CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
WRITTEN ASSIGNMENT 1 ANSWER KEY
1. Problem 1-1)
In general, this problem requires f(n) = some time period be solve for a value n. This can
be done for all case expect n lg n and n!, which will be evaluated numerically. For
example,
F(n) = lg n microsec = 1 sec => 2lg n = 21E6 => n = 21E6
lg n
n0.5
n
n lg n
n2
n3
2n
n!
1
second
21E6
1.00E12
1E6
62746
1.00E03
100
19
9
1
minute
26E7
3.60E15
6E7
2.8E6
7.75E03
391
25
10
1
hour
23.6E9
1.30E19
3.6E9
1.33E8
6.00E04
1532
31
12
1
day
28.64E10
7.46E21
8.64E10
2.75E9
2.94E05
4420
36
13
1
month
22.59E12
6.72E24
2.59E12
7.18E10
1.61E06
13736
41
15
1
year
23.15E13
9.95E26
3.15E13
7.97E11
5.62E06
31593
44
16
1
century
23.15E15
9.95E30
3.15E15
6.86E13
5.62E07
146645
51
17
In solving for problems like n! and n lg n, numerically solving in Excel is an acceptable
approach. For example, let cell A1 equal n and cell A2 equal “=FACT(A1)” Next, we
recognize 1 second equals 1x106 microseconds. So, increase cell A1 until cell A2 equals
1x106. Rather than manually changing cell A1, you could use Excel’s Goal Seek
function.
2. Exercise 2.2-2)
n ← length[A]
for j ← 1 to n-1
do smallest ← j
for i ← j + 1 to n
do if A[i] < A[smallest]
Cost
c1
c2
c3
c4
Time
1
n
n-1
n
n
n(n + 1)
tj = ! j =
"1
!
2
j =2
j =2
c5
n
" (t
! 1) =
j
j =2
n
n !1
" ( j ! 1) = " j =
j =2
then smallest ← i
c6
j =1
n
" (t
j
n(n ! 1)
2
! 1)
j =2
exchange A[j] ↔ A[smallest]
c7
n-1
1
CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
The algorithm maintains the loop invariant that at the start of each iteration of the outer
for loop, the subarray A[1, …, j-1] consists of the j-1 smallest elements in the array A[1,
…, n], and this subarray is the sorted order. After the first n-1 elements, the subarray
A[1, …, n-1] contains the smallest n-1 elements, sorted, and therefore element A[n] must
be the largest element.
Best Case) Best-case will occur when the array is already sorted, thus c6 will never be
executed. Hence the running time is:
T(n) = c1(1) + c2(n) + c3(n-1) + c4(n2+n-2)/2 + c5(n2-n)/2 + c7(n-1)
T(n) = (c4/2 + c5/2)n2 + (c2 + c3 + c4/2 – c5/2 + c7)n + (c1 – c3 – c4 – c7)
T(n) = a n2 + b n + c
T(n) = Θ(n2)
Worse Case) Worst-case will occur when the array is unsorted in reverse order (i.e. <5, 4,
3, 2, 1>), thus c6 will occur n(n-1)/2 time. Hence the running time is:
T(n) = c1(1) + c2(n) + c3(n-1) + c4(n2+n-2)/2 + c5(n2-n)/2 + c6(n2-n)/2 + c7(n-1)
T(n) = (c4/2 + c5/2 + c6/2)n2 + (c2 + c3 + c4/2 – c5/2 - c6/2 + c7)n + (c1 – c3 – c4 – c7)
T(n) = a n2 + b n + c
T(n) = Θ(n2)
The running time for the algorithm is Θ(n2) for all cases.
3. Exercise 2.2-4)
Modify the algorithm so it tests whether the input satisfies some special-case condition
and, if it does, output a pre-computed answer. The best-case running time is generally
not a good measure of an algorithm.
4. Exercise 2.3-1)
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CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
3 9 26 38 41 49 52 57
3 26 41 52
3 41
3
9 38 49 57
26 52
41
52
38 57
26
38
9 49
57
9
49
5. Exercise 2.3-3)
Prove by induction that
2
if n = 2
#
T ( n) = "
k
!2T (n / 2) + n if n = 2 , for k > 1
is
T (n) = n lg n
To prove by induction, we must 1) show it is true for some base case, 2) hypothesize for
some value of k, and 3) show that it holds for k+1.
Base Case:
when k=1 => n = 2
T(n) = n lg n = 2 lg 2 = 2 x 1 = 2.
