Algorithm in O(2n)
Presently we can solve in our hypothetical machine problem
instances of size 100 in 1 minute using algorithm A, which
is a O(2n). We would like to solve instances of size 200 in
1 minute using algorithm A on a new machine.
What is the speed of the new machine should be?
Faster Computers
Suppose you have a computer that requires 1 minute to solve
problem instances of size 1000. What instance sizes can be
run in 1 minute if you buy a new computer that runs 1000
times faster than the old one, assuming the following time
complexities T(n) for our algorithm?
(a) T(n) = O(n).
(b) T(n) = O(n3).
(c) T(n) = O(10n).
In A Book "Foundation Of Algorithems Using C Psudocode"
Min-Max
Write an algorithm that finds both the smallest and
largest numbers in a list of n numbers and with complexity
T(n) is at most about (1.5)n comparisons.
Complexity T(n)
Write a linear-time algorithm that sorts n distinct
integers, each of which is between 1 and 500.
Hint: Use a 500-element array. (Linear-time means your
algorithm runs in time c*n + b, where c and b are any
constants that do not depend on n.
For example, your algorithm can run in time n, or time 2n +
1, or time 5n + 10, or time 100n + 6, but not time c*n*n =
c*n?.)
Performance
Algorithm A performs 10n2 basic operations and algorithm B
performs 300 lg n basic operations. For what value of n does
algorithm B start to show its better performance?
Coin Problem
You are given 9 gold coins that look identical. One is
counterfeit and weighs a bit greater than the others, but
the difference is very small that only a balance scale can
tell it from the real one. You have a balance scale that
costs 25 USD per weighing.
Give an algorithm that finds the counterfeit coin with as
little weighting as possible. Of primary importance is that
your algorithm is correct; of secondary importance is that
your algorithm truly uses the minimum number of weightings
possible.
HINT: THE BEST ALGORITHM USES ONLY 2 WEIGHINGS!!!
What is the time complexity T(n) of the following program?
a)
int n, d, i, j;
cin >> n;
for (d=1; d<=n; d++)
for (i=1; i<=d; i++)
for (j=1; j<=n; j += n/10)
cout << d << " " << i << " " << j << endl;
b)
void main()
{ int n, s, t;
cin >> n;
for (s = 1; s <= n/4; s++)
{t = s;
while (t >= 1)
{ cout << s << " " << t << endl;
t--;
}
}
}
c)
void main()
{ int n, r, s, t;
cin >> n;
for (r = 2; r <= n; r = r * 2)
for (s = 1; s <= n/4; s++)
{
t = s;
while (t >= 1)
{
cout << s << " " << t << endl;
t--;
}
}
}
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