Lab of
CS103 Data Structure and C++
by
Han Young Ryoo,
Joseph Gomes
Big-O Notation

Definition:
T(n) is O(f(n))
iff (if and only if) there exist positive constants
c and n0 such that
T(n) <= c f(n), for all n >= n0
Examples


T(n) = 3n + 2 is O(n)
since 3n + 2 <= 4n for all n >= 2 where c=4
and n0 = 2.
T(n) = 10n2 + 4n + 2 is O(n2)
since 10n2 + 4n + 2 <= 11n2 for all n >= 5
where c=11 and n0 = 5.
Why do we need Big O



To decide whether an algorithm is adequate,
For a better implementation
* The algorithm will always be too slow on a big enough input.
Quicksort vs. Bubble sort (O(n log n) vs. O(n2)
Quicksort running on a small desktop computer can beat
bubble sort running on a super-computer if there are a lot of
numbers to sort.
To sort 1,000,000 numbers, the quicksort takes 20,000,000
steps on average, while the bubble sort takes
1,000,000,000,000 steps!
How to determine

Sequence of statements:
statement 1;
statement 2;
...
statement k;
The total time is found by adding the times for
all statements:
total time = time(statement 1) + time(statement 2) + ... + time(statement k)
Rules for using big-O

Ignoring constant factors
O(c f(N)) = O(f(N)), where c is a constant
Example) O(20 N3) = O(N3)
Rules for using big-O

Ignoring smaller terms
If a<b then O(a+b) = O(b)
Example) O(N2+N) = O(N2)
Rules for using big-O

Upper bound only
If a<b then an O(a) algorithm is also an O(b)
algorithm.
Example) an O(N) algorithm is also an O(N2)
algorithm (but not vice versa).
Rules for using big-O

N and log N are "bigger" than any constant, from an
asymptotic view (that means for large enough N).
So if k is a constant, an O(N + k) algorithm is also
O(N), by ignoring smaller terms.
Similarly, an O(log N + k) algorithm is also O(log N).
Rules for using big-O

An O(N log N + N) algorithm, which is
O(N(log N + 1)), can be simplified to O(N log
N).
Simple Statements

If each statement is "simple" (only involves
basic operations) then the time for each
statement is constant and the total time is also
constant: O(1).
if-then-else statements

if (condition)
{ sequence of statements 1 }
else
{ sequence of statements 2 }

Either sequence 1 will execute, or sequence 2 will execute.
The worst-case time: the slowest of the two possibilities:
max(time(sequence 1), time(sequence 2)).
If sequence 1 is O(N) and sequence 2 is O(1), what is the
worst-case time for the whole if-then-else statement?


loops

for (i = 0; i < N; i++)
{ sequence of statements }

The loop executes N times, so the sequence of
statements also executes N times.
If the statements are O(1), what is the total
time for the for loop?

Nested loops

for (i = 0; i < N; i++)
{
for (j = 0; j < M; j++)
{ sequence of statements }
}

The complexity is O(N * M).
Need Generalization:
In a common special case where the stopping
condition of the inner loop is j < N instead of j < M
(i.e., the inner loop also executes N times), the total
complexity for the two loops is O(N2).

Function/Method calls


When a statement involves a method call, the complexity of
the statement includes the complexity of the method call.
Assume that you know that method f takes constant time, and
that method g takes time proportional to (linear in) the value of
its parameter k. Then the statements below have the time
complexities indicated.
f(k); // O(1)
g(k); // O(k)
Example:
for (j = 0; j < N; j++)
{ g(N);}
What is the complexity?
A program that calculates
execution time in milliseconds
#include <sys/timeb.h>
#include "stdio.h"
#include "math.h"
int main(int argc, char* argv[])
{
struct _timeb time_start,time_end;
_ftime( &time_start );
for (int i=1;i<=10000;i++)
{
int m = 1;
for(int j=1;j<100000;j++)
{
m *= 2;
}
}
_ftime( &time_end );
printf("elapsed time = %d milliseconds\n",time_end.millitm-time_start.millitm);
return 0;
}
Exercises

Two loops in a row:
for (i = 0; i < N; i++)
{ sequence of statements }
for (j = 0; j < M; j++)
{ sequence of statements }
How would the complexity change if the
second loop went to N instead of M?
Exercises

A nested loop followed by a non-nested loop:
for (i = 0; i < N; i++)
{
for (j = 0; j < N; j++)
{ sequence of statements }
}
for (k = 0; k < N; k++)
{ sequence of statements }
Exercises

A nested loop with dependent loop index:
for (i = 0; i < N; i++)
{
for (j = i; j < N; j++)
{ sequence of statements }
}


A nested loop in which the number of times the inner
loop executes depends on the value of the outer loop
index
Quiz

for (i = 0; i < N; i++) {
for (j = 0; j < N; j++) {
C[i][j] = 0;
for (k = 0; k < N; k++) {
C[i][j] = C[i][j] + A[i][k] * B[k][j];
}
}
}
Quiz

Sorting
for (i = N-1; i > 0; i--)
for (j = 0; j < i; i++)
if (a[j] > a[j+1]) swap a[j] and a[j+1];
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CS103-31 Lab of Data Structure and C++

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