Hypothesis Step:
T (n) = (n) lg(n)
T (2 k ) = (2 k ) lg(2 k )
Inductive Step k+1:
T (2 k +1 ) = 2T (2 k +1 / 2) + 2 k +1 = 2T (2 k ) + 2 k +1
[
]
T (2 k +1 ) = 2 2 k lg 2 k + 2 k +1 = 2 k +1 lg 2 k + 2 k +1
T (2 k +1 ) = 2 k +1 (lg 2 k + 1)
T (2 k +1 ) = 2 k +1 lg 2 k +1
Q.E.D.
6. Exercise 2.3-5)
3
CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
The binary search procedures takes a sorted array A, a value v, and a range [low … high]
of the array A, in which to search for the value v.
ITERATIVE-BINARY-SEARCH(A, v, low,
high)
while low ≤ high
do mid !(low + high) / 2"
If v = A[mid]
then return mid
if v > A[mid]
then low ← mid + 1
else high ← mid - 1
return NIL
RECURSIVE-BINARY-SEARCH(A, v,
low, high)
if low > high
then return NIL
mid ← !(low + high) / 2"
if v = A[mid]
then return mid
if v > A[mid]
then return RECURSIVE-BINARYSEARCH(A, v, mid+1, high)
else return RECURSIVE-BINARYSEARCH(A, v, low, mid-1)
Based on the comparison of v to the middle element in the search range, the search
continues with the range halved.
4
CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
n=0
n=1
n=2
lg n
T(n/2)
c
c
c
c
T(n/2)
c
c
c
T(n/4)
c
c
lg n
...
T(1)
c
Total:c lg n
The depth of the tree is computed by noting that T(1) occurs when T(n/22)=T(1), or n/2j =
1. This means n = 2j, or j = lg n.
7. Exercise 3.1-4)
(a) Is 2n+1 = O(2n)?
f(n) = O(g(n))
0 ≤ f(n) ≤ c g(n) where c > 0 and n > n0
Hence,
0 ≤ 2n+1 ≤ c 2n where n > n0
0 ≤ 2n+1 / 2n ≤ c where n > n0
0 ≤ 2 ≤ c where n > 0
c ≥ 2 and n > 0
Therefore,
2n+1 = O(2n) Q.E.D.
(b) Is 22n = O(2n)?
f(n) = O(g(n))
0 ≤ f(n) ≤ c g(n) where n > n0
5
CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
Hence,
0 ≤ 22n ≤ c 2n where n > n0
0 ≤ 22n / 2n ≤ c where n > n0
0 ≤ 2n ≤ c where n > n0
0 ≤ 2no ≤ c
Now,
0 ≤ 22n ≤ 2no 2n where n > n0
0 ≤ 22n ≤ 2n+no where n > n0
2n ≤ n+ n0 for n > n0
n ≤ n0 for n > n0
Is a contradiction so,
22n ≠O(2n) Q.E.D.
8. Show that n(n-1)/2 is O(n2)
f(n) = O(g(n))
0 ≤ f(n) ≤ c g(n) where c > 0 and n > n0
Hence,
0 ≤ n(n-1)/2 ≤ c n2 where c > 0 and n > n0
0 ≤ 1/2-1/2n ≤ c where c > 0 and n > n0
0 ≤ 1/2-1/4 ≤ c where n ≥ 2
c = 1/4
Hence,
0 ≤ n(n-1)/2 ≤ n2/4 where n ≥ 2
Therefore,
n(n-1)/2 = O(n2)
9. Using "big-O" notation, give the worst case running times of the following procedures
as a function of n.
procedure mystery (n: integer);
var
i,j,k: integer;
begin
for i := 1 to n-1 do
for j := i + 1 to n do
6
CISC 615
Introduction Design and Analysis of Algorithms
Dr. Jerry A. Smith
for k := 1 to j do
... a statement requiring O(1) time
end
Line
1
2
Pseudo code
for i := 1 to n-1 do
for j := i + 1 to n do
Cost
c1
c2
Times
n
c3
n +1 n +1
n
!
j=
j =2
3
for k := 1 to j do
n2 + n " 2
2
"" j =
k =3 j = k
4
... a statement requiring O(1) time
c4
n
n
"" j =
k =2 j =k
n(2n + 5)(n ! 1)
6
n(n + 1)(n ! 1)
3
T(n) = c1n + c2(n2 + n –2)/2 + c3(n)(2n+5)(n-1)/6 + c4(n)(n+1)(n-1)/3
T(n) = (c3+ c4)n3/3 + (c2 + c3)n2/2 + (n1 + c2/2 – 5c3/6 – c4/3)n – c2
T(n) an3 + bn2 + cn + d
T(n) = O(n3)
7