Tutorial Exercise (Dynamic Programming)
1.
CS3381 Design and Analysis of Algorithms
Helena Wong, 2001
For the equations of the assembly-line scheduling problem:
f *= min (f1[n] + x1, f2[n] + x2)
e1+a1,1
f 1(j) =
min (f 1[j -1] + a1,j, f2[j-1] + t 2,j-1+ a1,j)
e2+a2,1
f 2(j) =
min (f 2[j-1] + a2,j, f1[j -1] + t 1,j-1+ a 2,j)
if j=1
if j>1
if j=1
if j>1
Let ri(j) = number of references made to fi[j] in a recursive algorithm, We have
r1(n) = r2(n) = 1
r1(j) = r2(j) = r1(j+1) + r2(j+1)
Show that ri(j) = 2n-j. [Hint: You may use substitution method to prove ri(n-k) = 2k first]
2. Using the result of question 1, show that the total number of references to all f i[j] values, is exactly 2n+1-2.
What does this imply to the running time of a recursive algorithm for the Assembly-Line Scheduling
Problem? Compare it with the running time of the “FASTEST-WAY” algorithm.
3. For the “MATRIX-CHAIN-ORDER” algorithm used in our Matrix-Chain Multiplication problem
a. By observation, state the tight upper bound of the running time.
b. Show that the total number of times line 9 being executed is (n3-n)/6. [Given 1+22+32+..n2 =
n(n+1)(2n+1)/6.]
c. Make a conclusion about the asymptotic tight bound of the running time of the algorithm.
4.
Draw the recursion tree for the merge-sort procedure on an array of 16 elements. Explain why memoization
is ineffective in speeding up a good divide-and-conquer algorithm such as MERGE-SORT.
Tutorial Exercise (Dynamic Programming)
CS3381 Design and Analysis of Algorithms
Helena Wong, 2001
5.
Show that the running time of the MEMOIZED-MATRIX-CHAIN algorithm is O(n3)
(text book pp. 348-349)
6.
A dynamic programming solution to the LCS
problem, LCS-LENGTH, is illustrated below.
What is the asymptotic upper bound of its running
time?
Write a Memoized version of LCS-LENGTH.
Compare the performances of the 2 solutions.
7.
Repeat question 6 for the OPTIMAL-BST algorithm
LCS-LENGTH(X,Y)
1
m = length of X
2
n = length of Y
3
for i = 1 to m
4
c[i,0] = 0
5
for j = 1 to n
6
c[0,j] = 0
7
for i = 1 to m
8
for j = 1 to n
9
if xi=yj
10
c[i,j] = c[i-1,j-1] + 1
11
b[i,j] = “use result of c[i,j] plus xi”
12
else if c[i-1,j] >= c[i,j-1]
13
c[i,j] = c[i-1,j]
14
b[i,j] = “use result of c[i-1,j]”
12
else
13
c[i,j] = c[i,j-1]
14
b[i,j] = “use result of c[i,j-1]”
15 return c and b
OPTIMAL-BST()
1
for i = 1 to n+1
2
e[i,i-1] = qi-1
3
w[i,i-1] = qi-1
4
for len = 1 to n
5
for i = 1 to n-len + 1
6
j = i + len - 1
7
e[i,j] = 
8
w[i,j] = w[i,j-1] + pj + qj
9
for r = i to j
10
t = e[i,r-1] + e[r+1,j] + w[i,j]
11
if t < e[i,j]
12
e[i,j] = t
13
root[i,j] = r
14 return e and root