Solution of assignment 1
R-1.10
Algorithm Loop1(n):
s0
for i  1 to n do
1 (assign a value)
(n+1) assignments & (n+1) comparisons & n increments
of i
ss+i
n * (one addition + one assignment)
(5n+3) total operations
f(n)=5n+3, which is in O(n) running time.
Proof: Let g(n)=n. We can find a c = 6, and n0 = 2, such that for every n >= n0, f(n) <=
cg(n). So, the running time of Loop1 method is O(n).
R-1.14
Algorithm Loop5(n):
s0
for i  1 to n2 do
for j  1 to i do
ss+i
1 (assign a value)
(n2+1) assignments & (n2+1) comparisons & n2 increments
of i
n2
3
7
 3i  2   n4  n2 operations

2
2
i 1
n2
 2i  n
4
 n 2 operations
i 1
(5n4/2+15n2/2+3) total operations
f(n) = 5n4/2+15n2/2+3. By omitting the lower-order term and the constant factor, we will
have that this algorithm is in O(n4).
Proof: Let g(n) = n4. We can find a c = 3 and n0 = 4, such that for every n >= n0, f(n) <=
cg(n). So, the running time of Loop5 method is O(n4).
R-1.15
Proof:
Step1: f(n) is O(g(n))  there exist constants c1 and n1, such that for every n >= n1, f(n)
<=c1g(n). Similarly, d(n) is O(h(n))  there exist constants c2 and n2, such that for every
n >= n2,d(n) <=c2h(n).
Step2: let c = max (c1, c2), and n0 = max (n1, n2).
Then for every n >= n0, f(n) + d(n) <= c1g(n) + c2h(n) <= c (g(n) + h(n)), so, f(n) + d(n) is
O(g(n)+h(n)).
R-1.19
Proof:
Let f(n) = (n+1)5and g(n) = O(n5).
We need to find constants c and n0, such that for every n >= n0 , f(n) <= cg(n).
The condition is equivalent to
(n+1)5 <= cn5
1
n+1 <= c1/5 n
(c1/5-1)n >= 1
Let c=2, then the condition is equivalent to
n >= 6.725
Thus, we can pick n0 = 7.
So, we have c = 2 and n0 = 7, such that for every n>=n0, f(n) <= cg(n).
So, (n+1)5 is O(n5).
R-1.23
Proof:
Let f(n) = n3logn, g(n) = n3 .
We need to find constants c and n0, such that for every n >= n0, f(n) >= cg(n).
The condition is equivalent to
n3logn >= cn3
logn >= c
Let c = 1 and n0 = 2.
Then for every n >= n0, we have logn >= c, that is, f(n) >= cg(n). Thus, n3logn is Ω(n3).
R-1.25
Proof:
Step1: f(n) is O(g(n))  there exist constants c1 and n1, such that for every n >= n1, f(n)
<=c1g(n). Similarly, d(n) is O(h(n))  there exist constants c2 and n2, such that for every
n >= n2,d(n) <=c2h(n).
Step2: let c = c1* c2 and n0 = max (n1, n2).
Then for every n>= n0, f(n) <=c1g(n) and d(n) <=c2h(n).
Because all terms>=0, we have f(n)* d(n)<= c1* c2* g(n)*f(n)
f(n)* d(n)<= c* g(n)*f(n).
Thus for every n >= n0, f(n)* d(n)<= c* g(n)*f(n), so, f(n)* d(n)is O(g(n)*f(n)).
R-5.4
a. T (n)  2T (n / 2)  log n
n logb a  n log2 2  n , f(n)=logn. Let , f(n) is O(n0.5), that is, f(n)is O(n logb a  )
Thus, we are in case 1. This means T ( n ) is (n) by the master method.
b. T (n)  8T (n / 2)  n2
n logb a  n log2 8  n3 , f(n)=n2. Let   1 , n 2 is O(n3 ) , that is, f(n)is O(n logb a  ) .
Thus, we are in case 1. This means T ( n ) is (n3 ) by the master method.
c. T (n)  16T (n / 2)  (n log n)4
2
n logb a  n log2 16  n 4 . F(n)= (nlogn)4. Let k  4 , (n log n) 4 is (n 4 log k n) , that is,
f (n)  (n logb a log k n) . Thus we are in case 2.
This means T ( n ) is (n4 log5 n)
d. T (n)  7T (n / 3)  n
n logb a  nlog3 7 , f(n)=n. Let   log3 7  1 , then n is O(nlog3 7  ) , that is, f(n)is O(n logb a  ) .
Thus, we are in case 1.
This means T ( n ) is (nlog3 7 )
e. T (n)  9T (n / 3)  n3 log n
n logb a  n log3 9  n 2 , f(n)=n3logn. let   1 , then, n3 log n is (n 2 ) , that is f(n) is
(n logb a  ).
af (n / b)  9 f (n / 3)  9((n / 3)3 log(n / 3))  (n3 / 3) log(n / 3)
let   1/ 3 , d=1, then, af (n / b)  (n3 / 3) log(n / 3)  (n3 / 3) log(n)   f (n) for all
n>=d. Thus, we are in case 3.
This means T ( n ) is (n3 log n)
Question 8
log n; n; nlogn; n2; n2 + log n; n− n3 + 7n5; 2n ; n!; nn;
Actually n2 and n2 + log n are the same.
3