b.1) (4 points). How the items are to be sorted? Show the items after sorting
They shall be sorted by their relative value decreasing. After sorting (relative value in
parenthesis):
Item4(4.5), Item3(3),Item1(2), Item2(4/3),Item5(2/3)
b.2) (4 points).What is the total value of the items placed in the knapsack?
125+40/3=138.33
b.3) (2 points). What items are placed in the knapsack?
4, 3, 1, and 1/3 of item 2
3
Q3 (30 points). Use Dynamic programming to find the Longest Common Subsequence of X=<a
b c a d e f >, Y=<c d f e a b d>
a) (5 points).Write out recursive formula for the problem solving
0, 𝑖𝑓⁡𝑖 = 0⁡𝑜𝑟⁡𝑗 = 0
𝑙𝑐𝑠𝑖−1,𝑗−1 + 1, 𝑖𝑓⁡𝑥𝑖 = 𝑦𝑗
𝑙𝑐𝑠𝑖𝑗 = {
max(𝑙𝑐𝑠𝑖−1,𝑗 , 𝑙𝑐𝑠𝑖,𝑗−1 ) , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
b) (5 points). What is the size of the matrix to be constructed? How its rows and columns
shall be labeled? What is the contents of the matrix?
Matrix is to have 8 rows (0..7) labeled by X characters and 8 columns (0..7) labeled by Y
characters. It will contain lcs(I,i) values and arrows allowing reconstructing LCS
c) (5 points). Fill the matrix
x\y
0
c
d
0
0
0
0
A
0
0↑
0↑
B
0
0↑
0↑
C
0
1\
1←
A
0
1↑
1↑
D
0
1↑
2\
E
0
1↑
2↑
F
0
1↑
2↑
f
0
0↑
0↑
1←
1↑
2←
2↑
3\
e
0
0↑
0↑
1←
1↑
2←
3\
3↑
a
0
1\
1↑
1↑
2\
2↑
3←
3↑
b
0
1←
2\
2↑
2↑
2↑
3←
3↑
D
0
1←
2←
2↑
2↑
3\
3↑
3↑
d) (5 points). What is the length of the longest common subsequence?
3
e) (5 points). How the longest common subsequence is to be constructed using contents of
the matrix? What is the meaning of signs”↑ “, “← “, “∖” used when filling the matrix?
It is constructed from the end. We go out from the write bottom corner first by ↑ go up, second
turn by ← go left until coming to a cell with \. A character marking that row and column is
inserted before the last selected character, or just placed if it is the first selected. Then we go by
diagonal to the left and above cell and the procedure repeats until we come to the first row or the
first column.
f) (5 points). Write out the longest common subsequence. Note: When constructing the
longest common subsequence, use ”↑ “ first turn, and “← “ second turn
Abd
4
Q4 (10 points). Consider the set of five activities
Activity name a1
a2
a3
a4
Start
0
6
5
4
Finish
4
12
9
8
Use a greedy algorithm to select a maximal set of non-overlapping activities
a5
4
6
a) (3 points). How the activities are to be sorted?
They are sorted by the finish time
b) (4 points). Write out sorted activities
Activity name a1
Start
0
Finish
4
a5
4
6
a4
4
8
a3
5
9
a2
6
12
c) (3 points). What activities are selected? List their names as they are shown in the table
above
A1, a5, a2
5